# [plt-scheme] multiply-by-pi problem

 From: arnuld (arnuld3 at gmail.com) Date: Sat Jan 28 02:04:06 EST 2006 Previous message: [plt-scheme] about SICP & PLT scheme Next message: [plt-scheme] bytes vs u8vector Messages sorted by: [date] [thread] [subject] [author]

```hai there,

again i am on section-11 of HtDP i did create 2 programmes (ex 11.5.1
& 11.5.2) whch work as desired but failed to create a programme based
on those 2. i created "add & multiply" which add and multiply 2
numbers without using + and * but i am not able to create
"multiply-by-pi". i am providing here the full programmes with problem
statements except for failed one for which i am able to provide
data-definition and template only.

i will appreciate any help from anybody.

thanks

"arnuld"

----------------------------------------------------
Exercise 11.5.1.   Define add, which consumes two natural numbers, n
and x, and produces n + x without using Scheme's +.

Exercise 11.5.2.   Develop the function multiply-by-pi, which consumes
a natural number and multiplies it by 3.14 without using *. For
example,

(= (multiply-by-pi 0) 0)
(= (multiply-by-pi 2) 6.28)
(= (multiply-by-pi 3) 9.42)

Define multiply, which consumes two natural numbers, n and x, and
produces n * x without using Scheme's *. Eliminate + from this
definition, too.

Hint: Recall that multiplying x by n means adding x to itself n times.

;;------------ failed attempt of ex 11.5.2's 1st part (multiply-by-pi)
-------------------------------

;; a natural number is either:
;;           1. 0, or
;;           2. (add1 n), where n is a natural number

;; multiply-by-pi : N -> number
;; to multiply n by pi without using *.
;; (define (multiply-by-pi n)..)

#| TEMPLATE
(define (multiply-by-pi n)
(cond
[(zero? n)....]
[else...(multiply-by-pi (sub1 n))....]))

;; examples
(= (multiply-by-pi 0) 0)
(= (multiply-by-pi 2) 6.28)
(= (multiply-by-pi 3) 9.42) |#

;;-------------- solution to ex 11.5.2's 2nd part (multiply two
numbers) -------------------------------

;; multiply : N, N -> N
;; to multiply n & x natural numbers without using *.
;; (define (multiply n x)..)

#| TEMPLATE
(define (multiply n x)
(cond
[(zero? n)...]
[else ...(multiply (sub1 n) x)...]))

;; examples
(= 30 (multiply 6 5))
(= 0 (multiply 6 0))
(= 0 (multiply 0 5))
(= 0 (multiply  0))
(= 63 (multiply 9 7)) |#

(define (multiply n x)
(cond
[(zero? n) 0]
[else (add x (multiply (sub1 n) x))]))

;; tests
(= 30 (multiply 6 5))
(= 0 (multiply 6 0))
(= 0 (multiply 0 5))
(= 0 (multiply  0))
(= 63 (multiply 9 7))

;;----------------------- solutionto ex 11.5.1 (add two numbers)
--------------------------------------
;; add : N, N  ->  number
;; to compute n + x without using +

;; TEMPLATE
#| (define (add n x)
(cond
[(zero? n)...]
[else....(add (sub1 n) x)...]))

but i did not understand one thing when < hellos > uses one input and
< add > uses 2 inputs then why templates in both
cases match? |#

(define (add n x)
(cond
[(zero? n) x]
[else (add1 (add (sub1 n) x))]))

;; tests
(= 8 (add 0 8))
(= 3 (add 3 0))
(= 11 (add 3 8))
(= 0 (add 0 0))
(= (+ 20 67) (add 20 67))

;; ------------- default programme from HtDP (add-to-pi)
-----------------------------

;; add-to-pi : N  ->  number
;; to compute n + 3.14 without using +

#| TEMPLATE
(define (add-to-pi n)
(cond
[(zero? n) ...]
[else ... (add-to-pi (sub1 n)) ... ]))) |#

(define (add-to-pi n)
(cond
[(zero? n) 3.14]
[else (add1 (add-to-pi (sub1 n)))])

;; tests
(= (add-to-pi 0) 3.14)
(= (add-to-pi 2) 5.14)
(= (add-to-pi 6) 9.14)

;;---------------------- end of programme
----------------------------------------
--
"the great intellectuals"

```

 Posted on the users mailing list. Previous message: [plt-scheme] about SICP & PLT scheme Next message: [plt-scheme] bytes vs u8vector Messages sorted by: [date] [thread] [subject] [author]