Advent of Racket 2023/04 - Scratchcards

Day four starts out very simple. We're given an input where we have two lists of numbers per line. Per line (or "card"), we get one point for the first number in the second list that is also found in the first, and the score doubles for every subsequent match. The example input looks like this:

Card 1: 41 48 83 86 17 | 83 86  6 31 17  9 48 53
Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19
Card 3:  1 21 53 59 44 | 69 82 63 72 16 21 14  1
Card 4: 41 92 73 84 69 | 59 84 76 51 58  5 54 83
Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36
Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11

Since part one seemed a little too easy, I decided to parse the input into a struct, in anticipation of a harder part two.

;; id : int
;; winning : listof int
;; have : listof int
(struct card (id winning have)

Parsing the cards is straightforward enough, though I did spend a minute scratching my head because I forgot to escape the | character:

(define cards
  (call-with-input-file "day04.txt"
    (lambda (in)
      (for/vector ([line (in-lines in)])
        (match-define (regexp #rx"Card +([0-9]+): ([^|]+) \\| (.+)"
                              (list _ (app string->number id) winning-str have-str))
         (map string->number (string-split winning-str))
         (map string->number (string-split have-str)))))))

I had originally stored the set of cards as a list, but changed it to a vector for part two. We'll see why in a bit. In the mean time, computing part one is just a matter of determining the number of matches in each card:

(define (card-matches c)
  (for/sum ([n (in-list (card-have c))]
            #:when (memv n (card-winning c)))

And computing the score per card:

(define (card-score c)
  (define matches
    (card-matches c))
    [(zero? matches) 0]
    [else (expt 2 (sub1 matches))]))

Putting those functions together, we get:

(define part1
  (for/sum ([c (in-vector cards)])
    (card-score c)))

For part two, the problem goes exponential. For every card, the number of matches that we find represents subsequent cards that we have to check for matches. We have to recursively add up all the cards we see.

We need a function that returns the won cards for any given card:

(define (card-wins c)
  (match-define (card id _winning _have) c)
  (for/list ([i (in-range 0 (card-matches c))])
    (vector-ref cards (+ id i))))

This function is the reason why I stored the cards as a vector earlier on. With this function in hand, we can now write a function to compute the number of cards seen when starting from any given card:

(define (add-counts cs)
  (apply hash-union cs #:combine +))

(define (card-counts c)
    (hasheqv (card-id c) 1)
    (map card-counts (card-wins c)))))

Running card-counts on the first card in the example input yields:

> (card-counts (vector-ref cards 0))
'#hasheqv((1 . 1) (2 . 1) (3 . 2) (4 . 4) (5 . 7))

To compute the result for part two we just have to add up all the counts for all the cards we have:

(apply + (hash-values
           (for/list ([c (in-vector cards)])
             (card-counts c)))))

This works fine for the example input, but the real input is much larger and requires many more iterations. The trick to notice here is that calling card-counts on an individual card will always return the same result, so we can simply memoize the result for every card and greatly reduce the number of iterations required to compute the solution.

All we have to do is change card-counts to:

(define card-counts
  (let ([memo (make-hasheqv)])
    (lambda (c)
       memo (card-id c)
       (lambda ()
           (hasheqv (card-id c) 1)
           (map card-counts (card-wins c)))))))))

The memo hash keeps a mapping of card ids to the number of cards seen when starting from that card. If a card's id is already in the hash, we return its associated value, otherwise we compute the result and store it in the hash for subsequent lookups.

That's it for day four!