165 lines
10 KiB
Plaintext
165 lines
10 KiB
Plaintext
# Advent of Code
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--- Day 16: Packet Decoder ---
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As you leave the cave and reach open waters, you receive a transmission from the Elves back on the ship.
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The transmission was sent using the Buoyancy Interchange Transmission System (BITS), a method of packing numeric expressions into a binary sequence. Your submarine's computer has saved the transmission in hexadecimal (your puzzle
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input).
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The first step of decoding the message is to convert the hexadecimal representation into binary. Each character of hexadecimal corresponds to four bits of binary data:
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0 = 0000
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1 = 0001
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2 = 0010
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3 = 0011
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4 = 0100
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5 = 0101
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6 = 0110
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7 = 0111
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8 = 1000
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9 = 1001
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A = 1010
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B = 1011
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C = 1100
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D = 1101
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E = 1110
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F = 1111
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The BITS transmission contains a single packet at its outermost layer which itself contains many other packets. The hexadecimal representation of this packet might encode a few extra 0 bits at the end; these are not part of the
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transmission and should be ignored.
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Every packet begins with a standard header: the first three bits encode the packet version, and the next three bits encode the packet type ID. These two values are numbers; all numbers encoded in any packet are represented as
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binary with the most significant bit first. For example, a version encoded as the binary sequence 100 represents the number 4.
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Packets with type ID 4 represent a literal value. Literal value packets encode a single binary number. To do this, the binary number is padded with leading zeroes until its length is a multiple of four bits, and then it is broken
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into groups of four bits. Each group is prefixed by a 1 bit except the last group, which is prefixed by a 0 bit. These groups of five bits immediately follow the packet header. For example, the hexadecimal string D2FE28 becomes:
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110100101111111000101000
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VVVTTTAAAAABBBBBCCCCC
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Below each bit is a label indicating its purpose:
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• The three bits labeled V (110) are the packet version, 6.
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• The three bits labeled T (100) are the packet type ID, 4, which means the packet is a literal value.
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• The five bits labeled A (10111) start with a 1 (not the last group, keep reading) and contain the first four bits of the number, 0111.
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• The five bits labeled B (11110) start with a 1 (not the last group, keep reading) and contain four more bits of the number, 1110.
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• The five bits labeled C (00101) start with a 0 (last group, end of packet) and contain the last four bits of the number, 0101.
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• The three unlabeled 0 bits at the end are extra due to the hexadecimal representation and should be ignored.
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So, this packet represents a literal value with binary representation 011111100101, which is 2021 in decimal.
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Every other type of packet (any packet with a type ID other than 4) represent an operator that performs some calculation on one or more sub-packets contained within. Right now, the specific operations aren't important; focus on
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parsing the hierarchy of sub-packets.
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An operator packet contains one or more packets. To indicate which subsequent binary data represents its sub-packets, an operator packet can use one of two modes indicated by the bit immediately after the packet header; this is
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called the length type ID:
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• If the length type ID is 0, then the next 15 bits are a number that represents the total length in bits of the sub-packets contained by this packet.
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• If the length type ID is 1, then the next 11 bits are a number that represents the number of sub-packets immediately contained by this packet.
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Finally, after the length type ID bit and the 15-bit or 11-bit field, the sub-packets appear.
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For example, here is an operator packet (hexadecimal string 38006F45291200) with length type ID 0 that contains two sub-packets:
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00111000000000000110111101000101001010010001001000000000
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VVVTTTILLLLLLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBBBBBB
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• The three bits labeled V (001) are the packet version, 1.
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• The three bits labeled T (110) are the packet type ID, 6, which means the packet is an operator.
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• The bit labeled I (0) is the length type ID, which indicates that the length is a 15-bit number representing the number of bits in the sub-packets.
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• The 15 bits labeled L (000000000011011) contain the length of the sub-packets in bits, 27.
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• The 11 bits labeled A contain the first sub-packet, a literal value representing the number 10.
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• The 16 bits labeled B contain the second sub-packet, a literal value representing the number 20.
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After reading 11 and 16 bits of sub-packet data, the total length indicated in L (27) is reached, and so parsing of this packet stops.
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As another example, here is an operator packet (hexadecimal string EE00D40C823060) with length type ID 1 that contains three sub-packets:
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11101110000000001101010000001100100000100011000001100000
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VVVTTTILLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBCCCCCCCCCCC
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• The three bits labeled V (111) are the packet version, 7.
