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Valeriano A.R.
2023-12-02 18:27:00 +01:00
parent 4d8bfbb377
commit 4b3d0fd0b6
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186
AdventOfCode2018/Day01.cs
View File

@@ -1,71 +1,97 @@
using System.Collections.Generic;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 1: Chronal Calibration ---
"We've detected some temporal anomalies," one of Santa's Elves at the Temporal Anomaly Research and Detection Instrument Station tells you. She sounded pretty worried when she called you down here. "At 500-year intervals into the past, someone has been changing Santa's history!"
"The good news is that the changes won't propagate to our time stream for another 25 days, and we have a device" - she attaches something to your wrist - "that will let you fix the changes with no such propagation delay. It's configured to send you 500 years further into the past every few days; that was the best we could do on such short notice."
"The bad news is that we are detecting roughly fifty anomalies throughout time; the device will indicate fixed anomalies with stars. The other bad news is that we only have one device and you're the best person for the job! Good lu--" She taps a button on the device and you suddenly feel like you're falling. To save Christmas, you need to get all fifty stars by December 25th.
Collect stars by solving puzzles. Two puzzles will be made available on each day in the advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
After feeling like you've been falling for a few minutes, you look at the device's tiny screen. "Error: Device must be calibrated before first use. Frequency drift detected. Cannot maintain destination lock." Below the message, the device shows a sequence of changes in frequency (your puzzle input). A value like +6 means the current frequency increases by 6; a value like -3 means the current frequency decreases by 3.
For example, if the device displays frequency changes of +1, -2, +3, +1, then starting from a frequency of zero, the following changes would occur:
Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.
In this example, the resulting frequency is 3.
Here are other example situations:
+1, +1, +1 results in 3
+1, +1, -2 results in 0
-1, -2, -3 results in -6
Starting with a frequency of zero, what is the resulting frequency after all of the changes in frequency have been applied?
--- Part Two ---
You notice that the device repeats the same frequency change list over and over. To calibrate the device, you need to find the first frequency it reaches twice.
For example, using the same list of changes above, the device would loop as follows:
Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.
(At this point, the device continues from the start of the list.)
Current frequency 3, change of +1; resulting frequency 4.
Current frequency 4, change of -2; resulting frequency 2, which has already been seen.
In this example, the first frequency reached twice is 2. Note that your device might need to repeat its list of frequency changes many times before a duplicate frequency is found, and that duplicates might be found while in the middle of processing the list.
Here are other examples:
+1, -1 first reaches 0 twice.
+3, +3, +4, -2, -4 first reaches 10 twice.
-6, +3, +8, +5, -6 first reaches 5 twice.
+7, +7, -2, -7, -4 first reaches 14 twice.
What is the first frequency your device reaches twice?
*/
public class Day01 : IDay
{
/*
--- Day 1: Chronal Calibration ---
"We've detected some temporal anomalies," one of Santa's Elves at the Temporal Anomaly Research and Detection Instrument Station tells you. She sounded pretty worried when she called you down here. "At 500-year intervals into the past, someone has been changing Santa's history!"
"The good news is that the changes won't propagate to our time stream for another 25 days, and we have a device" - she attaches something to your wrist - "that will let you fix the changes with no such propagation delay. It's configured to send you 500 years further into the past every few days; that was the best we could do on such short notice."
"The bad news is that we are detecting roughly fifty anomalies throughout time; the device will indicate fixed anomalies with stars. The other bad news is that we only have one device and you're the best person for the job! Good lu--" She taps a button on the device and you suddenly feel like you're falling. To save Christmas, you need to get all fifty stars by December 25th.
Collect stars by solving puzzles. Two puzzles will be made available on each day in the advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
After feeling like you've been falling for a few minutes, you look at the device's tiny screen. "Error: Device must be calibrated before first use. Frequency drift detected. Cannot maintain destination lock." Below the message, the device shows a sequence of changes in frequency (your puzzle input). A value like +6 means the current frequency increases by 6; a value like -3 means the current frequency decreases by 3.
For example, if the device displays frequency changes of +1, -2, +3, +1, then starting from a frequency of zero, the following changes would occur:
Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.
In this example, the resulting frequency is 3.
Here are other example situations:
+1, +1, +1 results in 3
+1, +1, -2 results in 0
-1, -2, -3 results in -6
Starting with a frequency of zero, what is the resulting frequency after all of the changes in frequency have been applied?
--- Part Two ---
You notice that the device repeats the same frequency change list over and over. To calibrate the device, you need to find the first frequency it reaches twice.
For example, using the same list of changes above, the device would loop as follows:
Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.
(At this point, the device continues from the start of the list.)
Current frequency 3, change of +1; resulting frequency 4.
Current frequency 4, change of -2; resulting frequency 2, which has already been seen.
In this example, the first frequency reached twice is 2. Note that your device might need to repeat its list of frequency changes many times before a duplicate frequency is found, and that duplicates might be found while in the middle of processing the list.
Here are other examples:
+1, -1 first reaches 0 twice.
+3, +3, +4, -2, -4 first reaches 10 twice.
-6, +3, +8, +5, -6 first reaches 5 twice.
+7, +7, -2, -7, -4 first reaches 14 twice.
What is the first frequency your device reaches twice?
*/
public class Day01 : IDay
public string ResolvePart1(string[] inputs)
{
public string ResolvePart1(string[] inputs)
int accumulator = 0;
foreach (string input in inputs)
{
int intInput;
if (int.TryParse(input.Substring(1), out intInput))
{
if (input[0] == '-')
{
accumulator -= intInput;
}
if (input[0] == '+')
{
accumulator += intInput;
}
}
}
return accumulator.ToString();
}
public string ResolvePart2(string[] inputs)
{
int accumulator = 0;
List<int> accumulatorHistory = new();
int? repeatedAccumulator = null;
while (repeatedAccumulator == null)
{
int accumulator = 0;
foreach (string input in inputs)
{
accumulatorHistory.Add(accumulator);
int intInput;
if (int.TryParse(input.Substring(1), out intInput))
{
@@ -77,43 +103,15 @@ namespace AdventOfCode2018
{
accumulator += intInput;
}
}
}
return accumulator.ToString();
}
public string ResolvePart2(string[] inputs)
{
int accumulator = 0;
List<int> accumulatorHistory = new List<int>();
int? repeatedAccumulator = null;
while (repeatedAccumulator == null)
{
foreach (string input in inputs)
{
accumulatorHistory.Add(accumulator);
int intInput;
if (int.TryParse(input.Substring(1), out intInput))
if (accumulatorHistory.Contains(accumulator))
{
if (input[0] == '-')
{
accumulator -= intInput;
}
if (input[0] == '+')
{
accumulator += intInput;
}
if (accumulatorHistory.Contains(accumulator))
{
repeatedAccumulator = accumulator;
break;
}
repeatedAccumulator = accumulator;
break;
}
}
}
return repeatedAccumulator.ToString();
}
return repeatedAccumulator.ToString();
}
}
}

220
AdventOfCode2018/Day02.cs
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@@ -2,117 +2,115 @@
using System.Linq;
using System.Text;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 2: Inventory Management System ---
You stop falling through time, catch your breath, and check the screen on the device. "Destination reached. Current Year: 1518. Current Location: North Pole Utility Closet 83N10." You made it! Now, to find those anomalies.
Outside the utility closet, you hear footsteps and a voice. "...I'm not sure either. But now that so many people have chimneys, maybe he could sneak in that way?" Another voice responds, "Actually, we've been working on a new kind of suit that would let him fit through tight spaces like that. But, I heard that a few days ago, they lost the prototype fabric, the design plans, everything! Nobody on the team can even seem to remember important details of the project!"
"Wouldn't they have had enough fabric to fill several boxes in the warehouse? They'd be stored together, so the box IDs should be similar. Too bad it would take forever to search the warehouse for two similar box IDs..." They walk too far away to hear any more.
Late at night, you sneak to the warehouse - who knows what kinds of paradoxes you could cause if you were discovered - and use your fancy wrist device to quickly scan every box and produce a list of the likely candidates (your puzzle input).
To make sure you didn't miss any, you scan the likely candidate boxes again, counting the number that have an ID containing exactly two of any letter and then separately counting those with exactly three of any letter. You can multiply those two counts together to get a rudimentary checksum and compare it to what your device predicts.
For example, if you see the following box IDs:
abcdef contains no letters that appear exactly two or three times.
bababc contains two a and three b, so it counts for both.
abbcde contains two b, but no letter appears exactly three times.
abcccd contains three c, but no letter appears exactly two times.
aabcdd contains two a and two d, but it only counts once.
abcdee contains two e.
ababab contains three a and three b, but it only counts once.
Of these box IDs, four of them contain a letter which appears exactly twice, and three of them contain a letter which appears exactly three times. Multiplying these together produces a checksum of 4 * 3 = 12.
What is the checksum for your list of box IDs?
--- Part Two ---
Confident that your list of box IDs is complete, you're ready to find the boxes full of prototype fabric.
The boxes will have IDs which differ by exactly one character at the same position in both strings. For example, given the following box IDs:
abcde
fghij
klmno
pqrst
fguij
axcye
wvxyz
The IDs abcde and axcye are close, but they differ by two characters (the second and fourth). However, the IDs fghij and fguij differ by exactly one character, the third (h and u). Those must be the correct boxes.
What letters are common between the two correct box IDs? (In the example above, this is found by removing the differing character from either ID, producing fgij.)
*/
public class Day02 : IDay
{
/*
--- Day 2: Inventory Management System ---
You stop falling through time, catch your breath, and check the screen on the device. "Destination reached. Current Year: 1518. Current Location: North Pole Utility Closet 83N10." You made it! Now, to find those anomalies.
Outside the utility closet, you hear footsteps and a voice. "...I'm not sure either. But now that so many people have chimneys, maybe he could sneak in that way?" Another voice responds, "Actually, we've been working on a new kind of suit that would let him fit through tight spaces like that. But, I heard that a few days ago, they lost the prototype fabric, the design plans, everything! Nobody on the team can even seem to remember important details of the project!"
"Wouldn't they have had enough fabric to fill several boxes in the warehouse? They'd be stored together, so the box IDs should be similar. Too bad it would take forever to search the warehouse for two similar box IDs..." They walk too far away to hear any more.
Late at night, you sneak to the warehouse - who knows what kinds of paradoxes you could cause if you were discovered - and use your fancy wrist device to quickly scan every box and produce a list of the likely candidates (your puzzle input).
To make sure you didn't miss any, you scan the likely candidate boxes again, counting the number that have an ID containing exactly two of any letter and then separately counting those with exactly three of any letter. You can multiply those two counts together to get a rudimentary checksum and compare it to what your device predicts.
For example, if you see the following box IDs:
abcdef contains no letters that appear exactly two or three times.
bababc contains two a and three b, so it counts for both.
abbcde contains two b, but no letter appears exactly three times.
abcccd contains three c, but no letter appears exactly two times.
aabcdd contains two a and two d, but it only counts once.
abcdee contains two e.
ababab contains three a and three b, but it only counts once.
Of these box IDs, four of them contain a letter which appears exactly twice, and three of them contain a letter which appears exactly three times. Multiplying these together produces a checksum of 4 * 3 = 12.
What is the checksum for your list of box IDs?
--- Part Two ---
Confident that your list of box IDs is complete, you're ready to find the boxes full of prototype fabric.
The boxes will have IDs which differ by exactly one character at the same position in both strings. For example, given the following box IDs:
abcde
fghij
klmno
pqrst
fguij
axcye
wvxyz
The IDs abcde and axcye are close, but they differ by two characters (the second and fourth). However, the IDs fghij and fguij differ by exactly one character, the third (h and u). Those must be the correct boxes.
What letters are common between the two correct box IDs? (In the example above, this is found by removing the differing character from either ID, producing fgij.)
*/
public class Day02 : IDay
private int CountOccurrencesOfLetter(string text, char letter)
{
private int CountOccurrencesOfLetter(string text, char letter)
{
return text.Count(c => (c == letter));
}
private Tuple<bool, bool> HasPairsAndTriplets(string text)
{
bool hasPair = false;
bool hasTriplet = false;
foreach (char c in text)
{
int count = CountOccurrencesOfLetter(text, c);
if (count == 2) { hasPair = true; }
if (count == 3) { hasTriplet = true; }
if (hasPair && hasTriplet) { break; }
}
return new Tuple<bool, bool>(hasPair, hasTriplet);
}
public string ResolvePart1(string[] inputs)
{
int pairsCount = 0;
int tripletsCount = 0;
foreach (string input in inputs)
{
var hasPairsAndTriplets = HasPairsAndTriplets(input);
if (hasPairsAndTriplets.Item1) { pairsCount++; }
if (hasPairsAndTriplets.Item2) { tripletsCount++; }
}
long result = pairsCount * tripletsCount;
return result.ToString();
}
private Tuple<int, string> CompareIDPair(string id1, string id2)
{
if (id1.Length != id2.Length) { throw new ArgumentException("id1 and id2 parameters must be of same length"); }
int diffCount = 0;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < id1.Length; i++)
{
if (id1[i] != id2[i]) { diffCount++; }
else { sb.Append(id1[i]); }
}
return new Tuple<int, string>(diffCount, sb.ToString());
}
public string ResolvePart2(string[] inputs)
{
for (int i = 0; i < (inputs.Length - 1); i++)
{
for (int j = (i + 1); j < inputs.Length; j++)
{
var result = CompareIDPair(inputs[i], inputs[j]);
if (result.Item1 == 1) { return result.Item2; }
}
}
return string.Empty;
}
return text.Count(c => (c == letter));
}
}
private Tuple<bool, bool> HasPairsAndTriplets(string text)
{
bool hasPair = false;
bool hasTriplet = false;
foreach (char c in text)
{
int count = CountOccurrencesOfLetter(text, c);
if (count == 2) { hasPair = true; }
if (count == 3) { hasTriplet = true; }
if (hasPair && hasTriplet) { break; }
}
return new Tuple<bool, bool>(hasPair, hasTriplet);
}
public string ResolvePart1(string[] inputs)
{
int pairsCount = 0;
int tripletsCount = 0;
foreach (string input in inputs)
{
var hasPairsAndTriplets = HasPairsAndTriplets(input);
if (hasPairsAndTriplets.Item1) { pairsCount++; }
if (hasPairsAndTriplets.Item2) { tripletsCount++; }
}
long result = pairsCount * tripletsCount;
return result.ToString();
}
private Tuple<int, string> CompareIDPair(string id1, string id2)
{
if (id1.Length != id2.Length) { throw new ArgumentException("id1 and id2 parameters must be of same length"); }
int diffCount = 0;
StringBuilder sb = new();
for (int i = 0; i < id1.Length; i++)
{
if (id1[i] != id2[i]) { diffCount++; }
else { sb.Append(id1[i]); }
}
return new Tuple<int, string>(diffCount, sb.ToString());
}
public string ResolvePart2(string[] inputs)
{
for (int i = 0; i < (inputs.Length - 1); i++)
{
for (int j = (i + 1); j < inputs.Length; j++)
{
var result = CompareIDPair(inputs[i], inputs[j]);
if (result.Item1 == 1) { return result.Item2; }
}
}
return string.Empty;
}
}

304
AdventOfCode2018/Day03.cs
View File

@@ -2,180 +2,176 @@
using System.Collections.Generic;
using System.Linq;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 3: No Matter How You Slice It ---
The Elves managed to locate the chimney-squeeze prototype fabric for Santa's suit (thanks to someone who helpfully wrote its box IDs on the wall of the warehouse in the middle of the night). Unfortunately, anomalies are still affecting them - nobody can even agree on how to cut the fabric.
The whole piece of fabric they're working on is a very large square - at least 1000 inches on each side.
Each Elf has made a claim about which area of fabric would be ideal for Santa's suit. All claims have an ID and consist of a single rectangle with edges parallel to the edges of the fabric. Each claim's rectangle is defined as follows:
The number of inches between the left edge of the fabric and the left edge of the rectangle.
The number of inches between the top edge of the fabric and the top edge of the rectangle.
The width of the rectangle in inches.
The height of the rectangle in inches.
A claim like #123 @ 3,2: 5x4 means that claim ID 123 specifies a rectangle 3 inches from the left edge, 2 inches from the top edge, 5 inches wide, and 4 inches tall. Visually, it claims the square inches of fabric represented by # (and ignores the square inches of fabric represented by .) in the diagram below:
...........
...........
...#####...
...#####...
...#####...
...#####...
...........
...........
...........
The problem is that many of the claims overlap, causing two or more claims to cover part of the same areas. For example, consider the following claims:
#1 @ 1,3: 4x4
#2 @ 3,1: 4x4
#3 @ 5,5: 2x2
Visually, these claim the following areas:
........
...2222.
...2222.
.11XX22.
.11XX22.
.111133.
.111133.
........
The four square inches marked with X are claimed by both 1 and 2. (Claim 3, while adjacent to the others, does not overlap either of them.)
If the Elves all proceed with their own plans, none of them will have enough fabric. How many square inches of fabric are within two or more claims?
--- Part Two ---
Amidst the chaos, you notice that exactly one claim doesn't overlap by even a single square inch of fabric with any other claim. If you can somehow draw attention to it, maybe the Elves will be able to make Santa's suit after all!
For example, in the claims above, only claim 3 is intact after all claims are made.
What is the ID of the only claim that doesn't overlap?
*/
public class Day03 : IDay
{
/*
--- Day 3: No Matter How You Slice It ---
The Elves managed to locate the chimney-squeeze prototype fabric for Santa's suit (thanks to someone who helpfully wrote its box IDs on the wall of the warehouse in the middle of the night). Unfortunately, anomalies are still affecting them - nobody can even agree on how to cut the fabric.
The whole piece of fabric they're working on is a very large square - at least 1000 inches on each side.
Each Elf has made a claim about which area of fabric would be ideal for Santa's suit. All claims have an ID and consist of a single rectangle with edges parallel to the edges of the fabric. Each claim's rectangle is defined as follows:
The number of inches between the left edge of the fabric and the left edge of the rectangle.
The number of inches between the top edge of the fabric and the top edge of the rectangle.
The width of the rectangle in inches.
The height of the rectangle in inches.
A claim like #123 @ 3,2: 5x4 means that claim ID 123 specifies a rectangle 3 inches from the left edge, 2 inches from the top edge, 5 inches wide, and 4 inches tall. Visually, it claims the square inches of fabric represented by # (and ignores the square inches of fabric represented by .) in the diagram below:
...........
...........
...#####...
...#####...
...#####...
...#####...
...........
...........
...........
The problem is that many of the claims overlap, causing two or more claims to cover part of the same areas. For example, consider the following claims:
#1 @ 1,3: 4x4
#2 @ 3,1: 4x4
#3 @ 5,5: 2x2
Visually, these claim the following areas:
........
...2222.
...2222.
.11XX22.
.11XX22.
.111133.
.111133.
........
The four square inches marked with X are claimed by both 1 and 2. (Claim 3, while adjacent to the others, does not overlap either of them.)
If the Elves all proceed with their own plans, none of them will have enough fabric. How many square inches of fabric are within two or more claims?
--- Part Two ---
Amidst the chaos, you notice that exactly one claim doesn't overlap by even a single square inch of fabric with any other claim. If you can somehow draw attention to it, maybe the Elves will be able to make Santa's suit after all!
For example, in the claims above, only claim 3 is intact after all claims are made.
What is the ID of the only claim that doesn't overlap?
*/
public class Day03 : IDay
public string ResolvePart1(string[] inputs)
{
public string ResolvePart1(string[] inputs)
List<Claim> claims = inputs.Select(i => Claim.FromString(i)).ToList();
const int edgeSize = 1000;
int[,] cells = new int[edgeSize, edgeSize];
foreach (Claim claim in claims)
{
List<Claim> claims = inputs.Select(i => Claim.FromString(i)).ToList();
const int edgeSize = 1000;
int[,] cells = new int[edgeSize, edgeSize];
foreach (Claim claim in claims)
for (int j = 0; j < claim.Height; j++)
{
for (int j = 0; j < claim.Height; j++)
for (int i = 0; i < claim.Width; i++)
{
for (int i = 0; i < claim.Width; i++)
{
cells[claim.Left + i, claim.Top + j]++;
}
cells[claim.Left + i, claim.Top + j]++;
}
}
int overlappedArea = 0;
for (int j = 0; j < edgeSize; j++)
{
for (int i = 0; i < edgeSize; i++)
{
if (cells[i, j] > 1)
{
overlappedArea++;
}
}
}
return overlappedArea.ToString();
}
public string ResolvePart2(string[] inputs)
int overlappedArea = 0;
for (int j = 0; j < edgeSize; j++)
{
List<Claim> claims = inputs.Select(i => Claim.FromString(i)).ToList();
Claim unoverlappingClaim = null;
for (int i = 0; i < claims.Count; i++)
for (int i = 0; i < edgeSize; i++)
{
bool overlaps = false;
for (int j = 0; j < claims.Count; j++)
if (cells[i, j] > 1)
{
if (i == j) { continue; }
if (Claim.Overlaps(claims[i], claims[j]))
{
overlaps = true;
break;
}
overlappedArea++;
}
if (overlaps == false)
}
}
return overlappedArea.ToString();
}
public string ResolvePart2(string[] inputs)
{
List<Claim> claims = inputs.Select(i => Claim.FromString(i)).ToList();
Claim unoverlappingClaim = null;
for (int i = 0; i < claims.Count; i++)
{
bool overlaps = false;
for (int j = 0; j < claims.Count; j++)
{
if (i == j) { continue; }
if (Claim.Overlaps(claims[i], claims[j]))
{
unoverlappingClaim = claims[i];
overlaps = true;
break;
}
}
return unoverlappingClaim.ID.ToString();
}
public class Claim
{
public int ID { get; set; }
public int Left { get; set; }
public int Top { get; set; }
public int Width { get; set; }
public int Height { get; set; }
public int MinX { get { return Left; } }
public int MaxX { get { return Left + Width; } }
public int MinY { get { return Top; } }
public int MaxY { get { return Top + Height; } }
public int GetArea()
if (overlaps == false)
{
return Width * Height;
}
public static Claim FromString(string strClaim)
{
Claim claim = new Claim();
string[] parts = strClaim.Split(new string[] { " @ ", ",", ": ", "x", }, StringSplitOptions.None);
claim.ID = Convert.ToInt32(parts[0].Substring(1));
claim.Left = Convert.ToInt32(parts[1]);
claim.Top = Convert.ToInt32(parts[2]);
claim.Width = Convert.ToInt32(parts[3]);
claim.Height = Convert.ToInt32(parts[4]);
return claim;
}
public static bool Overlaps(Claim claim1, Claim claim2)
{
if (claim1.MinX <= claim2.MaxX &&
claim2.MinX <= claim1.MaxX &&
claim1.MinY <= claim2.MaxY &&
claim2.MinY <= claim1.MaxY)
{
int minX = Math.Max(claim1.MinX, claim2.MinX);
int maxX = Math.Min(claim1.MaxX, claim2.MaxX);
int minY = Math.Max(claim1.MinY, claim2.MinY);
int maxY = Math.Min(claim1.MaxY, claim2.MaxY);
int width = maxX - minX;
int height = maxY - minY;
if (width <= 0 || height <= 0) { return false; }
return true;
}
return false;
unoverlappingClaim = claims[i];
break;
}
}
return unoverlappingClaim.ID.ToString();
}
public class Claim
{
public int ID { get; set; }
}
public int Left { get; set; }
public int Top { get; set; }
public int Width { get; set; }
public int Height { get; set; }
public int MinX { get { return Left; } }
public int MaxX { get { return Left + Width; } }
public int MinY { get { return Top; } }
public int MaxY { get { return Top + Height; } }
public int GetArea()
{
return Width * Height;
}
public static Claim FromString(string strClaim)
{
Claim claim = new();
string[] parts = strClaim.Split(new[] { " @ ", ",", ": ", "x", }, StringSplitOptions.None);
claim.ID = Convert.ToInt32(parts[0].Substring(1));
claim.Left = Convert.ToInt32(parts[1]);
claim.Top = Convert.ToInt32(parts[2]);
claim.Width = Convert.ToInt32(parts[3]);
claim.Height = Convert.ToInt32(parts[4]);
return claim;
}
public static bool Overlaps(Claim claim1, Claim claim2)
{
if (claim1.MinX <= claim2.MaxX &&
claim2.MinX <= claim1.MaxX &&
claim1.MinY <= claim2.MaxY &&
claim2.MinY <= claim1.MaxY)
{
int minX = Math.Max(claim1.MinX, claim2.MinX);
int maxX = Math.Min(claim1.MaxX, claim2.MaxX);
int minY = Math.Max(claim1.MinY, claim2.MinY);
int maxY = Math.Min(claim1.MaxY, claim2.MaxY);
int width = maxX - minX;
int height = maxY - minY;
if (width <= 0 || height <= 0) { return false; }
return true;
}
return false;
}
}
}

