day8, example3 new approach

day8/notes.org

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+#+title: Notes
+* ok, for part 2 i think i'm supposed to find cicles.
+so for each starting point?
+calculate how long it takes to get to first Z, let's call is initialN
+and then how long it takes to get to Z again, call it cicleN
+
+i have initial0 + n0 * cicle0 = k
+
+now i need to find minimal number K ( steps ), so that exists solution for each starting point
+
+(k - initial0) % cycle0 == 0
+
+so. get both these numbers for each starting point.
+if initialN are equal, that's the answer.
+
+if they are not equal?
+then i have what? huh
+
+i guess then i need to add 1 cycle length to all.
+if they are equal, nice.
+
+but if they are not equal?
+i guess i have one that's longest?
+then i need to advance all paths, up to and overshoot if necessary by their cycle lengths.
+so take the difference, if divisible by cycle - yay. just multiply, if not divisible - overshoot with +1 cycle
+
+if all are same, that's solution.
+if one is farthest, repeat
+* i stopped bruteforce at 4080000000