SOLO-026 · Hamiltonian path
EXPERTThe One-Way Crossing
Number all boxes from 1 to 36 to form a single orthogonal path without repetition.
This page contains the solution.
The answer
The full numbering below determines the only compatible route.
| L / C | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 29 | 30 | 31 | 32 | 33 | 36 |
| 2 | 28 | 27 | 26 | 25 | 34 | 35 |
| 3 | 21 | 22 | 23 | 24 | 1 | 2 |
| 4 | 20 | 19 | 18 | 17 | 16 | 3 |
| 5 | 11 | 12 | 13 | 14 | 15 | 4 |
| 6 | 10 | 9 | 8 | 7 | 6 | 5 |
The reasoning
- From marker 1 to marker 4, the portion retained is: 1 (L3C5) → 2 (L3C6) → 3 (L4C6) → 4 (L5C6). Check each orthogonal shift, then mark the boxes already consumed before continuing.
- From marker 4 to marker 6, the portion retained is: 4 (L5C6) → 5 (L6C6) → 6 (L6C5). Check each orthogonal shift, then mark the boxes already consumed before continuing.
- From marker 6 to marker 9, the portion retained is: 6 (L6C5) → 7 (L6C4) → 8 (L6C3) → 9 (L6C2). Check each orthogonal shift, then mark the boxes already consumed before continuing.
- From marker 9 to marker 12, the portion retained is: 9 (L6C2) → 10 (L6C1) → 11 (L5C1) → 12 (L5C2). Check each orthogonal displacement, then mark the boxes already consumed before continuing.
- From marker 12 to marker 21, the portion retained is: 12 (L5C2) → 13 (L5C3) → 14 (L5C4) → 15 (L5C5) → 16 (L4C5) → 17 (L4C4) → 18 (L4C3) → 19 (L4C2) → 20 (L4C1) → 21 (L3C1). Check each orthogonal shift, then mark the boxes already consumed before continuing.
- From marker 21 to marker 28, the portion retained is: 21 (L3C1) → 22 (L3C2) → 23 (L3C3) → 24 (L3C4) → 25 (L2C4) → 26 (L2C3) → 27 (L2C2) → 28 (L2C1). Check each orthogonal shift, then mark the boxes already consumed before continuing.
- From marker 28 to marker 36, the portion retained is: 28 (L2C1) → 29 (L1C1) → 30 (L1C2) → 31 (L1C3) → 32 (L1C4) → 33 (L1C5) → 34 (L2C5) → 35 (L2C6) → 36 (L1C6). Check each orthogonal shift, then mark the boxes already consumed before continuing.
- The exploration of all compatible continuations gives 423 states. It rejects 93 separations of the remaining cells and 83 violations of distance or parity. Only one complete visit remains.
- To reproduce this proof on paper, starting from 1, try each free neighbour in the order high, low, left, right, and keep a tree. Immediately reject a path that reaches a marker in the wrong rank, isolate an unvisited area or cannot reach the next marker in as many steps as required. The last complete path is that displayed; the validator provided performs exactly this exploration.
Common mistakes and false leads
- Local continuity is not enough: a detour can cut the remaining boxes in half.
- Two boxes of the same color on a checker require an even number of movements; check this parity between markers.
- Reaching a landmark too early is as forbidden as missing it.
How we verified this answer
DFS comprehensive simple paths, with necessary controls of connection, benchmarks, distance and parity; comparison of all solutions up to two.
{
"nodes": 423,
"dead_ends": 17,
"connectivity_rejections": 93,
"distance_rejections": 83
}Coordinate the intervals between markers, the separations of regions and the constraints of parity. The solver must explore 423 prefixes; the level reflects this combined exploration, without automatically absorbing time of calculation and human difficulty.