485
AdventOfCode2018/Day04.cs
View File

@@ -2,274 +2,273 @@
using System.Collections.Generic;
using System.Linq;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 4: Repose Record ---
You've sneaked into another supply closet - this time, it's across from the prototype suit manufacturing lab. You need to sneak inside and fix the issues with the suit, but there's a guard stationed outside the lab, so this is as close as you can safely get.
As you search the closet for anything that might help, you discover that you're not the first person to want to sneak in. Covering the walls, someone has spent an hour starting every midnight for the past few months secretly observing this guard post! They've been writing down the ID of the one guard on duty that night - the Elves seem to have decided that one guard was enough for the overnight shift - as well as when they fall asleep or wake up while at their post (your puzzle input).
For example, consider the following records, which have already been organized into chronological order:
[1518-11-01 00:00] Guard #10 begins shift
[1518-11-01 00:05] falls asleep
[1518-11-01 00:25] wakes up
[1518-11-01 00:30] falls asleep
[1518-11-01 00:55] wakes up
[1518-11-01 23:58] Guard #99 begins shift
[1518-11-02 00:40] falls asleep
[1518-11-02 00:50] wakes up
[1518-11-03 00:05] Guard #10 begins shift
[1518-11-03 00:24] falls asleep
[1518-11-03 00:29] wakes up
[1518-11-04 00:02] Guard #99 begins shift
[1518-11-04 00:36] falls asleep
[1518-11-04 00:46] wakes up
[1518-11-05 00:03] Guard #99 begins shift
[1518-11-05 00:45] falls asleep
[1518-11-05 00:55] wakes up
Timestamps are written using year-month-day hour:minute format. The guard falling asleep or waking up is always the one whose shift most recently started. Because all asleep/awake times are during the midnight hour (00:00 - 00:59), only the minute portion (00 - 59) is relevant for those events.
Visually, these records show that the guards are asleep at these times:
Date ID Minute
000000000011111111112222222222333333333344444444445555555555
012345678901234567890123456789012345678901234567890123456789
11-01 #10 .....####################.....#########################.....
11-02 #99 ........................................##########..........
11-03 #10 ........................#####...............................
11-04 #99 ....................................##########..............
11-05 #99 .............................................##########.....
The columns are Date, which shows the month-day portion of the relevant day; ID, which shows the guard on duty that day; and Minute, which shows the minutes during which the guard was asleep within the midnight hour. (The Minute column's header shows the minute's ten's digit in the first row and the one's digit in the second row.) Awake is shown as ., and asleep is shown as #.
Note that guards count as asleep on the minute they fall asleep, and they count as awake on the minute they wake up. For example, because Guard #10 wakes up at 00:25 on 1518-11-01, minute 25 is marked as awake.
If you can figure out the guard most likely to be asleep at a specific time, you might be able to trick that guard into working tonight so you can have the best chance of sneaking in. You have two strategies for choosing the best guard/minute combination.
Strategy 1: Find the guard that has the most minutes asleep. What minute does that guard spend asleep the most?
In the example above, Guard #10 spent the most minutes asleep, a total of 50 minutes (20+25+5), while Guard #99 only slept for a total of 30 minutes (10+10+10). Guard #10 was asleep most during minute 24 (on two days, whereas any other minute the guard was asleep was only seen on one day).
While this example listed the entries in chronological order, your entries are in the order you found them. You'll need to organize them before they can be analyzed.
What is the ID of the guard you chose multiplied by the minute you chose? (In the above example, the answer would be 10 * 24 = 240.)
--- Part Two ---
Strategy 2: Of all guards, which guard is most frequently asleep on the same minute?
In the example above, Guard #99 spent minute 45 asleep more than any other guard or minute - three times in total. (In all other cases, any guard spent any minute asleep at most twice.)
What is the ID of the guard you chose multiplied by the minute you chose? (In the above example, the answer would be 99 * 45 = 4455.)
*/
public class Day04 : IDay
{
/*
--- Day 4: Repose Record ---
You've sneaked into another supply closet - this time, it's across from the prototype suit manufacturing lab. You need to sneak inside and fix the issues with the suit, but there's a guard stationed outside the lab, so this is as close as you can safely get.
As you search the closet for anything that might help, you discover that you're not the first person to want to sneak in. Covering the walls, someone has spent an hour starting every midnight for the past few months secretly observing this guard post! They've been writing down the ID of the one guard on duty that night - the Elves seem to have decided that one guard was enough for the overnight shift - as well as when they fall asleep or wake up while at their post (your puzzle input).
For example, consider the following records, which have already been organized into chronological order:
[1518-11-01 00:00] Guard #10 begins shift
[1518-11-01 00:05] falls asleep
[1518-11-01 00:25] wakes up
[1518-11-01 00:30] falls asleep
[1518-11-01 00:55] wakes up
[1518-11-01 23:58] Guard #99 begins shift
[1518-11-02 00:40] falls asleep
[1518-11-02 00:50] wakes up
[1518-11-03 00:05] Guard #10 begins shift
[1518-11-03 00:24] falls asleep
[1518-11-03 00:29] wakes up
[1518-11-04 00:02] Guard #99 begins shift
[1518-11-04 00:36] falls asleep
[1518-11-04 00:46] wakes up
[1518-11-05 00:03] Guard #99 begins shift
[1518-11-05 00:45] falls asleep
[1518-11-05 00:55] wakes up
Timestamps are written using year-month-day hour:minute format. The guard falling asleep or waking up is always the one whose shift most recently started. Because all asleep/awake times are during the midnight hour (00:00 - 00:59), only the minute portion (00 - 59) is relevant for those events.
Visually, these records show that the guards are asleep at these times:
Date ID Minute
000000000011111111112222222222333333333344444444445555555555
012345678901234567890123456789012345678901234567890123456789
11-01 #10 .....####################.....#########################.....
11-02 #99 ........................................##########..........
11-03 #10 ........................#####...............................
11-04 #99 ....................................##########..............
11-05 #99 .............................................##########.....
The columns are Date, which shows the month-day portion of the relevant day; ID, which shows the guard on duty that day; and Minute, which shows the minutes during which the guard was asleep within the midnight hour. (The Minute column's header shows the minute's ten's digit in the first row and the one's digit in the second row.) Awake is shown as ., and asleep is shown as #.
Note that guards count as asleep on the minute they fall asleep, and they count as awake on the minute they wake up. For example, because Guard #10 wakes up at 00:25 on 1518-11-01, minute 25 is marked as awake.
If you can figure out the guard most likely to be asleep at a specific time, you might be able to trick that guard into working tonight so you can have the best chance of sneaking in. You have two strategies for choosing the best guard/minute combination.
Strategy 1: Find the guard that has the most minutes asleep. What minute does that guard spend asleep the most?
In the example above, Guard #10 spent the most minutes asleep, a total of 50 minutes (20+25+5), while Guard #99 only slept for a total of 30 minutes (10+10+10). Guard #10 was asleep most during minute 24 (on two days, whereas any other minute the guard was asleep was only seen on one day).
While this example listed the entries in chronological order, your entries are in the order you found them. You'll need to organize them before they can be analyzed.
What is the ID of the guard you chose multiplied by the minute you chose? (In the above example, the answer would be 10 * 24 = 240.)
--- Part Two ---
Strategy 2: Of all guards, which guard is most frequently asleep on the same minute?
In the example above, Guard #99 spent minute 45 asleep more than any other guard or minute - three times in total. (In all other cases, any guard spent any minute asleep at most twice.)
What is the ID of the guard you chose multiplied by the minute you chose? (In the above example, the answer would be 99 * 45 = 4455.)
*/
public class Day04 : IDay
public string ResolvePart1(string[] inputs)
{
public string ResolvePart1(string[] inputs)
{
List<GuardEvent> guardEvents = GuardEvent.FromStringArray(inputs);
Dictionary<int, GuardSleepHistogram> dictFullHistogram = BuildFullHistorgram(guardEvents);
List<GuardEvent> guardEvents = GuardEvent.FromStringArray(inputs);
Dictionary<int, GuardSleepHistogram> dictFullHistogram = BuildFullHistorgram(guardEvents);
// Find sleepier guard
GuardSleepHistogram highestSleeperHistogram = null;
long highestTotalSleep = long.MinValue;
// Find sleepier guard
GuardSleepHistogram highestSleeperHistogram = null;
long highestTotalSleep = long.MinValue;
foreach (GuardSleepHistogram guardHistogram in dictFullHistogram.Values)
{
int totalSleep = guardHistogram.SleepOnMunute.Sum();
if (totalSleep > highestTotalSleep)
{
highestSleeperHistogram = guardHistogram;
highestTotalSleep = totalSleep;
}
}
// Find sleepier minute
int maxSleepMinute = int.MinValue;
int maxSleepMinuteValue = int.MinValue;
for (int i = 0; i < GuardSleepHistogram.MinutesOnHour; i++)
{
if (highestSleeperHistogram.SleepOnMunute[i] > maxSleepMinuteValue)
{
maxSleepMinute = i;
maxSleepMinuteValue = highestSleeperHistogram.SleepOnMunute[i];
}
}
int result = highestSleeperHistogram.ID * maxSleepMinute;
return result.ToString();
}
public string ResolvePart2(string[] inputs)
{
List<GuardEvent> guardEvents = GuardEvent.FromStringArray(inputs);
Dictionary<int, GuardSleepHistogram> dictFullHistogram = BuildFullHistorgram(guardEvents);
int selectedGuardID = int.MinValue;
int selectedMinute = int.MinValue;
int maxSleepMinuteValue = int.MinValue;
for (int i = 0; i < GuardSleepHistogram.MinutesOnHour; i++)
{
foreach (GuardSleepHistogram guardHistogram in dictFullHistogram.Values)
{
int totalSleep = guardHistogram.SleepOnMunute.Sum();
if (totalSleep > highestTotalSleep)
if (guardHistogram.SleepOnMunute[i] > maxSleepMinuteValue)
{
highestSleeperHistogram = guardHistogram;
highestTotalSleep = totalSleep;
maxSleepMinuteValue = guardHistogram.SleepOnMunute[i];
selectedGuardID = guardHistogram.ID;
selectedMinute = i;
}
}
// Find sleepier minute
int maxSleepMinute = int.MinValue;
int maxSleepMinuteValue = int.MinValue;
for (int i = 0; i < GuardSleepHistogram.MinutesOnHour; i++)
{
if (highestSleeperHistogram.SleepOnMunute[i] > maxSleepMinuteValue)
{
maxSleepMinute = i;
maxSleepMinuteValue = highestSleeperHistogram.SleepOnMunute[i];
}
}
int result = highestSleeperHistogram.ID * maxSleepMinute;
return result.ToString();
}
public string ResolvePart2(string[] inputs)
{
List<GuardEvent> guardEvents = GuardEvent.FromStringArray(inputs);
Dictionary<int, GuardSleepHistogram> dictFullHistogram = BuildFullHistorgram(guardEvents);
int selectedGuardID = int.MinValue;
int selectedMinute = int.MinValue;
int maxSleepMinuteValue = int.MinValue;
for (int i = 0; i < GuardSleepHistogram.MinutesOnHour; i++)
{
foreach (GuardSleepHistogram guardHistogram in dictFullHistogram.Values)
{
if (guardHistogram.SleepOnMunute[i] > maxSleepMinuteValue)
{
maxSleepMinuteValue = guardHistogram.SleepOnMunute[i];
selectedGuardID = guardHistogram.ID;
selectedMinute = i;
}
}
}
int result = selectedGuardID * selectedMinute;
return result.ToString();
}
private static Dictionary<int, GuardSleepHistogram> BuildFullHistorgram(List<GuardEvent> guardEvents)
{
Dictionary<int, GuardSleepHistogram> dictFullHistogram = new Dictionary<int, GuardSleepHistogram>();
foreach (IGrouping<int, GuardEvent> group in guardEvents.GroupBy(guardEvent => guardEvent.Date.DayOfYear))
{
Dictionary<int, GuardSleepHistogram> dictDayHistogram = new Dictionary<int, GuardSleepHistogram>();
foreach (GuardEvent guardEvent in group)
{
if (guardEvent.ID == null) { continue; }
GuardSleepHistogram dayGuardHistogram = null;
if (dictDayHistogram.ContainsKey((int)guardEvent.ID))
{
dayGuardHistogram = dictDayHistogram[(int)guardEvent.ID];
}
else
{
dayGuardHistogram = new GuardSleepHistogram { ID = (int)guardEvent.ID };
dictDayHistogram.Add(dayGuardHistogram.ID, dayGuardHistogram);
}
if (guardEvent.Type == GuardEventType.FallSleep)
{
dayGuardHistogram.FallSleep(guardEvent.Date.Minute);
}
if (guardEvent.Type == GuardEventType.WakeUp)
{
dayGuardHistogram.WakeUp(guardEvent.Date.Minute);
}
}
foreach (GuardSleepHistogram dayGuardHistogram in dictDayHistogram.Values)
{
GuardSleepHistogram guardHistogram = null;
if (dictFullHistogram.ContainsKey(dayGuardHistogram.ID))
{
guardHistogram = dictFullHistogram[dayGuardHistogram.ID];
guardHistogram.AddHistogram(dayGuardHistogram);
}
else
{
dictFullHistogram.Add(dayGuardHistogram.ID, dayGuardHistogram);
}
}
}
return dictFullHistogram;
}
int result = selectedGuardID * selectedMinute;
return result.ToString();
}
public enum GuardEventType
private static Dictionary<int, GuardSleepHistogram> BuildFullHistorgram(List<GuardEvent> guardEvents)
{
ShiftBegin,
FallSleep,
WakeUp,
}
public class GuardEvent
{
public DateTime Date { get; set; }
public int? ID { get; set; }
public GuardEventType Type { get; set; }
public static GuardEvent FromString(string strEvent)
Dictionary<int, GuardSleepHistogram> dictFullHistogram = new();
foreach (IGrouping<int, GuardEvent> group in guardEvents.GroupBy(guardEvent => guardEvent.Date.DayOfYear))
{
GuardEvent guardEvent = new GuardEvent();
string[] parts = strEvent.Split(new string[] { "[", "-", " ", ":", "]", "#", }, StringSplitOptions.RemoveEmptyEntries);
guardEvent.Date = new DateTime(
Convert.ToInt32(parts[0]),
Convert.ToInt32(parts[1]),
Convert.ToInt32(parts[2]),
Convert.ToInt32(parts[3]),
Convert.ToInt32(parts[4]),
0
);
if (parts[5] == "Guard")
Dictionary<int, GuardSleepHistogram> dictDayHistogram = new();
foreach (GuardEvent guardEvent in group)
{
guardEvent.ID = Convert.ToInt32(parts[6]);
guardEvent.Type = GuardEventType.ShiftBegin;
}
if (parts[5] == "falls")
{
guardEvent.Type = GuardEventType.FallSleep;
}
if (parts[5] == "wakes")
{
guardEvent.Type = GuardEventType.WakeUp;
}
return guardEvent;
}
public static List<GuardEvent> FromStringArray(string[] strEvents)
{
List<GuardEvent> guardEvents = strEvents
.Select(strEvent => FromString(strEvent))
.OrderBy(guardEvent => guardEvent.Date)
.ToList();
int? guardID = null;
foreach (GuardEvent guardEvent in guardEvents)
{
if (guardEvent.Type == GuardEventType.ShiftBegin)
if (guardEvent.ID == null) { continue; }
GuardSleepHistogram dayGuardHistogram;
if (dictDayHistogram.ContainsKey((int)guardEvent.ID))
{
guardID = guardEvent.ID;
dayGuardHistogram = dictDayHistogram[(int)guardEvent.ID];
}
else
{
guardEvent.ID = guardID;
dayGuardHistogram = new GuardSleepHistogram { ID = (int)guardEvent.ID };
dictDayHistogram.Add(dayGuardHistogram.ID, dayGuardHistogram);
}
if (guardEvent.Type == GuardEventType.FallSleep)
{
dayGuardHistogram.FallSleep(guardEvent.Date.Minute);
}
if (guardEvent.Type == GuardEventType.WakeUp)
{
dayGuardHistogram.WakeUp(guardEvent.Date.Minute);
}
}
return guardEvents;
}
}
public class GuardSleepHistogram
{
public const int MinutesOnHour = 60;
public int ID { get; set; }
public int[] SleepOnMunute { get; } = new int[MinutesOnHour];
public void FallSleep(int minute)
{
for (int i = minute; i < MinutesOnHour; i++)
foreach (GuardSleepHistogram dayGuardHistogram in dictDayHistogram.Values)
{
SleepOnMunute[i] = 1;
GuardSleepHistogram guardHistogram;
if (dictFullHistogram.ContainsKey(dayGuardHistogram.ID))
{
guardHistogram = dictFullHistogram[dayGuardHistogram.ID];
guardHistogram.AddHistogram(dayGuardHistogram);
}
else
{
dictFullHistogram.Add(dayGuardHistogram.ID, dayGuardHistogram);
}
}
}
public void WakeUp(int minute)
{
for (int i = minute; i < MinutesOnHour; i++)
{
SleepOnMunute[i] = 0;
}
}
public void AddHistogram(GuardSleepHistogram histogram)
{
for (int i = 0; i < MinutesOnHour; i++)
{
SleepOnMunute[i] += histogram.SleepOnMunute[i];
}
}
return dictFullHistogram;
}
}
public enum GuardEventType
{
ShiftBegin,
FallSleep,
WakeUp,
}
public class GuardEvent
{
public DateTime Date { get; set; }
public int? ID { get; set; }
public GuardEventType Type { get; set; }
public static GuardEvent FromString(string strEvent)
{
GuardEvent guardEvent = new();
string[] parts = strEvent.Split(new[] { "[", "-", " ", ":", "]", "#", }, StringSplitOptions.RemoveEmptyEntries);
guardEvent.Date = new DateTime(
Convert.ToInt32(parts[0]),
Convert.ToInt32(parts[1]),
Convert.ToInt32(parts[2]),
Convert.ToInt32(parts[3]),
Convert.ToInt32(parts[4]),
0
);
if (parts[5] == "Guard")
{
guardEvent.ID = Convert.ToInt32(parts[6]);
guardEvent.Type = GuardEventType.ShiftBegin;
}
if (parts[5] == "falls")
{
guardEvent.Type = GuardEventType.FallSleep;
}
if (parts[5] == "wakes")
{
guardEvent.Type = GuardEventType.WakeUp;
}
return guardEvent;
}
public static List<GuardEvent> FromStringArray(string[] strEvents)
{
List<GuardEvent> guardEvents = strEvents
.Select(strEvent => FromString(strEvent))
.OrderBy(guardEvent => guardEvent.Date)
.ToList();
int? guardID = null;
foreach (GuardEvent guardEvent in guardEvents)
{
if (guardEvent.Type == GuardEventType.ShiftBegin)
{
guardID = guardEvent.ID;
}
else
{
guardEvent.ID = guardID;
}
}
return guardEvents;
}
}
public class GuardSleepHistogram
{
public const int MinutesOnHour = 60;
public int ID { get; set; }
public int[] SleepOnMunute { get; } = new int[MinutesOnHour];
public void FallSleep(int minute)
{
for (int i = minute; i < MinutesOnHour; i++)
{
SleepOnMunute[i] = 1;
}
}
public void WakeUp(int minute)
{
for (int i = minute; i < MinutesOnHour; i++)
{
SleepOnMunute[i] = 0;
}
}
public void AddHistogram(GuardSleepHistogram histogram)
{
for (int i = 0; i < MinutesOnHour; i++)
{
SleepOnMunute[i] += histogram.SleepOnMunute[i];
}
}
}

232
AdventOfCode2018/Day05.cs
View File

@@ -2,130 +2,128 @@
using System.Linq;
using System.Text;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 5: Alchemical Reduction ---
You've managed to sneak in to the prototype suit manufacturing lab. The Elves are making decent progress, but are still struggling with the suit's size reduction capabilities.
While the very latest in 1518 alchemical technology might have solved their problem eventually, you can do better. You scan the chemical composition of the suit's material and discover that it is formed by extremely long polymers (one of which is available as your puzzle input).
The polymer is formed by smaller units which, when triggered, react with each other such that two adjacent units of the same type and opposite polarity are destroyed. Units' types are represented by letters; units' polarity is represented by capitalization. For instance, r and R are units with the same type but opposite polarity, whereas r and s are entirely different types and do not react.
For example:
In aA, a and A react, leaving nothing behind.
In abBA, bB destroys itself, leaving aA. As above, this then destroys itself, leaving nothing.
In abAB, no two adjacent units are of the same type, and so nothing happens.
In aabAAB, even though aa and AA are of the same type, their polarities match, and so nothing happens.
Now, consider a larger example, dabAcCaCBAcCcaDA:
dabAcCaCBAcCcaDA The first 'cC' is removed.
dabAaCBAcCcaDA This creates 'Aa', which is removed.
dabCBAcCcaDA Either 'cC' or 'Cc' are removed (the result is the same).
dabCBAcaDA No further actions can be taken.
After all possible reactions, the resulting polymer contains 10 units.
How many units remain after fully reacting the polymer you scanned? (Note: in this puzzle and others, the input is large; if you copy/paste your input, make sure you get the whole thing.)
--- Part Two ---
Time to improve the polymer.
One of the unit types is causing problems; it's preventing the polymer from collapsing as much as it should. Your goal is to figure out which unit type is causing the most problems, remove all instances of it (regardless of polarity), fully react the remaining polymer, and measure its length.
For example, again using the polymer dabAcCaCBAcCcaDA from above:
Removing all A/a units produces dbcCCBcCcD. Fully reacting this polymer produces dbCBcD, which has length 6.
Removing all B/b units produces daAcCaCAcCcaDA. Fully reacting this polymer produces daCAcaDA, which has length 8.
Removing all C/c units produces dabAaBAaDA. Fully reacting this polymer produces daDA, which has length 4.
Removing all D/d units produces abAcCaCBAcCcaA. Fully reacting this polymer produces abCBAc, which has length 6.
In this example, removing all C/c units was best, producing the answer 4.
What is the length of the shortest polymer you can produce by removing all units of exactly one type and fully reacting the result?
*/
public class Day05 : IDay
{
/*
--- Day 5: Alchemical Reduction ---
You've managed to sneak in to the prototype suit manufacturing lab. The Elves are making decent progress, but are still struggling with the suit's size reduction capabilities.
While the very latest in 1518 alchemical technology might have solved their problem eventually, you can do better. You scan the chemical composition of the suit's material and discover that it is formed by extremely long polymers (one of which is available as your puzzle input).
The polymer is formed by smaller units which, when triggered, react with each other such that two adjacent units of the same type and opposite polarity are destroyed. Units' types are represented by letters; units' polarity is represented by capitalization. For instance, r and R are units with the same type but opposite polarity, whereas r and s are entirely different types and do not react.
For example:
In aA, a and A react, leaving nothing behind.
In abBA, bB destroys itself, leaving aA. As above, this then destroys itself, leaving nothing.
In abAB, no two adjacent units are of the same type, and so nothing happens.
In aabAAB, even though aa and AA are of the same type, their polarities match, and so nothing happens.
Now, consider a larger example, dabAcCaCBAcCcaDA:
dabAcCaCBAcCcaDA The first 'cC' is removed.
dabAaCBAcCcaDA This creates 'Aa', which is removed.
dabCBAcCcaDA Either 'cC' or 'Cc' are removed (the result is the same).
dabCBAcaDA No further actions can be taken.
After all possible reactions, the resulting polymer contains 10 units.
How many units remain after fully reacting the polymer you scanned? (Note: in this puzzle and others, the input is large; if you copy/paste your input, make sure you get the whole thing.)
--- Part Two ---
Time to improve the polymer.
One of the unit types is causing problems; it's preventing the polymer from collapsing as much as it should. Your goal is to figure out which unit type is causing the most problems, remove all instances of it (regardless of polarity), fully react the remaining polymer, and measure its length.
For example, again using the polymer dabAcCaCBAcCcaDA from above:
Removing all A/a units produces dbcCCBcCcD. Fully reacting this polymer produces dbCBcD, which has length 6.
Removing all B/b units produces daAcCaCAcCcaDA. Fully reacting this polymer produces daCAcaDA, which has length 8.
Removing all C/c units produces dabAaBAaDA. Fully reacting this polymer produces daDA, which has length 4.
Removing all D/d units produces abAcCaCBAcCcaA. Fully reacting this polymer produces abCBAc, which has length 6.
In this example, removing all C/c units was best, producing the answer 4.
What is the length of the shortest polymer you can produce by removing all units of exactly one type and fully reacting the result?
*/
public class Day05 : IDay
public string ReducePolymer(string polymer)
{
public string ReducePolymer(string polymer)
{
if (polymer.Length <= 1) { return polymer; }
StringBuilder sb = new StringBuilder();
if (polymer.Length <= 1) { return polymer; }
StringBuilder sb = new();
int i;
for (i = 1; i < polymer.Length; i++)
int i;
for (i = 1; i < polymer.Length; i++)
{
char previousCharacter = polymer[i - 1];
char character = polymer[i];
if (previousCharacter != character && char.ToLower(previousCharacter) == char.ToLower(character))
{
char previousCharacter = polymer[i - 1];
char character = polymer[i];
if (previousCharacter != character && char.ToLower(previousCharacter) == char.ToLower(character))
{
i++;
}
else
{
sb.Append(previousCharacter);
}
i++;
}
if (i == polymer.Length) { sb.Append(polymer[i - 1]); }
return sb.ToString();
}
public string FullyReducePolymer(string input)
{
string previousPolymer = null;
string polymer = input;
do
else
{
previousPolymer = polymer;
polymer = ReducePolymer(polymer);
} while (previousPolymer.Length > polymer.Length);
return polymer;
}
public string ResolvePart1(string[] inputs)
{
string reducedPolymer = FullyReducePolymer(inputs[0]);
return reducedPolymer.Length.ToString();
}
public string RemoveUnitTypeFromPolymer(string polymer, char unitType)
{
StringBuilder sb = new StringBuilder();
unitType = char.ToLower(unitType);
foreach (char c in polymer)
{
if (char.ToLower(c) != unitType)
{
sb.Append(c);
}
sb.Append(previousCharacter);
}
return sb.ToString();
}
if (i == polymer.Length) { sb.Append(polymer[i - 1]); }
public string ResolvePart2(string[] inputs)
{
string input = inputs[0];
List<char> allUnitTypes = input.Select(c => char.ToLower(c)).Distinct().ToList();
int minPolymerLenght = int.MaxValue;
foreach (char unitType in allUnitTypes)
{
string fixedPolymer = RemoveUnitTypeFromPolymer(input, unitType);
string fixedReducedPolymer = FullyReducePolymer(fixedPolymer);
int fixedReducedPolymerLenght = fixedReducedPolymer.Length;
if (minPolymerLenght > fixedReducedPolymerLenght)
{
minPolymerLenght = fixedReducedPolymerLenght;
}
}
return minPolymerLenght.ToString();
}
return sb.ToString();
}
}
public string FullyReducePolymer(string input)
{
string previousPolymer;
string polymer = input;
do
{
previousPolymer = polymer;
polymer = ReducePolymer(polymer);
} while (previousPolymer.Length > polymer.Length);
return polymer;
}
public string ResolvePart1(string[] inputs)
{
string reducedPolymer = FullyReducePolymer(inputs[0]);
return reducedPolymer.Length.ToString();
}
public string RemoveUnitTypeFromPolymer(string polymer, char unitType)
{
StringBuilder sb = new();
unitType = char.ToLower(unitType);
foreach (char c in polymer)
{
if (char.ToLower(c) != unitType)
{
sb.Append(c);
}
}
return sb.ToString();
}
public string ResolvePart2(string[] inputs)
{
string input = inputs[0];
List<char> allUnitTypes = input.Select(c => char.ToLower(c)).Distinct().ToList();
int minPolymerLenght = int.MaxValue;
foreach (char unitType in allUnitTypes)
{
string fixedPolymer = RemoveUnitTypeFromPolymer(input, unitType);
string fixedReducedPolymer = FullyReducePolymer(fixedPolymer);
int fixedReducedPolymerLenght = fixedReducedPolymer.Length;
if (minPolymerLenght > fixedReducedPolymerLenght)
{
minPolymerLenght = fixedReducedPolymerLenght;
}
}
return minPolymerLenght.ToString();
}
}

