Why is modulo operator slow?

Why is modulo operator slow?

Integer division and modulo are relatively slow because there is no direct hardware support (they compile to multiple instruction sequences). Floating point modulo is fast. Integer modulo is also slow on CPUs for the same reason. you can get a improvement by replacing the modulo op by the actual formula.

Is modulo expensive operation?

Orange: division and modulus Division and modulus are more than twice as expensive as multiplication (a weight 10). The division by two or a multiple of two is always a trick, but not much more can be done without having side-effects.

What is faster modulo or division?

Branchless non-power-of-two modulus is possible by precomputing magic constants at run-time, to implement division using a multiply-add-shift. This is roughly 2x faster than the built-in modulo operator % on my Intel Core i5.

What is the complexity of modulo operation?

Modulo/remainder is a O(1) operation (it’s essentially just a variation on division, which takes constant time on fixed-sized numbers). Therefore, the inside of the loop is an O(1) operation, which makes the total complexity O(√n) .

How do you reduce modulus?

The reduction modulo d of an integer is, loosely speaking, its remainder in the division by d. Definition Let d be a non-zero integer. Two integers a and b are congruent modulo d if d | (a − b). Two integers a and b are congruent modulo d if and only if they have the same remainder in the division by d.

Is modulo slow in Java?

Modulo is usually implemented by division which is normally the slowest hardware operation on a CPU.

Is modulus faster than if?

A modulo-operation is very slow. An if is most likely to be faster than a modulo and more readable.

Is mod slower than division?

So in simple terms, this should give you a feel for why division and hence modulo is slower: computers still have to do long division in the same stepwise fashion tha you did in grade school.

Is modulo faster than bitwise?

The result was consistent with the idea that bitwise is faster than modulo operator.

What is the time complexity of modular exponentiation?

Modular exponentiation: [xy mod N]. Directly calculate xy with y-1 multiplications would take too long. The algorithm in Figure 1.4 does O(n) recursive calls, and each of them takes O(n2) time, so the complexity is O(n3).

How does modulo operator work?

The modulus operator is added in the arithmetic operators in C, and it works between two available operands. It divides the given numerator by the denominator to find a result. In simpler words, it produces a remainder for the integer division. Thus, the remainder is also always an integer number only.

How do you fix mod problems?

How to calculate the modulo – an example

  1. Start by choosing the initial number (before performing the modulo operation).
  2. Choose the divisor.
  3. Divide one number by the other, rounding down: 250 / 24 = 10 .
  4. Multiply the divisor by the quotient.
  5. Subtract this number from your initial number (dividend).

How do you calculate modulo fast?

How can we calculate A^B mod C quickly for any B?

  1. Step 1: Divide B into powers of 2 by writing it in binary. Start at the rightmost digit, let k=0 and for each digit:
  2. Step 2: Calculate mod C of the powers of two ≤ B. 5^1 mod 19 = 5.
  3. Step 3: Use modular multiplication properties to combine the calculated mod C values.

How can modular multiplication prevent overflow?

We can multiply recursively to overcome the difficulty of overflow. To multiply a*b, first calculate a*b/2 then add it twice. For calculating a*b/2 calculate a*b/4 and so on (similar to log n exponentiation algorithm).

What does modulo operator do?

Is bit shift faster than addition?

On most older microprocessors, bitwise operations are slightly faster than addition and subtraction operations and usually significantly faster than multiplication and division operations.

Is bit shift fast?

Bit-shifting is still faster, but for non-power-of-two mul/div by the time you do all your shifts and add the results it’s slower again.

Why do we need modular exponentiation?

Modular exponentiation is efficient to compute, even for very large integers. On the other hand, computing the modular discrete logarithm – that is, finding the exponent e when given b, c, and m – is believed to be difficult.

How do you do fast exponents?

Why is modulo useful?

The modulus operator is useful in a variety of circumstances. It is commonly used to take a randomly generated number and reduce that number to a random number on a smaller range, and it can also quickly tell you if one number is a factor of another.