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Lec 2: Buzz Project

This demonstrates a function with the correct definition of:

  1. 0 or more inputs
  2. No free variables
#lang racket
(require (planet dyoo/simply-scheme:2:2))

;Buzz game
;  Rules

;  Count upwards
;    If your number is divisible by 7 you have to say buzz.
;    If your number contains a 7 you have to say buzz.
;    Buzz(1) = 1
;    Buzz(7) = Buzz
;    Buzz(8) = 8
;    Buzz(17) = Buzz

; Test
(define (is_equal? n x)
  (if (equal? n x)
      (sentence n x '=> '✅)
      (sentence n x '=> '❌)))

; Game function
(define (Buzz n)
  (cond ((equal? (modulo n 7) 0) (word 'Buzz))
        ((member? 7 n)           (word 'Buzz))
        (else n)))

;Tests
(is_equal? (Buzz 1) 1)
(is_equal? (Buzz 2) 2)
(is_equal? (Buzz 3) 3)
(is_equal? (Buzz 4) 4)
(is_equal? (Buzz 5) 5)
(is_equal? (Buzz 6) 6)
(is_equal? (Buzz 7) 'Buzz)
(is_equal? (Buzz 8) 8)
(is_equal? (Buzz 9) 9)
(is_equal? (Buzz 10) 10)
(is_equal? (Buzz 11) 11)
(is_equal? (Buzz 12) 12)
(is_equal? (Buzz 13) 13)
(is_equal? (Buzz 14) 'Buzz)
(is_equal? (Buzz 15) 15)
(is_equal? (Buzz 16) 16)
(is_equal? (Buzz 17) 'Buzz)
(is_equal? (Buzz 18) 18)
(is_equal? (Buzz 71) 'Buzz)

I had initially only been checking the last digit of the input to see if it contains a 7. This of course only checks the last digit not the rest of them. So, '781' for instance, would fail.

Assuming my test cases are correct, this works.

'(1 1 => ✅)
'(2 2 => ✅)
'(3 3 => ✅)
'(4 4 => ✅)
'(5 5 => ✅)
'(6 6 => ✅)
'("Buzz" Buzz => ✅)
'(8 8 => ✅)
'(9 9 => ✅)
'(10 10 => ✅)
'(11 11 => ✅)
'(12 12 => ✅)
'(13 13 => ✅)
'("Buzz" Buzz => ✅)
'(15 15 => ✅)
'(16 16 => ✅)
'("Buzz" Buzz => ✅)
'(18 18 => ✅)
'("Buzz" Buzz => ✅)