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#lang racket
(require advent-of-code
(only-in relation ->number ->symbol))
(struct monkey (name op) #:transparent)
(struct op (f first second) #:transparent)
(define (parse-monkey str)
(match (string-split str " ")
[(list (app (curryr string-trim ":") name) name1 (app ->symbol f) name2)
(monkey name (op f name1 name2))]
[(list (app (curryr string-trim ":") name) (app ->number int))
(monkey name (op 'constant int #f))]))
(define raw-monkeys (port->lines (open-aoc-input (find-session) 2022 21 #:cache #true)))
(define monkey-table
(for/hash ([m raw-monkeys] #:do [(match-define (monkey name op) (parse-monkey m))])
(values name op)))
;; part 1
(define (evaluate-monkey m-name [guess #f])
(match-define (op f name1 name2)
(if (and guess (equal? m-name "humn")) (op 'constant guess #f) (hash-ref monkey-table m-name)))
(match f
['constant name1]
['+ (+ (evaluate-monkey name1 guess) (evaluate-monkey name2 guess))]
['- (- (evaluate-monkey name1 guess) (evaluate-monkey name2 guess))]
['* (* (evaluate-monkey name1 guess) (evaluate-monkey name2 guess))]
['/ (/ (evaluate-monkey name1 guess) (evaluate-monkey name2 guess))]))
(evaluate-monkey "root")
;; part 2
;; since humn only ever appears once, and it's never the divisor in a division operation,
;; the difference of the branches is linearly proportional to humn
;; therefore, if we find two points we can calculate the root directly
(match-define (op _ branch-1 branch-2) (hash-ref monkey-table "root"))
(define known-side (evaluate-monkey branch-2))
(define humn-zero (- known-side (evaluate-monkey branch-1 0)))
(define humn-one (- known-side (evaluate-monkey branch-1 1)))
(- (/ humn-zero (- humn-one humn-zero)))
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