Why is modulo slow?
It is not exactly slower than other operators, but typically \% operator is equivalent to three operations . Suppose x = y \% z is equivalent to x = x – y *(x/y ) . So number of instruction cycle will obviously much more for its computation , so it is slower.
Is modulo slow C++?
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 division?
It works, but a modulo reduction involves a division, and divisions are expensive. Much more expensive than multiplications. A single 32-bit division on a recent x64 processor has a throughput of one instruction every six cycles with a latency of 26 cycles….A fast alternative to the modulo reduction.
| modulo reduction | fast range |
|---|---|
| 8.1 | 2.2 |
Why modulo operator is costly?
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. A side-effect is that errors (especially floats) multiply too and you can end up with less precision.
Is modulo efficient?
Modular exponentiation can be performed with a negative exponent e by finding the modular multiplicative inverse d of b modulo m using the extended Euclidean algorithm. That is: c = be mod m = d−e mod m, where e < 0 and b ⋅ d ≡ 1 (mod m). Modular exponentiation is efficient to compute, even for very large integers.
How long does modulo operation take?
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 modulo?
a = q × d + r where q is the quotient, d is the divisor and r is the remainder. 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).
Can modulo result be negative?
Can a modulus be negative? \% can be negative as it is the remainder operator, the remainder after division, not after Euclidean_division. Since C99 the result may be 0, negative or positive. The modulo OP wanted is a classic Euclidean modulo, not \% .
What is modulo reduction?
If you divide an integer a by a non-zero integer d, you get. a = q × d + r where q is the quotient, d is the divisor and r is the remainder. There are d possible remainders: 0,1,2,…,d − 1. The reduction modulo d of an integer is, loosely speaking, its remainder in the division by d.
Is the modulo operator constant time?
Modulo/remainder is a O(1) operation (it’s essentially just a variation on division, which takes constant time on fixed-sized numbers).
How does the modulo operator work?
The modulus operator, sometimes also called the remainder operator or integer remainder operator works on integers (and integer expressions) and yields the remainder when the first operand is divided by the second. The syntax is the same as for other operators.
What does the modulo operator do in C?
The modulo operator, denoted by \%, is an arithmetic operator. The modulo division operator produces the remainder of an integer division. Syntax: If x and y are integers, then the expression: x \% y. produces the remainder when x is divided by y. Return Value: If y completely divides x, the result of the expression is 0.
How does the modulus operator work in between 2 operands?
This modulus operator works in between 2 operands. The modulus operator finds the division with numerator by denominator which results in the remainder of the number. Remainder always integer number only. If no remainder is there, then it gives you 0 (zero) as the remainder. Let’s consider a and b are 2 integers then the modulus expression becomes
Why is the modulo of a number so slow?
Feels like an O(N^3) with variable N 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.
Why is my modulo operator not working with floating-point variables?
If you try to use the modulo operator with floating-point constants or variables, the compiler will produce a error: The sign of the result for modulo operator is machine-dependent for negative operands, as the action takes as a result of underflow or overflow.