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knot_hash_part_two.py
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knot_hash_part_two.py
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"""
--- Part Two ---
The logic you've constructed forms a single round of the Knot Hash algorithm;
running the full thing requires many of these rounds. Some input and output
processing is also required.
First, from now on, your input should be taken not as a list of numbers, but as
a string of bytes instead. Unless otherwise specified, convert characters to
bytes using their ASCII codes. This will allow you to handle arbitrary ASCII
strings, and it also ensures that your input lengths are never larger than 255.
For example, if you are given 1,2,3, you should convert it to the ASCII codes
for each character: 49,44,50,44,51.
Once you have determined the sequence of lengths to use, add the following
lengths to the end of the sequence: 17, 31, 73, 47, 23. For example, if you are
given 1,2,3, your final sequence of lengths should be 49,44,50,44,51,17,31,73,
47,23 (the ASCII codes from the input string combined with the standard length
suffix values).
Second, instead of merely running one round like you did above, run a total of
64 rounds, using the same length sequence in each round. The current position
and skip size should be preserved between rounds. For example, if the previous
example was your first round, you would start your second round with the same
length sequence (3, 4, 1, 5, 17, 31, 73, 47, 23, now assuming they came from
ASCII codes and include the suffix), but start with the previous round's
current position (4) and skip size (4).
Once the rounds are complete, you will be left with the numbers from 0 to 255
in some order, called the sparse hash. Your next task is to reduce these to a
list of only 16 numbers called the dense hash. To do this, use numeric bitwise
XOR to combine each consecutive block of 16 numbers in the sparse hash (there
are 16 such blocks in a list of 256 numbers). So, the first element in the
dense hash is the first sixteen elements of the sparse hash XOR'd together, the
second element in the dense hash is the second sixteen elements of the sparse
hash XOR'd together, etc.
For example, if the first sixteen elements of your sparse hash are as shown
below, and the XOR operator is ^, you would calculate the first output number
like this:
65 ^ 27 ^ 9 ^ 1 ^ 4 ^ 3 ^ 40 ^ 50 ^ 91 ^ 7 ^ 6 ^ 0 ^ 2 ^ 5 ^ 68 ^ 22 = 64
Perform this operation on each of the sixteen blocks of sixteen numbers in your
sparse hash to determine the sixteen numbers in your dense hash.
Finally, the standard way to represent a Knot Hash is as a single hexadecimal
string; the final output is the dense hash in hexadecimal notation. Because
each number in your dense hash will be between 0 and 255 (inclusive), always
represent each number as two hexadecimal digits (including a leading zero as
necessary). So, if your first three numbers are 64, 7, 255, they correspond to
the hexadecimal numbers 40, 07, ff, and so the first six characters of the hash
would be 4007ff. Because every Knot Hash is sixteen such numbers, the
hexadecimal representation is always 32 hexadecimal digits (0-f) long.
Here are some example hashes:
The empty string becomes a2582a3a0e66e6e86e3812dcb672a272.
AoC 2017 becomes 33efeb34ea91902bb2f59c9920caa6cd.
1,2,3 becomes 3efbe78a8d82f29979031a4aa0b16a9d.
1,2,4 becomes 63960835bcdc130f0b66d7ff4f6a5a8e.
Treating your puzzle input as a string of ASCII characters, what is the Knot
Hash of your puzzle input? Ignore any leading or trailing whitespace you might
encounter.
"""
import sys
from functools import reduce
from utils.round import run_round
lengths = []
if len(sys.argv) > 1:
lengths += [ord(i) for i in sys.argv[1]]
lengths += [17, 31, 73, 47, 23]
pos = 0
skip = 0
sparse_hash = list(range(0, 256))
for i in range(0, 64):
sparse_hash, pos, skip = run_round(
lengths, sparse_hash, pos=pos, skip=skip)
dense_hash = ''
for i in range(0, int(len(sparse_hash)/16)):
chunk = sparse_hash[(i * 16):(i * 16 + 16)]
dense_hash += '%0.2X'.lower() % reduce(lambda j, k: j ^ k, chunk)
print(dense_hash)