1 --- Day 16: Packet Decoder ---
3 As you leave the cave and reach open waters, you receive a transmission from
4 the Elves back on the ship.
6 The transmission was sent using the Buoyancy Interchange Transmission System (
7 BITS), a method of packing numeric expressions into a binary sequence. Your
8 submarine's computer has saved the transmission in hexadecimal (your puzzle
11 The first step of decoding the message is to convert the hexadecimal
12 representation into binary. Each character of hexadecimal corresponds to four
32 The BITS transmission contains a single packet at its outermost layer which
33 itself contains many other packets. The hexadecimal representation of this
34 packet might encode a few extra 0 bits at the end; these are not part of the
35 transmission and should be ignored.
37 Every packet begins with a standard header: the first three bits encode the
38 packet version, and the next three bits encode the packet type ID. These two
39 values are numbers; all numbers encoded in any packet are represented as binary
40 with the most significant bit first. For example, a version encoded as the
41 binary sequence 100 represents the number 4.
43 Packets with type ID 4 represent a literal value. Literal value packets encode
44 a single binary number. To do this, the binary number is padded with leading
45 zeroes until its length is a multiple of four bits, and then it is broken into
46 groups of four bits. Each group is prefixed by a 1 bit except the last group,
47 which is prefixed by a 0 bit. These groups of five bits immediately follow the
48 packet header. For example, the hexadecimal string D2FE28 becomes:
50 110100101111111000101000
53 Below each bit is a label indicating its purpose:
55 • The three bits labeled V (110) are the packet version, 6.
56 • The three bits labeled T (100) are the packet type ID, 4, which means the
57 packet is a literal value.
58 • The five bits labeled A (10111) start with a 1 (not the last group, keep
59 reading) and contain the first four bits of the number, 0111.
60 • The five bits labeled B (11110) start with a 1 (not the last group, keep
61 reading) and contain four more bits of the number, 1110.
62 • The five bits labeled C (00101) start with a 0 (last group, end of packet)
63 and contain the last four bits of the number, 0101.
64 • The three unlabeled 0 bits at the end are extra due to the hexadecimal
65 representation and should be ignored.
67 So, this packet represents a literal value with binary representation
68 011111100101, which is 2021 in decimal.
70 Every other type of packet (any packet with a type ID other than 4) represent
71 an operator that performs some calculation on one or more sub-packets contained
72 within. Right now, the specific operations aren't important; focus on parsing
73 the hierarchy of sub-packets.
75 An operator packet contains one or more packets. To indicate which subsequent
76 binary data represents its sub-packets, an operator packet can use one of two
77 modes indicated by the bit immediately after the packet header; this is called
80 • If the length type ID is 0, then the next 15 bits are a number that
81 represents the total length in bits of the sub-packets contained by this
83 • If the length type ID is 1, then the next 11 bits are a number that
84 represents the number of sub-packets immediately contained by this packet.
86 Finally, after the length type ID bit and the 15-bit or 11-bit field, the
89 For example, here is an operator packet (hexadecimal string 38006F45291200)
90 with length type ID 0 that contains two sub-packets:
92 00111000000000000110111101000101001010010001001000000000
93 VVVTTTILLLLLLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBBBBBB
95 • The three bits labeled V (001) are the packet version, 1.
96 • The three bits labeled T (110) are the packet type ID, 6, which means the
97 packet is an operator.
98 • The bit labeled I (0) is the length type ID, which indicates that the
99 length is a 15-bit number representing the number of bits in the
101 • The 15 bits labeled L (000000000011011) contain the length of the
102 sub-packets in bits, 27.
103 • The 11 bits labeled A contain the first sub-packet, a literal value
104 representing the number 10.
105 • The 16 bits labeled B contain the second sub-packet, a literal value
106 representing the number 20.
108 After reading 11 and 16 bits of sub-packet data, the total length indicated in
109 L (27) is reached, and so parsing of this packet stops.
111 As another example, here is an operator packet (hexadecimal string
112 EE00D40C823060) with length type ID 1 that contains three sub-packets:
114 11101110000000001101010000001100100000100011000001100000
115 VVVTTTILLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBCCCCCCCCCCC
117 • The three bits labeled V (111) are the packet version, 7.
