Here is the complete, deterministic solution for the 12 Coins Puzzle.
Bonus first: Why 3 weighings is the theoretical minimum
Each coin can be either Heavy (H) or Light (L), giving 12 × 2 = 24 possible states.
A balance scale has 3 possible outcomes per weighing: Left down, Balanced, Right down (a ternary digit, $\log_3$).
With $k$ weighings, the maximum number of distinguishable states is $3^k$:
- $k = 2 \implies 3^2 = 9 < 24$ (mathematically impossible).
- $k = 3 \implies 3^3 = 27 \ge 24$.
Therefore, 3 weighings is the absolute information-theoretic minimum.
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The Strategy
Label coins 1 to 12.
Weighing 1: Weigh [1, 2, 3, 4] against [5, 6, 7, 8]
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CASE A: Scales Balance ([1..4] == [5..8])
The counterfeit is in [9, 10, 11, 12], and coins 1..8 are proven genuine (G).
Possible states: 9H, 9L, 10H, 10L, 11H, 11L, 12H, 12L (8 states).
- Weighing 2: Weigh [9, 10, 11] against [G, G, G] (e.g. 1, 2, 3).
- Balanced: The counterfeit is 12.
- Weighing 3: Weigh [12] against [G]. If 12 goes down $\to$ 12 Heavy. If up $\to$ 12 Light.
- Left is Heavy: The counterfeit is in [9, 10, 11] and is Heavy.
- Weighing 3: Weigh [9] against [10]. If 9 down $\to$ 9 Heavy; if 10 down $\to$ 10 Heavy; if balanced $\to$ 11 Heavy.
- Right is Heavy (Left up): The counterfeit is in [9, 10, 11] and is Light.
- Weighing 3: Weigh [9] against [10]. If 9 up $\to$ 9 Light; if 10 up $\to$ 10 Light; if balanced $\to$ 11 Light.
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CASE B: Left is Heavy ([1, 2, 3, 4] > [5, 6, 7, 8])
(Note: If Right is heavy, the logic is identical by inverting sides).
Coins 9..12 are proven genuine (G).
The counterfeit is either in [1, 2, 3, 4] and Heavy, OR in [5, 6, 7, 8] and Light (8 candidate states: 1H, 2H, 3H, 4H, 5L, 6L, 7L, 8L).
- Weighing 2: Rearrange coins:
- Left pan: [1, 2, 5] (keep two heavy-candidates, bring one light-candidate 5)
- Right pan: [3, 6, G] (bring one heavy-candidate 3, keep one light-candidate 6, add genuine G)
- On the table: [4, 7, 8]
- Subcase B1: Scales Balance
The counterfeit was left on the table: candidates are 4H, 7L, 8L.
- Weighing 3: Weigh [7] against [8].
- If balanced $\to$ 4 Heavy.
- If 7 goes up $\to$ 7 Light.
- If 8 goes up $\to$ 8 Light.
- Subcase B2: Left stays Heavy ([1, 2, 5] > [3, 6, G])
The tilt stayed the same. The counterfeit cannot be 5 (a light candidate on the heavy side would make it go up), cannot be 3 (a heavy candidate on the light side would make it go down).
Remaining candidates: 1H, 2H, 6L.
- Weighing 3: Weigh [1] against [2].
- If 1 goes down $\to$ 1 Heavy.
- If 2 goes down $\to$ 2 Heavy.
- If balanced $\to$ 6 Light.
- Subcase B3: Right becomes Heavy ([1, 2, 5] < [3, 6, G])
The tilt reversed! The only coins whose relocation could cause this are 5 (light candidate moved left) or 3 (heavy candidate moved right).
Remaining candidates: 3H, 5L.
- Weighing 3: Weigh [3] against [G].
- If 3 goes down $\to$ 3 Heavy.
- If balanced $\to$ 5 Light.
Every single leaf of the decision tree terminates in $\le 3$ weighings with exact coin identification and heavy/light determination.