When do you need a fast modular exponentiation?

When do you need a fast modular exponentiation?

This has given us a method to calculate A^B mod C quickly provided that B is a power of 2. However, we also need a method for fast modular exponentiation when B is not a power of 2.

How to make the execution time less in C + +?

Increment by one make the number non-divisible. Next increment – number still non-divisible. Third increment makes the number again divisible. First optimization one can try is to not to let number grow above 2: if during increment number goes from 2 to 3, set it back to zero.

Is it possible to eliminate the modulus operator?

Presumably this is because the ARM has native support for the modulus operation. Notwithstanding the ARM results, it’s clear that at least in this example, it’s possible to significantly speed up an algorithm by eliminating the modulus operator. I could of course just stop at this point.

Is it possible to make a modulus faster in C #?

If you are dividing by literals that are powers of two, then the answer is probably No: Any decent compiler will automatically turn such expressions into a variation of an AND operation, which is pretty close to optimal. Thanks for contributing an answer to Stack Overflow!

What happens if base is zero in modular exponentiation?

If a is zero, no code executes since this effectively multiplies the running total by one. If a instead is one, the variable base (containing the value b2i mod m of the original base) is simply multiplied in. In this example, the base b is raised to the exponent e = 13 .

When to use modular exponentiation in cryptography?

Modular exponentiation is a type of exponentiation performed over a modulus. It is useful in computer science, especially in the field of public-key cryptography . The operation of modular exponentiation calculates the remainder when an integer b (the base) raised to the e th power (the exponent), be,…

How to calculate the power of modular arithmetic?

Below is the fundamental modular property that is used for efficiently computing power under modular arithmetic. (ab) mod p = ( (a mod p) (b mod p) ) mod p For example a = 50, b = 100, p = 13 50 mod 13 = 11 100 mod 13 = 9 (50 * 100) mod 13 = ( (50 mod 13) * (100 mod 13) ) mod 13 or (5000) mod 13 = ( 11 * 9 ) mod 13 or 8 = 8

Why is power evaluated under modulo of modular arithmetic?

The problem with above solutions is, overflow may occur for large value of n or x. Therefore, power is generally evaluated under modulo of a large number. Below is the fundamental modular property that is used for efficiently computing power under modular arithmetic.

When do we use modular exponentiation in cryptography?

It should be noted that when we perform modular exponentiation in cryptography, it is not unusual to use exponents for B > 1000 bits. What is modular arithmetic? This is the currently selected item. Posted 8 years ago. Direct link to JMGClark’s post “Is there a tutorial/challenge for writing numbers …”

How can we calculate a ^ b mod c quickly?

This has given us a method to calculate A^B mod C quickly provided that B is a power of 2. However, we also need a method for fast modular exponentiation when B is not a power of 2. How can we calculate A^B mod C quickly for any B ?