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• The three bits labeled T (011) are the packet type ID, 3, which means the packet is an operator.
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• The bit labeled I (1) is the length type ID, which indicates that the length is a 11-bit number representing the number of sub-packets.
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• The 11 bits labeled L (00000000011) contain the number of sub-packets, 3.
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• The 11 bits labeled A contain the first sub-packet, a literal value representing the number 1.
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• The 11 bits labeled B contain the second sub-packet, a literal value representing the number 2.
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• The 11 bits labeled C contain the third sub-packet, a literal value representing the number 3.
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After reading 3 complete sub-packets, the number of sub-packets indicated in L (3) is reached, and so parsing of this packet stops.
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For now, parse the hierarchy of the packets throughout the transmission and add up all of the version numbers.
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Here are a few more examples of hexadecimal-encoded transmissions:
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• 8A004A801A8002F478 represents an operator packet (version 4) which contains an operator packet (version 1) which contains an operator packet (version 5) which contains a literal value (version 6); this packet has a version sum
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of 16.
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• 620080001611562C8802118E34 represents an operator packet (version 3) which contains two sub-packets; each sub-packet is an operator packet that contains two literal values. This packet has a version sum of 12.
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• C0015000016115A2E0802F182340 has the same structure as the previous example, but the outermost packet uses a different length type ID. This packet has a version sum of 23.
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• A0016C880162017C3686B18A3D4780 is an operator packet that contains an operator packet that contains an operator packet that contains five literal values; it has a version sum of 31.
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Decode the structure of your hexadecimal-encoded BITS transmission; what do you get if you add up the version numbers in all packets?
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Your puzzle answer was 999.
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--- Part Two ---
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Now that you have the structure of your transmission decoded, you can calculate the value of the expression it represents.
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Literal values (type ID 4) represent a single number as described above. The remaining type IDs are more interesting:
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• Packets with type ID 0 are sum packets - their value is the sum of the values of their sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
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• Packets with type ID 1 are product packets - their value is the result of multiplying together the values of their sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
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• Packets with type ID 2 are minimum packets - their value is the minimum of the values of their sub-packets.
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• Packets with type ID 3 are maximum packets - their value is the maximum of the values of their sub-packets.
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• Packets with type ID 5 are greater than packets - their value is 1 if the value of the first sub-packet is greater than the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two
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sub-packets.
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• Packets with type ID 6 are less than packets - their value is 1 if the value of the first sub-packet is less than the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two
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sub-packets.
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• Packets with type ID 7 are equal to packets - their value is 1 if the value of the first sub-packet is equal to the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two sub-packets.
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Using these rules, you can now work out the value of the outermost packet in your BITS transmission.
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For example:
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• C200B40A82 finds the sum of 1 and 2, resulting in the value 3.
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• 04005AC33890 finds the product of 6 and 9, resulting in the value 54.
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• 880086C3E88112 finds the minimum of 7, 8, and 9, resulting in the value 7.
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• CE00C43D881120 finds the maximum of 7, 8, and 9, resulting in the value 9.
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• D8005AC2A8F0 produces 1, because 5 is less than 15.
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• F600BC2D8F produces 0, because 5 is not greater than 15.
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• 9C005AC2F8F0 produces 0, because 5 is not equal to 15.
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• 9C0141080250320F1802104A08 produces 1, because 1 + 3 = 2 * 2.
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What do you get if you evaluate the expression represented by your hexadecimal-encoded BITS transmission?
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Your puzzle answer was 3408662834145.
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Both parts of this puzzle are complete! They provide two gold stars: **
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At this point, all that is left is for you to admire your Advent calendar.
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If you still want to see it, you can get your puzzle input.
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References
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Visible links
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. https://adventofcode.com/
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. https://adventofcode.com/2021/about
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. https://adventofcode.com/2021/events
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. https://adventofcode.com/2021/settings
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. https://adventofcode.com/2021/auth/logout
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. Advent of Code Supporter
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https://adventofcode.com/2021/support
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. https://adventofcode.com/2021
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. https://adventofcode.com/2021
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. https://adventofcode.com/2021/support
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. https://adventofcode.com/2021/sponsors
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. https://adventofcode.com/2021/leaderboard
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. https://adventofcode.com/2021/stats
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. https://adventofcode.com/2021/sponsors
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. https://en.wikipedia.org/wiki/Hexadecimal
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. https://adventofcode.com/2021
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. https://adventofcode.com/2021/day/16/input
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