6.3.90.900
8 Day 8
| (require aoc-racket/day08) | package: aoc-racket |
The puzzle. Our input consists of a list of seemingly random strings within quotation marks.
<day08> ::=
8.1 What’s the difference between the literal length of the strings, and their length in memory?
The puzzle relies the fact that within strings, certain single characters — like the backslash \ and double-quote mark " — are described with more than one character. Thus, the question asks us to compare the two lengths.
The literal length of the string is trivial — use string-length. The memory length requires interpreting a string as a Racket value, which (as seen in Day 7) simply means using read.
<day08-setup> ::=
(require racket rackunit) (provide (all-defined-out))
<day08-q1> ::=
(define (memory-length str) (string-length (read (open-input-string str)))) (define (q1 strs) (- (apply + (map string-length strs)) (apply + (map memory-length strs))))
8.2 What’s the difference between the re-encoded length of the literal string, and the original length?
This question simply comes down to — do you know how to use the string-formatting functions in your programming language?
In Racket, a string can be re-encoded with ~v. Not a very puzzling puzzle overall.
<day08-q2> ::=
(define (encoded-length str) (string-length (~v str))) (define (q2 strs) (- (apply + (map encoded-length strs)) (apply + (map string-length strs))))
8.3 Testing Day 8
<day08-test> ::=
(module+ test (define input-strs (file->lines "day08-input.txt")) (check-equal? (q1 input-strs) 1333) (check-equal? (q2 input-strs) 2046))