121 lines
4.4 KiB
Plaintext
121 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
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your presence. You need to overwrite some disks with random-looking data to
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cover your tracks and update the local security system with a new checksum
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for those disks.
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For the data to not be suspiscious, it needs to have certain properties;
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purely random data will be detected as tampering. To generate appropriate
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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
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as you don't have enough data yet to fill the disk, repeat the following
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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
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data. Calculate the checksum only for the data that fits on the disk, even
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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
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non-overlapping pair of characters in the input data. If the two characters
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match (00 or 11), the next checksum character is a 1. If the characters do
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not match (01 or 10), the next checksum character is a 0. This should
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produce a new string which is exactly half as long as the original. If the
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length of the checksum is even, repeat the process until you end up with a
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checksum with an odd length.
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For example, suppose we want to fill a disk of length 12, and when we
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finally generate a string of at least length 12, the first 12 characters are
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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
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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
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length 20 using an initial state of 10000:
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• Because 10000 is too short, we first use the modified dragon curve to
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make it longer.
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• After one round, it becomes 10000011110 (11 characters), still too
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short.
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• After two rounds, it becomes 10000011110010000111110 (23 characters),
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which is enough.
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• Since we only need 20, but we have 23, we get rid of all but the first
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20 characters: 10000011110010000111.
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• Next, we start calculating the checksum; after one round, we have
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0111110101, which 10 characters long (even), so we continue.
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• After two rounds, we have 01100, which is 5 characters long (odd), so we
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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
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your puzzle input, what is the correct checksum?
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Your puzzle answer 10011010010010010.
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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: 10101011110100011.
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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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