413
AdventOfCode2018/Day06.cs
View File

@@ -2,224 +2,221 @@
using System.Collections.Generic;
using System.Linq;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 6: Chronal Coordinates ---
The device on your wrist beeps several times, and once again you feel like you're falling.
"Situation critical," the device announces. "Destination indeterminate. Chronal interference detected. Please specify new target coordinates."
The device then produces a list of coordinates (your puzzle input). Are they places it thinks are safe or dangerous? It recommends you check manual page 729. The Elves did not give you a manual.
If they're dangerous, maybe you can minimize the danger by finding the coordinate that gives the largest distance from the other points.
Using only the Manhattan distance, determine the area around each coordinate by counting the number of integer X,Y locations that are closest to that coordinate (and aren't tied in distance to any other coordinate).
Your goal is to find the size of the largest area that isn't infinite. For example, consider the following list of coordinates:
1, 1
1, 6
8, 3
3, 4
5, 5
8, 9
If we name these coordinates A through F, we can draw them on a grid, putting 0,0 at the top left:
..........
.A........
..........
........C.
...D......
.....E....
.B........
..........
..........
........F.
This view is partial - the actual grid extends infinitely in all directions. Using the Manhattan distance, each location's closest coordinate can be determined, shown here in lowercase:
aaaaa.cccc
aAaaa.cccc
aaaddecccc
aadddeccCc
..dDdeeccc
bb.deEeecc
bBb.eeee..
bbb.eeefff
bbb.eeffff
bbb.ffffFf
Locations shown as . are equally far from two or more coordinates, and so they don't count as being closest to any.
In this example, the areas of coordinates A, B, C, and F are infinite - while not shown here, their areas extend forever outside the visible grid. However, the areas of coordinates D and E are finite: D is closest to 9 locations, and E is closest to 17 (both including the coordinate's location itself). Therefore, in this example, the size of the largest area is 17.
What is the size of the largest area that isn't infinite?
--- Part Two ---
On the other hand, if the coordinates are safe, maybe the best you can do is try to find a region near as many coordinates as possible.
For example, suppose you want the sum of the Manhattan distance to all of the coordinates to be less than 32. For each location, add up the distances to all of the given coordinates; if the total of those distances is less than 32, that location is within the desired region. Using the same coordinates as above, the resulting region looks like this:
..........
.A........
..........
...###..C.
..#D###...
..###E#...
.B.###....
..........
..........
........F.
In particular, consider the highlighted location 4,3 located at the top middle of the region. Its calculation is as follows, where abs() is the absolute value function:
Distance to coordinate A: abs(4-1) + abs(3-1) = 5
Distance to coordinate B: abs(4-1) + abs(3-6) = 6
Distance to coordinate C: abs(4-8) + abs(3-3) = 4
Distance to coordinate D: abs(4-3) + abs(3-4) = 2
Distance to coordinate E: abs(4-5) + abs(3-5) = 3
Distance to coordinate F: abs(4-8) + abs(3-9) = 10
Total distance: 5 + 6 + 4 + 2 + 3 + 10 = 30
Because the total distance to all coordinates (30) is less than 32, the location is within the region.
This region, which also includes coordinates D and E, has a total size of 16.
Your actual region will need to be much larger than this example, though, instead including all locations with a total distance of less than 10000.
What is the size of the region containing all locations which have a total distance to all given coordinates of less than 10000?
*/
public class Day06 : IDay
{
/*
--- Day 6: Chronal Coordinates ---
The device on your wrist beeps several times, and once again you feel like you're falling.
"Situation critical," the device announces. "Destination indeterminate. Chronal interference detected. Please specify new target coordinates."
The device then produces a list of coordinates (your puzzle input). Are they places it thinks are safe or dangerous? It recommends you check manual page 729. The Elves did not give you a manual.
If they're dangerous, maybe you can minimize the danger by finding the coordinate that gives the largest distance from the other points.
Using only the Manhattan distance, determine the area around each coordinate by counting the number of integer X,Y locations that are closest to that coordinate (and aren't tied in distance to any other coordinate).
Your goal is to find the size of the largest area that isn't infinite. For example, consider the following list of coordinates:
1, 1
1, 6
8, 3
3, 4
5, 5
8, 9
If we name these coordinates A through F, we can draw them on a grid, putting 0,0 at the top left:
..........
.A........
..........
........C.
...D......
.....E....
.B........
..........
..........
........F.
This view is partial - the actual grid extends infinitely in all directions. Using the Manhattan distance, each location's closest coordinate can be determined, shown here in lowercase:
aaaaa.cccc
aAaaa.cccc
aaaddecccc
aadddeccCc
..dDdeeccc
bb.deEeecc
bBb.eeee..
bbb.eeefff
bbb.eeffff
bbb.ffffFf
Locations shown as . are equally far from two or more coordinates, and so they don't count as being closest to any.
In this example, the areas of coordinates A, B, C, and F are infinite - while not shown here, their areas extend forever outside the visible grid. However, the areas of coordinates D and E are finite: D is closest to 9 locations, and E is closest to 17 (both including the coordinate's location itself). Therefore, in this example, the size of the largest area is 17.
What is the size of the largest area that isn't infinite?
--- Part Two ---
On the other hand, if the coordinates are safe, maybe the best you can do is try to find a region near as many coordinates as possible.
For example, suppose you want the sum of the Manhattan distance to all of the coordinates to be less than 32. For each location, add up the distances to all of the given coordinates; if the total of those distances is less than 32, that location is within the desired region. Using the same coordinates as above, the resulting region looks like this:
..........
.A........
..........
...###..C.
..#D###...
..###E#...
.B.###....
..........
..........
........F.
In particular, consider the highlighted location 4,3 located at the top middle of the region. Its calculation is as follows, where abs() is the absolute value function:
Distance to coordinate A: abs(4-1) + abs(3-1) = 5
Distance to coordinate B: abs(4-1) + abs(3-6) = 6
Distance to coordinate C: abs(4-8) + abs(3-3) = 4
Distance to coordinate D: abs(4-3) + abs(3-4) = 2
Distance to coordinate E: abs(4-5) + abs(3-5) = 3
Distance to coordinate F: abs(4-8) + abs(3-9) = 10
Total distance: 5 + 6 + 4 + 2 + 3 + 10 = 30
Because the total distance to all coordinates (30) is less than 32, the location is within the region.
This region, which also includes coordinates D and E, has a total size of 16.
Your actual region will need to be much larger than this example, though, instead including all locations with a total distance of less than 10000.
What is the size of the region containing all locations which have a total distance to all given coordinates of less than 10000?
*/
public class Day06 : IDay
private List<ChronoPoint> InputsToPoints(string[] inputs)
{
private List<ChronoPoint> InputsToPoints(string[] inputs)
{
return inputs
.Where(input => string.IsNullOrEmpty(input) == false)
.Select(input => ChronoPoint.FromString(input))
.ToList();
}
public string ResolvePart1(string[] inputs)
{
List<ChronoPoint> points = InputsToPoints(inputs);
Dictionary<int, int> pointsAreas = new Dictionary<int, int>();
for (int i = 0; i < points.Count; i++)
{
pointsAreas.Add(i, 0);
}
int minX = points.Min(p => p.X) - 1;
int maxX = points.Max(p => p.X) + 1;
int minY = points.Min(p => p.Y) - 1;
int maxY = points.Max(p => p.Y) + 1;
ChronoPoint samplingPoint = new ChronoPoint();
for(int i=minX; i <= maxX; i++)
{
for(int j = minY; j <= maxY; j++)
{
samplingPoint.X = i;
samplingPoint.Y = j;
bool isEdge = i == minX || i == maxX || j == minY || j == maxY;
int idxMin = -1;
int distanceMin = int.MaxValue;
for (int idx = 0; idx < points.Count; idx++)
{
int distance = ChronoPoint.ManhattanDistance(samplingPoint, points[idx]);
if (distance == distanceMin)
{
idxMin = -1;
}
else if (distance < distanceMin)
{
distanceMin = distance;
idxMin = idx;
}
}
if (idxMin < 0) { continue; }
if (isEdge)
{
pointsAreas[idxMin] = -1;
}
else
{
int previousArea = pointsAreas[idxMin];
if (previousArea >= 0)
{
pointsAreas[idxMin] = previousArea + 1;
}
}
}
}
int maxArea = pointsAreas.Max(p => p.Value);
return maxArea.ToString();
}
private int AreaInThresold(List<ChronoPoint> points, int distanceThresold)
{
int minX = points.Min(p => p.X) - 1;
int maxX = points.Max(p => p.X) + 1;
int minY = points.Min(p => p.Y) - 1;
int maxY = points.Max(p => p.Y) + 1;
int areaInRange = 0;
ChronoPoint samplingPoint = new ChronoPoint();
for (int i = minX; i <= maxX; i++)
{
for (int j = minY; j <= maxY; j++)
{
samplingPoint.X = i;
samplingPoint.Y = j;
int distanceSum = points.Sum(p => ChronoPoint.ManhattanDistance(samplingPoint, p));
if (distanceSum < distanceThresold) { areaInRange++; }
}
}
return areaInRange;
}
public int DistanceThresold { get; set; } = 10000;
public string ResolvePart2(string[] inputs)
{
List<ChronoPoint> points = InputsToPoints(inputs);
int areaInRange = AreaInThresold(points, DistanceThresold);
return areaInRange.ToString();
}
return inputs
.Where(input => string.IsNullOrEmpty(input) == false)
.Select(input => ChronoPoint.FromString(input))
.ToList();
}
public class ChronoPoint
public string ResolvePart1(string[] inputs)
{
public int X { get; set; }
public int Y { get; set; }
public static ChronoPoint FromString(string strPoint)
List<ChronoPoint> points = InputsToPoints(inputs);
Dictionary<int, int> pointsAreas = new();
for (int i = 0; i < points.Count; i++)
{
if (string.IsNullOrEmpty(strPoint)) { return null; }
string[] parts = strPoint.Split(new string[] { ", ", }, StringSplitOptions.RemoveEmptyEntries);
if (parts.Length < 2) { return null; }
ChronoPoint point = new ChronoPoint
pointsAreas.Add(i, 0);
}
int minX = points.Min(p => p.X) - 1;
int maxX = points.Max(p => p.X) + 1;
int minY = points.Min(p => p.Y) - 1;
int maxY = points.Max(p => p.Y) + 1;
ChronoPoint samplingPoint = new();
for(int i=minX; i <= maxX; i++)
{
for(int j = minY; j <= maxY; j++)
{
X = Convert.ToInt32(parts[0]),
Y = Convert.ToInt32(parts[1]),
};
return point;
samplingPoint.X = i;
samplingPoint.Y = j;
bool isEdge = i == minX || i == maxX || j == minY || j == maxY;
int idxMin = -1;
int distanceMin = int.MaxValue;
for (int idx = 0; idx < points.Count; idx++)
{
int distance = ChronoPoint.ManhattanDistance(samplingPoint, points[idx]);
if (distance == distanceMin)
{
idxMin = -1;
}
else if (distance < distanceMin)
{
distanceMin = distance;
idxMin = idx;
}
}
if (idxMin < 0) { continue; }
if (isEdge)
{
pointsAreas[idxMin] = -1;
}
else
{
int previousArea = pointsAreas[idxMin];
if (previousArea >= 0)
{
pointsAreas[idxMin] = previousArea + 1;
}
}
}
}
public static int ManhattanDistance(ChronoPoint p0, ChronoPoint p1)
int maxArea = pointsAreas.Max(p => p.Value);
return maxArea.ToString();
}
private int AreaInThresold(List<ChronoPoint> points, int distanceThresold)
{
int minX = points.Min(p => p.X) - 1;
int maxX = points.Max(p => p.X) + 1;
int minY = points.Min(p => p.Y) - 1;
int maxY = points.Max(p => p.Y) + 1;
int areaInRange = 0;
ChronoPoint samplingPoint = new();
for (int i = minX; i <= maxX; i++)
{
int distance = Math.Abs(p1.X - p0.X) + Math.Abs(p1.Y - p0.Y);
return distance;
for (int j = minY; j <= maxY; j++)
{
samplingPoint.X = i;
samplingPoint.Y = j;
int distanceSum = points.Sum(p => ChronoPoint.ManhattanDistance(samplingPoint, p));
if (distanceSum < distanceThresold) { areaInRange++; }
}
}
return areaInRange;
}
public int DistanceThresold { get; set; } = 10000;
public string ResolvePart2(string[] inputs)
{
List<ChronoPoint> points = InputsToPoints(inputs);
int areaInRange = AreaInThresold(points, DistanceThresold);
return areaInRange.ToString();
}
}
public class ChronoPoint
{
public int X { get; set; }
public int Y { get; set; }
public static ChronoPoint FromString(string strPoint)
{
if (string.IsNullOrEmpty(strPoint)) { return null; }
string[] parts = strPoint.Split(new[] { ", ", }, StringSplitOptions.RemoveEmptyEntries);
if (parts.Length < 2) { return null; }
ChronoPoint point = new() {
X = Convert.ToInt32(parts[0]),
Y = Convert.ToInt32(parts[1]),
};
return point;
}
public static int ManhattanDistance(ChronoPoint p0, ChronoPoint p1)
{
int distance = Math.Abs(p1.X - p0.X) + Math.Abs(p1.Y - p0.Y);
return distance;
}
}

492
AdventOfCode2018/Day07.cs
View File

@@ -3,283 +3,281 @@ using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 7: The Sum of Its Parts ---
You find yourself standing on a snow-covered coastline; apparently, you landed a little off course. The region is too hilly to see the North Pole from here, but you do spot some Elves that seem to be trying to unpack something that washed ashore. It's quite cold out, so you decide to risk creating a paradox by asking them for directions.
"Oh, are you the search party?" Somehow, you can understand whatever Elves from the year 1018 speak; you assume it's Ancient Nordic Elvish. Could the device on your wrist also be a translator? "Those clothes don't look very warm; take this." They hand you a heavy coat.
"We do need to find our way back to the North Pole, but we have higher priorities at the moment. You see, believe it or not, this box contains something that will solve all of Santa's transportation problems - at least, that's what it looks like from the pictures in the instructions." It doesn't seem like they can read whatever language it's in, but you can: "Sleigh kit. Some assembly required."
"'Sleigh'? What a wonderful name! You must help us assemble this 'sleigh' at once!" They start excitedly pulling more parts out of the box.
The instructions specify a series of steps and requirements about which steps must be finished before others can begin (your puzzle input). Each step is designated by a single letter. For example, suppose you have the following instructions:
Step C must be finished before step A can begin.
Step C must be finished before step F can begin.
Step A must be finished before step B can begin.
Step A must be finished before step D can begin.
Step B must be finished before step E can begin.
Step D must be finished before step E can begin.
Step F must be finished before step E can begin.
Visually, these requirements look like this:
-->A--->B--
/ \ \
C -->D----->E
\ /
---->F-----
Your first goal is to determine the order in which the steps should be completed. If more than one step is ready, choose the step which is first alphabetically. In this example, the steps would be completed as follows:
Only C is available, and so it is done first.
Next, both A and F are available. A is first alphabetically, so it is done next.
Then, even though F was available earlier, steps B and D are now also available, and B is the first alphabetically of the three.
After that, only D and F are available. E is not available because only some of its prerequisites are complete. Therefore, D is completed next.
F is the only choice, so it is done next.
Finally, E is completed.
So, in this example, the correct order is CABDFE.
In what order should the steps in your instructions be completed?
--- Part Two ---
As you're about to begin construction, four of the Elves offer to help. "The sun will set soon; it'll go faster if we work together." Now, you need to account for multiple people working on steps simultaneously. If multiple steps are available, workers should still begin them in alphabetical order.
Each step takes 60 seconds plus an amount corresponding to its letter: A=1, B=2, C=3, and so on. So, step A takes 60+1=61 seconds, while step Z takes 60+26=86 seconds. No time is required between steps.
To simplify things for the example, however, suppose you only have help from one Elf (a total of two workers) and that each step takes 60 fewer seconds (so that step A takes 1 second and step Z takes 26 seconds). Then, using the same instructions as above, this is how each second would be spent:
Second Worker 1 Worker 2 Done
0 C .
1 C .
2 C .
3 A F C
4 B F CA
5 B F CA
6 D F CAB
7 D F CAB
8 D F CAB
9 D . CABF
10 E . CABFD
11 E . CABFD
12 E . CABFD
13 E . CABFD
14 E . CABFD
15 . . CABFDE
Each row represents one second of time. The Second column identifies how many seconds have passed as of the beginning of that second. Each worker column shows the step that worker is currently doing (or . if they are idle). The Done column shows completed steps.
Note that the order of the steps has changed; this is because steps now take time to finish and multiple workers can begin multiple steps simultaneously.
In this example, it would take 15 seconds for two workers to complete these steps.
With 5 workers and the 60+ second step durations described above, how long will it take to complete all of the steps?
*/
public class Day07 : IDay
{
/*
--- Day 7: The Sum of Its Parts ---
You find yourself standing on a snow-covered coastline; apparently, you landed a little off course. The region is too hilly to see the North Pole from here, but you do spot some Elves that seem to be trying to unpack something that washed ashore. It's quite cold out, so you decide to risk creating a paradox by asking them for directions.
"Oh, are you the search party?" Somehow, you can understand whatever Elves from the year 1018 speak; you assume it's Ancient Nordic Elvish. Could the device on your wrist also be a translator? "Those clothes don't look very warm; take this." They hand you a heavy coat.
"We do need to find our way back to the North Pole, but we have higher priorities at the moment. You see, believe it or not, this box contains something that will solve all of Santa's transportation problems - at least, that's what it looks like from the pictures in the instructions." It doesn't seem like they can read whatever language it's in, but you can: "Sleigh kit. Some assembly required."
"'Sleigh'? What a wonderful name! You must help us assemble this 'sleigh' at once!" They start excitedly pulling more parts out of the box.
The instructions specify a series of steps and requirements about which steps must be finished before others can begin (your puzzle input). Each step is designated by a single letter. For example, suppose you have the following instructions:
Step C must be finished before step A can begin.
Step C must be finished before step F can begin.
Step A must be finished before step B can begin.
Step A must be finished before step D can begin.
Step B must be finished before step E can begin.
Step D must be finished before step E can begin.
Step F must be finished before step E can begin.
Visually, these requirements look like this:
-->A--->B--
/ \ \
C -->D----->E
\ /
---->F-----
Your first goal is to determine the order in which the steps should be completed. If more than one step is ready, choose the step which is first alphabetically. In this example, the steps would be completed as follows:
Only C is available, and so it is done first.
Next, both A and F are available. A is first alphabetically, so it is done next.
Then, even though F was available earlier, steps B and D are now also available, and B is the first alphabetically of the three.
After that, only D and F are available. E is not available because only some of its prerequisites are complete. Therefore, D is completed next.
F is the only choice, so it is done next.
Finally, E is completed.
So, in this example, the correct order is CABDFE.
In what order should the steps in your instructions be completed?
--- Part Two ---
As you're about to begin construction, four of the Elves offer to help. "The sun will set soon; it'll go faster if we work together." Now, you need to account for multiple people working on steps simultaneously. If multiple steps are available, workers should still begin them in alphabetical order.
Each step takes 60 seconds plus an amount corresponding to its letter: A=1, B=2, C=3, and so on. So, step A takes 60+1=61 seconds, while step Z takes 60+26=86 seconds. No time is required between steps.
To simplify things for the example, however, suppose you only have help from one Elf (a total of two workers) and that each step takes 60 fewer seconds (so that step A takes 1 second and step Z takes 26 seconds). Then, using the same instructions as above, this is how each second would be spent:
Second Worker 1 Worker 2 Done
0 C .
1 C .
2 C .
3 A F C
4 B F CA
5 B F CA
6 D F CAB
7 D F CAB
8 D F CAB
9 D . CABF
10 E . CABFD
11 E . CABFD
12 E . CABFD
13 E . CABFD
14 E . CABFD
15 . . CABFDE
Each row represents one second of time. The Second column identifies how many seconds have passed as of the beginning of that second. Each worker column shows the step that worker is currently doing (or . if they are idle). The Done column shows completed steps.
Note that the order of the steps has changed; this is because steps now take time to finish and multiple workers can begin multiple steps simultaneously.
In this example, it would take 15 seconds for two workers to complete these steps.
With 5 workers and the 60+ second step durations described above, how long will it take to complete all of the steps?
*/
public class Day07 : IDay
private static Instructions BuildInstructions(string[] inputs, int baseCost)
{
private static Instructions BuildInstructions(string[] inputs, int baseCost)
Instructions instructions = new();
foreach (string input in inputs)
{
Instructions instructions = new Instructions();
foreach (string input in inputs)
{
if (string.IsNullOrEmpty(input)) { continue; }
string[] parts = input.Split(new string[] {
"Step ",
" must be finished before step ",
" can begin.",
}, StringSplitOptions.RemoveEmptyEntries);
instructions.AddNodeRelation(parts[1].ToUpper(), parts[0].ToUpper());
}
foreach (InstructionNode node in instructions.Nodes.Values)
{
char nodeID = node.NodeID[0];
int nodeCost = baseCost + (nodeID - 'A') + 1;
node.Cost = nodeCost;
}
return instructions;
if (string.IsNullOrEmpty(input)) { continue; }
string[] parts = input.Split(new[] {
"Step ",
" must be finished before step ",
" can begin.",
}, StringSplitOptions.RemoveEmptyEntries);
instructions.AddNodeRelation(parts[1].ToUpper(), parts[0].ToUpper());
}
foreach (InstructionNode node in instructions.Nodes.Values)
{
char nodeID = node.NodeID[0];
int nodeCost = baseCost + (nodeID - 'A') + 1;
node.Cost = nodeCost;
}
public string ResolvePart1(string[] inputs)
return instructions;
}
public string ResolvePart1(string[] inputs)
{
Instructions instructions = BuildInstructions(inputs, 0);
List<InstructionNode> finalInstructions = instructions.SortInstructions();
StringBuilder sbInstructions = new();
foreach (InstructionNode node in finalInstructions)
{
Instructions instructions = BuildInstructions(inputs, 0);
List<InstructionNode> finalInstructions = instructions.SortInstructions();
StringBuilder sbInstructions = new StringBuilder();
foreach (InstructionNode node in finalInstructions)
{
sbInstructions.Append(node.NodeID);
}
return sbInstructions.ToString();
sbInstructions.Append(node.NodeID);
}
return sbInstructions.ToString();
}
public int BaseCost { get; set; } = 60;
public int NumberOfWorkers { get; set; } = 5;
public string ResolvePart2(string[] inputs)
{
Instructions instructions = BuildInstructions(inputs, BaseCost);
int totalElapsedTime = instructions.SimulateInstructionsUsage(NumberOfWorkers);
return totalElapsedTime.ToString();
}
}
public class InstructionNode
{
public string NodeID { get; set; }
public List<string> PreviousNodeIDs { get; } = new();
public int Cost { get; set; }
public bool Running { get; set; }
public bool Used { get; set; }
public bool CanBeUsed(Dictionary<string, InstructionNode> allNodes)
{
if (PreviousNodeIDs.Count == 0) { return true; }
bool allPreviousUsed = PreviousNodeIDs.All(nodeID => allNodes[nodeID].Used);
return allPreviousUsed;
}
}
public class Instructions
{
public Dictionary<string, InstructionNode> Nodes { get; } = new();
public InstructionNode GetNode(string nodeID)
{
InstructionNode node = null;
if (Nodes.ContainsKey(nodeID))
{
node = Nodes[nodeID];
}
else
{
node = new InstructionNode { NodeID = nodeID, };
Nodes.Add(nodeID, node);
}
return node;
}
public void AddNodeRelation(string nodeID, string previousNodeID)
{
InstructionNode node = GetNode(nodeID);
InstructionNode previousNode = GetNode(previousNodeID);
node.PreviousNodeIDs.Add(previousNode.NodeID);
}
public List<InstructionNode> SortInstructions()
{
List<InstructionNode> finalNodes = new();
foreach (InstructionNode node in Nodes.Values)
{
node.Used = false;
}
public int BaseCost { get; set; } = 60;
public int NumberOfWorkers { get; set; } = 5;
public string ResolvePart2(string[] inputs)
List<InstructionNode> unusedNodes;
do
{
Instructions instructions = BuildInstructions(inputs, BaseCost);
int totalElapsedTime = instructions.SimulateInstructionsUsage(NumberOfWorkers);
return totalElapsedTime.ToString();
unusedNodes = Nodes.Values
.Where(n =>
n.Used == false &&
n.CanBeUsed(Nodes))
.OrderBy(n => n.NodeID)
.ToList();
if (unusedNodes.Count > 0)
{
InstructionNode node = unusedNodes.FirstOrDefault();
finalNodes.Add(node);
node.Used = true;
}
} while (unusedNodes.Count > 0);
return finalNodes;
}
private class SimulatedWorker
{
public InstructionNode CurrentInstruction { get; set; }
public int ElapsedTime { get; set; }
public void SetInstruction(InstructionNode instruction)
{
CurrentInstruction = instruction;
ElapsedTime = 0;
instruction.Running = true;
}
public bool Work()
{
if (CurrentInstruction == null) { return false; }
ElapsedTime++;
if (CurrentInstruction.Cost <= ElapsedTime)
{
CurrentInstruction.Running = false;
CurrentInstruction.Used = true;
CurrentInstruction = null;
}
return true;
}
}
public class InstructionNode
public int SimulateInstructionsUsage(int numberOfWorkers)
{
public string NodeID { get; set; }
public List<string> PreviousNodeIDs { get; } = new List<string>();
public int Cost { get; set; }
public bool Running { get; set; } = false;
public bool Used { get; set; } = false;
public bool CanBeUsed(Dictionary<string, InstructionNode> allNodes)
int totalElapsedTime = 0;
foreach (InstructionNode node in Nodes.Values)
{
if (PreviousNodeIDs.Count == 0) { return true; }
bool allPreviousUsed = PreviousNodeIDs.All(nodeID => allNodes[nodeID].Used);
return allPreviousUsed;
node.Used = false;
node.Running = false;
}
}
public class Instructions
{
public Dictionary<string, InstructionNode> Nodes { get; } = new Dictionary<string, InstructionNode>();
public InstructionNode GetNode(string nodeID)
List<SimulatedWorker> workers = new(numberOfWorkers);
for (int i = 0; i < numberOfWorkers; i++)
{
InstructionNode node = null;
if (Nodes.ContainsKey(nodeID))
{
node = Nodes[nodeID];
}
else
{
node = new InstructionNode { NodeID = nodeID, };
Nodes.Add(nodeID, node);
}
return node;
workers.Add(new SimulatedWorker());
}
public void AddNodeRelation(string nodeID, string previousNodeID)
bool anyWorkerWitoutWork;
do
{
InstructionNode node = GetNode(nodeID);
InstructionNode previousNode = GetNode(previousNodeID);
node.PreviousNodeIDs.Add(previousNode.NodeID);
}
public List<InstructionNode> SortInstructions()
{
List<InstructionNode> finalNodes = new List<InstructionNode>();
foreach (InstructionNode node in Nodes.Values)
bool anyWorkDone = false;
foreach (SimulatedWorker worker in workers)
{
node.Used = false;
if (worker.Work())
{
anyWorkDone = true;
}
}
if (anyWorkDone) { totalElapsedTime++; }
List<InstructionNode> unusedNodes = null;
do
anyWorkerWitoutWork = workers.Any(w => w.CurrentInstruction == null);
if (anyWorkerWitoutWork)
{
unusedNodes = Nodes.Values
List<InstructionNode> unusedNodes = Nodes.Values
.Where(n =>
n.Used == false &&
n.Used == false && n.Running == false &&
n.CanBeUsed(Nodes))
.OrderBy(n => n.NodeID)
.ToList();
if (unusedNodes.Count > 0)
{
InstructionNode node = unusedNodes.FirstOrDefault();
finalNodes.Add(node);
node.Used = true;
}
} while (unusedNodes.Count > 0);
return finalNodes;
}
private class SimulatedWorker
{
public InstructionNode CurrentInstruction { get; set; }
public int ElapsedTime { get; set; }
public void SetInstruction(InstructionNode instruction)
{
CurrentInstruction = instruction;
ElapsedTime = 0;
instruction.Running = true;
}
public bool Work()
{
if (CurrentInstruction == null) { return false; }
ElapsedTime++;
if (CurrentInstruction.Cost <= ElapsedTime)
{
CurrentInstruction.Running = false;
CurrentInstruction.Used = true;
CurrentInstruction = null;
}
return true;
}
}
public int SimulateInstructionsUsage(int numberOfWorkers)
{
int totalElapsedTime = 0;
foreach (InstructionNode node in Nodes.Values)
{
node.Used = false;
node.Running = false;
}
List<SimulatedWorker> workers = new List<SimulatedWorker>(numberOfWorkers);
for (int i = 0; i < numberOfWorkers; i++)
{
workers.Add(new SimulatedWorker());
}
bool anyWorkerWitoutWork;
do
{
bool anyWorkDone = false;
foreach (SimulatedWorker worker in workers)
{
if (worker.Work())
List<SimulatedWorker> workersWithoutWork = workers.Where(w => w.CurrentInstruction == null).ToList();
for (int i = 0; i < workersWithoutWork.Count && i < unusedNodes.Count; i++)
{
anyWorkDone = true;
workersWithoutWork[i].SetInstruction(unusedNodes[i]);
}
}
if (anyWorkDone) { totalElapsedTime++; }
anyWorkerWitoutWork = workers.Any(w => w.CurrentInstruction == null);
if (anyWorkerWitoutWork)
{
List<InstructionNode> unusedNodes = Nodes.Values
.Where(n =>
n.Used == false && n.Running == false &&
n.CanBeUsed(Nodes))
.OrderBy(n => n.NodeID)
.ToList();
if (unusedNodes.Count > 0)
{
List<SimulatedWorker> workersWithoutWork = workers.Where(w => w.CurrentInstruction == null).ToList();
for (int i = 0; i < workersWithoutWork.Count && i < unusedNodes.Count; i++)
{
workersWithoutWork[i].SetInstruction(unusedNodes[i]);
}
}
}
} while (workers.Any(w => w.CurrentInstruction != null));
return totalElapsedTime;
}
}
} while (workers.Any(w => w.CurrentInstruction != null));
return totalElapsedTime;
}
}
}

