Java inverse modulo 2**64

问题

Given an odd long x, I'm looking for long y such that their product modulo 2**64 (i.e., using the normal overflowing arithmetic) equals to 1. To make clear what I mean: This could be computed in a few thousand year this way:

for (long y=1; ; y+=2) {
    if (x*y == 1) return y;
}

I know that this can be solved quickly using the extended Euclidean algorithm, but it requires the ability to represent all the involved numbers (ranging up to 2**64, so even unsigned arithmetic wouldn't help). Using BigInteger would surely help, but I wonder if there's a simpler way, possibly using the extended Euclidean algorithm implemented for positive longs.

回答1:

Here's one way of doing it. This uses the extended Euclidean algorithm to find an inverse of abs(x) modulo 2⁶², and at the end it 'extends' the answer up to an inverse modulo 2⁶⁴ and applies a sign change if necessary:

public static long longInverse(long x) {

    if (x % 2 == 0) { throw new RuntimeException("must be odd"); }

    long power = 1L << 62;

    long a = Math.abs(x);
    long b = power;
    long sign = (x < 0) ? -1 : 1;

    long c1 = 1;
    long d1 = 0;
    long c2 = 0;
    long d2 = 1;

    // Loop invariants:
    // c1 * abs(x) + d1 * 2^62 = a
    // c2 * abs(x) + d2 * 2^62 = b 

    while (b > 0) {
        long q = a / b;
        long r = a % b;
        // r = a - qb.

        long c3 = c1 - q*c2;
        long d3 = d1 - q*d2;

        // Now c3 * abs(x) + d3 * 2^62 = r, with 0 <= r < b.

        c1 = c2;
        d1 = d2;
        c2 = c3;
        d2 = d3;
        a = b;
        b = r;
    }

    if (a != 1) { throw new RuntimeException("gcd not 1 !"); }

    // Extend from modulo 2^62 to modulo 2^64, and incorporate sign change
    // if necessary.
    for (int i = 0; i < 4; ++i) {
        long possinv = sign * (c1 + (i * power));
        if (possinv * x == 1L) { return possinv; }
    }

    throw new RuntimeException("failed");
}

I found it easier to work with 2⁶² than 2⁶³, mainly because it avoids problems with negative numbers: 2⁶³ as a Java long is negative.

回答2:

In the meantime I've recalled/reinvented a very simple solution:

public static int inverseOf(int x) {
    Preconditions.checkArgument((x&1)!=0, "Only odd numbers have an inverse, got " + x);
    int y = 1;
    for (int mask=2; mask!=0; mask<<=1) {
        final int product = x * y;
        final int delta = product & mask;
        y |= delta;
    }
    return y;
}

It works because of two things:

in each iteration if the corresponding bit of product is 1, then it's wrong, and the only way to fix is is by changing the corresponding bit of y
no bit of y influences any less significant bit of product, so no previous work gets undone

I started with int since for long it must work too, and for int I could run an exhaustive test.

Another idea: there must a a number n>0 such that x**n == 1, and therefore y == x**(n-1). This should probably be faster, I just can't recall enough math to compute n.

来源：https://stackoverflow.com/questions/11702215/java-inverse-modulo-264

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