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aoc-racket/day25.rkt

81 lines
2.3 KiB
Racket

9 years ago
#lang scribble/lp2
@(require scribble/manual aoc-racket/helper)
@aoc-title[25]
@defmodule[aoc-racket/day25]
@link["http://adventofcode.com/day/25"]{The puzzle}. Our @link-rp["day25-input.txt"]{input} is a row and column number for a grid.
@chunk[<day25>
<day25-setup>
<day25-q1>
<day25-test>]
@section{What code do you give the machine?}
The puzzle involves a series of codes in a grid. The first code is @racket[20151125] and subsequent codes are computed by multiplying by @racket[252533] and calculating the remainder after dividing by @racket[33554393]. Then the codes are put into a grid in diagonal fashion, in the order shown:
@tt{
  |  1  2  3  4  5  6
@(linebreak)
++++++++++++++++++++++ @(linebreak)
1 |  1  3  6 10 15 21 @(linebreak)
2 |  2  5  9 14 20 @(linebreak)
3 |  4  8 13 19 @(linebreak)
4 |  7 12 18 @(linebreak)
5 | 11 17 @(linebreak)
6 | 16}
The grid is described as infinite. But this doesn't play a role in the puzzle, since our job is to find the number at the finite coordinate given in our input. So we need a function for finding the @italic{n}th code in the series, and a function for finding which @italic{n} resides at a given grid location. Short & sweet.
@chunk[<day25-setup>
(require racket rackunit)
(provide (all-defined-out))
]
@chunk[<day25-q1>
(define first-code 20151125)
(define (next-code code)
(modulo (* code 252533) 33554393))
(define (nth-code n)
(for/fold ([code-so-far first-code])
([i (in-range (sub1 n))])
(next-code code-so-far)))
(define (rc->n row col)
(define first-col-val (add1 (apply + (range row))))
(define col-offset-val (apply + (range (add1 row) (+ row col))))
(+ first-col-val col-offset-val))
(define (q1 input-str)
(match-define (list _ row col)
(map string->number
(regexp-match #px"row (\\d+), column (\\d+)" input-str)))
(nth-code (rc->n row col)))
]
@section{Testing Day 25}
@chunk[<day25-test>
(module+ test
(define input-str (file->string "day25-input.txt"))
(check-equal? (q1 input-str) 19980801))]