DECISION STUDY · peg-07
Read a finish backwards
Imagine undoing the final jump: one peg splits into two with an empty landing left behind. Use that idea to explain a complete forward finish.
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Select a peg, then one of its available empty landing squares. A jump removes the peg in between.
Keyboard: Tab to an enabled control, then press Enter or Space. The study starts from the diagram shown; reload can restore your own saved moves.
Try the task before inspecting the answer.
Show the starting-position answer and evidence
Every jump removes exactly one peg, so a one-peg finish needs 6 jumps. Every legal first jump is evaluated. A successful continuation is verified by a complete finishing route; a no-finish conclusion exhausts all remaining legal branches. The route below is one verified finish; its last peg may be anywhere on this classic board. Read the final jump backward: the finish has its only peg at R3C3. Remove that peg and restore pegs at R5C3 and R4C3. That is exactly the two-peg position before the last jump. Continue reversing the written route from bottom to top: each undo restores its source and middle, and empties its landing. These reverse operations explain construction; the playable board permits only forward jumps.
| First move | Immediate effect | Checked consequence |
|---|---|---|
| R3C2 → R1C2 | Remove the jumped peg at R2C2. 6 pegs remain. | No one-peg finish exists after this jump; every remaining legal branch was checked. |
| R3C2 → R5C2 | Remove the jumped peg at R4C2. 6 pegs remain. | No one-peg finish exists after this jump; every remaining legal branch was checked. |
| R4C2 → R4C4 | Remove the jumped peg at R4C3. 6 pegs remain. | No one-peg finish exists after this jump; every remaining legal branch was checked. |
| R4C3 → R4C1 | Remove the jumped peg at R4C2. 6 pegs remain. | Can finish with one peg in 5 further jumps. |
| R5C3 → R3C3 | Remove the jumped peg at R4C3. 6 pegs remain. | No one-peg finish exists after this jump; every remaining legal branch was checked. |
| R5C3 → R5C5 | Remove the jumped peg at R5C4. 6 pegs remain. | No one-peg finish exists after this jump; every remaining legal branch was checked. |
| R5C4 → R5C2 | Remove the jumped peg at R5C3. 6 pegs remain. | Can finish with one peg in 5 further jumps. |
One complete checked route
These moves start from the original diagram, not from any different choices you made above. Read coordinates at the moment each move is made.
- Jump R4C3 → R4C1 over R4C2.
- Jump R2C2 → R4C2 over R3C2.
- Jump R4C1 → R4C3 over R4C2.
- Jump R5C4 → R5C2 over R5C3.
- Jump R5C1 → R5C3 over R5C2.
- Jump R5C3 → R3C3 over R4C3.
After the first move
Verified finish
How this position was checked
Built by reversing legal jumps from one centre peg; the recorded forward construction is verified.
Explain what you learned
Choose one first move from the comparison table. Explain its effect without rereading the answer. Then change the first move and identify which part of the explanation changes. A checked route answers this starting position; it does not automatically apply to a new one.
Rules and limits
Playfield's classic version uses a full five-by-five grid of holes, not the traditional cross-shaped board. Jump one peg horizontally or vertically over exactly one adjacent peg into the empty hole immediately beyond. The jumped peg is removed. Finish with exactly one peg anywhere on the board; the classic version does not require a centre finish. Diagonal jumps and jumps over longer runs are not allowed.
Every classic jump removes exactly one peg, so a successful finish from N pegs uses N minus 1 jumps. That arithmetic does not prove a successful sequence exists. An available continuation is not proof of eventual solvability. Exact winning or losing claims require a complete checked sequence or exhaustive search of the stated position. The live generator uses reverse jumps to build reachable layouts, but a poor forward choice can still strand the pegs.