# TADM2E 4.46

Question: You are given 12 coins. One of them is heavier or lighter than the rest. Identify this coin in just three weighings

Solution: Number the coins 1 through 12 and divide them coins into 4 sets of 3...

There are multiple comparison sets possible. This is an acceptable template to find a few of them. (This template is NOT definitive, there are other solutions that don't follow this template)

Compare (set 1 & 1st coin from set 4) against (set 2 + 2nd coin from set 4) Compare (set 1 & 2nd coin from set 4) against (set 3 + 1st coin from set 4) Compare (1st coin from each set) against (3rd coin from each set)

A more concise example:

Compare 1 2 3 10 against 4 5 6 11 Compare 1 2 3 11 against 7 8 9 10 Compare 1 4 7 10 against 3 6 9 12

Each weighing can have 3 possible outcomes: Left Heavy, Right Heavy, or Balanced (L,R or B)

Build a truth table to interpret outcomes...many outcomes are not possible. Note: THE TABLE VALUES ARE DERIVED FROM THE CHOSEN COMPARISON SETS!

outcome: fake coin: l l l 1 is heavy r r r 1 is light l l b 2 is heavy r r b 2 is light l l r 3 is heavy r r l 3 is light r b l 4 is heavy l b r 4 is light r b b 5 is heavy l b b 5 is light r b r 6 is heavy l b l 6 is light b r l 7 is heavy b l r 7 is light b r b 8 is heavy b l b 8 is light b r r 9 is heavy b l l 9 is light r l r 10 is heavy l r l 10 is light l r b 11 is heavy r l b 11 is light b b r 12 is heavy b b l 12 is light

There are multiple comparison set possibilities, each with their own comparison table solution.

A simpler solution #2:

- put 6 coins on each side of the scale, one side will be heavier.
- use the heavier side from the first weighing and put 3 coins on each side of the scale.
- using the heavier side from the 2nd weighing, pick 2 coins and put 1 on each side of the scale.

If the scale is balanced then the coin you didn't weigh is the heavier one. Otherwise, the scale will show which one of the other 2 is the heavy coin.

Since we do not know if the faulty coin is heavier or lighter , the soluition #2 is not correct

The only solution is solution nr 1, for which we can also use a binary tree