298
AdventOfCode2018/Day08.cs
View File

@@ -2,160 +2,158 @@
using System.Collections.Generic;
using System.Linq;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 8: Memory Maneuver ---
The sleigh is much easier to pull than you'd expect for something its weight. Unfortunately, neither you nor the Elves know which way the North Pole is from here.
You check your wrist device for anything that might help. It seems to have some kind of navigation system! Activating the navigation system produces more bad news: "Failed to start navigation system. Could not read software license file."
The navigation system's license file consists of a list of numbers (your puzzle input). The numbers define a data structure which, when processed, produces some kind of tree that can be used to calculate the license number.
The tree is made up of nodes; a single, outermost node forms the tree's root, and it contains all other nodes in the tree (or contains nodes that contain nodes, and so on).
Specifically, a node consists of:
A header, which is always exactly two numbers:
The quantity of child nodes.
The quantity of metadata entries.
Zero or more child nodes (as specified in the header).
One or more metadata entries (as specified in the header).
Each child node is itself a node that has its own header, child nodes, and metadata. For example:
2 3 0 3 10 11 12 1 1 0 1 99 2 1 1 2
A----------------------------------
B----------- C-----------
D-----
In this example, each node of the tree is also marked with an underline starting with a letter for easier identification. In it, there are four nodes:
A, which has 2 child nodes (B, C) and 3 metadata entries (1, 1, 2).
B, which has 0 child nodes and 3 metadata entries (10, 11, 12).
C, which has 1 child node (D) and 1 metadata entry (2).
D, which has 0 child nodes and 1 metadata entry (99).
The first check done on the license file is to simply add up all of the metadata entries. In this example, that sum is 1+1+2+10+11+12+2+99=138.
What is the sum of all metadata entries?
--- Part Two ---
The second check is slightly more complicated: you need to find the value of the root node (A in the example above).
The value of a node depends on whether it has child nodes.
If a node has no child nodes, its value is the sum of its metadata entries. So, the value of node B is 10+11+12=33, and the value of node D is 99.
However, if a node does have child nodes, the metadata entries become indexes which refer to those child nodes. A metadata entry of 1 refers to the first child node, 2 to the second, 3 to the third, and so on. The value of this node is the sum of the values of the child nodes referenced by the metadata entries. If a referenced child node does not exist, that reference is skipped. A child node can be referenced multiple time and counts each time it is referenced. A metadata entry of 0 does not refer to any child node.
For example, again using the above nodes:
Node C has one metadata entry, 2. Because node C has only one child node, 2 references a child node which does not exist, and so the value of node C is 0.
Node A has three metadata entries: 1, 1, and 2. The 1 references node A's first child node, B, and the 2 references node A's second child node, C. Because node B has a value of 33 and node C has a value of 0, the value of node A is 33+33+0=66.
So, in this example, the value of the root node is 66.
What is the value of the root node?
*/
public class Day08 : IDay
{
/*
--- Day 8: Memory Maneuver ---
The sleigh is much easier to pull than you'd expect for something its weight. Unfortunately, neither you nor the Elves know which way the North Pole is from here.
You check your wrist device for anything that might help. It seems to have some kind of navigation system! Activating the navigation system produces more bad news: "Failed to start navigation system. Could not read software license file."
The navigation system's license file consists of a list of numbers (your puzzle input). The numbers define a data structure which, when processed, produces some kind of tree that can be used to calculate the license number.
The tree is made up of nodes; a single, outermost node forms the tree's root, and it contains all other nodes in the tree (or contains nodes that contain nodes, and so on).
Specifically, a node consists of:
A header, which is always exactly two numbers:
The quantity of child nodes.
The quantity of metadata entries.
Zero or more child nodes (as specified in the header).
One or more metadata entries (as specified in the header).
Each child node is itself a node that has its own header, child nodes, and metadata. For example:
2 3 0 3 10 11 12 1 1 0 1 99 2 1 1 2
A----------------------------------
B----------- C-----------
D-----
In this example, each node of the tree is also marked with an underline starting with a letter for easier identification. In it, there are four nodes:
A, which has 2 child nodes (B, C) and 3 metadata entries (1, 1, 2).
B, which has 0 child nodes and 3 metadata entries (10, 11, 12).
C, which has 1 child node (D) and 1 metadata entry (2).
D, which has 0 child nodes and 1 metadata entry (99).
The first check done on the license file is to simply add up all of the metadata entries. In this example, that sum is 1+1+2+10+11+12+2+99=138.
What is the sum of all metadata entries?
--- Part Two ---
The second check is slightly more complicated: you need to find the value of the root node (A in the example above).
The value of a node depends on whether it has child nodes.
If a node has no child nodes, its value is the sum of its metadata entries. So, the value of node B is 10+11+12=33, and the value of node D is 99.
However, if a node does have child nodes, the metadata entries become indexes which refer to those child nodes. A metadata entry of 1 refers to the first child node, 2 to the second, 3 to the third, and so on. The value of this node is the sum of the values of the child nodes referenced by the metadata entries. If a referenced child node does not exist, that reference is skipped. A child node can be referenced multiple time and counts each time it is referenced. A metadata entry of 0 does not refer to any child node.
For example, again using the above nodes:
Node C has one metadata entry, 2. Because node C has only one child node, 2 references a child node which does not exist, and so the value of node C is 0.
Node A has three metadata entries: 1, 1, and 2. The 1 references node A's first child node, B, and the 2 references node A's second child node, C. Because node B has a value of 33 and node C has a value of 0, the value of node A is 33+33+0=66.
So, in this example, the value of the root node is 66.
What is the value of the root node?
*/
public class Day08 : IDay
public string ResolvePart1(string[] inputs)
{
public string ResolvePart1(string[] inputs)
{
IntStream values = new IntStream(inputs[0]);
ChronoLicenceNode licenceTree = ChronoLicenceNode.BuildFromIntStream(values);
int result = licenceTree.GetChecksum();
return result.ToString();
}
public string ResolvePart2(string[] inputs)
{
IntStream values = new IntStream(inputs[0]);
ChronoLicenceNode licenceTree = ChronoLicenceNode.BuildFromIntStream(values);
int result = licenceTree.GetValue();
return result.ToString();
}
IntStream values = new(inputs[0]);
ChronoLicenceNode licenceTree = ChronoLicenceNode.BuildFromIntStream(values);
int result = licenceTree.GetChecksum();
return result.ToString();
}
public class IntStream
public string ResolvePart2(string[] inputs)
{
private int[] values;
private int index;
public IntStream(string strValues)
{
values = strValues
.Split(new string[] { " ", }, StringSplitOptions.RemoveEmptyEntries)
.Select(strVal => Convert.ToInt32(strVal))
.ToArray();
index = 0;
}
public int Get()
{
int value = values[index];
index++;
return value;
}
}
public class ChronoLicenceNode
{
public List<ChronoLicenceNode> Childs { get; } = new List<ChronoLicenceNode>();
public List<int> Metadata { get; } = new List<int>();
public static ChronoLicenceNode BuildFromIntStream(IntStream stream)
{
ChronoLicenceNode node = new ChronoLicenceNode();
int numChilds = stream.Get();
int numMetadata = stream.Get();
for (int i = 0; i < numChilds; i++)
{
ChronoLicenceNode childNode = BuildFromIntStream(stream);
node.Childs.Add(childNode);
}
for (int i = 0; i < numMetadata; i++)
{
node.Metadata.Add(stream.Get());
}
return node;
}
public int GetChecksum()
{
int checksum = Metadata.Sum();
foreach (ChronoLicenceNode child in Childs)
{
checksum += child.GetChecksum();
}
return checksum;
}
public int GetValue()
{
int value = 0;
if (Childs.Count == 0)
{
value = Metadata.Sum();
}
else
{
foreach (int metadata in Metadata)
{
int index = metadata - 1;
if (index < 0 || index >= Childs.Count) { continue; }
value += Childs[index].GetValue();
}
}
return value;
}
IntStream values = new(inputs[0]);
ChronoLicenceNode licenceTree = ChronoLicenceNode.BuildFromIntStream(values);
int result = licenceTree.GetValue();
return result.ToString();
}
}
public class IntStream
{
private int[] _values;
private int _index;
public IntStream(string strValues)
{
_values = strValues
.Split(new[] { " ", }, StringSplitOptions.RemoveEmptyEntries)
.Select(strVal => Convert.ToInt32(strVal))
.ToArray();
_index = 0;
}
public int Get()
{
int value = _values[_index];
_index++;
return value;
}
}
public class ChronoLicenceNode
{
public List<ChronoLicenceNode> Childs { get; } = new();
public List<int> Metadata { get; } = new();
public static ChronoLicenceNode BuildFromIntStream(IntStream stream)
{
ChronoLicenceNode node = new();
int numChilds = stream.Get();
int numMetadata = stream.Get();
for (int i = 0; i < numChilds; i++)
{
ChronoLicenceNode childNode = BuildFromIntStream(stream);
node.Childs.Add(childNode);
}
for (int i = 0; i < numMetadata; i++)
{
node.Metadata.Add(stream.Get());
}
return node;
}
public int GetChecksum()
{
int checksum = Metadata.Sum();
foreach (ChronoLicenceNode child in Childs)
{
checksum += child.GetChecksum();
}
return checksum;
}
public int GetValue()
{
int value = 0;
if (Childs.Count == 0)
{
value = Metadata.Sum();
}
else
{
foreach (int metadata in Metadata)
{
int index = metadata - 1;
if (index < 0 || index >= Childs.Count) { continue; }
value += Childs[index].GetValue();
}
}
return value;
}
}

318
AdventOfCode2018/Day09.cs
View File

@@ -2,173 +2,171 @@
using System.Collections.Generic;
using System.Linq;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 9: Marble Mania ---
You talk to the Elves while you wait for your navigation system to initialize. To pass the time, they introduce you to their favorite marble game.
The Elves play this game by taking turns arranging the marbles in a circle according to very particular rules. The marbles are numbered starting with 0 and increasing by 1 until every marble has a number.
First, the marble numbered 0 is placed in the circle. At this point, while it contains only a single marble, it is still a circle: the marble is both clockwise from itself and counter-clockwise from itself. This marble is designated the current marble.
Then, each Elf takes a turn placing the lowest-numbered remaining marble into the circle between the marbles that are 1 and 2 marbles clockwise of the current marble. (When the circle is large enough, this means that there is one marble between the marble that was just placed and the current marble.) The marble that was just placed then becomes the current marble.
However, if the marble that is about to be placed has a number which is a multiple of 23, something entirely different happens. First, the current player keeps the marble they would have placed, adding it to their score. In addition, the marble 7 marbles counter-clockwise from the current marble is removed from the circle and also added to the current player's score. The marble located immediately clockwise of the marble that was removed becomes the new current marble.
For example, suppose there are 9 players. After the marble with value 0 is placed in the middle, each player (shown in square brackets) takes a turn. The result of each of those turns would produce circles of marbles like this, where clockwise is to the right and the resulting current marble is in parentheses:
[-] (0)
[1] 0 (1)
[2] 0 (2) 1
[3] 0 2 1 (3)
[4] 0 (4) 2 1 3
[5] 0 4 2 (5) 1 3
[6] 0 4 2 5 1 (6) 3
[7] 0 4 2 5 1 6 3 (7)
[8] 0 (8) 4 2 5 1 6 3 7
[9] 0 8 4 (9) 2 5 1 6 3 7
[1] 0 8 4 9 2(10) 5 1 6 3 7
[2] 0 8 4 9 2 10 5(11) 1 6 3 7
[3] 0 8 4 9 2 10 5 11 1(12) 6 3 7
[4] 0 8 4 9 2 10 5 11 1 12 6(13) 3 7
[5] 0 8 4 9 2 10 5 11 1 12 6 13 3(14) 7
[6] 0 8 4 9 2 10 5 11 1 12 6 13 3 14 7(15)
[7] 0(16) 8 4 9 2 10 5 11 1 12 6 13 3 14 7 15
[8] 0 16 8(17) 4 9 2 10 5 11 1 12 6 13 3 14 7 15
[9] 0 16 8 17 4(18) 9 2 10 5 11 1 12 6 13 3 14 7 15
[1] 0 16 8 17 4 18 9(19) 2 10 5 11 1 12 6 13 3 14 7 15
[2] 0 16 8 17 4 18 9 19 2(20)10 5 11 1 12 6 13 3 14 7 15
[3] 0 16 8 17 4 18 9 19 2 20 10(21) 5 11 1 12 6 13 3 14 7 15
[4] 0 16 8 17 4 18 9 19 2 20 10 21 5(22)11 1 12 6 13 3 14 7 15
[5] 0 16 8 17 4 18(19) 2 20 10 21 5 22 11 1 12 6 13 3 14 7 15
[6] 0 16 8 17 4 18 19 2(24)20 10 21 5 22 11 1 12 6 13 3 14 7 15
[7] 0 16 8 17 4 18 19 2 24 20(25)10 21 5 22 11 1 12 6 13 3 14 7 15
The goal is to be the player with the highest score after the last marble is used up. Assuming the example above ends after the marble numbered 25, the winning score is 23+9=32 (because player 5 kept marble 23 and removed marble 9, while no other player got any points in this very short example game).
Here are a few more examples:
10 players; last marble is worth 1618 points: high score is 8317
13 players; last marble is worth 7999 points: high score is 146373
17 players; last marble is worth 1104 points: high score is 2764
21 players; last marble is worth 6111 points: high score is 54718
30 players; last marble is worth 5807 points: high score is 37305
What is the winning Elf's score?
--- Part Two ---
Amused by the speed of your answer, the Elves are curious:
What would the new winning Elf's score be if the number of the last marble were 100 times larger?
*/
public class Day09 : IDay
{
/*
--- Day 9: Marble Mania ---
You talk to the Elves while you wait for your navigation system to initialize. To pass the time, they introduce you to their favorite marble game.
The Elves play this game by taking turns arranging the marbles in a circle according to very particular rules. The marbles are numbered starting with 0 and increasing by 1 until every marble has a number.
First, the marble numbered 0 is placed in the circle. At this point, while it contains only a single marble, it is still a circle: the marble is both clockwise from itself and counter-clockwise from itself. This marble is designated the current marble.
Then, each Elf takes a turn placing the lowest-numbered remaining marble into the circle between the marbles that are 1 and 2 marbles clockwise of the current marble. (When the circle is large enough, this means that there is one marble between the marble that was just placed and the current marble.) The marble that was just placed then becomes the current marble.
However, if the marble that is about to be placed has a number which is a multiple of 23, something entirely different happens. First, the current player keeps the marble they would have placed, adding it to their score. In addition, the marble 7 marbles counter-clockwise from the current marble is removed from the circle and also added to the current player's score. The marble located immediately clockwise of the marble that was removed becomes the new current marble.
For example, suppose there are 9 players. After the marble with value 0 is placed in the middle, each player (shown in square brackets) takes a turn. The result of each of those turns would produce circles of marbles like this, where clockwise is to the right and the resulting current marble is in parentheses:
[-] (0)
[1] 0 (1)
[2] 0 (2) 1
[3] 0 2 1 (3)
[4] 0 (4) 2 1 3
[5] 0 4 2 (5) 1 3
[6] 0 4 2 5 1 (6) 3
[7] 0 4 2 5 1 6 3 (7)
[8] 0 (8) 4 2 5 1 6 3 7
[9] 0 8 4 (9) 2 5 1 6 3 7
[1] 0 8 4 9 2(10) 5 1 6 3 7
[2] 0 8 4 9 2 10 5(11) 1 6 3 7
[3] 0 8 4 9 2 10 5 11 1(12) 6 3 7
[4] 0 8 4 9 2 10 5 11 1 12 6(13) 3 7
[5] 0 8 4 9 2 10 5 11 1 12 6 13 3(14) 7
[6] 0 8 4 9 2 10 5 11 1 12 6 13 3 14 7(15)
[7] 0(16) 8 4 9 2 10 5 11 1 12 6 13 3 14 7 15
[8] 0 16 8(17) 4 9 2 10 5 11 1 12 6 13 3 14 7 15
[9] 0 16 8 17 4(18) 9 2 10 5 11 1 12 6 13 3 14 7 15
[1] 0 16 8 17 4 18 9(19) 2 10 5 11 1 12 6 13 3 14 7 15
[2] 0 16 8 17 4 18 9 19 2(20)10 5 11 1 12 6 13 3 14 7 15
[3] 0 16 8 17 4 18 9 19 2 20 10(21) 5 11 1 12 6 13 3 14 7 15
[4] 0 16 8 17 4 18 9 19 2 20 10 21 5(22)11 1 12 6 13 3 14 7 15
[5] 0 16 8 17 4 18(19) 2 20 10 21 5 22 11 1 12 6 13 3 14 7 15
[6] 0 16 8 17 4 18 19 2(24)20 10 21 5 22 11 1 12 6 13 3 14 7 15
[7] 0 16 8 17 4 18 19 2 24 20(25)10 21 5 22 11 1 12 6 13 3 14 7 15
The goal is to be the player with the highest score after the last marble is used up. Assuming the example above ends after the marble numbered 25, the winning score is 23+9=32 (because player 5 kept marble 23 and removed marble 9, while no other player got any points in this very short example game).
Here are a few more examples:
10 players; last marble is worth 1618 points: high score is 8317
13 players; last marble is worth 7999 points: high score is 146373
17 players; last marble is worth 1104 points: high score is 2764
21 players; last marble is worth 6111 points: high score is 54718
30 players; last marble is worth 5807 points: high score is 37305
What is the winning Elf's score?
--- Part Two ---
Amused by the speed of your answer, the Elves are curious:
What would the new winning Elf's score be if the number of the last marble were 100 times larger?
*/
public class Day09 : IDay
public string ResolvePart1(string[] inputs)
{
public string ResolvePart1(string[] inputs)
{
return CalculateHighScore(inputs[0], 1);
}
public string ResolvePart2(string[] inputs)
{
return CalculateHighScore(inputs[0], 100);
}
private static string CalculateHighScore(string input, long factor)
{
string[] parts = input.Split(new string[] { " players; last marble is worth ", " points" }, StringSplitOptions.RemoveEmptyEntries);
long numberOfPlayers = Convert.ToInt32(parts[0]);
long lastMarble = Convert.ToInt32(parts[1]) * factor;
MarbleGame marbleGame = new MarbleGame();
marbleGame.PlayGame(numberOfPlayers, lastMarble);
long result = marbleGame.GetHighScore();
return result.ToString();
}
return CalculateHighScore(inputs[0], 1);
}
public class Marble
public string ResolvePart2(string[] inputs)
{
public long Value { get; set; }
public Marble Previous { get; set; }
public Marble Next { get; set; }
return CalculateHighScore(inputs[0], 100);
}
public class MarbleGame
private static string CalculateHighScore(string input, long factor)
{
public Dictionary<long, long> Scores { get; } = new Dictionary<long, long>();
private Marble firstMarble;
private Marble currentMarble;
private long currentPlayer = 0;
private const long PointValueMultiple = 23;
public void PlayGame(long numPlayers, long lastMarble, bool showStatus = false)
{
Scores.Clear();
firstMarble = new Marble { Value = 0 };
firstMarble.Previous = firstMarble;
firstMarble.Next = firstMarble;
currentMarble = firstMarble;
for (long i = 1; i <= numPlayers; i++) { Scores.Add(i, 0); }
for (long i = 0; i <= lastMarble; i++)
{
if (showStatus) { PrintStatus(); }
currentPlayer = (i % numPlayers) + 1;
Marble newMarble = new Marble { Value = i + 1 };
if ((newMarble.Value % PointValueMultiple) > 0)
{
Marble previousMarble = currentMarble.Next;
Marble nextMarble = previousMarble.Next;
newMarble.Previous = previousMarble;
newMarble.Next = nextMarble;
previousMarble.Next = newMarble;
nextMarble.Previous = newMarble;
currentMarble = newMarble;
}
else
{
Marble marbleToRemove = currentMarble.Previous.Previous.Previous.Previous.Previous.Previous.Previous;
currentMarble = marbleToRemove.Next;
marbleToRemove.Previous.Next = marbleToRemove.Next;
marbleToRemove.Next.Previous = marbleToRemove.Previous;
long currentPlayerScore = Scores[currentPlayer] + (newMarble.Value + marbleToRemove.Value);
Scores[currentPlayer] = currentPlayerScore;
}
}
}
public void PrintStatus()
{
Console.Write("[{0}] ", currentPlayer);
Marble marble = firstMarble;
do
{
if (currentMarble.Value == marble.Value)
{
Console.Write("({0}) ", marble.Value);
}
else
{
Console.Write("{0} ", marble.Value);
}
marble = marble.Next;
} while (marble.Value != 0);
Console.WriteLine();
}
public long GetHighScore()
{
return Scores.Values.Max();
}
string[] parts = input.Split(new[] { " players; last marble is worth ", " points" }, StringSplitOptions.RemoveEmptyEntries);
long numberOfPlayers = Convert.ToInt32(parts[0]);
long lastMarble = Convert.ToInt32(parts[1]) * factor;
MarbleGame marbleGame = new();
marbleGame.PlayGame(numberOfPlayers, lastMarble);
long result = marbleGame.GetHighScore();
return result.ToString();
}
}
public class Marble
{
public long Value { get; set; }
public Marble Previous { get; set; }
public Marble Next { get; set; }
}
public class MarbleGame
{
public Dictionary<long, long> Scores { get; } = new();
private Marble _firstMarble;
private Marble _currentMarble;
private long _currentPlayer;
private const long PointValueMultiple = 23;
public void PlayGame(long numPlayers, long lastMarble, bool showStatus = false)
{
Scores.Clear();
_firstMarble = new Marble { Value = 0 };
_firstMarble.Previous = _firstMarble;
_firstMarble.Next = _firstMarble;
_currentMarble = _firstMarble;
for (long i = 1; i <= numPlayers; i++) { Scores.Add(i, 0); }
for (long i = 0; i <= lastMarble; i++)
{
if (showStatus) { PrintStatus(); }
_currentPlayer = (i % numPlayers) + 1;
Marble newMarble = new() { Value = i + 1 };
if ((newMarble.Value % PointValueMultiple) > 0)
{
Marble previousMarble = _currentMarble.Next;
Marble nextMarble = previousMarble.Next;
newMarble.Previous = previousMarble;
newMarble.Next = nextMarble;
previousMarble.Next = newMarble;
nextMarble.Previous = newMarble;
_currentMarble = newMarble;
}
else
{
Marble marbleToRemove = _currentMarble.Previous.Previous.Previous.Previous.Previous.Previous.Previous;
_currentMarble = marbleToRemove.Next;
marbleToRemove.Previous.Next = marbleToRemove.Next;
marbleToRemove.Next.Previous = marbleToRemove.Previous;
long currentPlayerScore = Scores[_currentPlayer] + (newMarble.Value + marbleToRemove.Value);
Scores[_currentPlayer] = currentPlayerScore;
}
}
}
public void PrintStatus()
{
Console.Write("[{0}] ", _currentPlayer);
Marble marble = _firstMarble;
do
{
if (_currentMarble.Value == marble.Value)
{
Console.Write("({0}) ", marble.Value);
}
else
{
Console.Write("{0} ", marble.Value);
}
marble = marble.Next;
} while (marble.Value != 0);
Console.WriteLine();
}
public long GetHighScore()
{
return Scores.Values.Max();
}
}

