96 lines
4.4 KiB
Plaintext
96 lines
4.4 KiB
Plaintext
Advent of Code
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--- Day 16: Dragon Checksum ---
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You're done scanning this part of the network, but you've left traces of your presence. You need to
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overwrite some disks with random-looking data to cover your tracks and update the local security system
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with a new checksum for those disks.
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For the data to not be suspiscious, it needs to have certain properties; purely random data will be
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detected as tampering. To generate appropriate random data, you'll need to use a modified dragon curve.
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Start with an appropriate initial state (your puzzle input). Then, so long as you don't have enough data
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yet to fill the disk, repeat the following steps:
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• Call the data you have at this point "a".
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• Make a copy of "a"; call this copy "b".
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• Reverse the order of the characters in "b".
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• In "b", replace all instances of 0 with 1 and all 1s with 0.
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• The resulting data is "a", then a single 0, then "b".
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For example, after a single step of this process,
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• 1 becomes 100.
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• 0 becomes 001.
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• 11111 becomes 11111000000.
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• 111100001010 becomes 1111000010100101011110000.
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Repeat these steps until you have enough data to fill the desired disk.
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Once the data has been generated, you also need to create a checksum of that data. Calculate the checksum
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only for the data that fits on the disk, even if you generated more data than that in the previous step.
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The checksum for some given data is created by considering each non-overlapping pair of characters in the
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input data. If the two characters match (00 or 11), the next checksum character is a 1. If the characters
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do not match (01 or 10), the next checksum character is a 0. This should produce a new string which is
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exactly half as long as the original. If the length of the checksum is even, repeat the process until you
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end up with a checksum with an odd length.
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For example, suppose we want to fill a disk of length 12, and when we finally generate a string of at
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least length 12, the first 12 characters are 110010110100. To generate its checksum:
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• Consider each pair: 11, 00, 10, 11, 01, 00.
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• These are same, same, different, same, different, same, producing 110101.
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• The resulting string has length 6, which is even, so we repeat the process.
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• The pairs are 11 (same), 01 (different), 01 (different).
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• This produces the checksum 100, which has an odd length, so we stop.
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Therefore, the checksum for 110010110100 is 100.
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Combining all of these steps together, suppose you want to fill a disk of length 20 using an initial state
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of 10000:
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• Because 10000 is too short, we first use the modified dragon curve to make it longer.
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• After one round, it becomes 10000011110 (11 characters), still too short.
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• After two rounds, it becomes 10000011110010000111110 (23 characters), which is enough.
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• Since we only need 20, but we have 23, we get rid of all but the first 20 characters:
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10000011110010000111.
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• Next, we start calculating the checksum; after one round, we have 0111110101, which 10 characters long
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(even), so we continue.
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• After two rounds, we have 01100, which is 5 characters long (odd), so we are done.
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In this example, the correct checksum would therefore be 01100.
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The first disk you have to fill has length 272. Using the initial state in your puzzle input, what is the
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correct checksum?
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Your puzzle answer _____________________.
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The first half of this puzzle is complete! It provides one gold star: *
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--- Part Two ---
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The second disk you have to fill has length 35651584. Again using the initial state in your puzzle input,
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what is the correct checksum for this disk?
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Your puzzle input is still 00111101111101000.
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Answer: _____________________
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References
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Visible links
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. https://en.wikipedia.org/wiki/Dragon_curve
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