Question

I have an array which tells whether a card is in use:

int used[52];

This is a terrible way to pick a random card if I have many used cards:

do {
  card = rand() % 52;
} while (used[card]);

since if I have only 3-4 unused cards, it’ll take forever to find them.

I came up with this:

 int card;
 int k = 0;
 int numUsed = 0;
 for (k=0; k < 52; ++k) {
   if (used[k]) numUsed += 1;
 }
 if (numUsed == 52) return -1;
 card = rand() % (52 - numUsed);

 for (k=0; k < 52; ++k) {
   if (used[k]) continue;
   if (card == 0) return k;
   card -= 1;
 }

which I guess works better if the deck is full, but works worse when the deck is empty since I have to go through two for loops.

What’s the most efficient way to do this?

Answer 1

I think your two-pass algorithm is likely to be the best you can do, given the constraint you added in a comment that you don’t know in advance which cards are eligible for a given draw.

You could try the cunning “select at random from a list of unknown size in a single pass” algorithm:

int sofar = 0;
int selected = -1;
for (i = 0; i < 52; ++i) {
    if (used[i]) continue;
    ++sofar;
    if ((rand() % sofar) == 0) selected = i;
}
if (selected == -1) panic; // there were no usable cards 
else used[selected] = 1;   // we have selected a card

Then if (as you say in a comment) different draws have different criteria, you can replace used[i] with whatever the actual criteria are.

The way it works is that you select the first card. Then you replace it with the second card with probability 1/2. Replace the result with the third card with probability 1/3, etc. It’s easy to prove by induction that after n steps, the probability of each of the preceding cards being the selected one, is 1/n.

This method uses lots of random numbers, so it’s probably slower than your two-pass version unless getting each item is slow, or evaluating the criteria is slow. It’d normally be used e.g. for selecting a random line from a file, where you really don’t want to run over the data twice. It’s also sensitive to bias in the random numbers.

It’s good and simple, though.

[Edit: proof

Let p(j,k) be the probability that card number j is the currently-selected card after step k.

Required to prove: for all n, p(j,n) = 1/n for all 1 <= j <= n

For n = 1, obviously p(1,1) = 1, since the first card is selected at the first step with probability 1/1 = 1.

Suppose that p(j,k) = 1/k for all 1 <= j <= k.

Then we select the (k+1)th card at step (k+1) with probability 1/(k+1), i.e p(k+1,k+1) = 1/(k+1).

We retain the existing selection with probability k/(k+1), so for any j < k+1:

p(j,k+1) = p(j,k) * k/(k+1)
         = 1/k    * k/(k+1)   // by the inductive hypothesis
         = 1/(k+1)

So p(j,k+1) = 1/(k+1) for all 1 <= k <= k+1

Hence, by induction, for all n: p(j,n) = 1/n for all 1 <= j <= n]