How do you randomly zero a bit in an integer?

Question

Updated with newer answer and better test

Let\'s say I have the number 382 which is 101111110.

How could I randomly turn a bit which is not 0 to

Answer 1

static Random random = new Random();

public static int Perturb(int data)
{
    if (data == 0) return 0;

    // attempt to pick a more narrow search space
    int minBits = (data & 0xFFFF0000) == 0 ? 16 : 32;

    // int used = 0; // Uncomment for more-bounded performance
    int newData = data;
    do
    {
        // Unbounded performance guarantees
        newData &= ~(1 << random.Next(minBits));

        // // More-bounded performance:
        // int bit = 1 << random.Next(minBits);
        // if ((used & bit) == bit) continue;
        // used |= bit;
        // newData &= ~bit;
    } while (newData == data); // XXX: we know we've inverted at least one 1
                               // when the new value differs

    return newData;
}

Update: added code above which can be used for more-bounded performance guarantees (or less unbounded if you want to think of it that way). Interestingly enough this performs better than the original uncommented version.

Below is an alternate approach that is fast but without bounded performance guarantees:

public static int FastPerturb(int data)
{
    if (data == 0) return 0;

    int bit = 0;
    while (0 == (data & (bit = 1 << random.Next(32))));

    return data & ~bit;
}