Thinking about day 3, solving the Manhattan distance seems to be the easy part. The naïve solution might be generating a matrix but it’s probably better to find some shortcut for calculating the position of any number in the spiral. 🤔

Plugging the sequence of the first few lower-right corners (1, 9, 25, 49) into Wolfram Alpha gives a closed form of $a_n = 4 n^2 - 4 n + 1$, or in Lisp:

(λ (n)
(+ (- (* 4 (* n n))
(* 4 n))
1))

Not sure where I’m going with this, though. Maybe other corners or edges share a similar closed form.

@amdt I just took the naïve approach of generating the grid completely. It made part 2 very simple at the end though.

Also for the audience at home, Part 2 is an Ulam spiral.

my (unfortunately imperative) solution: https://github.com/a-red-christmas/aoc2017-fs/blob/master/day3/Program.fs

@amdt errrrr i'm a moron, part 1 is the ulam spiral

@calvin I hadn’t heard of the Ulam spiral (thanks!) but if part 2 isn’t particularly about primes I might just go with the naïve solution.