Day 10: Elves Look, Elves Say
Picat: Both parts (only a single digit needs to change)
Count the length of the look and say sequence iterated N times over a given input.
I'd read about this sequence before and remembered that it grew exponentially, so I figured part 2 might be something like "now iterate 10,000 times", requiring a solution that doesn't take linear space. But I also remembered that only the digits 1, 2 and 3 appear in the sequence (unless the input string contains other digits), which means a sort of run-length encoded representation (which is really what the sequence produces) would be useless. So I tried the simple implementation, using pattern matching on the input string to check first for a run of 3 identical characters, then 2, then 1.
This worked just fine for 40 iterations.
...And it also looked perfectly fine, requiring 1.7 seconds compared to the 50ms or so needed by part 1. I could have easily beaten my record for fastest split, but I wasted a few seconds staring at the console.
Another nice one for Picat, only let down a little by the syntax for pattern matching successive list elements. Haskell lets you do something like a:b:c:d:xs rather than Picat's [A|[A|[A|[A[Xs]]]]], but on the other hand, the trick of using the variable A in multiple parts of the pattern is not legal in Haskell, so you have to add a guard: | a == b && a == c && ....
None.