of n = 2¹? [Hint: When input n-1, the number of lines will be printed: P(1) -0. Identify the recurrence relation and find the solution using the iteration method.] function fun(n) if n > 1 print.line('good day') fun (n/2) fun (n/2) end if
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How many lines does the following
![of n = 2¹? [Hint: When input n = 1, the number of lines will be printed: P(1) = 0. Identify the
recurrence relation and find the solution using the iteration method.]
function fun (n)
if n > 1
print.line('good day')
fun (n/2)
fun (n/2)
end if](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F324f16c4-c4e5-4b06-b074-f261b39e021a%2Fca772a11-4a00-465a-a6e9-e6284b39b993%2Fukleudk_processed.png&w=3840&q=75)
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- 1. Consider the following NFA N: :+ nfa N { Q={q1,q2,q3,q4,q5} S={a,b} q0=q1 d(q1,a)={q2} d(q2,\e)={q3} d(q2,a)={q1} d(q3,\e)={q1} d(q3,b)={q4} d(q4,b)={q5} d(q5,b)={q3} F={q3} } Question: Use !{the algorithm from the notes} to construct an NFA N_2 that does not use \e-transitions and that is !{@eqnt} to N.Find the runtime of the following recurrence relation using a substitution method T(n) = T(n/5) + T(7n/10) + O(n)Each time you enter a guess, the program will tell you whether the secret word is alphabetically before your guess, alphabetically after your guess, or exactly matches your guess. Each secret word is randomly chosen from a dictionary with exactly 267, 751 words. Let T(n) be the maximum number of guesses required to correctly identify a secret word that is randomly chosen from a dictionary with exactly n words. Determine a recurrence relation for T(n), explain why the recurrence relation is true, and then apply the Master Theorem to show that T (n) = Θ(log n).
- Q1 The periodic function sin(2x) has multiple roots between x values of -5π and 5π. If xL = -15 and xU = 15, which of the following statements is true using a bracketed method? Select one: a. All roots will be returned b. The middle root will be returned c. The chosen bracket is invalid for bracketed methods d. A single root will be returned e. The algorithm will be stuck in an infinite loop Q2 Consider x and y to represent data points (xi,yi), where i = 1, 2, 3, … n. What is the length of pafter running the following command? p = polyval(x,y) Select one: a. n b. n - 1 c. n + 1 d. Empty variable e. 1 Q3 Consider a system of linear equations in the form of AX = B, where X is the unknown vector. Which of the following can be used to solve for X? Select one: a. X = A\B b. X = B./A c. X = inv(B)*A d. X = inv(A)./B e. X = B\ADrag all else please.a)Write a recursive definition for the set of odd positive integers. b)Use master theorem to find the solution to the recurrence relation f(n) = 4f(n/2) + 2? ! ,when n = 2" , where k is a positive integer and f(1) = 1.
- ALGO(A)x = A[0]p = -1r = nwhile (True) { do { p = p + 1 } while (A[p] < x) do { r = r - 1 } while (A[r] > x) if (p >= r ){ return r } exchange A[p] and A[r]} From the above algorithm, we have to state whether the following statements are True or False(With explanation) (A) At the end of each iteration of the while loop of Line 4, each index of A smaller than p contains an integer smaller than x. (B) At the end of each iteration of the while loop of Line 4, each index of A larger than r contains an integer larger than or equal to x. (C) When ALGO terminates, the integer at index r is x.discrete structure Consider the balanced parentheses problem from the tutorial 5. True or False: We could have equivalently proven the property P(n) that every balanced parentheses string S of length n has the same number of [ as ], for every n ≥ 0, by strong induction. Question 5 options: a) True b) FalseIN STO a IN STO b test LDA a SUB b BRP count LDA c OUT LDA d OUT HLT count STO a LDA c ADD one STO c LDA d SUB b STO d BR test a DAT b DAT c DAT 0 one DAT 1 d DAT 0 Using this LMC program to solve a÷b=cRd where a = 20 b = 4 what is c and remainder d. IF a = 20 b = 4 c should be 5 and d should be 0. IS there reason to why i keep getting the remainder as 980
- 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).Given a positive integer 'n', find and return the minimum number of steps that 'n' has to take to get reduced to 1. You can perform any one of the following 3 steps:1.) Subtract 1 from it. (n = n - 1) ,2.) If its divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,3.) If its divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ). Write brute-force recursive solution for this.Input format :The first and the only line of input contains an integer value, 'n'.Output format :Print the minimum number of steps.Constraints :1 <= n <= 200 Time Limit: 1 secSample Input 1 :4Sample Output 1 :2 Explanation of Sample Output 1 :For n = 4Step 1 : n = 4 / 2 = 2Step 2 : n = 2 / 2 = 1 Sample Input 2 :7Sample Output 2 :3Explanation of Sample Output 2 :For n = 7Step 1 : n = 7 - 1 = 6Step 2 : n = 6 / 3 = 2 Step 3 : n = 2 / 2 = 1 SolutionDp///.in this code: def insertSort(A): for i in range(1,len(A)): j = i-1 while A[j]>A[j+1] and j>=0: A[j],A[j+1] = A[j+1],A[j] j -= 1 A = [3,1,11,9,10,2] insertSort(A) print(A) 1) What is the Hypothesis? 2) What is Your Induction