# Joe Ganley

I make software and sometimes other things.

Due to Blogger's termination of support for FTP, this blog is no longer active.

It is possible that some links from here, particularly those within the site, are now broken. If you encounter one of those, your best bet is to go to the new front page and hunt for it from there.

Most, but not all, of the blog's posts are on this page; the archives are here.

Consider the problem of choosing a random element from a linked list whose length you do not know. Obviously, you could count the length of the list, then pick a random number between 1 and length, and then walk to that element, but there is a clever way to do this in one pass.

The algorithm is as follows:

```Element choice; int count = 0; for (Element e = head; e != NULL; e = e->next) {    if (rand01() <= (1.0 / ++count)) {       choice = e;    } }```

You can easily show that when this finishes, choice is equal to any given element with probability 1/n, where n is the number of elements: The probability of setting the ith element to be the new choice is 1/i. The probability of not selecting any of the subsequent elements to be the new choice is the product as j goes from i+1 to n of (1 - 1/j), which if you work out the math ends up as i/n. Multiply the two probabilities together and you get 1/n.

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