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Discussion on: Daily Challenge #108 - Find the Counterfeit Coin

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Daan Wilmer

While one might be tempted at first to weigh all coins in pairs, requiring up to n/2 weightings, we can also weigh multiple coins at once. If we split the number of coins in two equal piles of coins, the pile with the counterfeit coin will weigh less. We can then split that pile and weigh those piles, until we are down to two coins.
One might wonder, as I did at first, "what happens when you have an odd number of coins?" In this case, you can take one coin out, and you have an even number of coins. Either the counterfeit coin is in one of the two piles you are weighing and one pile is therefore lighter, or the counterfeit coin is the one you took out and the two piles are equally weighted.
While you could simulate this very easily with the following function:

function numWeighings($numCoins) {
    if ($numCoins <= 1) {
        return 0;

    return numWeighings(floor($numCoins / 2)) + 1;

this is, in fact, the logarithm of the number of coins in base two, which can be coded as such:

function numWeighings($numCoins) {
    return floor(log($numCoins, 2));

(and yes I had to double check if it was, indeed, the logarithm, by testing both functions up to 10000000)