Day 5 is wrapped in a story about plane ticket seat finding, but really it boils down to a simple binary to integer conversion, and then finding the difference of two sets and cleaning up what’s left based on some simple rules.

Challenge

The puzzle can be found here. I’m given seat numbers using a “binary space partitioning” strategy. The idea is that the first character, B for back or F for front identifies if the seat is in the back or front half of the plane. The second B or F identifies if the seat is in the back or front half of that previously chosen half. There are exactly 128 rows, so seven Bs and/or Fs determine the row number. Then there are three Ls and/or Rs to determine a column. To make a seat ID, just multiple the row by eight and add the column. One example input would be FBFBBFFRLR, which represents row 44, column 5, seat id 357.

Solution

Part 1

I’ve done enough CTFs working with binary data to immediately recognize this for what it is - binary.

  FBFBBFFRLR
      |
      v
  0101100101
      |
      v
 0101100 101
    |     |
    v     v
    44    5

I can also notice that the seat id is just original string as binary:

357 = 44 * 8 + 5
    = (44 << 3) | 5
    = (0101100 << 3) | 101
    = 0101100000 | 101
    = 0101100101

This is the given ticket! This makes finding the seat id super easy. Replace Fs and Ls with 0, Bs and Rs with 1s, and convert to an int.

#!/usr/bin/env python3

import sys


with open(sys.argv[1], "r") as f:
    seats = (
        f.read()
        .strip()
        .replace("B", "1")
        .replace("F", "0")
        .replace("R", "1")
        .replace("L", "0")
        .split("\n")
    )

print(f"Part 1: {max([int(s, 2) for s in seats])}")

For each seat, convert it to binary, and grab the max in the array.

Part 2

In part two, I’m asked to go through all the seats, and find mine. There will be 128 * 8 = 1024 seats on the plane, but my puzzle input has only 781 filled seats:

$ wc -l 05-puzzle_input.txt 
781 05-puzzle_input.txt

It turns out there will be seats at the front and back of the plain that aren’t filled. My seat will have someone in both the the id one higher and one lower.

I’ll adjust the code from above to create a set of claimed seats, and then create the set of all numbers 0-1023, and find the difference, leaving the open seats.

This will produce a list like this:

>>> openseats = allseats - claimed
>>> len(openseats)
243
>>> openseats
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 517, 833, 834, 835, 836, 837, 838, 839, 840, 841, 842, 843, 844, 845, 846, 847, 848, 849, 850, 851, 852, 853, 854, 855, 856, 857, 858, 859, 860, 861, 862, 863, 864, 865, 866, 867, 868, 869, 870, 871, 872, 873, 874, 875, 876, 877, 878, 879, 880, 881, 882, 883, 884, 885, 886, 887, 888, 889, 890, 891, 892, 893, 894, 895, 896, 897, 898, 899, 900, 901, 902, 903, 904, 905, 906, 907, 908, 909, 910, 911, 912, 913, 914, 915, 916, 917, 918, 919, 920, 921, 922, 923, 924, 925, 926, 927, 928, 929, 930, 931, 932, 933, 934, 935, 936, 937, 938, 939, 940, 941, 942, 943, 944, 945, 946, 947, 948, 949, 950, 951, 952, 953, 954, 955, 956, 957, 958, 959, 960, 961, 962, 963, 964, 965, 966, 967, 968, 969, 970, 971, 972, 973, 974, 975, 976, 977, 978, 979, 980, 981, 982, 983, 984, 985, 986, 987, 988, 989, 990, 991, 992, 993, 994, 995, 996, 997, 998, 999, 1000, 1001, 1002, 1003, 1004, 1005, 1006, 1007, 1008, 1009, 1010, 1011, 1012, 1013, 1014, 1015, 1016, 1017, 1018, 1019, 1020, 1021, 1022, 1023}

I could just look through that list and find the seat without it’s neighbors, but completeness I’ll use a list comprehension and have it pull the one number:

#!/usr/bin/env python3

import sys


with open(sys.argv[1], "r") as f:
    seats = (
        f.read()
        .strip()
        .replace("B", "1")
        .replace("F", "0")
        .replace("R", "1")
        .replace("L", "0")
        .split("\n")
    )

claimed = set([int(s, 2) for s in seats])
print(f"Part 1: {max(claimed)}")

allseats = set(range(128 * 8))
openseats = allseats - claimed
myseat = [
    seat
    for seat in openseats
    if seat + 1 not in openseats and seat - 1 not in openseats
]

print(f"Part 2: {myseat[0]}")

Time is still not yet a factor this year, so this also solves instantly:

$ time python3 day5.py 05-puzzle_input.txt 
Part 1: 832
Part 2: 517

real    0m0.061s
user    0m0.041s
sys     0m0.016s