Determine a function T(n) that relates input sizen to number of runtime steps and the Big-O Time complexity for this T(n): Example only: T(n) = an^2 +k Complexity = O(n) def f6(n): for i_1 in range(n): for i 2 in range (n): T(n) = Complexity = ... for i n in range (n):
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T(n) = _____
O(n) = _____
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- Please written by computer source Let x be a binary string. The minimal description of x, written d(x), is the shortest string ⟨M, w⟩ where TM M on input w halts with x on its tape. If several such strings exist, select the lexicographically first among them. The descriptive complexity, or Kolmogorov complexity, of x, written K(x), is K(x) = |d(x)|. Show that the function K(x) is not a computable function. HINTS: If K is a computable function, there is some TM which computes it. That TM can be used to find strings of large complexity. Try to design a program which outputs “complex” strings but which contradicts their supposed complexity, and even contradicting the supposed complexity of a single string suffices.Suppose that your implementation of a particular algorithm appears in C++ as for (int pass = 1; pass <= n; pass++) { for (int index = 0; index < n; index++) { for (int count = 1; count < 10; count++) { ... } // end for } // end for } // end for The previous code shows only the repetition in the algorithm, not the computations that occur within the loops. These computations, however, are independent of n. What is the Big O of the algorithm?What is the running time for this java code? I want the complexity and asymptotic upper bound using Big-Oh notation class lcs { static void LongestCommonArraySubsequence(int X[], int Y[], int n, int m) { // dp[][] array int[][] dp = new int[n + 1][m + 1]; // dp matrix initialization for (int i = 0; i <= n; i++) for (int j = 0; j <= m; j++) dp[i][j] = 0; for (int i = n - 1; i >= 0; i--) { for (int j = m - 1; j >= 0; j--) { if (X[i] == Y[j]) dp[i][j] = dp[i + 1][j + 1] + 1; // bottom up fill } } int max = 0; // maximum length int Xindex = 0; // starting index of array X int Yindex = 0; // starting index of array Y for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { // calculating the maximum value in dp matrix if (dp[i][j] > max) {…
- Given below is the implementation of the bellman ford and dijkstras algorithm. Please complete the code for the time_shortest_path_algs() function according to the instructions in the 1st screenshot provided. Done in python 3.10 or later please def bellman_ford(self,s) : """Bellman Ford Algorithm for single source shortest path. Keyword Arguments: s - The source vertex. """ distances = {v: float('inf') for v in self.adjacency_list} distances[s] = 0 parents = {v: None for v in self.adjacency_list} for _ in range(len(self.adjacency_list) - 1): for from_vertex in self.adjacency_list: for to_vertex in self.adjacency_list[from_vertex]: if distances[from_vertex] + self.weights[(from_vertex, to_vertex)] < distances[to_vertex]: distances[to_vertex] = distances[from_vertex] + self.weights[(from_vertex, to_vertex)] parents[to_vertex] =…What is the time complexity of the following function- 5T(n/5+50)+n,T(1)=1 (a) θ(50logn) (b) θ(n50logn) (c) θ(logn) (d) θ(nlogn) (e) θ(nlogn)Information is present in the screenshot and below. Based on that need help in solving the code for this problem in python. The time complexity has to be as less as possible (nlogn or n at best, no n^2). Apply divide-and-conquer algorithm in the problem. Make sure all test cases return expected outputs. Hint: Apply bisection method/modules Output FormatOutput contains a line with two space-separated integers W_a and W_b.- W_a is the maximum matchups won by Hamiltonia- W_b is the maximum matchups won by Burrgadia. Sample Input 03 554402410 Sample Output 03 0 Sample Input 15 4833485183 Sample Output 12 2 Sample Input 27 81028121912601319851 Sample Output 27 0 The actual code """Solves a test case Parameters:a : int - number of leaders in Hamiltoniab : int - number of leaders in Burrgadias_i : array-like - rap proficiencies of Hamiltonia's leadersr_j : array-like - rap proficiencies of Burrgadia's leaders Returns:win_a : int - number of…
- #4. Euler's totient function, also known as phi-function ϕ(n),counts the number of integers between 1 and n inclusive,which are coprime to n.(Two numbers are coprime if their greatest common divisor (GCD) equals 1)."""def euler_totient(n): """Euler's totient function or Phi function. Time Complexity: O(sqrt(n)).""" result = n for i in range(2, int(n ** 0.5) + 1): if n % i == 0: while n % i == 0: n //= i.Let, the time complexity of each of the following code snippets be T(n). Find out a tight bound for T(n) in Big-Theta () notation. for (i=n, i>=1; i=i/3) p=2 while(p<n) print(“hello”) p=p*p1. Input: G = (V, E)2. Output: MIS I of G3. I ← ∅4. V ← V5. while V = ∅6. assign a random number r(v) to each vertex v ∈ V7. for all v ∈ V in parallel8. if r(v) is minimum amongst all neighbors9. I ← I ∪ {v}10. V ← V \ {v ∪ N(v)}This algorithm terminates in O(log n)rounds with good probability [3]. The stepsof the algorithm between lines 7–10 can be performed in parallel which will providea speedup.Make a Python Implementation of given algorithm?
- Assume that each of the expressions below gives the processing time T(n) spent by an algorithm for solving a problem of size n. Select the dominant term(s) having the steepest increase in n and specify the lowest Big-Oh complexity of each algorithm. For example, the dominant term in 0.1n + 10n4 is 10n4 and it is O(n4). Expression Dominant term(s) O(. . .) 5 + 0.001n3 + 0.025n 500n + 100n1.5 + 50n log10 n 0.3n + 5n1.5 + 2.5 · n1.75 n2 log2 n + n(log2 n)2 n log3 n + n log2 n 100n + 0.01n2 0.01n + 100n2 2n + n0.5 + 0.5n1.25 0.01n log2 n + n(log2 n)2 100n log3 n + n3 + 100nThe given time complexity is: m T(m-1) + ca > 1 cb = 0 expanding using substitution: m T(m-1) + ca m[(m-1) T(m-2) + ca]+ ca m(m-1) T(m-2) + mca + ca m(m-1) [(m-2) T(m-3) + ca] + mca + ca m(m-1)(m-2) T(m-3) + m(m-1)ca + mca + ca what is the time complexity? form an expression for adding all the caAssume that you were given N cents (N is an integer) and you were asked to break up the N cents into coins consisting of 1 cent, 2 cents and 5 cents. Write a dynamicprogramming based recursive algorithm, which returns the smallest (optimal) number of coins needed to solve this problem. For example, if your algorithm is called A, and N = 13, then A(N) = A(13) returns 4, since 5+5+2+1 = 13 used the smallest (optimal) number of coins. In contrast, 5+5+1+1+1 is not an optimal answer.