Two sensors are set in the tunnel. The sensor closest to the cage is G1, the other G2. Only one lion can walk through the tunnel at a time. A lion is allowed to reverse in the tunnel.
As the lion starts walking out, (G1, G2) = (1, 0), the counter increments by 1. One of two scenarios then occurs: either the sensors read (G1, G2) = (0, 0) before it reads (G1, G2) = (0, 1). In that case, the counter will decrement as the lion must have gone back into the cage. Otherwise, the sensors will read (G1, G2) = (0, 1) before it reads (G1, G2) = (0, 0), thereby letting us know that it has passed (G1, G2) = (1, 1) as well. The counter will not change in this case.
If we read (G1, G2) = (1, 1) before reading (G1, G2) = (1, 0), we know a lion is moving from the enclosure to the cage. We then repeat the above two cases. This allows us to create a two state graph of the problem, transitioning from S0 to S1 on either (G1, G2) = (1, 0) or (G1, G2) = (1, 1) and returning back on (G1, G2) = (0, 1) or (G1, G2) = (0, 0), incrementing the counter if S0->S1 on (G1, G2) = (1, 0) and decrementing the counter if S1->S0 on (G1, G2) = (0, 0).
After reset, the counter should increase by 1 if a lion moves from the cage to the enclosure, and the opposite if vice versa.
| # | Input | Output | Bidirectional |
|---|---|---|---|
| 0 | G1, first sensor in tunnel | segment a | They do nothing. |
| 1 | G2, second sensor in tunnnel | segment b | |
| 2 | segment c | ||
| 3 | segment d | ||
| 4 | segment e | ||
| 5 | segment f | ||
| 6 | segment g | ||
| 7 |