X-Git-Url: https://git.sommitrealweird.co.uk/advent-of-code-2021.git/blobdiff_plain/9017e242041b51939c88e7e99221923965e919e1..44ccb9f2e0493b5aa32f10f00ab62a8ec3d860d9:/day16/task.txt diff --git a/day16/task.txt b/day16/task.txt new file mode 100644 index 0000000..00fdc60 --- /dev/null +++ b/day16/task.txt @@ -0,0 +1,213 @@ +--- Day 16: Packet Decoder --- + +As you leave the cave and reach open waters, you receive a transmission from +the Elves back on the ship. + +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 +input). + +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: + +0 = 0000 +1 = 0001 +2 = 0010 +3 = 0011 +4 = 0100 +5 = 0101 +6 = 0110 +7 = 0111 +8 = 1000 +9 = 1001 +A = 1010 +B = 1011 +C = 1100 +D = 1101 +E = 1110 +F = 1111 + +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 +transmission and should be ignored. + +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 binary +with the most significant bit first. For example, a version encoded as the +binary sequence 100 represents the number 4. + +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 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: + +110100101111111000101000 +VVVTTTAAAAABBBBBCCCCC + +Below each bit is a label indicating its purpose: + + • The three bits labeled V (110) are the packet version, 6. + • The three bits labeled T (100) are the packet type ID, 4, which means the + packet is a literal value. + • 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. + • 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. + • 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. + • The three unlabeled 0 bits at the end are extra due to the hexadecimal + representation and should be ignored. + +So, this packet represents a literal value with binary representation +011111100101, which is 2021 in decimal. + +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 parsing +the hierarchy of sub-packets. + +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 called +the length type ID: + + • 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. + • 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. + +Finally, after the length type ID bit and the 15-bit or 11-bit field, the +sub-packets appear. + +For example, here is an operator packet (hexadecimal string 38006F45291200) +with length type ID 0 that contains two sub-packets: + +00111000000000000110111101000101001010010001001000000000 +VVVTTTILLLLLLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBBBBBB + + • The three bits labeled V (001) are the packet version, 1. + • The three bits labeled T (110) are the packet type ID, 6, which means the + packet is an operator. + • 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. + • The 15 bits labeled L (000000000011011) contain the length of the + sub-packets in bits, 27. + • The 11 bits labeled A contain the first sub-packet, a literal value + representing the number 10. + • The 16 bits labeled B contain the second sub-packet, a literal value + representing the number 20. + +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. + +As another example, here is an operator packet (hexadecimal string +EE00D40C823060) with length type ID 1 that contains three sub-packets: + +11101110000000001101010000001100100000100011000001100000 +VVVTTTILLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBCCCCCCCCCCC + + • The three bits labeled V (111) are the packet version, 7. + • The three bits labeled T (011) are the packet type ID, 3, which means the + packet is an operator. + • 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. + • The 11 bits labeled L (00000000011) contain the number of sub-packets, 3. + • The 11 bits labeled A contain the first sub-packet, a literal value + representing the number 1. + • The 11 bits labeled B contain the second sub-packet, a literal value + representing the number 2. + • The 11 bits labeled C contain the third sub-packet, a literal value + representing the number 3. + +After reading 3 complete sub-packets, the number of sub-packets indicated in L +(3) is reached, and so parsing of this packet stops. + +For now, parse the hierarchy of the packets throughout the transmission and add +up all of the version numbers. + +Here are a few more examples of hexadecimal-encoded transmissions: + + • 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 of 16. + • 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. + • 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. + • 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. + +Decode the structure of your hexadecimal-encoded BITS transmission; what do you +get if you add up the version numbers in all packets? + +Your puzzle answer was 945. + +--- Part Two --- + +Now that you have the structure of your transmission decoded, you can calculate +the value of the expression it represents. + +Literal values (type ID 4) represent a single number as described above. The +remaining type IDs are more interesting: + + • 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. + • 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. + • Packets with type ID 2 are minimum packets - their value is the minimum of + the values of their sub-packets. + • Packets with type ID 3 are maximum packets - their value is the maximum of + the values of their sub-packets. + • 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 sub-packets. + • 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 sub-packets. + • 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. + +Using these rules, you can now work out the value of the outermost packet in +your BITS transmission. + +For example: + + • C200B40A82 finds the sum of 1 and 2, resulting in the value 3. + • 04005AC33890 finds the product of 6 and 9, resulting in the value 54. + • 880086C3E88112 finds the minimum of 7, 8, and 9, resulting in the value 7. + • CE00C43D881120 finds the maximum of 7, 8, and 9, resulting in the value 9. + • D8005AC2A8F0 produces 1, because 5 is less than 15. + • F600BC2D8F produces 0, because 5 is not greater than 15. + • 9C005AC2F8F0 produces 0, because 5 is not equal to 15. + • 9C0141080250320F1802104A08 produces 1, because 1 + 3 = 2 * 2. + +What do you get if you evaluate the expression represented by your +hexadecimal-encoded BITS transmission? + +Your puzzle answer was 10637009915279. + +Both parts of this puzzle are complete! They provide two gold stars: ** + +At this point, you should return to your Advent calendar and try another +puzzle. + +If you still want to see it, you can get your puzzle input. +