这里是使用 Racket 完成 HackerRank 函数式编程题目的第三章，包含函数关系和几何计算（22 - 26 题）。
仓库：13m0n4de/HackerRank-FP-Solutions
22-Functions or Not?
数学中函数的定义为：
对于每个定义域中的元素（x 值），有唯一的值域元素（y 值）与之对应。
如果测试用例中所有 x 值都唯一地映射到 y 值，那么这些关系可以代表一个有效的函数。
函数签名：
valid-function? : Listof(Integer) -> Boolean
#lang racket ;; valid-function? : Listof(Integer) -> Boolean (define (valid-function? pairs) (define mapping (make-vector 501 -1)) (for/and ([pair pairs]) (match-let ([(list x y) pair]) (let ([mapped-y (vector-ref mapping x)]) (cond [(= mapped-y -1) (vector-set! mapping x y) #t] [(= mapped-y y) #t] [else #f]))))) (define (read-test-case) (for/list ([_ (in-range (read))]) (list (read) (read)))) (define (read-test-cases) (for/list ([_ (in-range (read))]) (read-test-case))) (define test-cases (read-test-cases)) (for ([test-case test-cases]) (if (valid-function? test-case) (displayln "YES") (displayln "NO")))
for/and 会在条件失败时短路。
读取时用 read-line 会比较麻烦，因为测试用例中带有 \r，需要指定 'return-linefeed。用 read 简单些。
23-Compute the Perimeter of a Polygon
函数签名：
distance : (Listof Integer) (Listof Integer) -> Number perimeter : (Listof (Listof Integer)) -> Number
依次计算每个边的长度（两点距离），加在一起就是周长。
#lang racket ;; distance : (Listof Integer) (Listof Integer) -> Number (define (distance point-a point-b) (let ([ax (first point-a)] [ay (second point-a)] [bx (first point-b)] [by (second point-b)]) (sqrt (+ (expt (- bx ax) 2) (expt (- by ay) 2))))) ;; perimeter : (Listof (Listof Integer)) -> Number (define (perimeter points) (define first-point (first points)) (let perimeter-helper ([acc 0.0] [remaining-points (cdr points)] [prev-point first-point]) (if (empty? remaining-points) (+ acc (distance prev-point first-point)) (perimeter-helper (+ acc (distance prev-point (first remaining-points))) (cdr remaining-points) (first remaining-points))))) (define (read-points) (for/list ([_ (in-range (read))]) (list (read) (read)))) (displayln (perimeter (read-points)))
24-Compute the Area of a Polygon
计算多边形面积的鞋带公式 (Shoelace formula)：
\[ S={\frac {1}{2}}{\begin{vmatrix}x_{1}&x_{2}&x_{3}&{...}&x_{n}\\y_{1}&y_{2}&y_{3}&{...}&y_{n}\end{vmatrix}}={\frac {1}{2}}\left({\begin{vmatrix}x_{1}&x_{2}\\y_{1}&{y_{2}}\end{vmatrix}}+{\begin{vmatrix}x_{2}&x_{3}\\y_{2}&{y_{3}}\end{vmatrix}}+...+{\begin{vmatrix}x_{n-1}&x_{n}\\y_{n-1}&{y_{n}}\end{vmatrix}}+{\begin{vmatrix}x_{n}&x_{1}\\y_{n}&{y_{1}}\end{vmatrix}}\right) \]
函数签名：
cross-product : (Listof Integer) (Listof Integer) -> Number area : (Listof (Listof Integer)) -> Number
#lang racket ;; cross-product : (Listof Integer) (Listof Integer) -> Number (define (cross-product point-a point-b) (let ([ax (first point-a)] [ay (second point-a)] [bx (first point-b)] [by (second point-b)]) (- (* ax by) (* bx ay)))) ;; area : (Listof (Listof Integer)) -> Number (define (area points) (define first-point (first points)) (let shoelace ([acc 0.0] [remaining-points (cdr points)] [prev-point first-point]) (if (empty? remaining-points) (/ (abs (+ acc (cross-product prev-point first-point))) 2) (shoelace (+ acc (cross-product prev-point (first remaining-points))) (cdr remaining-points) (first remaining-points))))) (define (read-points) (for/list ([_ (in-range (read))]) (list (read) (read)))) (displayln (area (read-points)))
和计算周长时差不多，在每次计算时更新前后点和剩余点，并保存第一个点用于最后一个叉乘项的计算。
25-Computing the GCD
改进版的欧几里得算法：Euclidean_algorithm
函数签名：
gcd : Integer Integer -> Integer
#lang racket ;; gcd : Integer Integer -> Integer (define (gcd x y) (let [(r (remainder x y))] (if (= r 0) y (gcd y r)))) (displayln (gcd (read) (read)))
26-Fibonacci Numbers
函数签名：
fibonacci : Integer -> Integer
#lang racket ;; fibonacci : Integer -> Integer (define (fibonacci n) (define (fib-helper a b count) (if (= count 3) (+ a b) (fib-helper b (+ a b) (- count 1)))) (cond [(= n 1) 0] [(= n 2) 1] [else (fib-helper 0 1 n)])) (displayln (fibonacci (read)))