You're minding your own business on a ship at sea when the overboard alarm goes off! You rush to see if you can help. Apparently, one of the Elves tripped and accidentally sent the sleigh keys flying into the ocean! Before you know it, you're inside a submarine the Elves keep ready for situations like this. It's covered in Christmas lights (because of course it is), and it even has an experimental antenna that should be able to track the keys if you can boost its signal strength high enough; there's a little meter that indicates the antenna's signal strength by displaying 0-50 stars. Your instincts tell you that in order to save Christmas, you'll 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! As the submarine drops below the surface of the ocean, it automatically performs a sonar sweep of the nearby sea floor. On a small screen, the sonar sweep report (your puzzle input) appears: each line is a measurement of the sea floor depth as the sweep looks further and further away from the submarine. For example, suppose you had the following report: 199 200 208 210 200 207 240 269 260 263 This report indicates that, scanning outward from the submarine, the sonar sweep found depths of  199 ,  200 ,  208 ,  210 , and so on. The first order of business is to figure out how quickly the depth increases, just so you know what you're dealing with - you never know if the keys will get carried into deeper water by an ocean current or a fish or something. To do this, count the number of times a depth measurement increases from the previous measurement. (There is no measurement before the first measurement.) In the example above, the changes are as follows: 199 (N/A - no previous measurement) 200 (increased) 208 (increased) 210 (increased) 200 (decreased) 207 (increased) 240 (increased) 269 (increased) 260 (decreased) 263 (increased) In this example, there are  7  measurements that are larger than the previous measurement. How many measurements are larger than the previous measurement?   All credit for the above puzzle goes to Eric Wastl and Advent of Code. Don't forget to submit your solutions at adventofcode.com and join the KX leaderboard using code 1526329-bb0915a5. ... View more

There's not much to do as you slowly descend to the bottom of the ocean. The submarine computer challenges you to a nice game of Dirac Dice. This game consists of a single die, two pawns, and a game board with a circular track containing ten spaces marked  1  through  10  clockwise. Each player's starting space is chosen randomly (your puzzle input). Player 1 goes first. Players take turns moving. On each player's turn, the player rolls the die three times and adds up the results. Then, the player moves their pawn that many times forward around the track (that is, moving clockwise on spaces in order of increasing value, wrapping back around to  1  after  10 ). So, if a player is on space  7  and they roll  2 ,  2 , and  1 , they would move forward 5 times, to spaces  8 ,  9 ,  10 ,  1 , and finally stopping on  2 . After each player moves, they increase their score by the value of the space their pawn stopped on. Players' scores start at  0 . So, if the first player starts on space  7  and rolls a total of  5 , they would stop on space  2  and add  2  to their score (for a total score of  2 ). The game immediately ends as a win for any player whose score reaches at least  1000 . Since the first game is a practice game, the submarine opens a compartment labeled deterministic dice and a 100-sided die falls out. This die always rolls  1  first, then  2 , then  3 , and so on up to  100 , after which it starts over at  1  again. Play using this die. For example, given these starting positions: Player 1 starting position: 4 Player 2 starting position: 8 This is how the game would go: Player 1 rolls  1 + 2 + 3  and moves to space  10  for a total score of  10 . Player 2 rolls  4 + 5 + 6  and moves to space  3  for a total score of  3 . Player 1 rolls  7 + 8 + 9  and moves to space  4  for a total score of  14 . Player 2 rolls  10 + 11 + 12  and moves to space  6  for a total score of  9 . Player 1 rolls  13 + 14 + 15  and moves to space  6  for a total score of  20 . Player 2 rolls  16 + 17 + 18  and moves to space  7  for a total score of  16 . Player 1 rolls  19 + 20 + 21  and moves to space  6  for a total score of  26 . Player 2 rolls  22 + 23 + 24  and moves to space  6  for a total score of  22 . ...after many turns... Player 2 rolls  82 + 83 + 84  and moves to space  6  for a total score of  742 . Player 1 rolls  85 + 86 + 87  and moves to space  4  for a total score of  990 . Player 2 rolls  88 + 89 + 90  and moves to space  3  for a total score of  745 . Player 1 rolls  91 + 92 + 93  and moves to space  10  for a final score,  1000 . Since player 1 has at least  1000  points, player 1 wins and the game ends. At this point, the losing player had  745  points and the die had been rolled a total of  993  times;  745 * 993 = 739785 . Play a practice game using the deterministic 100-sided die. The moment either player wins, what do you get if you multiply the score of the losing player by the number of times the die was rolled during the game?   All credit for the above puzzle goes to Eric Wastl and Advent of Code. Don't forget to submit your solutions at adventofcode.com and join the KX leaderboard using code 1526329-bb0915a5. ... View more