118 • The three bits labeled T (011) are the packet type ID, 3, which means the
119 packet is an operator.
120 • The bit labeled I (1) is the length type ID, which indicates that the
121 length is a 11-bit number representing the number of sub-packets.
122 • The 11 bits labeled L (00000000011) contain the number of sub-packets, 3.
123 • The 11 bits labeled A contain the first sub-packet, a literal value
124 representing the number 1.
125 • The 11 bits labeled B contain the second sub-packet, a literal value
126 representing the number 2.
127 • The 11 bits labeled C contain the third sub-packet, a literal value
128 representing the number 3.
130 After reading 3 complete sub-packets, the number of sub-packets indicated in L
131 (3) is reached, and so parsing of this packet stops.
133 For now, parse the hierarchy of the packets throughout the transmission and add
134 up all of the version numbers.
136 Here are a few more examples of hexadecimal-encoded transmissions:
138 • 8A004A801A8002F478 represents an operator packet (version 4) which contains
139 an operator packet (version 1) which contains an operator packet (version
140 5) which contains a literal value (version 6); this packet has a version
142 • 620080001611562C8802118E34 represents an operator packet (version 3) which
143 contains two sub-packets; each sub-packet is an operator packet that
144 contains two literal values. This packet has a version sum of 12.
145 • C0015000016115A2E0802F182340 has the same structure as the previous
146 example, but the outermost packet uses a different length type ID. This
147 packet has a version sum of 23.
148 • A0016C880162017C3686B18A3D4780 is an operator packet that contains an
149 operator packet that contains an operator packet that contains five literal
150 values; it has a version sum of 31.
152 Decode the structure of your hexadecimal-encoded BITS transmission; what do you
153 get if you add up the version numbers in all packets?
155 Your puzzle answer was 945.
159 Now that you have the structure of your transmission decoded, you can calculate
160 the value of the expression it represents.
162 Literal values (type ID 4) represent a single number as described above. The
163 remaining type IDs are more interesting:
165 • Packets with type ID 0 are sum packets - their value is the sum of the
166 values of their sub-packets. If they only have a single sub-packet, their
167 value is the value of the sub-packet.
168 • Packets with type ID 1 are product packets - their value is the result of
169 multiplying together the values of their sub-packets. If they only have a
170 single sub-packet, their value is the value of the sub-packet.
171 • Packets with type ID 2 are minimum packets - their value is the minimum of
172 the values of their sub-packets.
173 • Packets with type ID 3 are maximum packets - their value is the maximum of
174 the values of their sub-packets.
175 • Packets with type ID 5 are greater than packets - their value is 1 if the
176 value of the first sub-packet is greater than the value of the second
177 sub-packet; otherwise, their value is 0. These packets always have exactly
179 • Packets with type ID 6 are less than packets - their value is 1 if the
180 value of the first sub-packet is less than the value of the second
181 sub-packet; otherwise, their value is 0. These packets always have exactly
183 • Packets with type ID 7 are equal to packets - their value is 1 if the value
184 of the first sub-packet is equal to the value of the second sub-packet;
185 otherwise, their value is 0. These packets always have exactly two
188 Using these rules, you can now work out the value of the outermost packet in
189 your BITS transmission.
193 • C200B40A82 finds the sum of 1 and 2, resulting in the value 3.
194 • 04005AC33890 finds the product of 6 and 9, resulting in the value 54.
195 • 880086C3E88112 finds the minimum of 7, 8, and 9, resulting in the value 7.
196 • CE00C43D881120 finds the maximum of 7, 8, and 9, resulting in the value 9.
197 • D8005AC2A8F0 produces 1, because 5 is less than 15.
198 • F600BC2D8F produces 0, because 5 is not greater than 15.
199 • 9C005AC2F8F0 produces 0, because 5 is not equal to 15.
200 • 9C0141080250320F1802104A08 produces 1, because 1 + 3 = 2 * 2.
202 What do you get if you evaluate the expression represented by your
203 hexadecimal-encoded BITS transmission?
205 Your puzzle answer was 10637009915279.
207 Both parts of this puzzle are complete! They provide two gold stars: **
209 At this point, you should return to your Advent calendar and try another
212 If you still want to see it, you can get your puzzle input.