procedure gcd(a,b: positive integer x := a y := b while y # 0 begin r := x mod y x := y y :=r end {gcd(a,b) is x when y = r = 0} (Try example acd(9. 6))
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Try to compute gcd or lcm ( gcd(9, 6) or lcm(9,6) ) by the following
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- Using C language, trace this: void trace(int x, int *y, int z) {x = 1; *y=2;z=4;printf("%2d %2d %2d\n", x, *y, z);}main() {int x=1, y=3,z=4;clrscr();printf("%2d %2d %2d\n",x,y,z);trace(y,&x,z);printf("%2d %2d %2d\n",x,y,z);trace(x,&z,y);printf("%2d %2d %2d\n",x,y,z);trace(z,&y,x);printf("%2d %2d %2d\n",x,y,z);getch();return 0; }(c)#include<sfunction iterFunc(x) {return x^3 + x - 1 }oldEst = 0 tol = 0.00005 error = 1 while (error > tol) newEst = iterFunc(oldEst) print(newEst)error = abs(newEst - oldEst) oldEst = newEst } Edit this psuedocode to make it work in python, the output should be 0.6823 as this is a method of root finding!Example: Enter an integer = 75 Smallest divisor is = 3
- Computer Science A code C runs T seconds to obtain results for an input of size S. How to find the order of runtime for the code Cfor given pairs. Hint the answer is a formula that uses logarithm.(a) (S1,T1), (S2,T2)?(b) (S1,T1), (S2,T2), (S3, T3)?A number of the form a + ib, in which i2 = -1 and a and b are real numbers, is called a complex number. We call the real part and b the imaginary part of a + ib. Complex numbers can also be represented as ordered pairs (a, b). The addition and multiplication of complex numbers are defined by the following rules: (a + ib) + (c + id) = (a + c) + i(b + d ) (a + ib) * (c + id) = (ac - bd) + i(ad + bc) Using the ordered pair notation, these rules are written as: (a, b) + (c, d) = ((a + c), (b + d )) (a, b) * (c, d) = ((ac - bd ), (ad + bc)) C++ has no built-in data type that allows us to manipulate complex numbers. Construct a data type, complex Type, that can be used to process complex numbers. Overload the stream insertion and stream extraction operators for easy input and output. We will also overload the operators + and * to perform addition and multiplication of complex numbers. If x and y are complex numbers, we can evaluate expressions such as x + y and x * y./* segvhunt.cFind and eliminate all code that generates Segmentation Fault*/#include <stdio.h>int main() {char **s;char foo[] = "Hello World";*s = foo;printf("s is %s\n",s);s[0] = foo;printf("s[0] is %s\n",s[0]);return(0);}
- // Description: Write a program that outputs a root of a given function f(x) by using a bi-section method. // f(x) = x3 + 5x2 +7x +2 // IN C++ Please! not using #define EP 0.01mplement a Java program that applies the Newton-Raphson's method xn+1 = xn – f(xn) / f '(xn) to search the roots for this polynomial function ax6 – bx5 + cx4 – dx3+ ex2 – fx + g = 0. Fill out a, b, c, d, e, f, and g using the first 7 digits of your ID, respectively. For example, if ID is 4759284, the polynomial function would be 4x6 – 7x5 + 5x4 – 9x3+ 2x2 – 8x + 4 = 0. The program terminates when the difference between the new solution and the previous one is smaller than 0.00001 within 2000 iterations. Otherwise, it shows Not Found as the final solution.Given an integer N and a base X, the task is to find the minimum number of operations required to represent N as a sum of the distinct powers of X. In each operation, you can either increment or decrement N. You are allowed to make the given operation any number of times Examples: Input: N = 7, X = 3 Output: 3.
- write a c++ program that asks user to enter a number and then print the number that is maximum time repeated in that number and count how mnay times its repeated. remember that if two numbers are repeated same times then you need to output both of them and total times they are repeated. Do this only using loops and functions. arrays pointers are not allowed. for example:input:122344, digit = 2,frequency=2 digit= 4, frequency = 2Factorial of a number is defined as: n! = n(n-1)(n-2)(n-3)...(2)(1) For example, 4! = 4*3*2*1 The n! can be written in terms of (n-1)! as: n! = n* (n-1)! (n-1)! = (n-1)*(n-2) ! and so forth. Thus, in order to compute n!, we need (n-1)!, to have (n-1)!, we need (n-2)! and so forth. As you may immediately notice, the base case for factorial is 1 because 1! = 1. Write a program that uses a recursive function called factorial that takes an integer n as its argument and returns n! to the main. C++ PLEASEWrite a Phyton code to compare these two divisions ;c=a//b and d=a/b. (// Integer division operator). Assign integer (like a=5 and b=3)and real values (like a=5.0 and b=3.0) to a and b and explain the division results.