With the scanners fully deployed, you turn their attention to mapping the floor of the ocean trench. When you get back the image from the scanners, it seems to just be random noise. Perhaps you can combine an image enhancement algorithm and the input image (your puzzle input) to clean it up a little. For example: ..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..## #..######.###...####..#..#####..##..#.#####...##.#.#..#.##..#.#......#.### .######.###.####...#.##.##..#..#..#####.....#.#....###..#.##......#.....#. .#..#..##..#...##.######.####.####.#.#...#.......#..#.#.#...####.##.#..... .#..#...##.#.##..#...##.#.##..###.#......#.#.......#.#.#.####.###.##...#.. ...####.#..#..#.##.#....##..#.####....##...##..#...#......#.#.......#..... ..##..####..#...#.#.#...##..#.#..###..#####........#..####......#..# #..#. #.... ##..# ..#.. ..### The first section is the image enhancement algorithm. It is normally given on a single line, but it has been wrapped to multiple lines in this example for legibility. The second section is the input image, a two-dimensional grid of light pixels ( # ) and dark pixels ( . ). The image enhancement algorithm describes how to enhance an image by simultaneously converting all pixels in the input image into an output image. Each pixel of the output image is determined by looking at a 3x3 square of pixels centered on the corresponding input image pixel. So, to determine the value of the pixel at (5,10) in the output image, nine pixels from the input image need to be considered: (4,9), (4,10), (4,11), (5,9), (5,10), (5,11), (6,9), (6,10), and (6,11). These nine input pixels are combined into a single binary number that is used as an index in the image enhancement algorithm string. For example, to determine the output pixel that corresponds to the very middle pixel of the input image, the nine pixels marked by  [...]  would need to be considered: # . . # . #[. . .]. #[# . .]# .[. # .]. . . # # # Starting from the top-left and reading across each row, these pixels are  ... , then  #.. , then  .#. ; combining these forms  ...#...#. . By turning dark pixels ( . ) into  0  and light pixels ( # ) into  1 , the binary number  000100010  can be formed, which is  34  in decimal. The image enhancement algorithm string is exactly 512 characters long, enough to match every possible 9-bit binary number. The first few characters of the string (numbered starting from zero) are as follows: 0 10 20 30 34 40 50 60 70 | | | | | | | | | ..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..## In the middle of this first group of characters, the character at index 34 can be found:  # . So, the output pixel in the center of the output image should be  # , a light pixel. This process can then be repeated to calculate every pixel of the output image. Through advances in imaging technology, the images being operated on here are infinite in size. Every pixel of the infinite output image needs to be calculated exactly based on the relevant pixels of the input image. The small input image you have is only a small region of the actual infinite input image; the rest of the input image consists of dark pixels ( . ). For the purposes of the example, to save on space, only a portion of the infinite-sized input and output images will be shown. The starting input image, therefore, looks something like this, with more dark pixels ( . ) extending forever in every direction not shown here: ............... ............... ............... ............... ............... .....#..#...... .....#......... .....##..#..... .......#....... .......###..... ............... ............... ............... ............... ............... By applying the image enhancement algorithm to every pixel simultaneously, the following output image can be obtained: ............... ............... ............... ............... .....##.##..... ....#..#.#..... ....##.#..#.... ....####..#.... .....#..##..... ......##..#.... .......#.