The function f(x) = x^3 is bijection from R to R. Is it True or False? O a. True O b. False
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- SUCCESSIVE-OVER-RELAXATION Method TROUBLESHOOTING: Fix the errors in the code below and run the script with your modified code and screenshot your output and codes with answer function [x,numIter,omega] = gaussSeidel(func,x,maxIter,epsilon) % Solves Ax = b by Gauss-Seidel method with relaxation. % USAGE: [x,numIter,omega] = gaussSeidel(func,x,maxIter,epsilon) % INPUT: % func = handle of function that returns improved x using % x = starting solution vector % maxIter = allowable number of iterations (default is 500) % epsilon = error tolerance (default is 1.0e-9) % OUTPUT: % x = solution vector % numIter = number of iterations carried out % omega = computed relaxation factor if nargin < 4; epsilon = 1.0e-9; end if nargin < 3; maxIter = 500; end k = 10; p = 1; omega = 1; for numIter = 1:maxIter xOld = x; x = feval(func,x,omega); dx = sqrt(dot(x - xOld,x - xOld)); if dx < epsilon; return; end if numIter == k; dx1 = dx; end if numIter == k + p omega = 2/(1 + sqrt(1 -…Write a C++ function, bool ok(int q[]), which takes an array q and returns true if the array represents a valid configuration of the eight queens and returns false otherwise.def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab' >>> longest_unique_substring('abcabcbb')'abc'"""
- def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab' >>> longest_unique_substring('abcabcbb')'abc'""" RESTRICTIONS: - Do not add any imports and do it on python .Do not use recursion. Do not use break/continue.Do not use try-except statements.def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab'""" RESTRICTIONS: - Do not add any imports and do it on python .Do not use recursion. Do not use break/continue.Do not use try-except statements.Take a sentence and as well as a word/character(s) and return True if the word/character(s) can be found in the sentence. Sample Parameter(s): "chicken", "z" Sample Return: False
- SUCCESSIVE-OVER-RELAXATION Method TROUBLESHOOTING: Fix the errors in the code below and run the script with your modified code function [x,numIter,omega] = gaussSeidel(func,x,maxIter,epsilon) % Solves Ax = b by Gauss-Seidel method with relaxation. % USAGE: [x,numIter,omega] = gaussSeidel(func,x,maxIter,epsilon) % INPUT: % func = handle of function that returns improved x using % x = starting solution vector % maxIter = allowable number of iterations (default is 500) % epsilon = error tolerance (default is 1.0e-9) % OUTPUT: % x = solution vector % numIter = number of iterations carried out % omega = computed relaxation factor if nargin < 4; epsilon = 1.0e-9; end if nargin < 3; maxIter = 500; end k = 10; p = 1; omega = 1; for numIter = 1:maxIter xOld = x; x = feval(func,x,omega); dx = sqrt(dot(x - xOld,x - xOld)); if dx < epsilon; return; end if numIter == k; dx1 = dx; end if numIter == k + p omega = 2/(1 + sqrt(1 - (dx/dx1)ˆ(1/p))); end end error(’Too many iterations’)Write a program to implement linear search.Let f,g:Z+⟶R, and f(n)=nlog2(n) forn∈Z+. For which function g(n)below is f∉O(g)
- write a code c++ to represnt polynomial function like this: (explain every line on code) Px=a(n) X(n)+a(n-1)X(n-1)+a(n-2) X(n-2)+…+ a1x+a0 by using method for ist datastructure lilke : L.insert or L.remove or . L.retrieve or L.replace or L.empty or L.size()Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] """ def generate_parenthesis_v1(n): defadd_pair(res, s, left, right): ifleft==0andright==0: res.append(s) return ifright>0: add_pair(res, s+")", left, right-1) ifleft>0: add_pair(res, s+"(", left-1, right+1) res= [] add_pair(res, "", n, 0) returnres def generate_parenthesis_v2(n): defadd_pair(res, s, left, right): ifleft==0andright==0: res.append(s) ifleft>0: add_pair(res, s+"(", left-1, right) ifright>0andleft<right: add_pair(res, s+")", left, right-1).Write a menu-driven program to read and display a p¥q¥r matrix. Also, find the sum, transpose, and product of the two p*q*r matrices