Implement the modular exponentiation (a.k.a. fast exponentiation) function mod_exp (b, n, m) to compute b n (mod m) more efficiently. (Hint: to read n bit-by-bit, use / and % operations repeatedly) a) Test your function for b = 3, n = 231 – 2, m = 231 – 1. b) Report the result and the time (in seconds) it takes to find the result. in pahton

Implement the modular exponentiation (a.k.a. fast exponentiation) function mod_exp (b, n, m) to compute b n (mod m) more efficiently. (Hint: to read n bit-by-bit, use / and % operations repeatedly) a) Test your function for b = 3, n = 231 – 2, m = 231 – 1. b) Report the result and the time (in seconds) it takes to find the result. in pahton

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter5: Repetition Statements

Section5.5: A Closer Look: Loop Programming Techniques

Problem 12E: (Program) Write a program that tests the effectiveness of the rand() library function. Start by...

See similar textbooks

Similar questions

Implement the quadratic_formula() function. The function takes 3 arguments, a, b, and c, and computes the two results of the quadratic formula: x1=−b+b2−4ac2a x2=−b−b2−4ac2a The quadratic_formula() function returns the tuple (x1, x2). Ex: When a = 1, b = -5, and c = 6, quadratic_formula() returns (3, 2). Code provided in main.py reads a single input line containing values for a, b, and c, separated by spaces. Each input is converted to a float and passed to the quadratic_formula() function. Ex: If the input is: 2 -3 -77 the output is: Solutions to 2x^2 + -3x + -77 = 0 x1 = 7 x2 = -5.50 code: # TODO: Import math module def quadratic_formula(a, b, c):# TODO: Compute the quadratic formula results in variables x1 and x2return (x1, x2) def print_number(number, prefix_str):if float(int(number)) == number:print("{}{:.0f}".format(prefix_str, number))else:print("{}{:.2f}".format(prefix_str, number)) if __name__ == "__main__":input_line = input()split_line = input_line.split(" ")a =…
Write a function that takes an unsigned integer andreturns the number of '1' bits it has(also known as the Hamming weight).For example, the 32-bit integer '11' has binaryrepresentation 00000000000000000000000000001011,so the function should return 3.T(n)- O(k) : k is the number of 1s present in binary representation.NOTE: this complexity is better than O(log n).e.g. for n = 00010100000000000000000000000000only 2 iterations are required.Number of loops isequal to the number of 1s in the binary representation."""def count_ones_recur(n):.
Write a function that takes an unsigned integer andreturns the number of '1' bits it has(also known as the Hamming weight).For example, the 32-bit integer '11' has binaryrepresentation 00000000000000000000000000001011,so the function should return 3.T(n)- O(k) : k is the number of 1s present in binary representation.NOTE: this complexity is better than O(log n).e.g. for n = 00010100000000000000000000000000only 2 iterations are required..
What is the complexity of the following pseudo-code * Write in two line.
Answer questions (a) and (b) below: (a) How many times exactly is the code block below executed? For (i = 1, n) { For (j = 1, i) { For (k = 1, j) { code block } } } Hint: You have to start with n=1, then make assumption what you make expect for any given n = N, and check if the formula you found works for n =N+1. This is what we call prove by induction. (b) What is the theta value of this code segment?
Recall the naïve divide and conquer O(n2)-time multiplication algorithm, call it NaiveD&Cmult, to multiply two n-digit numbers. Design a 8-bit multiplier using NaiveD&Cmult algorithm with basic building blocks of 1-bit multiplier and m-bit adders and shifters, m>0. Note that in class we did not consider the cases when n is not a power of two, you may have to modify the algorithm to take care of these cases. Also, we discussed algorithm using decimal digits, obviously it can be easily extended to any radix, so for this problem radix is 2.
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.
implement bitcount function Count the number of 1’s in x. you are onlyallowed to use the following eight operators:! ~ & ^ | + << >> “Max ops” field gives the maximumnumber of operators you are allowed to use to implement each function /** bitCount - returns count of number of 1's in x* Examples: bitCount(5) = 2, bitCount(7) = 3* Legal ops: ! ~ & ^ | + << >>* Max ops: 40*/int bitCount(int x) {return 2;}
Simplify the complement of the following function: F(A,B,C,D)=(0,2,4,5,8,9,10,11) Your answer: F=((A'B'D)' (BC)'(AB)')' F=((A'BD)'(BC)'(AB)')' F=((A'B'D)'(B'C)'(AB)') F=((A'B'D')' (BC)'(AB)')
implement anyEvenBit(x) Return 1 if any even bit in x is set to 1 you are only allowed to use the following eight operators: ! ~ & ^ | + << >> “Max ops” field gives the maximum number of operators you are allowed to use to implement each function /* * anyEvenBit - return 1 if any even-numbered bit in word set to 1* Examples anyEvenBit(0xA) = 0, anyEvenBit(0xE) = 1* Legal ops: ! ~ & ^ | + << >>* Max ops: 12*/int anyEvenBit(int x) {return 2;}
Consider the following code which includes a parity check for the third digit: C=(000,011,101,110). You get the third digit by adding the first two and then using the remainder on division by 2. Can this code detect and count single errors? how can this be implemented?
This exercise is about the bit-wise operators in C. Complete each functionskeleton using only straight-line code (i.e., no loops, conditionals, or functioncalls) and limited of C arithmetic and logical C operators. Specifically, youare only allowed to use the following eight operators: ! ∼, &,ˆ|, + <<>>.For more details on the Bit-Level Integer Coding Rules on p. 128/129 of thetext. A specific problem may restrict the list further: For example, to writea function to compute the bitwise xor of x and y, only using & |, ∼int bitXor(int x, int y){ return ((x&∼y) | (∼x & y));}(a) /*copyLSbit:Set all bits of result to least significant bit of x* Example: copyLSB(5) = 0xFFFFFFFF, copyLSB(6) = 0x00000000* Legal ops: ! ∼ & ˆ| + <<>>*/int copyLSbit(int x) {return 2;}(b) /* negate - return -x* Example: negate(1) = -1.* Legal ops:! ∼ & ˆ| + <<>>*/ int negate(int x) { return 2; }(c) /* isEqual - return 1 if x == y, and 0 otherwise*…