代写CSI2120、代做Programming Paradigms、代写Java代写、Python/c++程序代做

日期：2019-03-30 10:52

Assignment 3

CSI2120 Programming Paradigms

Winter 2019

Due on April 5th before 11:00 pm in Virtual Campus

6 marks

There are [10 points] in this assignment. The assignment is worth 6% of your final mark.

All code must be submitted in a scm file. Screenshots, files in a format of a word editor, pdfs,

handwritten solutions, etc. will not be marked and receive an automatic 0.

Question 1. [1 point]

Use the built-in function map to replace every number by its reciprocal value in the list. Define a global

1over and use 0 as the reciprocal value for 0, i.e., 0 = 1/0.

(1over '(0 2 3 4 12 0 0 1 0)) '(0 1/2 1/3 1/4 1/12 0 0 1 0))

Question 2. [2 points]

Implement Newton-Rhapson's method for root finding of function in one dimension. Newton-Rhapson

is defined by the iteration ;

where is the current estimate of the root,  is the function evaluated atlobal definition of the function must take three

arguments, the current estimate of the root, the function and the derivative of the function, i.e.,

; (newtonRhap x f fx)

Your routine must stop the iterations if the change in the solution is less than a tolerance. For this

purpose use a global define, i.e.,

(define TOL 1e-6)

See next page for examples.

CSI 2120 page 2

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Examples:

(newtonRhap 0.1 sin cos) 0

(newtonRhap 2.0 (lambda (x) (- (* x x) x 6))

(lambda (x) (- (* 2 x) 1))) 3.0

(newtonRhap -20.0 (lambda (x) (- (* x x) x 6))

(lambda (x) (- (* 2 x) 1))) -2.000000000000118

Question 3. [3 points]

Implement a routine p_cos which calculates the cosine of an angle in radians. Use the following

product approximation of cosine

Use as many terms until the change with the next term is less than a tolerance. For this purpose use

again the global define,

(define TOL 1e-6)

Examples:

(p_cos 0)1

(p_cos (/ pi 2)) 0.0

You are allowed extra global helper functions.

CSI 2120 page 3

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Question 4. [5 points]

You are not allowed to use any of the built-in string processing function for this question. You can

assume that all lists only contain characters.

a) Write a predicate separator? that returns true if a character is a space, tab or newline and

false otherwise. In Scheme this characters are written #\space, #\tab and #\newline

respectively. The built-in predicate char=? compares two characters for equality.

Example:

(separator #\space) #t

(separator #\b) #f

b) Write a function cpy that copies all characters from an input list into an output list until a

separator is encountered.

(cpy '(#\H #\e #\l #\l #\o #\space #\W #\o #\r #\l #\d)) '(#\H #\e #\l #\l #\o)

c) Write a function drop that removes all characters from an input list until a separator is

encountered and returns the remaining list.

(drop '(#\H #\e #\l #\l #\o #\newline #\W #\o #\r #\l #\d)) '(#\W #\o #\r #\l #\d)

d) Write a predicate same? that compares two list of characters and return true if the characters in

the first input list up to a separator are the same as in the second list.

(same '(#\H #\e #\l #\l #\o #\tab #\W #\o #\r #\l #\d)

'(#\H #\e #\l #\l #\o)) #t

(same '(#\H #\e #\l #\l #\o #\space #\W #\o #\r #\l #\d)

'(#\W #\o #\r #\l #\d)) #f

continued on next page.

CSI 2120 page 4

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e) Write a function replace that replaces a list of characters in the input list between separators with

another set of characters.

(replace

'(#\a #\space #\b #\i #\r #\d #\space #\e #\a #\t #\s #\space

#\a #\space #\t #\o #\m #\a #\t #\o)

‘(#\a)

‘(#\t #\h #\e)) '(#\t #\h #\e #\space #\b #\i #\r #\d #\space #\e #\a #\t #\s

#\space #\t #\h #\e #\space #\t #\o #\m #\a #\t #\o)

Or more readable:

(list->string (replace (string->list "a bird eats a tomato")

(string->list "a") (string->list "the"))) "the bird eats the tomato"