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| author | Louis Burda <quent.burda@gmail.com> | 2023-04-08 12:40:30 -0400 |
|---|---|---|
| committer | Louis Burda <quent.burda@gmail.com> | 2023-04-09 10:21:36 -0400 |
| commit | 9282e95e8844afe856ba76ceb6d2c3010df8bb1a (patch) | |
| tree | e35affc89b20324371381e079f7cb5f8a06aa81b /src/09/part2 | |
| parent | 2b5d4232879dc74491dabf54a0ddc958d66ebcec (diff) | |
| download | aoc2020-zig-master.tar.gz aoc2020-zig-master.zip | |
Diffstat (limited to 'src/09/part2')
| -rw-r--r-- | src/09/part2 | 21 |
1 files changed, 21 insertions, 0 deletions
diff --git a/src/09/part2 b/src/09/part2 new file mode 100644 index 0000000..674e22d --- /dev/null +++ b/src/09/part2 @@ -0,0 +1,21 @@ +--- Part Two --- + +The final step in breaking the XMAS encryption relies on the invalid number you just found: you must +[1m[37mfind a contiguous set of at least two numbers[0m in your list which sum to the invalid +number from step 1. + +Again consider the above example: + +35 20 [1m[37m15[0m [1m[37m25[0m [1m[37m47[0m [1m[37m40[0m 62 55 65 95 102 117 150 182 +127 219 299 277 309 576 + +In this list, adding up all of the numbers from 15 through 40 produces the invalid number from step +1, 127. (Of course, the contiguous set of numbers in your actual list might be much longer.) + +To find the [1m[37mencryption weakness[0m, add together the [1m[37msmallest[0m and +[1m[37mlargest[0m number in this contiguous range; in this example, these are 15 and 47, +producing [1m[37m62[0m. + +[1m[37mWhat is the encryption weakness in your XMAS-encrypted list of numbers?[0m + + |