557
AdventOfCode2018/Day10.cs
View File

@@ -3,308 +3,305 @@ using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 10: The Stars Align ---
It's no use; your navigation system simply isn't capable of providing walking directions in the arctic circle, and certainly not in 1018.
The Elves suggest an alternative. In times like these, North Pole rescue operations will arrange points of light in the sky to guide missing Elves back to base. Unfortunately, the message is easy to miss: the points move slowly enough that it takes hours to align them, but have so much momentum that they only stay aligned for a second. If you blink at the wrong time, it might be hours before another message appears.
You can see these points of light floating in the distance, and record their position in the sky and their velocity, the relative change in position per second (your puzzle input). The coordinates are all given from your perspective; given enough time, those positions and velocities will move the points into a cohesive message!
Rather than wait, you decide to fast-forward the process and calculate what the points will eventually spell.
For example, suppose you note the following points:
position=< 9, 1> velocity=< 0, 2>
position=< 7, 0> velocity=<-1, 0>
position=< 3, -2> velocity=<-1, 1>
position=< 6, 10> velocity=<-2, -1>
position=< 2, -4> velocity=< 2, 2>
position=<-6, 10> velocity=< 2, -2>
position=< 1, 8> velocity=< 1, -1>
position=< 1, 7> velocity=< 1, 0>
position=<-3, 11> velocity=< 1, -2>
position=< 7, 6> velocity=<-1, -1>
position=<-2, 3> velocity=< 1, 0>
position=<-4, 3> velocity=< 2, 0>
position=<10, -3> velocity=<-1, 1>
position=< 5, 11> velocity=< 1, -2>
position=< 4, 7> velocity=< 0, -1>
position=< 8, -2> velocity=< 0, 1>
position=<15, 0> velocity=<-2, 0>
position=< 1, 6> velocity=< 1, 0>
position=< 8, 9> velocity=< 0, -1>
position=< 3, 3> velocity=<-1, 1>
position=< 0, 5> velocity=< 0, -1>
position=<-2, 2> velocity=< 2, 0>
position=< 5, -2> velocity=< 1, 2>
position=< 1, 4> velocity=< 2, 1>
position=<-2, 7> velocity=< 2, -2>
position=< 3, 6> velocity=<-1, -1>
position=< 5, 0> velocity=< 1, 0>
position=<-6, 0> velocity=< 2, 0>
position=< 5, 9> velocity=< 1, -2>
position=<14, 7> velocity=<-2, 0>
position=<-3, 6> velocity=< 2, -1>
Each line represents one point. Positions are given as <X, Y> pairs: X represents how far left (negative) or right (positive) the point appears, while Y represents how far up (negative) or down (positive) the point appears.
At 0 seconds, each point has the position given. Each second, each point's velocity is added to its position. So, a point with velocity <1, -2> is moving to the right, but is moving upward twice as quickly. If this point's initial position were <3, 9>, after 3 seconds, its position would become <6, 3>.
Over time, the points listed above would move like this:
Initially:
........#.............
................#.....
.........#.#..#.......
......................
#..........#.#.......#
...............#......
....#.................
..#.#....#............
.......#..............
......#...............
...#...#.#...#........
....#..#..#.........#.
.......#..............
...........#..#.......
#...........#.........
...#.......#..........
After 1 second:
......................
......................
..........#....#......
........#.....#.......
..#.........#......#..
......................
......#...............
....##.........#......
......#.#.............
.....##.##..#.........
........#.#...........
........#...#.....#...
..#...........#.......
....#.....#.#.........
......................
......................
After 2 seconds:
......................
......................
......................
..............#.......
....#..#...####..#....
......................
........#....#........
......#.#.............
.......#...#..........
.......#..#..#.#......
....#....#.#..........
.....#...#...##.#.....
........#.............
......................
......................
......................
After 3 seconds:
......................
......................
......................
......................
......#...#..###......
......#...#...#.......
......#...#...#.......
......#####...#.......
......#...#...#.......
......#...#...#.......
......#...#...#.......
......#...#..###......
......................
......................
......................
......................
After 4 seconds:
......................
......................
......................
............#.........
........##...#.#......
......#.....#..#......
.....#..##.##.#.......
.......##.#....#......
...........#....#.....
..............#.......
....#......#...#......
.....#.....##.........
...............#......
...............#......
......................
......................
After 3 seconds, the message appeared briefly: HI. Of course, your message will be much longer and will take many more seconds to appear.
What message will eventually appear in the sky?
--- Part Two ---
Good thing you didn't have to wait, because that would have taken a long time - much longer than the 3 seconds in the example above.
Impressed by your sub-hour communication capabilities, the Elves are curious: exactly how many seconds would they have needed to wait for that message to appear?
*/
public class Day10 : IDay
{
/*
--- Day 10: The Stars Align ---
public int Width { get; set; } = 65;
public int Height { get; set; } = 12;
It's no use; your navigation system simply isn't capable of providing walking directions in the arctic circle, and certainly not in 1018.
The Elves suggest an alternative. In times like these, North Pole rescue operations will arrange points of light in the sky to guide missing Elves back to base. Unfortunately, the message is easy to miss: the points move slowly enough that it takes hours to align them, but have so much momentum that they only stay aligned for a second. If you blink at the wrong time, it might be hours before another message appears.
You can see these points of light floating in the distance, and record their position in the sky and their velocity, the relative change in position per second (your puzzle input). The coordinates are all given from your perspective; given enough time, those positions and velocities will move the points into a cohesive message!
Rather than wait, you decide to fast-forward the process and calculate what the points will eventually spell.
For example, suppose you note the following points:
position=< 9, 1> velocity=< 0, 2>
position=< 7, 0> velocity=<-1, 0>
position=< 3, -2> velocity=<-1, 1>
position=< 6, 10> velocity=<-2, -1>
position=< 2, -4> velocity=< 2, 2>
position=<-6, 10> velocity=< 2, -2>
position=< 1, 8> velocity=< 1, -1>
position=< 1, 7> velocity=< 1, 0>
position=<-3, 11> velocity=< 1, -2>
position=< 7, 6> velocity=<-1, -1>
position=<-2, 3> velocity=< 1, 0>
position=<-4, 3> velocity=< 2, 0>
position=<10, -3> velocity=<-1, 1>
position=< 5, 11> velocity=< 1, -2>
position=< 4, 7> velocity=< 0, -1>
position=< 8, -2> velocity=< 0, 1>
position=<15, 0> velocity=<-2, 0>
position=< 1, 6> velocity=< 1, 0>
position=< 8, 9> velocity=< 0, -1>
position=< 3, 3> velocity=<-1, 1>
position=< 0, 5> velocity=< 0, -1>
position=<-2, 2> velocity=< 2, 0>
position=< 5, -2> velocity=< 1, 2>
position=< 1, 4> velocity=< 2, 1>
position=<-2, 7> velocity=< 2, -2>
position=< 3, 6> velocity=<-1, -1>
position=< 5, 0> velocity=< 1, 0>
position=<-6, 0> velocity=< 2, 0>
position=< 5, 9> velocity=< 1, -2>
position=<14, 7> velocity=<-2, 0>
position=<-3, 6> velocity=< 2, -1>
Each line represents one point. Positions are given as <X, Y> pairs: X represents how far left (negative) or right (positive) the point appears, while Y represents how far up (negative) or down (positive) the point appears.
At 0 seconds, each point has the position given. Each second, each point's velocity is added to its position. So, a point with velocity <1, -2> is moving to the right, but is moving upward twice as quickly. If this point's initial position were <3, 9>, after 3 seconds, its position would become <6, 3>.
Over time, the points listed above would move like this:
Initially:
........#.............
................#.....
.........#.#..#.......
......................
#..........#.#.......#
...............#......
....#.................
..#.#....#............
.......#..............
......#...............
...#...#.#...#........
....#..#..#.........#.
.......#..............
...........#..#.......
#...........#.........
...#.......#..........
After 1 second:
......................
......................
..........#....#......
........#.....#.......
..#.........#......#..
......................
......#...............
....##.........#......
......#.#.............
.....##.##..#.........
........#.#...........
........#...#.....#...
..#...........#.......
....#.....#.#.........
......................
......................
After 2 seconds:
......................
......................
......................
..............#.......
....#..#...####..#....
......................
........#....#........
......#.#.............
.......#...#..........
.......#..#..#.#......
....#....#.#..........
.....#...#...##.#.....
........#.............
......................
......................
......................
After 3 seconds:
......................
......................
......................
......................
......#...#..###......
......#...#...#.......
......#...#...#.......
......#####...#.......
......#...#...#.......
......#...#...#.......
......#...#...#.......
......#...#..###......
......................
......................
......................
......................
After 4 seconds:
......................
......................
......................
............#.........
........##...#.#......
......#.....#..#......
.....#..##.##.#.......
.......##.#....#......
...........#....#.....
..............#.......
....#......#...#......
.....#.....##.........
...............#......
...............#......
......................
......................
After 3 seconds, the message appeared briefly: HI. Of course, your message will be much longer and will take many more seconds to appear.
What message will eventually appear in the sky?
--- Part Two ---
Good thing you didn't have to wait, because that would have taken a long time - much longer than the 3 seconds in the example above.
Impressed by your sub-hour communication capabilities, the Elves are curious: exactly how many seconds would they have needed to wait for that message to appear?
*/
public class Day10 : IDay
public string ResolvePart1(string[] inputs)
{
public int Width { get; set; } = 65;
public int Height { get; set; } = 12;
public string ResolvePart1(string[] inputs)
{
LightField lightField = new LightField(inputs);
int t = lightField.SearchSmallerBoundindBox();
string result = lightField.Render(t, Width, Height);
return result;
}
public string ResolvePart2(string[] inputs)
{
LightField lightField = new LightField(inputs);
int t = lightField.SearchSmallerBoundindBox();
return t.ToString();
}
LightField lightField = new(inputs);
int t = lightField.SearchSmallerBoundindBox();
string result = lightField.Render(t, Width, Height);
return result;
}
public class LightPoint
public string ResolvePart2(string[] inputs)
{
public int X { get; set; }
public int Y { get; set; }
public int VX { get; set; }
public int VY { get; set; }
LightField lightField = new(inputs);
int t = lightField.SearchSmallerBoundindBox();
return t.ToString();
}
}
public static LightPoint FromString(string strPoint)
{
string[] parts = strPoint.Split(new string[] { "position=<", " ", ",", "> velocity=<", ">" }, StringSplitOptions.RemoveEmptyEntries);
LightPoint point = new LightPoint
{
X = Convert.ToInt32(parts[0]),
Y = Convert.ToInt32(parts[1]),
VX = Convert.ToInt32(parts[2]),
VY = Convert.ToInt32(parts[3]),
};
return point;
}
public class LightPoint
{
public int X { get; set; }
public int Y { get; set; }
public int VX { get; set; }
public int VY { get; set; }
public int GetX(int t)
{
return X + (VX * t);
}
public int GetY(int t)
{
return Y + (VY * t);
}
public static LightPoint FromString(string strPoint)
{
string[] parts = strPoint.Split(new[] { "position=<", " ", ",", "> velocity=<", ">" }, StringSplitOptions.RemoveEmptyEntries);
LightPoint point = new() {
X = Convert.ToInt32(parts[0]),
Y = Convert.ToInt32(parts[1]),
VX = Convert.ToInt32(parts[2]),
VY = Convert.ToInt32(parts[3]),
};
return point;
}
public class LightField
public int GetX(int t)
{
private List<LightPoint> points;
return X + (VX * t);
}
public LightField(string[] strPoints)
public int GetY(int t)
{
return Y + (VY * t);
}
}
public class LightField
{
private readonly List<LightPoint> _points;
public LightField(string[] strPoints)
{
_points = strPoints.Select(strPoint => LightPoint.FromString(strPoint)).ToList();
}
public int SearchSmallerBoundindBox()
{
int minHeight = int.MaxValue;
int minT = 0;
int missCount = 0;
int t = 0;
do
{
points = strPoints.Select(strPoint => LightPoint.FromString(strPoint)).ToList();
int minY = int.MaxValue;
int maxY = int.MinValue;
foreach (LightPoint point in _points)
{
int y = point.GetY(t);
if (y < minY) { minY = y; }
if (y > maxY) { maxY = y; }
}
int height = (maxY - minY);
if (height < minHeight)
{
minHeight = height;
minT = t;
missCount = 0;
}
else
{
missCount++;
}
t++;
} while (missCount < 1000);
return minT;
}
public string Render(int t, int width, int height)
{
int minX = int.MaxValue;
int minY = int.MaxValue;
int maxX = int.MinValue;
int maxY = int.MinValue;
foreach (LightPoint point in _points)
{
int x = point.GetX(t);
int y = point.GetY(t);
if (x < minX) { minX = x; }
if (y < minY) { minY = y; }
if (x > maxX) { maxX = x; }
if (y > maxY) { maxY = y; }
}
minX--;
minY--;
maxX += 2;
maxY += 2;
double scaleX = (maxX - minX) / (double)width;
double scaleY = (maxY - minY) / (double)height;
int[,] field = new int[width, height];
foreach (LightPoint point in _points)
{
int x = point.GetX(t);
int y = point.GetY(t);
x = (int)(((x - minX) / scaleX) + 0.5);
y = (int)(((y - minY) / scaleY) + 0.5);
if (x < 0 || x >= width) { continue; }
if (y < 0 || y >= height) { continue; }
field[x, y]++;
}
public int SearchSmallerBoundindBox()
StringBuilder sb = new();
for (int j = 0; j < height; j++)
{
int minHeight = int.MaxValue;
int minT = 0;
int missCount = 0;
int t = 0;
do
sb.AppendLine();
for (int i = 0; i < width; i++)
{
int minY = int.MaxValue;
int maxY = int.MinValue;
foreach (LightPoint point in points)
if (field[i, j] > 0)
{
int y = point.GetY(t);
if (y < minY) { minY = y; }
if (y > maxY) { maxY = y; }
}
int height = (maxY - minY);
if (height < minHeight)
{
minHeight = height;
minT = t;
missCount = 0;
sb.Append("#");
}
else
{
missCount++;
}
t++;
} while (missCount < 1000);
return minT;
}
public string Render(int t, int width, int height)
{
int minX = int.MaxValue;
int minY = int.MaxValue;
int maxX = int.MinValue;
int maxY = int.MinValue;
foreach (LightPoint point in points)
{
int x = point.GetX(t);
int y = point.GetY(t);
if (x < minX) { minX = x; }
if (y < minY) { minY = y; }
if (x > maxX) { maxX = x; }
if (y > maxY) { maxY = y; }
}
minX--;
minY--;
maxX += 2;
maxY += 2;
double scaleX = (maxX - minX) / (double)width;
double scaleY = (maxY - minY) / (double)height;
int[,] field = new int[width, height];
foreach (LightPoint point in points)
{
int x = point.GetX(t);
int y = point.GetY(t);
x = (int)(((x - minX) / scaleX) + 0.5);
y = (int)(((y - minY) / scaleY) + 0.5);
if (x < 0 || x >= width) { continue; }
if (y < 0 || y >= height) { continue; }
field[x, y]++;
}
StringBuilder sb = new StringBuilder();
for (int j = 0; j < height; j++)
{
sb.AppendLine();
for (int i = 0; i < width; i++)
{
if (field[i, j] > 0)
{
sb.Append("#");
}
else
{
sb.Append(".");
}
sb.Append(".");
}
}
return sb.ToString();
}
return sb.ToString();
}
}
}

