What does modulo 3 mean?

1 mod 3 equals 1, since 1/3 = 0 with a remainder of 1. To find 1 mod 3 using the modulus method, we first find the highest multiple of the divisor, 3 that is equal to or less than the dividend, 1. Then, we subtract the highest multiple from the dividend to get the answer to 1 mod 3. Multiples of 3 are 0, 3, 6, 9, etc.

Which is faster modulo or division?

When the modulus m is constant, even where there is a hardware divide instruction, it can be faster to take the modulus directly than to use the divide instruction. These tricks become even more valuable on machines without a hardware divide instruction or where the numbers involved are out of range.

How do you calculate mod 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 is modulo calculated?

In mathematics, the modulo is the remainder or the number that’s left after a number is divided by another value. Modulo is also referred to as ‘mod.’ For example, you’re calculating 15 mod 4. When you divide 15 by 4, there’s a remainder.

Is modulo slow?

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.

What does modulo 8 mean?

The modulus method requires us to first find out what the highest common multiple of the Divisor (8) is that is equal to or less than the Dividend (8). We can see that multiples of 8 are 0, 8, 16, 24, etc. The highest multiple that is less than or equal to 8 is 8.

Why is modulo expensive?

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.

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.

Which is faster x * n Div 2 32 or modulo reduction?

Computing ( x * N) div 2 32 is very fast on a 64-bit processor. It is a multiplication followed by a shift. On a recent Intel processor, I expect that it has a latency of about 4 cycles and a throughput of at least on call every 2 cycles. So how fast is our map compared to a 32-bit modulo reduction?

Is there a way to precompute a modulo reduction?

There are fancy tricks to “precompute” a modulo reduction so that it can be transformed into a couple of multiplications as well as a few other operations, as long as N is known ahead of time. Your compiler will make use of them if N is known at compile time.

How can I tell that the modulo reduction is fair?

If N is small compared to 2 32, then this map could be considered as good as perfect. The common solution is to do a modulo reduction: x mod N. (Since we are computer scientists, we define the modulo reduction to be the remainder of the division, unless otherwise stated.) How can I tell that it is fair? Well.