The Dragon Guard
To pass the bridge, ensure the GUARD signal is active. The GUARD must trigger only if both FIRE and COLD are not present at the same time, but exactly one of them must be active.
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The truth table
Inputs on the left, the output on the right. A circuit solves the puzzle when it reproduces every row.
| FIRE | COLD | STORM | GUARD |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 |
What the table tells you
- GUARD is 1 in 2 of the 8 rows. Every input matters: for each one there is a row where flipping it changes GUARD. The rows that turn it on: FIRE=0, COLD=1, STORM=0; FIRE=1, COLD=0, STORM=0.
The smallest known circuit: 3 gates
As a formula, one line per output:
GUARD = ((FIRE XOR COLD) AND NOT STORM)
As a netlist, in the order you would wire it. Each g is one gate; the last line says which signal drives the output.
g1 = FIRE XOR COLDg2 = NOT STORMg3 = g1 AND g2GUARD = g3
The search ran out of budget before it could rule out every smaller circuit, so 3 is the best anyone found, not a proof. If you build it in fewer, the leaderboard records it. Gates that do not feed an output are never counted.
Leaderboard for the day
Nobody posted a result on the day. The circuit above is the reference.