# --- Day 10: Knot Hash ---

# You come across some programs that are trying to implement a software
# emulation of a hash based on knot-tying. The hash these programs are
# implementing isn't very strong, but you decide to help them anyway. You make
# a mental note to remind the Elves later not to invent their own cryptographic
# functions.

# This hash function simulates tying a knot in a circle of string with 256
# marks on it. Based on the input to be hashed, the function repeatedly selects
# a span of string, brings the ends together, and gives the span a half-twist
# to reverse the order of the marks within it. After doing this many times, the
# order of the marks is used to build the resulting hash.

#   4--5   pinch   4  5           4   1
#  /    \  5,0,1  / \/ \  twist  / \ / \
# 3      0  -->  3      0  -->  3   X   0
#  \    /         \ /\ /         \ / \ /
#   2--1           2  1           2   5

# To achieve this, begin with a list of numbers from 0 to 255, a current
# position which begins at 0 (the first element in the list), a skip size
# (which starts at 0), and a sequence of lengths (your puzzle input). Then, for
# each length:

#     Reverse the order of that length of elements in the list, starting with
# the element at the current position.
#     Move the current position forward by that length plus the skip size.
#     Increase the skip size by one.

# The list is circular; if the current position and the length try to reverse
# elements beyond the end of the list, the operation reverses using as many
# extra elements as it needs from the front of the list. If the current
# position moves past the end of the list, it wraps around to the front.
# Lengths larger than the size of the list are invalid.

# Here's an example using a smaller list:

# Suppose we instead only had a circular list containing five elements, 0, 1,
# 2, 3, 4, and were given input lengths of 3, 4, 1, 5.

#     The list begins as [0] 1 2 3 4 (where square brackets indicate the
# current position).
#     The first length, 3, selects ([0] 1 2) 3 4 (where parentheses indicate
# the sublist to be reversed).
#     After reversing that section (0 1 2 into 2 1 0), we get ([2] 1 0) 3 4.
#     Then, the current position moves forward by the length, 3, plus the skip
# size, 0: 2 1 0 [3] 4. Finally, the skip size increases to 1.

#     The second length, 4, selects a section which wraps: 2 1) 0 ([3] 4.
#     The sublist 3 4 2 1 is reversed to form 1 2 4 3: 4 3) 0 ([1] 2.
#     The current position moves forward by the length plus the skip size, a
# total of 5, causing it not to move because it wraps around: 4 3 0 [1] 2. The
# skip size increases to 2.

#     The third length, 1, selects a sublist of a single element, and so
# reversing it has no effect.
#     The current position moves forward by the length (1) plus the skip size
# (2): 4 [3] 0 1 2. The skip size increases to 3.

#     The fourth length, 5, selects every element starting with the second:
# 4) ([3] 0 1 2. Reversing this sublist (3 0 1 2 4 into 4 2 1 0 3) produces:
# 3) ([4] 2 1 0.
#     Finally, the current position moves forward by 8: 3 4 2 1 [0]. The skip
# size increases to 4.

# In this example, the first two numbers in the list end up being 3 and 4; to
# check the process, you can multiply them together to produce 12.

# However, you should instead use the standard list size of 256 (with values 0
# to 255) and the sequence of lengths in your puzzle input. Once this process
# is complete, what is the result of multiplying the first two numbers in the
# list?

from collections import deque
from typing import Deque, Tuple

with open("files/P10.txt") as f:
    lengths = [int(num) for num in f.read().strip().split(",")]


def simulated_hash(
    circle: Deque[int], sequence: list, curr_pos: int, skip: int
) -> Tuple[Deque[int], int, int]:
    for length in sequence:
        circle.rotate(-curr_pos)
        # creates a copy of the list
        rotated_l = list(circle)
        # reverse the order of the partial list
        rotated_l[:length] = reversed(rotated_l[:length])
        # as the list is circular, it wraps around when it has to
        circle = deque(rotated_l)
        circle.rotate(curr_pos)
        # move forward the current position
        curr_pos = (curr_pos + length + skip) % 256
        skip += 1
    return circle, curr_pos, skip


def part_1() -> None:
    list_of_numbers = deque(list(range(256)))
    curr_pos, skip = 0, 0
    simulated_hash_res, *_ = simulated_hash(
        list_of_numbers, lengths, curr_pos, skip
    )
    res = simulated_hash_res[0] * simulated_hash_res[1]

    print(f"The result is {res}")


# --- 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.


with open("files/P10.txt") as f:
    ascii_lengths = [ord(c) for c in f.read().strip()]
    ascii_lengths.extend([17, 31, 73, 47, 23])


def part_2():
    list_of_numbers = deque(list(range(256)))
    curr_pos, skip = 0, 0
    for _ in range(64):
        list_of_numbers, curr_pos, skip = simulated_hash(
            list_of_numbers, ascii_lengths, curr_pos, skip
        )

    sparse = list(list_of_numbers)
    dense = []

    for block in range(0, 256, 16):
        hashed = 0
        group = sparse[block : block + 16]
        for n in group:
            hashed ^= n
        dense.append(hashed)

    res = "".join(f"{n:02x}" for n in dense)
    print(f"The Knot Hash of the input is {res}")


if __name__ == "__main__":
    part_1()
    part_2()