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 \(1 + 3 = 2 + 2\).

Assignment

What do you get if you evaluate the expression represented by your hexadecimal-encoded BITS transmission? Determine this in the following way:

• Write a function decode that takes the pathname (str) of a text file containing a hexadecimal-encoded BITS transmission. The function must return the evaluation (int) the expression represented by the transmission.

Example

In this interactive session we assume the text files transmission01.txt¹, transmission02.txt², transmission03.txt³, transmission04.txt⁴ and transmission05.txt⁵ to be located in the current directory.

>>> decode('transmission01.txt')
15
>>> decode('transmission02.txt')
46
>>> decode('transmission03.txt')
46
>>> decode('transmission04.txt')
54
>>> decode('transmission05.txt')
10637009915279