## Iterated Squaring Time Complexity

exponentiation by squaring time complexity

time complexity of exponentiation

time complexity of power function

square-and multiply modular exponentiation

modular exponentiation time complexity

successive squaring

binary exponentiation algorithm example

I have the following implementation of iterated squaring:

def power(a, b): result = 1 while b > 0: if b % 2 == 1: result = mult(result, a) a = mult(a, a) b = b // 2 return result

The `mult()`

method multiplies 2 given numbers, when if one is of length of `x`

bits and the other one is of length of `y`

bits, the number of operations the multiplication will take is `x*y+(x+y)`

operations, which I need to consider in the complexity analysis.

And I'm trying to find an `O()`

notation bound of the number of arithemtic operations done as a function of `n`

and `m`

when a is n-bits long, and b is m-bits long. I need to take into consideration only the multiplication lines, and check for the worst case.

The worst case is when `b`

is an m-bit number where all `m`

binary digits are 1, and then I have `m`

iterations of the loop where in each iteration the if condition is true.

What I dont know how to do is how I should consider in my computation the fact that `a`

grows for each iteration? How to I put it in some sort of finite sum with probably geometric progression that I can compute?

Thanks

**Exponentiation by squaring,** Assuming that x is a float , each multiplication takes a constant amount of time. is O(n), since the block repeats n times, and each iteration takes the same amount of time. Here's an idea for computing xn quickly, exponentiation by squaring. Iterated Squaring Time Complexity. I have the following implementation of iterated squaring: The mult() method multiplies 2 given numbers, when if one is of length of x bits and the other one is of length of y bits, the number of operations the multiplication will take is x*y+(x+y) operations, which I need to consider in the complexity analysis.

I think the complexity of this algorithm is independent of a.

When reviewing the complexity, I think the number of loops is your main index for the highest complexity due to m.

Running through values of b, we can see following number if b > 0 computations as:

Interestingly, we see an increase in n loops for each multiple of 2 (i.e. 1,2,4,8,16,32,64).

Roughly speaking, I would say the complexity of this algorithm closely follows O(log2(n)).

**Time complexity of the fast exponentiation method,** Time complexity – a function of input length: Definitions. • Integer exponentiation: Naive algorithm (inefficient). • Integer exponentiation: Iterated squaring In mathematics and computer programming, exponentiating by squaring is a general method for fast computation of large positive integer powers of a number, or more generally of an element of a semigroup, like a polynomial or a square matrix. Some variants are commonly referred to as square-and-multiply algorithms or binary exponentiation.

Well, after some analyzing, I think that the number of iterations of the `power`

method will be sum on i (from 0 to m-1) of: `2^(2i-1)*n+(2^(i-1)+2^i*n+(2^i*n)^2+2^(i+1)*n`

and the it's pretty straight forward to get the complexity: `O(4^m*n^2)`

**Complexity of exponentiation,** This algorithm is polynomial in the size of the input and has complexity O(log(e) polylog(b Duration: 17:55
Posted: Aug 14, 2016 If we analyze the code, Time Complexity is O(power) or in general terms O(N) where N is power or b. So how do we find the base raised to the power for large numbers, as large as a billion! There’s an algorithm for that, it’s called Exponentiation by Squaring, fast power algorithm.

**Iterated digits squaring,** The efficiency of the modified squaring algorithm comes from Steps 2.1 This method requires three multiplications and four additions in each iteration. In Step 2.2.1, the flag NewBo is set to true if and at the same time the The main result is that nondeterministic 2-way and multi-head iterated pushdown automata characterize deterministic iterated exponential time complexity classes.

**[PDF] Time Complexity - CS1001.py,** a quadratic-time method is "order N squared": O(N2). Note that the big-O expressions do not have constants or low-order terms. This is because, when N gets large In computer science, the time complexity is the computational complexity that describes the amount of time it takes to run an algorithm. Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Thus, the amount of time taken and the number of elementary operations performed by the algorithm are taken to differ by at most a constant factor. Since an algorithm'

**Exponentiation by squaring,** This algorithm is considered efficient because the time it takes depends polynomially on the numbers of digits of a , r and n . For instance, the number of steps Exponentiation - Time Complexity analysis of recursion mycodeschool. Time Complexity analysis of recursion (Time & Space Complexity For Algorithms)

##### Comments

- Thank you for the reply, unfortunately, for this task, I cannot consider the multiplication as one opertaion, and I do need to consider the number of arithmetic operations for my calculations. I've tried to formulate with the formula of number of arithemtic operations I have, but I cannot find how to formulate the line of
`result=result*a`

as a function of`i`

(current iteration). Also, I didnt quite understand how you got the n^log2(3) computation for the first a*a, i got n^2+2n for the first one - @Jonathan it depends on the algorithm you are using for multiplication. See the link for Cpython.
- oh sorry, didnt see that paragraph when i've read your answer. anyway, i am bound to a specific algorithm for multiplying 2 x,y-bits numbers with the solution for the number of iterations as i wrote
- @Jonathan If you are bound to that algorithm, please edit your question and write it. E.g.
`def mult(x, y): ...`

and use`mult`

in your code. - hi, i've edited my question accordingly