This is a possible solution for the Pascal's triangle exploration at the end of chapter 5:
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#lang racket | |
(require rackunit) | |
(define (pascal-triangle-number row col) | |
(if (or (= col 0) | |
(= col row)) | |
1 | |
(+ (pascal-triangle-number (- row 1) col) | |
(pascal-triangle-number (- row 1) (- col 1))))) | |
(check-equal? | |
(pascal-triangle-number 0 0) 1) | |
(check-equal? | |
(pascal-triangle-number 1 0) 1) | |
(check-equal? | |
(pascal-triangle-number 1 1) 1) | |
(check-equal? | |
(pascal-triangle-number 2 0) 1) | |
(check-equal? | |
(pascal-triangle-number 2 1) 2) | |
(check-equal? | |
(pascal-triangle-number 2 2) 1) | |
(check-equal? | |
(pascal-triangle-number 3 1) 3) | |
(check-equal? | |
(pascal-triangle-number 3 2) 3) | |
(check-equal? | |
(pascal-triangle-number 4 1) 4) | |
(check-equal? | |
(pascal-triangle-number 4 2) 6) | |
(check-equal? | |
(pascal-triangle-number 4 3) 4) | |
(define (pascal-triangle-row n) | |
(define (helper n i) | |
(if (< i 0) | |
null | |
(cons (pascal-triangle-number n i) | |
(helper n (- i 1))))) | |
(helper n n)) | |
(check-equal? (pascal-triangle-row 0) (list 1)) | |
(check-equal? (pascal-triangle-row 1) (list 1 1)) | |
(check-equal? (pascal-triangle-row 2) (list 1 2 1)) | |
(define (range from to) | |
(if (> from to) | |
null | |
(cons from | |
(range (+ from 1) to)))) | |
(define (pascal-triangle n) | |
(map | |
pascal-triangle-row | |
(range 0 n))) | |
(check-equal? | |
(pascal-triangle 0) | |
(list (list 1))) | |
(check-equal? | |
(pascal-triangle 1) | |
(list (list 1) (list 1 1))) | |
(check-equal? | |
(pascal-triangle 2) | |
(list (list 1) (list 1 1) (list 1 2 1))) | |
(check-equal? | |
(pascal-triangle 4) | |
(list (list 1) (list 1 1) (list 1 2 1) (list 1 3 3 1) (list 1 4 6 4 1))) |
It's a nice exercise to practise recursion.
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