#..... ............... ............... ............... ............... Through further advances in imaging technology, the above output image can also be used as an input image! This allows it to be enhanced a second time: ............... ............... ............... ..........#.... ....#..#.#..... ...#.#...###... ...#...##.#.... ...#.....#.#... ....#.#####.... .....#.#####... ......##.##.... .......###..... ............... ............... ............... Truly incredible - now the small details are really starting to come through. After enhancing the original input image twice,  35  pixels are lit. Start with the original input image and apply the image enhancement algorithm twice, being careful to account for the infinite size of the images. How many pixels are lit in the resulting image?   All credit for the above puzzle goes to Eric Wastl and Advent of Code. Don't forget to submit your solutions at adventofcode.com and join the KX leaderboard using code 1526329-bb0915a5. ... View more

Now, you need to figure out how to pilot this thing. It seems like the submarine can take a series of commands like  forward 1 ,  down 2 , or  up 3 : forward X  increases the horizontal position by  X  units. down X  increases the depth by  X  units. up X  decreases the depth by  X  units. Note that since you're on a submarine,  down  and  up  affect your depth, and so they have the opposite result of what you might expect. The submarine seems to already have a planned course (your puzzle input). You should probably figure out where it's going. For example: forward 5 down 5 forward 8 up 3 down 8 forward 2 Your horizontal position and depth both start at  0 . The steps above would then modify them as follows: forward 5  adds  5  to your horizontal position, a total of  5 . down 5  adds  5  to your depth, resulting in a value of  5 . forward 8  adds  8  to your horizontal position, a total of  13 . up 3  decreases your depth by  3 , resulting in a value of  2 . down 8  adds  8  to your depth, resulting in a value of  10 . forward 2  adds  2  to your horizontal position, a total of  15 . After following these instructions, you would have a horizontal position of  15  and a depth of  10 . (Multiplying these together produces  150 .) Calculate the horizontal position and depth you would have after following the planned course.  What do you get if you multiply your final horizontal position by your final depth?   All credit for the above puzzle goes to Eric Wastl and Advent of Code. Don't forget to submit your solutions at adventofcode.com and join the KX leaderboard using code 1526329-bb0915a5. ... View more

A giant whale has decided your submarine is its next meal, and it's much faster than you are. There's nowhere to run! Suddenly, a swarm of crabs (each in its own tiny submarine - it's too deep for them otherwise) zooms in to rescue you! They seem to be preparing to blast a hole in the ocean floor; sensors indicate a massive underground cave system just beyond where they're aiming! The crab submarines all need to be aligned before they'll have enough power to blast a large enough hole for your submarine to get through. However, it doesn't look like they'll be aligned before the whale catches you! Maybe you can help? There's one major catch - crab submarines can only move horizontally. You quickly make a list of the horizontal position of each crab (your puzzle input). Crab submarines have limited fuel, so you need to find a way to make all of their horizontal positions match while requiring them to spend as little fuel as possible. For example, consider the following horizontal positions: 16,1,2,0,4,2,7,1,2,14 This means there's a crab with horizontal position  16 , a crab with horizontal position  1 , and so on. Each change of 1 step in horizontal position of a single crab costs 1 fuel. You could choose any horizontal position to align them all on, but the one that costs the least fuel is horizontal position  2 : Move from  16  to  2 :  14  fuel Move from  1  to  2 :  1  fuel Move from  2  to  2 :  0  fuel Move from  0  to  2 :  2  fuel Move from  4  to  2 :  2  fuel Move from  2  to  2 :  0  fuel Move from  7  to  2 :  5  fuel Move from  1  to  2 :  1  fuel Move from  2  to  2 :  0  fuel Move from  14  to  2 :  12  fuel This costs a total of  37  fuel. This is the cheapest possible outcome; more expensive outcomes include aligning at position  1  ( 41  fuel), position  3  ( 39  fuel), or position  10  ( 71  fuel). Determine the horizontal position that the crabs can align to using the least fuel possible. How much fuel must they spend to align to that position?   All credit for the above puzzle goes to Eric Wastl and Advent of Code. Don't forget to submit your solutions at adventofcode.com and join the KX leaderboard using code 1526329-bb0915a5. ... View more