396
AdventOfCode2018/Day11.cs
View File

@@ -1,226 +1,224 @@
using System;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 11: Chronal Charge ---
You watch the Elves and their sleigh fade into the distance as they head toward the North Pole.
Actually, you're the one fading. The falling sensation returns.
The low fuel warning light is illuminated on your wrist-mounted device. Tapping it once causes it to project a hologram of the situation: a 300x300 grid of fuel cells and their current power levels, some negative. You're not sure what negative power means in the context of time travel, but it can't be good.
Each fuel cell has a coordinate ranging from 1 to 300 in both the X (horizontal) and Y (vertical) direction. In X,Y notation, the top-left cell is 1,1, and the top-right cell is 300,1.
The interface lets you select any 3x3 square of fuel cells. To increase your chances of getting to your destination, you decide to choose the 3x3 square with the largest total power.
The power level in a given fuel cell can be found through the following process:
Find the fuel cell's rack ID, which is its X coordinate plus 10.
Begin with a power level of the rack ID times the Y coordinate.
Increase the power level by the value of the grid serial number (your puzzle input).
Set the power level to itself multiplied by the rack ID.
Keep only the hundreds digit of the power level (so 12345 becomes 3; numbers with no hundreds digit become 0).
Subtract 5 from the power level.
For example, to find the power level of the fuel cell at 3,5 in a grid with serial number 8:
The rack ID is 3 + 10 = 13.
The power level starts at 13 * 5 = 65.
Adding the serial number produces 65 + 8 = 73.
Multiplying by the rack ID produces 73 * 13 = 949.
The hundreds digit of 949 is 9.
Subtracting 5 produces 9 - 5 = 4.
So, the power level of this fuel cell is 4.
Here are some more example power levels:
Fuel cell at 122,79, grid serial number 57: power level -5.
Fuel cell at 217,196, grid serial number 39: power level 0.
Fuel cell at 101,153, grid serial number 71: power level 4.
Your goal is to find the 3x3 square which has the largest total power. The square must be entirely within the 300x300 grid. Identify this square using the X,Y coordinate of its top-left fuel cell. For example:
For grid serial number 18, the largest total 3x3 square has a top-left corner of 33,45 (with a total power of 29); these fuel cells appear in the middle of this 5x5 region:
-2 -4 4 4 4
-4 4 4 4 -5
4 3 3 4 -4
1 1 2 4 -3
-1 0 2 -5 -2
For grid serial number 42, the largest 3x3 square's top-left is 21,61 (with a total power of 30); they are in the middle of this region:
-3 4 2 2 2
-4 4 3 3 4
-5 3 3 4 -4
4 3 3 4 -3
3 3 3 -5 -1
What is the X,Y coordinate of the top-left fuel cell of the 3x3 square with the largest total power?
--- Part Two ---
You discover a dial on the side of the device; it seems to let you select a square of any size, not just 3x3. Sizes from 1x1 to 300x300 are supported.
Realizing this, you now must find the square of any size with the largest total power. Identify this square by including its size as a third parameter after the top-left coordinate: a 9x9 square with a top-left corner of 3,5 is identified as 3,5,9.
For example:
For grid serial number 18, the largest total square (with a total power of 113) is 16x16 and has a top-left corner of 90,269, so its identifier is 90,269,16.
For grid serial number 42, the largest total square (with a total power of 119) is 12x12 and has a top-left corner of 232,251, so its identifier is 232,251,12.
What is the X,Y,size identifier of the square with the largest total power?
*/
public class Day11 : IDay
{
/*
--- Day 11: Chronal Charge ---
You watch the Elves and their sleigh fade into the distance as they head toward the North Pole.
Actually, you're the one fading. The falling sensation returns.
The low fuel warning light is illuminated on your wrist-mounted device. Tapping it once causes it to project a hologram of the situation: a 300x300 grid of fuel cells and their current power levels, some negative. You're not sure what negative power means in the context of time travel, but it can't be good.
Each fuel cell has a coordinate ranging from 1 to 300 in both the X (horizontal) and Y (vertical) direction. In X,Y notation, the top-left cell is 1,1, and the top-right cell is 300,1.
The interface lets you select any 3x3 square of fuel cells. To increase your chances of getting to your destination, you decide to choose the 3x3 square with the largest total power.
The power level in a given fuel cell can be found through the following process:
Find the fuel cell's rack ID, which is its X coordinate plus 10.
Begin with a power level of the rack ID times the Y coordinate.
Increase the power level by the value of the grid serial number (your puzzle input).
Set the power level to itself multiplied by the rack ID.
Keep only the hundreds digit of the power level (so 12345 becomes 3; numbers with no hundreds digit become 0).
Subtract 5 from the power level.
For example, to find the power level of the fuel cell at 3,5 in a grid with serial number 8:
The rack ID is 3 + 10 = 13.
The power level starts at 13 * 5 = 65.
Adding the serial number produces 65 + 8 = 73.
Multiplying by the rack ID produces 73 * 13 = 949.
The hundreds digit of 949 is 9.
Subtracting 5 produces 9 - 5 = 4.
So, the power level of this fuel cell is 4.
Here are some more example power levels:
Fuel cell at 122,79, grid serial number 57: power level -5.
Fuel cell at 217,196, grid serial number 39: power level 0.
Fuel cell at 101,153, grid serial number 71: power level 4.
Your goal is to find the 3x3 square which has the largest total power. The square must be entirely within the 300x300 grid. Identify this square using the X,Y coordinate of its top-left fuel cell. For example:
For grid serial number 18, the largest total 3x3 square has a top-left corner of 33,45 (with a total power of 29); these fuel cells appear in the middle of this 5x5 region:
-2 -4 4 4 4
-4 4 4 4 -5
4 3 3 4 -4
1 1 2 4 -3
-1 0 2 -5 -2
For grid serial number 42, the largest 3x3 square's top-left is 21,61 (with a total power of 30); they are in the middle of this region:
-3 4 2 2 2
-4 4 3 3 4
-5 3 3 4 -4
4 3 3 4 -3
3 3 3 -5 -1
What is the X,Y coordinate of the top-left fuel cell of the 3x3 square with the largest total power?
--- Part Two ---
You discover a dial on the side of the device; it seems to let you select a square of any size, not just 3x3. Sizes from 1x1 to 300x300 are supported.
Realizing this, you now must find the square of any size with the largest total power. Identify this square by including its size as a third parameter after the top-left coordinate: a 9x9 square with a top-left corner of 3,5 is identified as 3,5,9.
For example:
For grid serial number 18, the largest total square (with a total power of 113) is 16x16 and has a top-left corner of 90,269, so its identifier is 90,269,16.
For grid serial number 42, the largest total square (with a total power of 119) is 12x12 and has a top-left corner of 232,251, so its identifier is 232,251,12.
What is the X,Y,size identifier of the square with the largest total power?
*/
public class Day11 : IDay
public string ResolvePart1(string[] inputs)
{
public string ResolvePart1(string[] inputs)
{
int serial = Convert.ToInt32(inputs[0]);
SearchBestRegionOfOneSize(300, 300, 3, serial, out int x, out int y);
return string.Format("{0},{1}", x, y);
}
int serial = Convert.ToInt32(inputs[0]);
SearchBestRegionOfOneSize(300, 300, 3, serial, out int x, out int y);
return $"{x},{y}";
}
public string ResolvePart2(string[] inputs)
{
int serial = Convert.ToInt32(inputs[0]);
SearchBestRegion(300, 300, serial, out int x, out int y, out int size);
return string.Format("{0},{1},{2}", x, y, size);
}
public string ResolvePart2(string[] inputs)
{
int serial = Convert.ToInt32(inputs[0]);
SearchBestRegion(300, 300, serial, out int x, out int y, out int size);
return $"{x},{y},{size}";
}
public static int CalculatePowerLevelOfCell(int x, int y, int serial)
{
int powerLevel = 0;
int rackID = x + 10;
powerLevel += rackID * y;
powerLevel += serial;
powerLevel = powerLevel * rackID;
powerLevel = (powerLevel / 100) % 10;
powerLevel -= 5;
return powerLevel;
}
public static int CalculatePowerLevelOfCell(int x, int y, int serial)
{
int powerLevel = 0;
int rackID = x + 10;
powerLevel += rackID * y;
powerLevel += serial;
powerLevel = powerLevel * rackID;
powerLevel = (powerLevel / 100) % 10;
powerLevel -= 5;
return powerLevel;
}
public static int CalculatePowerLevelOfRegion(int x, int y, int size, int serial)
public static int CalculatePowerLevelOfRegion(int x, int y, int size, int serial)
{
int powerLevel = 0;
for (int i = 0; i < size; i++)
{
int powerLevel = 0;
for (int i = 0; i < size; i++)
for (int j = 0; j < size; j++)
{
for (int j = 0; j < size; j++)
powerLevel += CalculatePowerLevelOfCell(x + i, y + j, serial);
}
}
return powerLevel;
}
public static void SearchBestRegionOfOneSize(int width, int height, int size, int serial, out int x, out int y)
{
int bestPowerLevel = int.MinValue;
int bestX = 0;
int bestY = 0;
for (int i = 0; i <= width - size; i++)
{
for (int j = 0; j <= height - size; j++)
{
int powerLevel = CalculatePowerLevelOfRegion(i, j, size, serial);
if (powerLevel > bestPowerLevel)
{
powerLevel += CalculatePowerLevelOfCell(x + i, y + j, serial);
bestPowerLevel = powerLevel;
bestX = i;
bestY = j;
}
}
return powerLevel;
}
x = bestX;
y = bestY;
}
public static void SearchBestRegionOfOneSize(int width, int height, int size, int serial, out int x, out int y)
public static int[,] GenerateSumationField(int width, int height, int serial)
{
int[,] field = new int[width, height];
for (int j = 0; j < height; j++)
{
int bestPowerLevel = int.MinValue;
int bestX = 0;
int bestY = 0;
for (int i = 0; i <= width - size; i++)
for (int i = 0; i < width; i++)
{
for (int j = 0; j <= height - size; j++)
int powerLevel = CalculatePowerLevelOfCell(i, j, serial);
field[i, j] = powerLevel;
}
}
int[,] summationFiled = new int[width, height];
for (int j = 0; j < height; j++)
{
for (int i = 0; i < width; i++)
{
int sum = field[i, j];
if (j > 0)
{
int powerLevel = CalculatePowerLevelOfRegion(i, j, size, serial);
sum += summationFiled[i, j - 1];
}
if (i > 0)
{
sum += summationFiled[i - 1, j];
}
if (i > 0 && j > 0)
{
sum -= summationFiled[i - 1, j - 1];
}
summationFiled[i, j] = sum;
}
}
return summationFiled;
}
public static int CalculatePowerLevelOfRegion(int x, int y, int size, int[,] summationFiled)
{
int powerLevel = summationFiled[x + (size - 1), y + (size - 1)];
if (x > 0 && y > 0)
{
powerLevel += summationFiled[x - 1, y - 1];
}
if (x > 0)
{
powerLevel -= summationFiled[x - 1, y + (size - 1)];
}
if (y > 0)
{
powerLevel -= summationFiled[x + (size - 1), y - 1];
}
return powerLevel;
}
public static void SearchBestRegion(int width, int height, int serial, out int x, out int y, out int size)
{
int[,] summationFiled = GenerateSumationField(width, height, serial);
int bestPowerLevel = int.MinValue;
int bestX = 0;
int bestY = 0;
int bestSize = 0;
int maxSize = Math.Min(width, height);
for (int k = 1; k <= maxSize; k++)
{
for (int i = 0; i <= width - k; i++)
{
for (int j = 0; j <= height - k; j++)
{
int powerLevel = CalculatePowerLevelOfRegion(i, j, k, summationFiled);
if (powerLevel > bestPowerLevel)
{
bestPowerLevel = powerLevel;
bestX = i;
bestY = j;
bestSize = k;
}
}
}
x = bestX;
y = bestY;
}
public static int[,] GenerateSumationField(int width, int height, int serial)
{
int[,] field = new int[width, height];
for (int j = 0; j < height; j++)
{
for (int i = 0; i < width; i++)
{
int powerLevel = CalculatePowerLevelOfCell(i, j, serial);
field[i, j] = powerLevel;
}
}
int[,] summationFiled = new int[width, height];
for (int j = 0; j < height; j++)
{
for (int i = 0; i < width; i++)
{
int sum = field[i, j];
if (j > 0)
{
sum += summationFiled[i, j - 1];
}
if (i > 0)
{
sum += summationFiled[i - 1, j];
}
if (i > 0 && j > 0)
{
sum -= summationFiled[i - 1, j - 1];
}
summationFiled[i, j] = sum;
}
}
return summationFiled;
}
public static int CalculatePowerLevelOfRegion(int x, int y, int size, int[,] summationFiled)
{
int powerLevel = summationFiled[x + (size - 1), y + (size - 1)];
if (x > 0 && y > 0)
{
powerLevel += summationFiled[x - 1, y - 1];
}
if (x > 0)
{
powerLevel -= summationFiled[x - 1, y + (size - 1)];
}
if (y > 0)
{
powerLevel -= summationFiled[x + (size - 1), y - 1];
}
return powerLevel;
}
public static void SearchBestRegion(int width, int height, int serial, out int x, out int y, out int size)
{
int[,] summationFiled = GenerateSumationField(width, height, serial);
int bestPowerLevel = int.MinValue;
int bestX = 0;
int bestY = 0;
int bestSize = 0;
int maxSize = Math.Min(width, height);
for (int k = 1; k <= maxSize; k++)
{
for (int i = 0; i <= width - k; i++)
{
for (int j = 0; j <= height - k; j++)
{
int powerLevel = CalculatePowerLevelOfRegion(i, j, k, summationFiled);
if (powerLevel > bestPowerLevel)
{
bestPowerLevel = powerLevel;
bestX = i;
bestY = j;
bestSize = k;
}
}
}
}
x = bestX;
y = bestY;
size = bestSize;
}
x = bestX;
y = bestY;
size = bestSize;
}
}
}

476
AdventOfCode2018/Day12.cs
View File

@@ -1,283 +1,281 @@
using System;
using System.Collections.Generic;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 12: Subterranean Sustainability ---
The year 518 is significantly more underground than your history books implied. Either that, or you've arrived in a vast cavern network under the North Pole.
After exploring a little, you discover a long tunnel that contains a row of small pots as far as you can see to your left and right. A few of them contain plants - someone is trying to grow things in these geothermally-heated caves.
The pots are numbered, with 0 in front of you. To the left, the pots are numbered -1, -2, -3, and so on; to the right, 1, 2, 3.... Your puzzle input contains a list of pots from 0 to the right and whether they do (#) or do not (.) currently contain a plant, the initial state. (No other pots currently contain plants.) For example, an initial state of #..##.... indicates that pots 0, 3, and 4 currently contain plants.
Your puzzle input also contains some notes you find on a nearby table: someone has been trying to figure out how these plants spread to nearby pots. Based on the notes, for each generation of plants, a given pot has or does not have a plant based on whether that pot (and the two pots on either side of it) had a plant in the last generation. These are written as LLCRR => N, where L are pots to the left, C is the current pot being considered, R are the pots to the right, and N is whether the current pot will have a plant in the next generation. For example:
A note like ..#.. => . means that a pot that contains a plant but with no plants within two pots of it will not have a plant in it during the next generation.
A note like ##.## => . means that an empty pot with two plants on each side of it will remain empty in the next generation.
A note like .##.# => # means that a pot has a plant in a given generation if, in the previous generation, there were plants in that pot, the one immediately to the left, and the one two pots to the right, but not in the ones immediately to the right and two to the left.
It's not clear what these plants are for, but you're sure it's important, so you'd like to make sure the current configuration of plants is sustainable by determining what will happen after 20 generations.
For example, given the following input:
initial state: #..#.#..##......###...###
...## => #
..#.. => #
.#... => #
.#.#. => #
.#.## => #
.##.. => #
.#### => #
#.#.# => #
#.### => #
##.#. => #
##.## => #
###.. => #
###.# => #
####. => #
For brevity, in this example, only the combinations which do produce a plant are listed. (Your input includes all possible combinations.) Then, the next 20 generations will look like this:
1 2 3
0 0 0 0
0: ...#..#.#..##......###...###...........
1: ...#...#....#.....#..#..#..#...........
2: ...##..##...##....#..#..#..##..........
3: ..#.#...#..#.#....#..#..#...#..........
4: ...#.#..#...#.#...#..#..##..##.........
5: ....#...##...#.#..#..#...#...#.........
6: ....##.#.#....#...#..##..##..##........
7: ...#..###.#...##..#...#...#...#........
8: ...#....##.#.#.#..##..##..##..##.......
9: ...##..#..#####....#...#...#...#.......
10: ..#.#..#...#.##....##..##..##..##......
11: ...#...##...#.#...#.#...#...#...#......
12: ...##.#.#....#.#...#.#..##..##..##.....
13: ..#..###.#....#.#...#....#...#...#.....
14: ..#....##.#....#.#..##...##..##..##....
15: ..##..#..#.#....#....#..#.#...#...#....
16: .#.#..#...#.#...##...#...#.#..##..##...
17: ..#...##...#.#.#.#...##...#....#...#...
18: ..##.#.#....#####.#.#.#...##...##..##..
19: .#..###.#..#.#.#######.#.#.#..#.#...#..
20: .#....##....#####...#######....#.#..##.
The generation is shown along the left, where 0 is the initial state. The pot numbers are shown along the top, where 0 labels the center pot, negative-numbered pots extend to the left, and positive pots extend toward the right. Remember, the initial state begins at pot 0, which is not the leftmost pot used in this example.
After one generation, only seven plants remain. The one in pot 0 matched the rule looking for ..#.., the one in pot 4 matched the rule looking for .#.#., pot 9 matched .##.., and so on.
In this example, after 20 generations, the pots shown as # contain plants, the furthest left of which is pot -2, and the furthest right of which is pot 34. Adding up all the numbers of plant-containing pots after the 20th generation produces 325.
After 20 generations, what is the sum of the numbers of all pots which contain a plant?
--- Part Two ---
You realize that 20 generations aren't enough. After all, these plants will need to last another 1500 years to even reach your timeline, not to mention your future.
After fifty billion (50000000000) generations, what is the sum of the numbers of all pots which contain a plant?
*/
public class Day12 : IDay
{
/*
--- Day 12: Subterranean Sustainability ---
The year 518 is significantly more underground than your history books implied. Either that, or you've arrived in a vast cavern network under the North Pole.
After exploring a little, you discover a long tunnel that contains a row of small pots as far as you can see to your left and right. A few of them contain plants - someone is trying to grow things in these geothermally-heated caves.
The pots are numbered, with 0 in front of you. To the left, the pots are numbered -1, -2, -3, and so on; to the right, 1, 2, 3.... Your puzzle input contains a list of pots from 0 to the right and whether they do (#) or do not (.) currently contain a plant, the initial state. (No other pots currently contain plants.) For example, an initial state of #..##.... indicates that pots 0, 3, and 4 currently contain plants.
Your puzzle input also contains some notes you find on a nearby table: someone has been trying to figure out how these plants spread to nearby pots. Based on the notes, for each generation of plants, a given pot has or does not have a plant based on whether that pot (and the two pots on either side of it) had a plant in the last generation. These are written as LLCRR => N, where L are pots to the left, C is the current pot being considered, R are the pots to the right, and N is whether the current pot will have a plant in the next generation. For example:
A note like ..#.. => . means that a pot that contains a plant but with no plants within two pots of it will not have a plant in it during the next generation.
A note like ##.## => . means that an empty pot with two plants on each side of it will remain empty in the next generation.
A note like .##.# => # means that a pot has a plant in a given generation if, in the previous generation, there were plants in that pot, the one immediately to the left, and the one two pots to the right, but not in the ones immediately to the right and two to the left.
It's not clear what these plants are for, but you're sure it's important, so you'd like to make sure the current configuration of plants is sustainable by determining what will happen after 20 generations.
For example, given the following input:
initial state: #..#.#..##......###...###
...## => #
..#.. => #
.#... => #
.#.#. => #
.#.## => #
.##.. => #
.#### => #
#.#.# => #
#.### => #
##.#. => #
##.## => #
###.. => #
###.# => #
####. => #
For brevity, in this example, only the combinations which do produce a plant are listed. (Your input includes all possible combinations.) Then, the next 20 generations will look like this:
1 2 3
0 0 0 0
0: ...#..#.#..##......###...###...........
1: ...#...#....#.....#..#..#..#...........
2: ...##..##...##....#..#..#..##..........
3: ..#.#...#..#.#....#..#..#...#..........
4: ...#.#..#...#.#...#..#..##..##.........
5: ....#...##...#.#..#..#...#...#.........
6: ....##.#.#....#...#..##..##..##........
7: ...#..###.#...##..#...#...#...#........
8: ...#....##.#.#.#..##..##..##..##.......
9: ...##..#..#####....#...#...#...#.......
10: ..#.#..#...#.##....##..##..##..##......
11: ...#...##...#.#...#.#...#...#...#......
12: ...##.#.#....#.#...#.#..##..##..##.....
13: ..#..###.#....#.#...#....#...#...#.....
14: ..#....##.#....#.#..##...##..##..##....
15: ..##..#..#.#....#....#..#.#...#...#....
16: .#.#..#...#.#...##...#...#.#..##..##...
17: ..#...##...#.#.#.#...##...#....#...#...
18: ..##.#.#....#####.#.#.#...##...##..##..
19: .#..###.#..#.#.#######.#.#.#..#.#...#..
20: .#....##....#####...#######....#.#..##.
The generation is shown along the left, where 0 is the initial state. The pot numbers are shown along the top, where 0 labels the center pot, negative-numbered pots extend to the left, and positive pots extend toward the right. Remember, the initial state begins at pot 0, which is not the leftmost pot used in this example.
After one generation, only seven plants remain. The one in pot 0 matched the rule looking for ..#.., the one in pot 4 matched the rule looking for .#.#., pot 9 matched .##.., and so on.
In this example, after 20 generations, the pots shown as # contain plants, the furthest left of which is pot -2, and the furthest right of which is pot 34. Adding up all the numbers of plant-containing pots after the 20th generation produces 325.
After 20 generations, what is the sum of the numbers of all pots which contain a plant?
--- Part Two ---
You realize that 20 generations aren't enough. After all, these plants will need to last another 1500 years to even reach your timeline, not to mention your future.
After fifty billion (50000000000) generations, what is the sum of the numbers of all pots which contain a plant?
*/
public class Day12 : IDay
public string ResolvePart1(string[] inputs)
{
public string ResolvePart1(string[] inputs)
Initialize(inputs);
Simulate(20, true);
return CalculateChecksum().ToString();
}
public string ResolvePart2(string[] inputs)
{
Initialize(inputs);
Simulate(500);
_offsetField -= (50000000000L - 500);
return CalculateChecksum().ToString();
}
private class PlantRule
{
public bool Minus2 { get; set; }
public bool Minus1 { get; set; }
public bool Current { get; set; }
public bool Plus1 { get; set; }
public bool Plus2 { get; set; }
public bool Result { get; set; }
}
private const int SideMargin = 5;
private const int SideProcessMargin = 2;
private List<bool> _initialState = new();
private List<PlantRule> _rules = new();
private long _offsetField;
private bool[] _field;
private bool[] _workField;
private void Initialize(string[] inputs)
{
_initialState.Clear();
foreach (char c in inputs[0].Substring("initial state: ".Length))
{
Initialize(inputs);
Simulate(20, true);
return CalculateChecksum().ToString();
_initialState.Add(c == '#');
}
public string ResolvePart2(string[] inputs)
_rules.Clear();
for (int i = 2; i < inputs.Length; i++)
{
Initialize(inputs);
Simulate(500, false);
_offsetField -= (50000000000L - 500);
return CalculateChecksum().ToString();
if (string.IsNullOrEmpty(inputs[i])) { continue; }
string[] parts = inputs[i].Split(new[] { " => " }, StringSplitOptions.RemoveEmptyEntries);
_rules.Add(new PlantRule
{
Minus2 = (parts[0][0] == '#'),
Minus1 = (parts[0][1] == '#'),
Current = (parts[0][2] == '#'),
Plus1 = (parts[0][3] == '#'),
Plus2 = (parts[0][4] == '#'),
Result = (parts[1][0] == '#'),
});
}
private class PlantRule
int maxSize = (SideMargin * 2) + _initialState.Count;
_offsetField = SideMargin;
_field = new bool[maxSize];
_workField = new bool[maxSize];
for (int i = 0; i < _initialState.Count; i++)
{
public bool Minus2 { get; set; }
public bool Minus1 { get; set; }
public bool Current { get; set; }
public bool Plus1 { get; set; }
public bool Plus2 { get; set; }
public bool Result { get; set; }
_field[i + _offsetField] = _initialState[i];
}
}
private const int SideMargin = 5;
private const int SideProcessMargin = 2;
private void SwapFields()
{
bool[] aux = _field;
_field = _workField;
_workField = aux;
}
private List<bool> _initialState = new List<bool>();
private List<PlantRule> _rules = new List<PlantRule>();
private long _offsetField = 0;
private bool[] _field;
private bool[] _workField;
private void Initialize(string[] inputs)
private void RecenterField()
{
long leftSpace = 0;
long rightSpace = 0;
for (long i = 0; i < _field.Length; i++)
{
_initialState.Clear();
foreach (char c in inputs[0].Substring("initial state: ".Length))
{
_initialState.Add(c == '#');
}
_rules.Clear();
for (int i = 2; i < inputs.Length; i++)
{
if (string.IsNullOrEmpty(inputs[i])) { continue; }
string[] parts = inputs[i].Split(new string[] { " => " }, StringSplitOptions.RemoveEmptyEntries);
_rules.Add(new PlantRule
{
Minus2 = (parts[0][0] == '#'),
Minus1 = (parts[0][1] == '#'),
Current = (parts[0][2] == '#'),
Plus1 = (parts[0][3] == '#'),
Plus2 = (parts[0][4] == '#'),
Result = (parts[1][0] == '#'),
});
}
int maxSize = (SideMargin * 2) + _initialState.Count;
_offsetField = SideMargin;
_field = new bool[maxSize];
_workField = new bool[maxSize];
for (int i = 0; i < _initialState.Count; i++)
{
_field[i + _offsetField] = _initialState[i];
}
if (_field[i]) { break; }
leftSpace++;
}
private void SwapFields()
for (long i = _field.Length - 1; i >= 0; i--)
{
bool[] aux = _field;
_field = _workField;
_workField = aux;
if (_field[i]) { break; }
rightSpace++;
}
if (leftSpace == SideMargin && rightSpace == SideMargin) { return; }
private void RecenterField()
long oldSize = _field.Length;
long newSize = oldSize + (SideMargin - leftSpace) + (SideMargin - rightSpace);
long diffOffset = SideMargin - leftSpace;
if (oldSize == newSize && diffOffset != 0)
{
long leftSpace = 0;
long rightSpace = 0;
for (long i = 0; i < _field.Length; i++)
if (diffOffset > 0)
{
if (_field[i]) { break; }
leftSpace++;
}
for (long i = _field.Length - 1; i >= 0; i--)
{
if (_field[i]) { break; }
rightSpace++;
}
if (leftSpace == SideMargin && rightSpace == SideMargin) { return; }
long oldSize = _field.Length;
long newSize = oldSize + (SideMargin - leftSpace) + (SideMargin - rightSpace);
long diffOffset = SideMargin - leftSpace;
if (oldSize == newSize && diffOffset != 0)
{
if (diffOffset > 0)
{
for (long i = 0; i < diffOffset; i++) { _workField[i] = false; }
}
else
{
for (long i = 0; i < Math.Abs(diffOffset); i++) { _workField[(_workField.Length - 1) - i] = false; }
}
for (long i = 0; i < newSize; i++)
{
long i2 = i - diffOffset;
if (i2 < 0 || i2 >= oldSize) { continue; }
_workField[i] = _field[i2];
}
SwapFields();
for (long i = 0; i < diffOffset; i++) { _workField[i] = false; }
}
else
{
bool[] tempField = new bool[newSize];
for (long i = 0; i < newSize; i++)
{
long i2 = i - diffOffset;
if (i2 < 0 || i2 >= oldSize) { continue; }
tempField[i] = _field[i2];
}
_field = tempField;
_workField = new bool[newSize];
for (long i = 0; i < Math.Abs(diffOffset); i++) { _workField[(_workField.Length - 1) - i] = false; }
}
_offsetField += diffOffset;
}
private void ShowField(long nGeneration)
{
Console.Write("({0:000}) [{1}]: ", nGeneration, _offsetField);
foreach (bool plant in _field)
for (long i = 0; i < newSize; i++)
{
Console.Write(plant ? "#" : ".");
long i2 = i - diffOffset;
if (i2 < 0 || i2 >= oldSize) { continue; }
_workField[i] = _field[i2];
}
Console.WriteLine();
SwapFields();
}
private static void SimulateGeneration(List<PlantRule> rules, bool[] field, bool[] finalField)
else
{
for (long i = SideProcessMargin; i < (field.Length - SideProcessMargin); i++)
bool[] tempField = new bool[newSize];
for (long i = 0; i < newSize; i++)
{
bool minus2 = field[i - 2];
bool minus1 = field[i - 1];
bool current = field[i];
bool plus1 = field[i + 1];
bool plus2 = field[i + 2];
bool result = false;
foreach (PlantRule rule in rules)
{
if (
rule.Minus2 == minus2 &&
rule.Minus1 == minus1 &&
rule.Current == current &&
rule.Plus1 == plus1 &&
rule.Plus2 == plus2 &&
true
)
{
result = rule.Result;
break;
}
}
finalField[i] = result;
long i2 = i - diffOffset;
if (i2 < 0 || i2 >= oldSize) { continue; }
tempField[i] = _field[i2];
}
_field = tempField;
_workField = new bool[newSize];
}
_offsetField += diffOffset;
}
private void Simulate(long nGenerations, bool showEvolution = false)
private void ShowField(long nGeneration)
{
Console.Write("({0:000}) [{1}]: ", nGeneration, _offsetField);
foreach (bool plant in _field)
{
for (int i = 0; i < nGenerations; i++)
{
RecenterField();
if (showEvolution)
{
ShowField(i);
}
SimulateGeneration(_rules, _field, _workField);
SwapFields();
}
ShowField(nGenerations);
Console.Write(plant ? "#" : ".");
}
Console.WriteLine();
}
private long CalculateChecksum()
private static void SimulateGeneration(List<PlantRule> rules, bool[] field, bool[] finalField)
{
for (long i = SideProcessMargin; i < (field.Length - SideProcessMargin); i++)
{
long sum = 0;
for (long i = 0; i < _field.Length; i++)
bool minus2 = field[i - 2];
bool minus1 = field[i - 1];
bool current = field[i];
bool plus1 = field[i + 1];
bool plus2 = field[i + 2];
bool result = false;
foreach (PlantRule rule in rules)
{
if (_field[i])
if (
rule.Minus2 == minus2 &&
rule.Minus1 == minus1 &&
rule.Current == current &&
rule.Plus1 == plus1 &&
rule.Plus2 == plus2 &&
true
)
{
sum += (i - _offsetField);
result = rule.Result;
break;
}
}
return sum;
finalField[i] = result;
}
}
}
private void Simulate(long nGenerations, bool showEvolution = false)
{
for (int i = 0; i < nGenerations; i++)
{
RecenterField();
if (showEvolution)
{
ShowField(i);
}
SimulateGeneration(_rules, _field, _workField);
SwapFields();
}
ShowField(nGenerations);
}
private long CalculateChecksum()
{
long sum = 0;
for (long i = 0; i < _field.Length; i++)
{
if (_field[i])
{
sum += (i - _offsetField);
}
}
return sum;
}
}

