How to do modulo multiplication for 64‑bit values without using BigInteger in Go

1 Answer

0 votes
package main

import (
    "fmt"
)

/*
===============================================================
    Modulo Multiplication (Slow and Fast Versions)
===============================================================

Purpose:
    Compute (a * b) % mod safely for large 64‑bit values.

Versions:
    1. Slow version:
        - Adds b to result a times.
        - Always correct.
        - Very slow for large numbers.
        - Useful as a correctness reference.

    2. Fast version:
        - Uses the classic "double‑and‑add" technique.
        - Runs in O(log b).
        - Avoids overflow by never multiplying large numbers.
        - Produces the same result as the slow version.

Notes:
    Go uint64 arithmetic wraps on overflow, so the fast version
    reduces modulo at every step to keep values safe.
*/


// ---------------------------------------------------------------
// SLOW VERSION (simple, correct, but extremely slow)
// ---------------------------------------------------------------
func mulModSlow(a, b, mod uint64) uint64 {
    /*
        Computes:
            (b + b + b + ... a times) % mod

        This avoids overflow because:
            - result stays below mod
            - b fits in uint64
            - addition is safe

        But it is O(a), which is too slow for large inputs.
    */

    if b < a {
        a, b = b, a // reduce loop count
    }

    var result uint64 = 0

    for i := uint64(0); i < a; i++ {
        result = (result + b) % mod
    }

    return result
}


// ---------------------------------------------------------------
// FAST VERSION (efficient and safe)
// ---------------------------------------------------------------
func mulModFast(a, b, mod uint64) uint64 {
    /*
        Uses the "double‑and‑add" method:

            - If the lowest bit of b is set, add a to result.
            - Double a each step.
            - Shift b right each step.

        This avoids overflow because:
            - We never compute a * b directly.
            - Doubling a is safe because we reduce modulo each step.

        This makes the algorithm:
            - Fast
            - Safe
            - Exact
    */

    var result uint64 = 0
    a %= mod

    for b > 0 {
        if b&1 == 1 {
            result = (result + a) % mod
        }

        a = (a << 1) % mod
        b >>= 1
    }

    return result
}


// ---------------------------------------------------------------
// MAIN PROGRAM
// ---------------------------------------------------------------
func main() {
    x := uint64(798345)
    y := uint64(20289473612815)
    mod := uint64(100000000000003)

    slow := mulModSlow(x, y, mod)
    fast := mulModFast(x, y, mod)

    fmt.Println("Slow result:", slow)
    fmt.Println("Fast result:", fast)
}



/*
run:

Slow result: 99811422305238
Fast result: 99811422305238

*/

 



answered Aug 16 by avibootz

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