The Lighthouse Beacon
To activate the BEACON, both FOG and WIND must be absent. However, if OIL is low, the BEACON will flicker even when conditions are clear.
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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.
| FOG | WIND | OIL | BEACON |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 |
What the table tells you
- BEACON is 1 in 1 of the 8 rows, so it is a single AND-shaped condition: one exact combination of the inputs. Every input matters: for each one there is a row where flipping it changes BEACON. The rows that turn it on: FOG=0, WIND=0, OIL=0.
The smallest known circuit: 5 gates
As a formula, one line per output:
BEACON = ((NOT FOG AND NOT WIND) AND NOT OIL)
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 = NOT FOGg2 = NOT WINDg3 = NOT OILg4 = g1 AND g2g5 = g4 AND g3BEACON = g5
The search ran out of budget before it could rule out every smaller circuit, so 5 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.