My solution to Advent of Code - Day 8

The trickiest part of Day 8 is Part 2 which I failed to solve by myself (I tried a bruteforce approach) and peeked at others solution.

Since I have not seen a very good explanation of why using the least common multiple of the number of steps to reach Z$ for all possible paths, here is my breakdown:

Advent Of Code - Day 8, Part 2 Explainer

The key is to notice three things:

a) There are cycles in both the sample graph (sample3.txt) and the puzzle input graph.

b) The exists only one cycle for each walk from node ending in A to node ending in Z and that cycle is the cycle that starts and ends at node Z. in other words: there is a unique path p from A to Z given the map instructions (L or R).

c) The number of steps n it takes to reach a node that ends in Z from a node that ends in A (i.e. the length of the path A->Z) is EQUAL to the number of steps of the cycle i.e. looping from node ending in Z back to node ending in Z.

For instance in the sample graph, the length of the path 11A -> 11Z is: |11A -> 11B -> 11Z| = 2 which is equal to the length of the cycle: |11Z -> 11B -> 11Z| = 2

Similarly for Path p' 22A -> 22Z: | 22A -> 22B -> 22C -> 22Z | = 3 which is equal to the length of the cycle: | 22Z -> 22B -> 22C -> 22Z | = 3

Where does the least common multiple come into play?

In the example sample graph, the 11Z node will always be reached in an 2 + 2*k steps | k = 1,2,3,.. 2 steps for path 11A -> 11Z + multiple of 2 for the cycle 11Z->11Z.

Similarly, the node 22Z will always be reached in 3*z steps | z = 1,2,3,...

To reach 22Z and 11Z simultaneously, the number of steps need to be equal i.e. 3*(z) == 2*(k)

Hence, the problem of finding the number of steps it takes to reach nodes ending in Z simultaneously is converted into finding the least common multiple between 2 and 3.

Generally, the solution is given by lcm(steps_to_reach_z_path1, steps_to_reach_z_path2, ..., steps_to_reach_z_pathN)