I have a question about Day 1 Part 2. I’ll put it in a thread to avoid spoilers.

Did anyone generalize to N, or did you just modify your solution for 2 to work specifically for 3?

I’ve added a <https://github.com/lojic/LearningRacket/tree/master/advent-of-code–2020|Advent of Code 2020 directory> to my <https://github.com/lojic/LearningRacket|Learning Racket repo>

I adapted mine. Now that I squint generalizing it shouldn’t be too hard.

I had to think a bit harder than I should have to create find-n - seems like a perfect problem for lisps.

I’m not super pleased with the end result, so if anyone comes up with a more elegant find-n, let me know: https://github.com/lojic/LearningRacket/blob/09b032f0ae7edcbb493954eec4e0bc53a417e24f/advent-of-code-2020/day1b.rkt#L28-L48

Here’s what I got: https://github.com/samdphillips/aoc-2020/blob/efa479d52b3edb5fd2c48cb40e2694aa1a65e026/01.rkt#L12-L22

Does macro-based generalization count? :stuck_out_tongue: https://github.com/Bogdanp/aoc2020/blob/8a43b3e6fc7c7bad6c4f050f4b05c542e41b9977/day01.rkt#L26-L38

I’m scared, and intrigued, to see what you have cooked up.

Looking at Racket’s cartesian-product: (define (cartesian-product . ls) (define (cp-2 as bs) (for*/list ([i (in-list as)] [j (in-list bs)]) (cons i j))) (foldr cp-2 (list (list)) ls)) I’m pretty sure there is something similarly elegant that combines elements in a way needed for this task.

I’m fine with not short circuiting - probably better to find all solutions anyway.

https://math.stackexchange.com/questions/531231/find-all-permutations-in-increasing-order

Ok, I feel more comfortable with the following solution to a slightly different problem: (define (ascending-permutations lst n) (reverse (let loop ([ lst lst ][ n n ][ stack '() ][ result '() ]) (if (= n 0) (cons (reverse stack) result) (if (null? lst) result (let ([ ans (loop (cdr lst) (sub1 n) (cons (car lst) stack) result) ]) (loop (cdr lst) n stack ans))))))) Produces all the “ascending permutations” - I’m really not sure what to call them.

Maybe making this into a generator would do the job.

Hmm… no need for the ans variable: (define (ascending-permutations lst n) (reverse (let loop ([ lst lst ][ n n ][ stack '() ][ result '() ]) (if (= n 0) (cons (reverse stack) result) (if (null? lst) result (loop (cdr lst) n stack (loop (cdr lst) (sub1 n) (cons (car lst) stack) result)))))))