16

2021, 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).

Part 1

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 999.

Part 2

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 3408662834145.

Solution Notes

There is nothing special about this puzzle: follow the instructions to the letter, and that's it. Turning this into a somewhat golfed solution was the greater challenge.

• Part 1, Python: 297 bytes, <100 ms
• Part 2, Python: 438 bytes, <100 ms