934
AdventOfCode2018/Day13.cs
View File

@@ -2,310 +2,330 @@
using System.Collections.Generic;
using System.Linq;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 13: Mine Cart Madness ---
A crop of this size requires significant logistics to transport produce, soil, fertilizer, and so on. The Elves are very busy pushing things around in carts on some kind of rudimentary system of tracks they've come up with.
Seeing as how cart-and-track systems don't appear in recorded history for another 1000 years, the Elves seem to be making this up as they go along. They haven't even figured out how to avoid collisions yet.
You map out the tracks (your puzzle input) and see where you can help.
Tracks consist of straight paths (| and -), curves (/ and \), and intersections (+). Curves connect exactly two perpendicular pieces of track; for example, this is a closed loop:
/----\
| |
| |
\----/
Intersections occur when two perpendicular paths cross. At an intersection, a cart is capable of turning left, turning right, or continuing straight. Here are two loops connected by two intersections:
/-----\
| |
| /--+--\
| | | |
\--+--/ |
| |
\-----/
Several carts are also on the tracks. Carts always face either up (^), down (v), left (<), or right (>). (On your initial map, the track under each cart is a straight path matching the direction the cart is facing.)
Each time a cart has the option to turn (by arriving at any intersection), it turns left the first time, goes straight the second time, turns right the third time, and then repeats those directions starting again with left the fourth time, straight the fifth time, and so on. This process is independent of the particular intersection at which the cart has arrived - that is, the cart has no per-intersection memory.
Carts all move at the same speed; they take turns moving a single step at a time. They do this based on their current location: carts on the top row move first (acting from left to right), then carts on the second row move (again from left to right), then carts on the third row, and so on. Once each cart has moved one step, the process repeats; each of these loops is called a tick.
For example, suppose there are two carts on a straight track:
| | | | |
v | | | |
| v v | |
| | | v X
| | ^ ^ |
^ ^ | | |
| | | | |
First, the top cart moves. It is facing down (v), so it moves down one square. Second, the bottom cart moves. It is facing up (^), so it moves up one square. Because all carts have moved, the first tick ends. Then, the process repeats, starting with the first cart. The first cart moves down, then the second cart moves up - right into the first cart, colliding with it! (The location of the crash is marked with an X.) This ends the second and last tick.
Here is a longer example:
/->-\
| | /----\
| /-+--+-\ |
| | | | v |
\-+-/ \-+--/
\------/
/-->\
| | /----\
| /-+--+-\ |
| | | | | |
\-+-/ \->--/
\------/
/---v
| | /----\
| /-+--+-\ |
| | | | | |
\-+-/ \-+>-/
\------/
/---\
| v /----\
| /-+--+-\ |
| | | | | |
\-+-/ \-+->/
\------/
/---\
| | /----\
| /->--+-\ |
| | | | | |
\-+-/ \-+--^
\------/
/---\
| | /----\
| /-+>-+-\ |
| | | | | ^
\-+-/ \-+--/
\------/
/---\
| | /----\
| /-+->+-\ ^
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /----<
| /-+-->-\ |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /---<\
| /-+--+>\ |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /--<-\
| /-+--+-v |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /-<--\
| /-+--+-\ |
| | | | v |
\-+-/ \-+--/
\------/
/---\
| | /<---\
| /-+--+-\ |
| | | | | |
\-+-/ \-<--/
\------/
/---\
| | v----\
| /-+--+-\ |
| | | | | |
\-+-/ \<+--/
\------/
/---\
| | /----\
| /-+--v-\ |
| | | | | |
\-+-/ ^-+--/
\------/
/---\
| | /----\
| /-+--+-\ |
| | | X | |
\-+-/ \-+--/
\------/
After following their respective paths for a while, the carts eventually crash. To help prevent crashes, you'd like to know the location of the first crash. Locations are given in X,Y coordinates, where the furthest left column is X=0 and the furthest top row is Y=0:
111
0123456789012
0/---\
1| | /----\
2| /-+--+-\ |
3| | | X | |
4\-+-/ \-+--/
5 \------/
In this example, the location of the first crash is 7,3.
--- Part Two ---
There isn't much you can do to prevent crashes in this ridiculous system. However, by predicting the crashes, the Elves know where to be in advance and instantly remove the two crashing carts the moment any crash occurs.
They can proceed like this for a while, but eventually, they're going to run out of carts. It could be useful to figure out where the last cart that hasn't crashed will end up.
For example:
/>-<\
| |
| /<+-\
| | | v
\>+</ |
| ^
\<->/
/---\
| |
| v-+-\
| | | |
\-+-/ |
| |
^---^
/---\
| |
| /-+-\
| v | |
\-+-/ |
^ ^
\---/
/---\
| |
| /-+-\
| | | |
\-+-/ ^
| |
\---/
After four very expensive crashes, a tick ends with only one cart remaining; its final location is 6,4.
What is the location of the last cart at the end of the first tick where it is the only cart left?
*/
public class Day13 : IDay
{
/*
private bool ShowProgress { get; set; } = false;
--- Day 13: Mine Cart Madness ---
A crop of this size requires significant logistics to transport produce, soil, fertilizer, and so on. The Elves are very busy pushing things around in carts on some kind of rudimentary system of tracks they've come up with.
Seeing as how cart-and-track systems don't appear in recorded history for another 1000 years, the Elves seem to be making this up as they go along. They haven't even figured out how to avoid collisions yet.
You map out the tracks (your puzzle input) and see where you can help.
Tracks consist of straight paths (| and -), curves (/ and \), and intersections (+). Curves connect exactly two perpendicular pieces of track; for example, this is a closed loop:
/----\
| |
| |
\----/
Intersections occur when two perpendicular paths cross. At an intersection, a cart is capable of turning left, turning right, or continuing straight. Here are two loops connected by two intersections:
/-----\
| |
| /--+--\
| | | |
\--+--/ |
| |
\-----/
Several carts are also on the tracks. Carts always face either up (^), down (v), left (<), or right (>). (On your initial map, the track under each cart is a straight path matching the direction the cart is facing.)
Each time a cart has the option to turn (by arriving at any intersection), it turns left the first time, goes straight the second time, turns right the third time, and then repeats those directions starting again with left the fourth time, straight the fifth time, and so on. This process is independent of the particular intersection at which the cart has arrived - that is, the cart has no per-intersection memory.
Carts all move at the same speed; they take turns moving a single step at a time. They do this based on their current location: carts on the top row move first (acting from left to right), then carts on the second row move (again from left to right), then carts on the third row, and so on. Once each cart has moved one step, the process repeats; each of these loops is called a tick.
For example, suppose there are two carts on a straight track:
| | | | |
v | | | |
| v v | |
| | | v X
| | ^ ^ |
^ ^ | | |
| | | | |
First, the top cart moves. It is facing down (v), so it moves down one square. Second, the bottom cart moves. It is facing up (^), so it moves up one square. Because all carts have moved, the first tick ends. Then, the process repeats, starting with the first cart. The first cart moves down, then the second cart moves up - right into the first cart, colliding with it! (The location of the crash is marked with an X.) This ends the second and last tick.
Here is a longer example:
/->-\
| | /----\
| /-+--+-\ |
| | | | v |
\-+-/ \-+--/
\------/
/-->\
| | /----\
| /-+--+-\ |
| | | | | |
\-+-/ \->--/
\------/
/---v
| | /----\
| /-+--+-\ |
| | | | | |
\-+-/ \-+>-/
\------/
/---\
| v /----\
| /-+--+-\ |
| | | | | |
\-+-/ \-+->/
\------/
/---\
| | /----\
| /->--+-\ |
| | | | | |
\-+-/ \-+--^
\------/
/---\
| | /----\
| /-+>-+-\ |
| | | | | ^
\-+-/ \-+--/
\------/
/---\
| | /----\
| /-+->+-\ ^
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /----<
| /-+-->-\ |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /---<\
| /-+--+>\ |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /--<-\
| /-+--+-v |
| | | | | |
\-+-/ \-+--/
\------/
/---\
| | /-<--\
| /-+--+-\ |
| | | | v |
\-+-/ \-+--/
\------/
/---\
| | /<---\
| /-+--+-\ |
| | | | | |
\-+-/ \-<--/
\------/
/---\
| | v----\
| /-+--+-\ |
| | | | | |
\-+-/ \<+--/
\------/
/---\
| | /----\
| /-+--v-\ |
| | | | | |
\-+-/ ^-+--/
\------/
/---\
| | /----\
| /-+--+-\ |
| | | X | |
\-+-/ \-+--/
\------/
After following their respective paths for a while, the carts eventually crash. To help prevent crashes, you'd like to know the location of the first crash. Locations are given in X,Y coordinates, where the furthest left column is X=0 and the furthest top row is Y=0:
111
0123456789012
0/---\
1| | /----\
2| /-+--+-\ |
3| | | X | |
4\-+-/ \-+--/
5 \------/
In this example, the location of the first crash is 7,3.
--- Part Two ---
There isn't much you can do to prevent crashes in this ridiculous system. However, by predicting the crashes, the Elves know where to be in advance and instantly remove the two crashing carts the moment any crash occurs.
They can proceed like this for a while, but eventually, they're going to run out of carts. It could be useful to figure out where the last cart that hasn't crashed will end up.
For example:
/>-<\
| |
| /<+-\
| | | v
\>+</ |
| ^
\<->/
/---\
| |
| v-+-\
| | | |
\-+-/ |
| |
^---^
/---\
| |
| /-+-\
| v | |
\-+-/ |
^ ^
\---/
/---\
| |
| /-+-\
| | | |
\-+-/ ^
| |
\---/
After four very expensive crashes, a tick ends with only one cart remaining; its final location is 6,4.
What is the location of the last cart at the end of the first tick where it is the only cart left?
*/
public class Day13 : IDay
public string ResolvePart1(string[] inputs)
{
private bool ShowProgress { get; set; } = false;
public string ResolvePart1(string[] inputs)
Initialize(inputs);
Train collidingTrain;
do
{
Initialize(inputs);
Train colidingTrain = null;
do
if (ShowProgress) { ShowGrid(); }
collidingTrain = SimulateForFirstCollision();
} while (collidingTrain == null);
return $"{collidingTrain.X},{collidingTrain.Y}";
}
public string ResolvePart2(string[] inputs)
{
Initialize(inputs);
Train lastCart;
do
{
if (ShowProgress) { ShowGrid(); }
lastCart = SimulateForLastCart();
} while (lastCart == null);
return $"{lastCart.X},{lastCart.Y}";
}
private enum TrainDirection
{
North,
South,
East,
West,
}
private enum TrainTurning
{
Left,
Right,
None,
}
private class Train
{
public int X { get; set; }
public int Y { get; set; }
public TrainDirection Direction { get; set; }
public TrainTurning NextIntersectionTurn { get; set; }
public void Simulate(char[,] grid)
{
if (Direction == TrainDirection.North)
{
if (ShowProgress) { ShowGrid(); }
colidingTrain = SimulateForFirstCollision();
} while (colidingTrain == null);
return string.Format("{0},{1}", colidingTrain.X, colidingTrain.Y);
}
public string ResolvePart2(string[] inputs)
{
Initialize(inputs);
Train lastCart = null;
do
Y--;
}
if (Direction == TrainDirection.South)
{
if (ShowProgress) { ShowGrid(); }
lastCart = SimulateForLastCart();
} while (lastCart == null);
return string.Format("{0},{1}", lastCart.X, lastCart.Y);
}
Y++;
}
if (Direction == TrainDirection.East)
{
X++;
}
if (Direction == TrainDirection.West)
{
X--;
}
private enum TrainDirection
{
North,
South,
East,
West,
};
private enum TrainTurning
{
Left,
Right,
None,
}
private class Train
{
public int X { get; set; }
public int Y { get; set; }
public TrainDirection Direction { get; set; }
public TrainTurning NextIntersectionTurn { get; set; }
public void Simulate(char[,] grid)
char cell = grid[X, Y];
if (cell == '/')
{
if (Direction == TrainDirection.North)
{
Y--;
Direction = TrainDirection.East;
}
if (Direction == TrainDirection.South)
else if (Direction == TrainDirection.South)
{
Y++;
Direction = TrainDirection.West;
}
if (Direction == TrainDirection.East)
else if (Direction == TrainDirection.East)
{
X++;
Direction = TrainDirection.North;
}
if (Direction == TrainDirection.West)
else if (Direction == TrainDirection.West)
{
X--;
Direction = TrainDirection.South;
}
char cell = grid[X, Y];
if (cell == '/')
}
if (cell == '\\')
{
if (Direction == TrainDirection.North)
{
if (Direction == TrainDirection.North)
{
Direction = TrainDirection.East;
}
else if (Direction == TrainDirection.South)
{
Direction = TrainDirection.West;
}
else if (Direction == TrainDirection.East)
{
Direction = TrainDirection.North;
}
else if (Direction == TrainDirection.West)
{
Direction = TrainDirection.South;
}
Direction = TrainDirection.West;
}
if (cell == '\\')
else if (Direction == TrainDirection.South)
{
Direction = TrainDirection.East;
}
else if (Direction == TrainDirection.East)
{
Direction = TrainDirection.South;
}
else if (Direction == TrainDirection.West)
{
Direction = TrainDirection.North;
}
}
if (cell == '+')
{
if (NextIntersectionTurn == TrainTurning.Left)
{
if (Direction == TrainDirection.North)
{
@@ -316,6 +336,30 @@ namespace AdventOfCode2018
Direction = TrainDirection.East;
}
else if (Direction == TrainDirection.East)
{
Direction = TrainDirection.North;
}
else if (Direction == TrainDirection.West)
{
Direction = TrainDirection.South;
}
NextIntersectionTurn = TrainTurning.None;
}
else if (NextIntersectionTurn == TrainTurning.None)
{
NextIntersectionTurn = TrainTurning.Right;
}
else if (NextIntersectionTurn == TrainTurning.Right)
{
if (Direction == TrainDirection.North)
{
Direction = TrainDirection.East;
}
else if (Direction == TrainDirection.South)
{
Direction = TrainDirection.West;
}
else if (Direction == TrainDirection.East)
{
Direction = TrainDirection.South;
}
@@ -323,190 +367,144 @@ namespace AdventOfCode2018
{
Direction = TrainDirection.North;
}
}
if (cell == '+')
{
if (NextIntersectionTurn == TrainTurning.Left)
{
if (Direction == TrainDirection.North)
{
Direction = TrainDirection.West;
}
else if (Direction == TrainDirection.South)
{
Direction = TrainDirection.East;
}
else if (Direction == TrainDirection.East)
{
Direction = TrainDirection.North;
}
else if (Direction == TrainDirection.West)
{
Direction = TrainDirection.South;
}
NextIntersectionTurn = TrainTurning.None;
}
else if (NextIntersectionTurn == TrainTurning.None)
{
NextIntersectionTurn = TrainTurning.Right;
}
else if (NextIntersectionTurn == TrainTurning.Right)
{
if (Direction == TrainDirection.North)
{
Direction = TrainDirection.East;
}
else if (Direction == TrainDirection.South)
{
Direction = TrainDirection.West;
}
else if (Direction == TrainDirection.East)
{
Direction = TrainDirection.South;
}
else if (Direction == TrainDirection.West)
{
Direction = TrainDirection.North;
}
NextIntersectionTurn = TrainTurning.Left;
}
NextIntersectionTurn = TrainTurning.Left;
}
}
}
private int _width;
private int _height;
private char[,] _grid = null;
private List<Train> _trains = new List<Train>();
private void Initialize(string[] inputs)
{
_width = inputs.Max(s => s.Length);
_height = inputs.Length;
_grid = new char[_width, _height];
_trains.Clear();
for (int j = 0; j < inputs.Length; j++)
{
for (int i = 0; i < inputs[j].Length; i++)
{
char cell = inputs[j][i];
if (cell == '^')
{
_trains.Add(new Train
{
X = i,
Y = j,
Direction = TrainDirection.North,
NextIntersectionTurn = TrainTurning.Left,
});
cell = '|';
}
if (cell == 'v')
{
_trains.Add(new Train
{
X = i,
Y = j,
Direction = TrainDirection.South,
NextIntersectionTurn = TrainTurning.Left,
});
cell = '|';
}
if (cell == '<')
{
_trains.Add(new Train
{
X = i,
Y = j,
Direction = TrainDirection.West,
NextIntersectionTurn = TrainTurning.Left,
});
cell = '-';
}
if (cell == '>')
{
_trains.Add(new Train
{
X = i,
Y = j,
Direction = TrainDirection.East,
NextIntersectionTurn = TrainTurning.Left,
});
cell = '-';
}
_grid[i, j] = cell;
}
}
}
private Train SimulateForFirstCollision()
{
Train collidingTrain = null;
IEnumerable<Train> orderedTrains = _trains.OrderBy(t => t.X + (t.Y * _width));
foreach (Train t in orderedTrains)
{
t.Simulate(_grid);
collidingTrain = _trains.FirstOrDefault(x => x.X == t.X && x.Y == t.Y && t != x);
if (collidingTrain != null) { break; }
}
return collidingTrain;
}
private Train SimulateForLastCart()
{
List<Train> orderedTrains = _trains.OrderBy(t => t.X + (t.Y * _width)).ToList();
foreach (Train t in orderedTrains)
{
t.Simulate(_grid);
Train collidingTrain = _trains.FirstOrDefault(x => x.X == t.X && x.Y == t.Y && t != x);
if (collidingTrain != null)
{
_trains.Remove(t);
_trains.Remove(collidingTrain);
}
}
if (_trains.Count == 1)
{
return _trains[0];
}
return null;
}
private void ShowGrid()
{
Console.WriteLine();
for (int j = 0; j < _height; j++)
{
for (int i = 0; i < _width; i++)
{
Train t = _trains.FirstOrDefault(x => x.X == i && x.Y == j);
if (t != null)
{
if (t.Direction == TrainDirection.North)
{
Console.Write("^");
}
if (t.Direction == TrainDirection.East)
{
Console.Write(">");
}
if (t.Direction == TrainDirection.South)
{
Console.Write("v");
}
if (t.Direction == TrainDirection.West)
{
Console.Write("<");
}
}
else
{
Console.Write(_grid[i, j]);
}
}
Console.WriteLine();
}
}
}
}
private int _width;
private int _height;
private char[,] _grid;
private List<Train> _trains = new();
private void Initialize(string[] inputs)
{
_width = inputs.Max(s => s.Length);
_height = inputs.Length;
_grid = new char[_width, _height];
_trains.Clear();
for (int j = 0; j < inputs.Length; j++)
{
for (int i = 0; i < inputs[j].Length; i++)
{
char cell = inputs[j][i];
if (cell == '^')
{
_trains.Add(new Train
{
X = i,
Y = j,
Direction = TrainDirection.North,
NextIntersectionTurn = TrainTurning.Left,
});
cell = '|';
}
if (cell == 'v')
{
_trains.Add(new Train
{
X = i,
Y = j,
Direction = TrainDirection.South,
NextIntersectionTurn = TrainTurning.Left,
});
cell = '|';
}
if (cell == '<')
{
_trains.Add(new Train
{
X = i,
Y = j,
Direction = TrainDirection.West,
NextIntersectionTurn = TrainTurning.Left,
});
cell = '-';
}
if (cell == '>')
{
_trains.Add(new Train
{
X = i,
Y = j,
Direction = TrainDirection.East,
NextIntersectionTurn = TrainTurning.Left,
});
cell = '-';
}
_grid[i, j] = cell;
}
}
}
private Train SimulateForFirstCollision()
{
Train collidingTrain = null;
IEnumerable<Train> orderedTrains = _trains.OrderBy(t => t.X + (t.Y * _width));
foreach (Train t in orderedTrains)
{
t.Simulate(_grid);
collidingTrain = _trains.FirstOrDefault(x => x.X == t.X && x.Y == t.Y && t != x);
if (collidingTrain != null) { break; }
}
return collidingTrain;
}
private Train SimulateForLastCart()
{
List<Train> orderedTrains = _trains.OrderBy(t => t.X + (t.Y * _width)).ToList();
foreach (Train t in orderedTrains)
{
t.Simulate(_grid);
Train collidingTrain = _trains.FirstOrDefault(x => x.X == t.X && x.Y == t.Y && t != x);
if (collidingTrain != null)
{
_trains.Remove(t);
_trains.Remove(collidingTrain);
}
}
if (_trains.Count == 1)
{
return _trains[0];
}
return null;
}
private void ShowGrid()
{
Console.WriteLine();
for (int j = 0; j < _height; j++)
{
for (int i = 0; i < _width; i++)
{
Train t = _trains.FirstOrDefault(x => x.X == i && x.Y == j);
if (t != null)
{
if (t.Direction == TrainDirection.North)
{
Console.Write("^");
}
if (t.Direction == TrainDirection.East)
{
Console.Write(">");
}
if (t.Direction == TrainDirection.South)
{
Console.Write("v");
}
if (t.Direction == TrainDirection.West)
{
Console.Write("<");
}
}
else
{
Console.Write(_grid[i, j]);
}
}
Console.WriteLine();
}
}
}

