Daily Challenge #108 - Find the Counterfeit Coin

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)

Randomly selected coin and lighter/heavier its value with Ruby:

# frozen_string_literal: true

# Save file as coins.rb and run `ruby coins.rb`
class Coins
  def initialize(place, value)
    @coins = Array.new(7, 1).insert(place, value)
    @count = 0
  end

  def weight
    return puts @count if @coins.size == 1

    @count += 1
    step = @coins.size / 2
    first = @coins.first(step)
    last = @coins.last(step)
    @coins = first.sum > last.sum ? first : last
    weight
  end
end

Coins.new(rand(0..7), [0, 2].sample).weight

Good one! This changes it to the following code:

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


    return numWeightings(ceil($numCoins / 3)) + 1;
}

which is equal to

function numWeightings($numCoins) {
    return ceil(log($numCoins, 3));
}

There's a modification where we don't know if the counterfeit is lighter or heavier, just that it will be different.
It is interesting because finding out if it is heavier or lighter is significant to picking the next pile to split.

This is a very old question. I remember being asked this question about 25 years ago.

The answer is to divide to three groups, as equal as possible. You have 9 coins. You start by measuring 3 against 3. If one is lighter than the other, that's where the coin is, and you again split to 3 piles of one coin each and measure. If both are the same, the counterfeit coin is the one left out - meaning you only need two weightings to find the counterfeit.

I've been thinking about this problem for a few days and each solution I can come up with is a tangled mess of comparative loops.