The following code implements Newton's algorithm for finding the square root of a number using repetition:
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In Python
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- calculate number of operations in this algorithm void my_dgemv(int n, double* A, double* x, double* y) { double alpha=1.0, beta=1.0; int lda=n, incx=1, incy=1; cblas_dgemv(CblasRowMajor, CblasNoTrans, n, n, alpha, A, lda, x, incx, beta, y, incy); }The greatest common divisor of two positive integers, A and B, is the largest number that can be evenly divided into both of them. Euclid’s algorithm can be used to find the greatest common divisor (GCD) of two positive integers. You can use this algorithm in the following manner: Compute the remainder of dividing the larger number by the smaller number. Replace the larger number with the smaller number and the smaller number with the remainder. Repeat this process until the smaller number is zero. The larger number at this point is the GCD of A and B. Write a program that lets the user enter two integers and then prints each step in the process of using the Euclidean algorithm to find their GCD. An example of the program input and output is shown below: Enter the smaller number: 5 Enter the larger number: 15 The greatest common divisor is 5calculate number of operations in this algorithm for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { y[i] += A[i * n + j] * x[j]; } }
- evenly divides num1gcd (num1, num2) is gcd(num2, num1) if num1 < num2gcd (num1, num2) is gcd(num2, num1%num2) otherwiseCreate and execute a program to find the greatest common factor of two positive integers using Euclid's method. The largest common divisor is the highest integer that divides both numbers without leaving a remainder. In the DivisorCalc class, define a static method called gcd that accepts two numbers, num1 and num2. Create a driver to put your program through its tests. The repeated procedure is as follows: gcd (num1, num2) is num2 if num2 <= num1 and num2 evenly divides num1gcd (num1, num2) is gcd(num2, num1) if num1 < num2gcd (num1, num2) is gcd(num2, num1%num2) otherwiseThe code below is given a sequence of positive integers to find the longest increasing and the longest decreasing subsequence of the sequence. write the algorithm for the python code below. def prints(arr): for x in arr: print(x, end=" ") print() def constructPrintLDS(arr, n): l = [[] for i in range(n)] l[0].append(arr[0]) for i in range(1, n): for j in range(i): if arr[i] < arr[j] and (len(l[i]) < len(l[j]) + 1): l[i] = l[j].copy() l[i].append(arr[i]) maxx = l[0] for x in l: if len(x) > len(maxx): maxx = x prints(maxx) def constructPrintLIS(arr, n): l = [[] for i in range(n)] l[0].append(arr[0]) for i in range(1, n): for j in range(i): if arr[i] > arr[j] and (len(l[i]) < len (l[j]) + 1): l[i] = l[j].copy() l[i].append(arr[i]) maxx = l[0]…Your goal is to guess a secret integer between 1 and N. You repeatedly guess integers between 1 and N. After each guess you learn if your guess equals thesecret integer (and the game stops). Otherwise, you learn if the guess is hotter (closer to)or colder (farther from) the secret number than your previous guess. Design an algorithm that finds the secret number in at most ~2 lg N guesses. Then design an algorithmthat finds the secret number in at most ~ 1 lg N guesses.
- Correct answer will be upvoted else downvoted. Computer science. You are given an integer n. Check if n has an odd divisor, more noteworthy than one (does there exist such a number x (x>1) that n is separable by x and x is odd). For instance, assuming n=6, there is x=3. Assuming n=4, such a number doesn't exist. Input The primary line contains one integer t (1≤t≤104) — the number of experiments. Then, at that point, t experiments follow. Each experiment contains one integer n (2≤n≤1014). If it's not too much trouble, note, that the input for some experiments will not squeeze into 32-cycle integer type, so you should use no less than 64-digit integer type in your programming language. Output For each experiment, output on a different line: "Indeed" if n has an odd divisor, more noteworthy than one; "NO" in any case. You can output "YES" and "NO" regardless (for instance, the strings yEs, indeed, Yes and YES will be perceived as certain).use JAVA to write the code. : Euclid’s algorithm for finding the greatest common divisor (gdc) of two numbers The algorithm: given two numbers, n1 and n2: Divide n1 by n2 and let r be the remainder. If the remainder r is 0, the algorithm is finished and the answer is n2. (If the remainder is 1, the numbers are mutually prime and we are done-see below.) Set n1 to the value of n2, set n2 to the value of r, and go back to step 1. Entering 0 for one of the values is bad. It should work for the other value, but you have to figure out which is OK and which is bad. Catch this problem as it happens and make the user enter another value until they enter an acceptable one. Give an appropriate error message if this happens.A Fibonacci sequence is a sequence of numbers where each successivenumber is the sum of the previous two. The classic Fibonacci sequencebegins: 1, 1, 2, 3, 5, 8, 13, .... Write a program that computes the nthFibonacci number where n is a value input by the user. For example, ifn = 6, then the result is 8.
- A positive integer is called a perfect number if it is equal to thesum of all of its positive divisors, excluding itself. For example, 6 is the first perfectnumber because 6 = 3 + 2 + 1. The next is 28 = 14 + 7 + 4 + 2 + 1. Thereare four perfect numbers < 10,000. Write a program to find all these four numbers.The following Python program solves Sudoku using backtracking. The method that starts the solution is "solve_sudoku(matrix)" and receives as input an n x n matrix where the empty inputs are represented by -1 from pprint import pprint def search_next_void(puzzle): for r in range(9): for c in range(9): if puzzle[r][c] == -1: return r, c return None, None def is_valid(puzzle, guess, row, col): row_vals = puzzle[row] if guess in row_vals: return False col_vars = [puzzle[i][col] for i in range(9)] if guess in col_vars: return False row_start = (row // 3) * 3 col_start = (col // 3) * 3 for r in range(row_start, row_start + 3): for c in range(col_start, col_start + 3): if puzzle[r][c] == guess: return False return True def solve_sudoku(puzzle): row, col = search_next_void(puzzle) if row is None: return True for guess in range(1, 10): if…The following Python program solves Sudoku using backtracking. The method that starts the solution is "solve_sudoku(matrix)" and receives as input an n x n matrix where the empty inputs are represented by -1 from pprint import pprint def search_next_void(puzzle): for r in range(9): for c in range(9): if puzzle[r][c] == -1: return r, c return None, None def is_valid(puzzle, guess, row, col): row_vals = puzzle[row] if guess in row_vals: return False col_vars = [puzzle[i][col] for i in range(9)] if guess in col_vars: return False row_start = (row // 3) * 3 col_start = (col // 3) * 3 for r in range(row_start, row_start + 3): for c in range(col_start, col_start + 3): if puzzle[r][c] == guess: return False return True def solve_sudoku(puzzle): row, col = search_next_void(puzzle) if row is None: return True for guess in range(1, 10): if…