标签：function Paradigms string space Programming list CSI2120 characters page
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,
代写CSI2120作业、代做Programming Paradigms作业
(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"
因为专业，所以值得信赖。如有需要，请加QQ：99515681 或邮箱：99515681@qq.com
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标签：function,Paradigms,string,space,Programming,list,CSI2120,characters,page 来源： https://www.cnblogs.com/bizhud/p/10632764.html
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