342
AdventOfCode2018/Day14.cs
View File

@@ -2,189 +2,187 @@
using System.Linq;
using System.Text;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
*
--- Day 14: Chocolate Charts ---
You finally have a chance to look at all of the produce moving around. Chocolate, cinnamon, mint, chili peppers, nutmeg, vanilla... the Elves must be growing these plants to make hot chocolate! As you realize this, you hear a conversation in the distance. When you go to investigate, you discover two Elves in what appears to be a makeshift underground kitchen/laboratory.
The Elves are trying to come up with the ultimate hot chocolate recipe; they're even maintaining a scoreboard which tracks the quality score (0-9) of each recipe.
Only two recipes are on the board: the first recipe got a score of 3, the second, 7. Each of the two Elves has a current recipe: the first Elf starts with the first recipe, and the second Elf starts with the second recipe.
To create new recipes, the two Elves combine their current recipes. This creates new recipes from the digits of the sum of the current recipes' scores. With the current recipes' scores of 3 and 7, their sum is 10, and so two new recipes would be created: the first with score 1 and the second with score 0. If the current recipes' scores were 2 and 3, the sum, 5, would only create one recipe (with a score of 5) with its single digit.
The new recipes are added to the end of the scoreboard in the order they are created. So, after the first round, the scoreboard is 3, 7, 1, 0.
After all new recipes are added to the scoreboard, each Elf picks a new current recipe. To do this, the Elf steps forward through the scoreboard a number of recipes equal to 1 plus the score of their current recipe. So, after the first round, the first Elf moves forward 1 + 3 = 4 times, while the second Elf moves forward 1 + 7 = 8 times. If they run out of recipes, they loop back around to the beginning. After the first round, both Elves happen to loop around until they land on the same recipe that they had in the beginning; in general, they will move to different recipes.
Drawing the first Elf as parentheses and the second Elf as square brackets, they continue this process:
(3)[7]
(3)[7] 1 0
3 7 1 [0](1) 0
3 7 1 0 [1] 0 (1)
(3) 7 1 0 1 0 [1] 2
3 7 1 0 (1) 0 1 2 [4]
3 7 1 [0] 1 0 (1) 2 4 5
3 7 1 0 [1] 0 1 2 (4) 5 1
3 (7) 1 0 1 0 [1] 2 4 5 1 5
3 7 1 0 1 0 1 2 [4](5) 1 5 8
3 (7) 1 0 1 0 1 2 4 5 1 5 8 [9]
3 7 1 0 1 0 1 [2] 4 (5) 1 5 8 9 1 6
3 7 1 0 1 0 1 2 4 5 [1] 5 8 9 1 (6) 7
3 7 1 0 (1) 0 1 2 4 5 1 5 [8] 9 1 6 7 7
3 7 [1] 0 1 0 (1) 2 4 5 1 5 8 9 1 6 7 7 9
3 7 1 0 [1] 0 1 2 (4) 5 1 5 8 9 1 6 7 7 9 2
The Elves think their skill will improve after making a few recipes (your puzzle input). However, that could take ages; you can speed this up considerably by identifying the scores of the ten recipes after that. For example:
If the Elves think their skill will improve after making 9 recipes, the scores of the ten recipes after the first nine on the scoreboard would be 5158916779 (highlighted in the last line of the diagram).
After 5 recipes, the scores of the next ten would be 0124515891.
After 18 recipes, the scores of the next ten would be 9251071085.
After 2018 recipes, the scores of the next ten would be 5941429882.
What are the scores of the ten recipes immediately after the number of recipes in your puzzle input?
--- Part Two ---
As it turns out, you got the Elves' plan backwards. They actually want to know how many recipes appear on the scoreboard to the left of the first recipes whose scores are the digits from your puzzle input.
51589 first appears after 9 recipes.
01245 first appears after 5 recipes.
92510 first appears after 18 recipes.
59414 first appears after 2018 recipes.
How many recipes appear on the scoreboard to the left of the score sequence in your puzzle input?
*/
public class Day14 : IDay
{
/*
*
--- Day 14: Chocolate Charts ---
private long _numRecipes;
private long _numRecipesAllocated;
private byte[] _recipes;
private long _idxA;
private long _idxB;
private long _searchSkip;
You finally have a chance to look at all of the produce moving around. Chocolate, cinnamon, mint, chili peppers, nutmeg, vanilla... the Elves must be growing these plants to make hot chocolate! As you realize this, you hear a conversation in the distance. When you go to investigate, you discover two Elves in what appears to be a makeshift underground kitchen/laboratory.
The Elves are trying to come up with the ultimate hot chocolate recipe; they're even maintaining a scoreboard which tracks the quality score (0-9) of each recipe.
Only two recipes are on the board: the first recipe got a score of 3, the second, 7. Each of the two Elves has a current recipe: the first Elf starts with the first recipe, and the second Elf starts with the second recipe.
To create new recipes, the two Elves combine their current recipes. This creates new recipes from the digits of the sum of the current recipes' scores. With the current recipes' scores of 3 and 7, their sum is 10, and so two new recipes would be created: the first with score 1 and the second with score 0. If the current recipes' scores were 2 and 3, the sum, 5, would only create one recipe (with a score of 5) with its single digit.
The new recipes are added to the end of the scoreboard in the order they are created. So, after the first round, the scoreboard is 3, 7, 1, 0.
After all new recipes are added to the scoreboard, each Elf picks a new current recipe. To do this, the Elf steps forward through the scoreboard a number of recipes equal to 1 plus the score of their current recipe. So, after the first round, the first Elf moves forward 1 + 3 = 4 times, while the second Elf moves forward 1 + 7 = 8 times. If they run out of recipes, they loop back around to the beginning. After the first round, both Elves happen to loop around until they land on the same recipe that they had in the beginning; in general, they will move to different recipes.
Drawing the first Elf as parentheses and the second Elf as square brackets, they continue this process:
(3)[7]
(3)[7] 1 0
3 7 1 [0](1) 0
3 7 1 0 [1] 0 (1)
(3) 7 1 0 1 0 [1] 2
3 7 1 0 (1) 0 1 2 [4]
3 7 1 [0] 1 0 (1) 2 4 5
3 7 1 0 [1] 0 1 2 (4) 5 1
3 (7) 1 0 1 0 [1] 2 4 5 1 5
3 7 1 0 1 0 1 2 [4](5) 1 5 8
3 (7) 1 0 1 0 1 2 4 5 1 5 8 [9]
3 7 1 0 1 0 1 [2] 4 (5) 1 5 8 9 1 6
3 7 1 0 1 0 1 2 4 5 [1] 5 8 9 1 (6) 7
3 7 1 0 (1) 0 1 2 4 5 1 5 [8] 9 1 6 7 7
3 7 [1] 0 1 0 (1) 2 4 5 1 5 8 9 1 6 7 7 9
3 7 1 0 [1] 0 1 2 (4) 5 1 5 8 9 1 6 7 7 9 2
The Elves think their skill will improve after making a few recipes (your puzzle input). However, that could take ages; you can speed this up considerably by identifying the scores of the ten recipes after that. For example:
If the Elves think their skill will improve after making 9 recipes, the scores of the ten recipes after the first nine on the scoreboard would be 5158916779 (highlighted in the last line of the diagram).
After 5 recipes, the scores of the next ten would be 0124515891.
After 18 recipes, the scores of the next ten would be 9251071085.
After 2018 recipes, the scores of the next ten would be 5941429882.
What are the scores of the ten recipes immediately after the number of recipes in your puzzle input?
--- Part Two ---
As it turns out, you got the Elves' plan backwards. They actually want to know how many recipes appear on the scoreboard to the left of the first recipes whose scores are the digits from your puzzle input.
51589 first appears after 9 recipes.
01245 first appears after 5 recipes.
92510 first appears after 18 recipes.
59414 first appears after 2018 recipes.
How many recipes appear on the scoreboard to the left of the score sequence in your puzzle input?
*/
public class Day14 : IDay
private void Init(long hintAllocation = 128)
{
private long _numRecipes = 0;
private long _numRecipesAllocated = 0;
private byte[] _recipes = null;
private long _idxA = 0;
private long _idxB = 0;
private long _searchSkip = 0;
_numRecipes = 2;
_numRecipesAllocated = hintAllocation;
_recipes = new byte[_numRecipesAllocated];
_recipes[0] = 3;
_recipes[1] = 7;
_idxA = 0;
_idxB = 1;
_searchSkip = 0;
}
private void Init(long hintAllocation = 128)
private void Step()
{
if (_numRecipes + 2 >= _numRecipesAllocated)
{
_numRecipes = 2;
_numRecipesAllocated = hintAllocation;
_recipes = new byte[_numRecipesAllocated];
_recipes[0] = 3;
_recipes[1] = 7;
_idxA = 0;
_idxB = 1;
_searchSkip = 0;
}
private void Step()
{
if (_numRecipes + 2 >= _numRecipesAllocated)
_numRecipesAllocated *= 2;
byte[] newRecipes = new byte[_numRecipesAllocated];
for (int i = 0; i < _numRecipes; i++)
{
_numRecipesAllocated *= 2;
byte[] newRecipes = new byte[_numRecipesAllocated];
for (int i = 0; i < _numRecipes; i++)
{
newRecipes[i] = _recipes[i];
}
_recipes = newRecipes;
newRecipes[i] = _recipes[i];
}
int newRecipe = _recipes[_idxA] + _recipes[_idxB];
if (newRecipe >= 10)
_recipes = newRecipes;
}
int newRecipe = _recipes[_idxA] + _recipes[_idxB];
if (newRecipe >= 10)
{
_recipes[_numRecipes] = 1;
_recipes[_numRecipes + 1] = (byte)(newRecipe % 10);
_numRecipes += 2;
}
else
{
_recipes[_numRecipes] = (byte)(newRecipe % 10);
_numRecipes += 1;
}
_idxA = (_idxA + _recipes[_idxA] + 1) % _numRecipes;
_idxB = (_idxB + _recipes[_idxB] + 1) % _numRecipes;
}
private void Print()
{
Console.WriteLine();
for (int i = 0; i < _numRecipes; i++)
{
if (i == _idxA)
{
_recipes[_numRecipes] = 1;
_recipes[_numRecipes + 1] = (byte)(newRecipe % 10);
_numRecipes += 2;
Console.Write(" ({0})", _recipes[i]);
}
else if (i == _idxB)
{
Console.Write(" [{0}]", _recipes[i]);
}
else
{
_recipes[_numRecipes] = (byte)(newRecipe % 10);
_numRecipes += 1;
Console.Write(" {0} ", _recipes[i]);
}
_idxA = (_idxA + _recipes[_idxA] + 1) % _numRecipes;
_idxB = (_idxB + _recipes[_idxB] + 1) % _numRecipes;
}
private void Print()
{
Console.WriteLine();
for (int i = 0; i < _numRecipes; i++)
{
if (i == _idxA)
{
Console.Write(" ({0})", _recipes[i]);
}
else if (i == _idxB)
{
Console.Write(" [{0}]", _recipes[i]);
}
else
{
Console.Write(" {0} ", _recipes[i]);
}
}
}
private long SearchPattern(byte[] pattern)
{
long i;
for (i = _searchSkip; i < (_numRecipes - pattern.Length); i++)
{
long j;
for (j = 0; j < pattern.Length; j++)
{
if (_recipes[i + j] != pattern[j])
{
break;
}
}
if (j == pattern.Length)
{
return i;
}
}
_searchSkip = i - 1;
if (_searchSkip < 0) { _searchSkip = 0; }
return -1;
}
private bool Show { get; set; } = false;
public string ResolvePart1(string[] inputs)
{
int numSkipRecipes = Convert.ToInt32(inputs[0]);
Init(numSkipRecipes + 20);
do
{
if (Show) { Print(); }
Step();
} while (_numRecipes < (numSkipRecipes + 10));
StringBuilder sb = new StringBuilder();
for (int i = 0; i < 10; i++)
{
sb.Append(_recipes[numSkipRecipes + i]);
}
return sb.ToString();
}
public string ResolvePart2(string[] inputs)
{
byte[] pattern = inputs[0].Select(c => (byte)(c - '0')).ToArray();
Init();
long position = -1;
do
{
if (Show) { Print(); }
Step();
position = SearchPattern(pattern);
} while (position < 0);
return position.ToString();
}
}
}
private long SearchPattern(byte[] pattern)
{
long i;
for (i = _searchSkip; i < (_numRecipes - pattern.Length); i++)
{
long j;
for (j = 0; j < pattern.Length; j++)
{
if (_recipes[i + j] != pattern[j])
{
break;
}
}
if (j == pattern.Length)
{
return i;
}
}
_searchSkip = i - 1;
if (_searchSkip < 0) { _searchSkip = 0; }
return -1;
}
private bool Show { get; set; } = false;
public string ResolvePart1(string[] inputs)
{
int numSkipRecipes = Convert.ToInt32(inputs[0]);
Init(numSkipRecipes + 20);
do
{
if (Show) { Print(); }
Step();
} while (_numRecipes < (numSkipRecipes + 10));
StringBuilder sb = new();
for (int i = 0; i < 10; i++)
{
sb.Append(_recipes[numSkipRecipes + i]);
}
return sb.ToString();
}
public string ResolvePart2(string[] inputs)
{
byte[] pattern = inputs[0].Select(c => (byte)(c - '0')).ToArray();
Init();
long position;
do
{
if (Show) { Print(); }
Step();
position = SearchPattern(pattern);
} while (position < 0);
return position.ToString();
}
}

1464
AdventOfCode2018/Day15.cs
View File

File diff suppressed because it is too large Load Diff

355
AdventOfCode2018/Day23.cs
View File

@@ -1,208 +1,203 @@
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace AdventOfCode2018
namespace AdventOfCode2018;
/*
--- Day 23: Experimental Emergency Teleportation ---
Using your torch to search the darkness of the rocky cavern, you finally locate the man's friend: a small reindeer.
You're not sure how it got so far in this cave. It looks sick - too sick to walk - and too heavy for you to carry all the way back. Sleighs won't be invented for another 1500 years, of course.
The only option is experimental emergency teleportation.
You hit the "experimental emergency teleportation" button on the device and push I accept the risk on no fewer than 18 different warning messages. Immediately, the device deploys hundreds of tiny nanobots which fly around the cavern, apparently assembling themselves into a very specific formation. The device lists the X,Y,Z position (pos) for each nanobot as well as its signal radius (r) on its tiny screen (your puzzle input).
Each nanobot can transmit signals to any integer coordinate which is a distance away from it less than or equal to its signal radius (as measured by Manhattan distance). Coordinates a distance away of less than or equal to a nanobot's signal radius are said to be in range of that nanobot.
Before you start the teleportation process, you should determine which nanobot is the strongest (that is, which has the largest signal radius) and then, for that nanobot, the total number of nanobots that are in range of it, including itself.
For example, given the following nanobots:
pos=<0,0,0>, r=4
pos=<1,0,0>, r=1
pos=<4,0,0>, r=3
pos=<0,2,0>, r=1
pos=<0,5,0>, r=3
pos=<0,0,3>, r=1
pos=<1,1,1>, r=1
pos=<1,1,2>, r=1
pos=<1,3,1>, r=1
The strongest nanobot is the first one (position 0,0,0) because its signal radius, 4 is the largest. Using that nanobot's location and signal radius, the following nanobots are in or out of range:
The nanobot at 0,0,0 is distance 0 away, and so it is in range.
The nanobot at 1,0,0 is distance 1 away, and so it is in range.
The nanobot at 4,0,0 is distance 4 away, and so it is in range.
The nanobot at 0,2,0 is distance 2 away, and so it is in range.
The nanobot at 0,5,0 is distance 5 away, and so it is not in range.
The nanobot at 0,0,3 is distance 3 away, and so it is in range.
The nanobot at 1,1,1 is distance 3 away, and so it is in range.
The nanobot at 1,1,2 is distance 4 away, and so it is in range.
The nanobot at 1,3,1 is distance 5 away, and so it is not in range.
In this example, in total, 7 nanobots are in range of the nanobot with the largest signal radius.
Find the nanobot with the largest signal radius. How many nanobots are in range of its signals?
--- Part Two ---
Now, you just need to figure out where to position yourself so that you're actually teleported when the nanobots activate.
To increase the probability of success, you need to find the coordinate which puts you in range of the largest number of nanobots. If there are multiple, choose one closest to your position (0,0,0, measured by manhattan distance).
For example, given the following nanobot formation:
pos=<10,12,12>, r=2
pos=<12,14,12>, r=2
pos=<16,12,12>, r=4
pos=<14,14,14>, r=6
pos=<50,50,50>, r=200
pos=<10,10,10>, r=5
Many coordinates are in range of some of the nanobots in this formation. However, only the coordinate 12,12,12 is in range of the most nanobots: it is in range of the first five, but is not in range of the nanobot at 10,10,10. (All other coordinates are in range of fewer than five nanobots.) This coordinate's distance from 0,0,0 is 36.
Find the coordinates that are in range of the largest number of nanobots. What is the shortest manhattan distance between any of those points and 0,0,0?
*/
public class Day23 : IDay
{
/*
--- Day 23: Experimental Emergency Teleportation ---
Using your torch to search the darkness of the rocky cavern, you finally locate the man's friend: a small reindeer.
You're not sure how it got so far in this cave. It looks sick - too sick to walk - and too heavy for you to carry all the way back. Sleighs won't be invented for another 1500 years, of course.
The only option is experimental emergency teleportation.
You hit the "experimental emergency teleportation" button on the device and push I accept the risk on no fewer than 18 different warning messages. Immediately, the device deploys hundreds of tiny nanobots which fly around the cavern, apparently assembling themselves into a very specific formation. The device lists the X,Y,Z position (pos) for each nanobot as well as its signal radius (r) on its tiny screen (your puzzle input).
Each nanobot can transmit signals to any integer coordinate which is a distance away from it less than or equal to its signal radius (as measured by Manhattan distance). Coordinates a distance away of less than or equal to a nanobot's signal radius are said to be in range of that nanobot.
Before you start the teleportation process, you should determine which nanobot is the strongest (that is, which has the largest signal radius) and then, for that nanobot, the total number of nanobots that are in range of it, including itself.
For example, given the following nanobots:
pos=<0,0,0>, r=4
pos=<1,0,0>, r=1
pos=<4,0,0>, r=3
pos=<0,2,0>, r=1
pos=<0,5,0>, r=3
pos=<0,0,3>, r=1
pos=<1,1,1>, r=1
pos=<1,1,2>, r=1
pos=<1,3,1>, r=1
The strongest nanobot is the first one (position 0,0,0) because its signal radius, 4 is the largest. Using that nanobot's location and signal radius, the following nanobots are in or out of range:
The nanobot at 0,0,0 is distance 0 away, and so it is in range.
The nanobot at 1,0,0 is distance 1 away, and so it is in range.
The nanobot at 4,0,0 is distance 4 away, and so it is in range.
The nanobot at 0,2,0 is distance 2 away, and so it is in range.
The nanobot at 0,5,0 is distance 5 away, and so it is not in range.
The nanobot at 0,0,3 is distance 3 away, and so it is in range.
The nanobot at 1,1,1 is distance 3 away, and so it is in range.
The nanobot at 1,1,2 is distance 4 away, and so it is in range.
The nanobot at 1,3,1 is distance 5 away, and so it is not in range.
In this example, in total, 7 nanobots are in range of the nanobot with the largest signal radius.
Find the nanobot with the largest signal radius. How many nanobots are in range of its signals?
--- Part Two ---
Now, you just need to figure out where to position yourself so that you're actually teleported when the nanobots activate.
To increase the probability of success, you need to find the coordinate which puts you in range of the largest number of nanobots. If there are multiple, choose one closest to your position (0,0,0, measured by manhattan distance).
For example, given the following nanobot formation:
pos=<10,12,12>, r=2
pos=<12,14,12>, r=2
pos=<16,12,12>, r=4
pos=<14,14,14>, r=6
pos=<50,50,50>, r=200
pos=<10,10,10>, r=5
Many coordinates are in range of some of the nanobots in this formation. However, only the coordinate 12,12,12 is in range of the most nanobots: it is in range of the first five, but is not in range of the nanobot at 10,10,10. (All other coordinates are in range of fewer than five nanobots.) This coordinate's distance from 0,0,0 is 36.
Find the coordinates that are in range of the largest number of nanobots. What is the shortest manhattan distance between any of those points and 0,0,0?
*/
public class Day23 : IDay
public string ResolvePart1(string[] inputs)
{
public string ResolvePart1(string[] inputs)
List<NanoBot> nanoBots = NanoBot.ListFromStrings(inputs);
NanoBot bestNanoBot = nanoBots.OrderBy(nanoBot => nanoBot.Range).LastOrDefault();
int countInRange = nanoBots.Where(nanoBot => bestNanoBot.InRange(nanoBot)).Count();
return countInRange.ToString();
}
public string ResolvePart2(string[] inputs)
{
List<NanoBot> nanoBots = NanoBot.ListFromStrings(inputs);
long maxX = long.MinValue;
long maxY = long.MinValue;
long maxZ = long.MinValue;
long minX = long.MaxValue;
long minY = long.MaxValue;
long minZ = long.MaxValue;
foreach(NanoBot nanoBot in nanoBots)
{
List<NanoBot> nanoBots = NanoBot.ListFromStrings(inputs);
NanoBot bestNanoBot = nanoBots.OrderBy(nanoBot => nanoBot.Range).LastOrDefault();
int countInRange = nanoBots.Where(nanoBot => bestNanoBot.InRange(nanoBot)).Count();
return countInRange.ToString();
if (nanoBot.X < minX) { minX = nanoBot.X; }
if (nanoBot.X > maxX) { maxX = nanoBot.X; }
if (nanoBot.Y < minY) { minY = nanoBot.Y; }
if (nanoBot.Y > maxY) { maxY = nanoBot.Y; }
if (nanoBot.Z < minZ) { minZ = nanoBot.Z; }
if (nanoBot.Z > maxZ) { maxZ = nanoBot.Z; }
}
public string ResolvePart2(string[] inputs)
{
List<NanoBot> nanoBots = NanoBot.ListFromStrings(inputs);
long maxX = long.MinValue;
long maxY = long.MinValue;
long maxZ = long.MinValue;
long minX = long.MaxValue;
long minY = long.MaxValue;
long minZ = long.MaxValue;
foreach(NanoBot nanoBot in nanoBots)
{
if (nanoBot.X < minX) { minX = nanoBot.X; }
if (nanoBot.X > maxX) { maxX = nanoBot.X; }
if (nanoBot.Y < minY) { minY = nanoBot.Y; }
if (nanoBot.Y > maxY) { maxY = nanoBot.Y; }
if (nanoBot.Z < minZ) { minZ = nanoBot.Z; }
if (nanoBot.Z > maxZ) { maxZ = nanoBot.Z; }
}
long sizeX = maxX - minX;
long sizeY = maxY - minY;
long sizeZ = maxZ - minZ;
long scale = Math.Min(sizeX, Math.Min(sizeY, sizeZ));
long sizeX = maxX - minX;
long sizeY = maxY - minY;
long sizeZ = maxZ - minZ;
long scale = Math.Min(sizeX, Math.Min(sizeY, sizeZ));
do
{
scale /= 2;
if (scale <= 0) { scale = 1; }
do
{
scale /= 2;
if (scale <= 0) { scale = 1; }
long bestX = 0;
long bestY = 0;
long bestZ = 0;
long bestCount = 0;
for (long k = minZ; k <= maxZ; k += scale)
long bestX = 0;
long bestY = 0;
long bestZ = 0;
long bestCount = 0;
for (long k = minZ; k <= maxZ; k += scale)
{
for (long j = minY; j <= maxY; j += scale)
{
for (long j = minY; j <= maxY; j += scale)
for (long i = minX; i <= maxX; i += scale)
{
for (long i = minX; i <= maxX; i += scale)
int count = 0;
foreach(NanoBot nanoBot in nanoBots)
{
int count = 0;
foreach(NanoBot nanoBot in nanoBots)
{
if (nanoBot.InRange(i, j, k, scale)) { count++; }
}
if(count> bestCount)
{
bestX = i;
bestY = j;
bestZ = k;
bestCount = count;
}
if (nanoBot.InRange(i, j, k, scale)) { count++; }
}
if(count> bestCount)
{
bestX = i;
bestY = j;
bestZ = k;
bestCount = count;
}
}
}
}
minX = bestX - scale;
maxX = bestX + scale;
minY = bestY - scale;
maxY = bestY + scale;
minZ = bestZ - scale;
maxZ = bestZ + scale;
minX = bestX - scale;
maxX = bestX + scale;
minY = bestY - scale;
maxY = bestY + scale;
minZ = bestZ - scale;
maxZ = bestZ + scale;
if(scale == 1)
{
long distance = bestX + bestY + bestZ;
return distance.ToString();
}
if(scale == 1)
{
long distance = bestX + bestY + bestZ;
return distance.ToString();
}
} while (true);
} while (true);
}
public class NanoBot
{
public long X { get; set; }
public long Y { get; set; }
public long Z { get; set; }
public long Range { get; set; }
public static NanoBot FromString(string strInput)
{
string[] parts = strInput.Split(new[] { "pos=<", ",", ">, r=", }, StringSplitOptions.RemoveEmptyEntries);
if (parts.Length != 4) { return null; }
NanoBot nanoBot = new() {
X = Convert.ToInt64(parts[0]),
Y = Convert.ToInt64(parts[1]),
Z = Convert.ToInt64(parts[2]),
Range = Convert.ToInt64(parts[3]),
};
return nanoBot;
}
public class NanoBot
public static List<NanoBot> ListFromStrings(string[] inputs)
{
public long X { get; set; }
public long Y { get; set; }
public long Z { get; set; }
public long Range { get; set; }
List<NanoBot> nanoBots = inputs
.Select(strInput => FromString(strInput))
.Where(nanoBot => nanoBot != null)
.ToList();
return nanoBots;
}
public static NanoBot FromString(string strInput)
{
string[] parts = strInput.Split(new string[] { "pos=<", ",", ">, r=", }, StringSplitOptions.RemoveEmptyEntries);
if (parts.Length != 4) { return null; }
NanoBot nanoBot = new NanoBot
{
X = Convert.ToInt64(parts[0]),
Y = Convert.ToInt64(parts[1]),
Z = Convert.ToInt64(parts[2]),
Range = Convert.ToInt64(parts[3]),
};
return nanoBot;
}
public long ManhattanDistance(NanoBot other)
{
return ManhattanDistance(other.X, other.Y, other.Z);
}
public static List<NanoBot> ListFromStrings(string[] inputs)
{
List<NanoBot> nanoBots = inputs
.Select(strInput => FromString(strInput))
.Where(nanoBot => nanoBot != null)
.ToList();
return nanoBots;
}
public long ManhattanDistance(long x, long y, long z)
{
long distance = Math.Abs(X - x) + Math.Abs(Y - y) + Math.Abs(Z - z);
return distance;
}
public long ManhattanDistance(NanoBot other)
{
return ManhattanDistance(other.X, other.Y, other.Z);
}
public bool InRange(NanoBot other)
{
long distance = ManhattanDistance(other);
return distance <= Range;
}
public long ManhattanDistance(long x, long y, long z)
{
long distance = Math.Abs(X - x) + Math.Abs(Y - y) + Math.Abs(Z - z);
return distance;
}
public bool InRange(NanoBot other)
{
long distance = ManhattanDistance(other);
return distance <= Range;
}
public bool InRange(long x, long y, long z, long scale)
{
long distance = (long)Math.Ceiling(ManhattanDistance(x, y, z) / (double)scale);
long range = (long)Math.Ceiling(Range / (double)scale);
return distance <= range;
}
public bool InRange(long x, long y, long z, long scale)
{
long distance = (long)Math.Ceiling(ManhattanDistance(x, y, z) / (double)scale);
long range = (long)Math.Ceiling(Range / (double)scale);
return distance <= range;
}
}
}
}

11
AdventOfCode2018/IDay.cs
View File

@@ -1,8 +1,7 @@
namespace AdventOfCode2018
namespace AdventOfCode2018;
public interface IDay
{
public interface IDay
{
string ResolvePart1(string[] inputs);
string ResolvePart2(string[] inputs);
}
string ResolvePart1(string[] inputs);
string ResolvePart2(string[] inputs);
}

67
AdventOfCode2018/Program.cs
View File

@@ -1,42 +1,41 @@
using System;
using System.IO;
namespace AdventOfCode2018
namespace AdventOfCode2018;
internal class Program
{
internal class Program
private static void Main()
{
private static void Main(string[] args)
int currentDayNumber = 15;
IDay currentDay = null;
switch (currentDayNumber)
{
int currentDayNumber = 15;
IDay currentDay = null;
switch (currentDayNumber)
{
case 1: currentDay = new Day01(); break;
case 2: currentDay = new Day02(); break;
case 3: currentDay = new Day03(); break;
case 4: currentDay = new Day04(); break;
case 5: currentDay = new Day05(); break;
case 6: currentDay = new Day06(); break;
case 7: currentDay = new Day07(); break;
case 8: currentDay = new Day08(); break;
case 9: currentDay = new Day09(); break;
case 10: currentDay = new Day10(); break;
case 11: currentDay = new Day11(); break;
case 12: currentDay = new Day12(); break;
case 13: currentDay = new Day13(); break;
case 14: currentDay = new Day14(); break;
case 15: currentDay = new Day15(); break;
case 23: currentDay = new Day23(); break;
}
string[] linesDay = File.ReadAllLines(string.Format("inputs/Day{0:00}.txt", currentDayNumber));
string resultPart1 = currentDay.ResolvePart1(linesDay);
Console.WriteLine("Day{1:00} Result Part1: {0}", resultPart1, currentDayNumber);
string resultPart2 = currentDay.ResolvePart2(linesDay);
Console.WriteLine("Day{1:00} Result Part2: {0}", resultPart2, currentDayNumber);
Console.Read();
case 1: currentDay = new Day01(); break;
case 2: currentDay = new Day02(); break;
case 3: currentDay = new Day03(); break;
case 4: currentDay = new Day04(); break;
case 5: currentDay = new Day05(); break;
case 6: currentDay = new Day06(); break;
case 7: currentDay = new Day07(); break;
case 8: currentDay = new Day08(); break;
case 9: currentDay = new Day09(); break;
case 10: currentDay = new Day10(); break;
case 11: currentDay = new Day11(); break;
case 12: currentDay = new Day12(); break;
case 13: currentDay = new Day13(); break;
case 14: currentDay = new Day14(); break;
case 15: currentDay = new Day15(); break;
case 23: currentDay = new Day23(); break;
}
string[] linesDay = File.ReadAllLines($"inputs/Day{currentDayNumber:00}.txt");
string resultPart1 = currentDay.ResolvePart1(linesDay);
Console.WriteLine("Day{1:00} Result Part1: {0}", resultPart1, currentDayNumber);
string resultPart2 = currentDay.ResolvePart2(linesDay);
Console.WriteLine("Day{1:00} Result Part2: {0}", resultPart2, currentDayNumber);
Console.Read();
}
}
}