The Porter’s One-Swap Claim

Case 26 · visual · Medium · About 4–7 minutes

Tally counts and their left-to-right order are clues; the railway background is decorative.

The museum’s luggage is waiting beside the night train. Each suitcase has a tally patch indicating its place in the loading order. The correct arrangement runs from one bar to four bars, left to right.

A porter claims the pictured row can be restored with just one exchange of neighboring cases. Vale checks the claim before approving the loading log.

The movement rule

• A move exchanges two suitcases currently next to one another.
• Only that pair changes position in the move. You cannot lift a case over another, reverse the whole row, or exchange a separated pair.
• All four cases must finish in increasing tally order: 1, 2, 3, 4.
• Each tally count occurs exactly once. The patches never change.

Picture evidence

Accessible evidence transcript: The four tally patches show, from left to right, two vertical bars, one bar, four bars, and three bars. In shorthand, the row is 2–1–4–3.

Is the porter’s claim possible? Find the minimum number of legal moves and name the pair exchanged in each. If more than one shortest move order exists, acknowledge it rather than claiming that a particular sequence is unique.

Only the tally patches define order. Suitcase size, station lighting, and the train behind them are not clues. Each exchange counts as one move even though two cases move at once.

Hint 1 - reveal when you want a nudge

There are two pairs currently in the wrong relative order. Consider 2 versus 1, and 4 versus 3.

Hint 2 - reveal when you want a nudge

One adjacent exchange reverses the relative order of only the two cases exchanged.

Reveal the solution and explanation

The claim is false. The minimum is two moves. Exchange the neighboring 2 and 1, then exchange the neighboring 4 and 3. This gives 1–2–3–4.

The two moves may be performed in either order. Swapping 4 and 3 first gives 2–1–3–4, after which swapping 2 and 1 finishes the task. Thus there are two shortest move sequences, with the same required exchanges.

For the lower bound, 2 must pass 1, and 4 must pass 3. An adjacent swap changes the relative order of just its exchanged pair; other pairs retain their relative order. One move cannot correct both reversed pairs.

At least two moves are necessary, and the constructions achieve two. That proves the unique minimum value while honestly retaining both valid shortest sequences.

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