Playfield Arcade

BRIDGES · INVESTIGATION

A connected network can still leave three answers

Enumerate a six-island ladder to separate three different tests: correct counts, one network and a unique solution.

Bridge networkA–B: 0 or 1 or 2 bridges; B–C: 0 or 1 or 2 bridges; D–E: 0 or 1 or 2 bridges; E–F: 0 or 1 or 2 bridges; A–D: 0 or 1 or 2 bridges; B–E: 0 or 1 or 2 bridges; C–F: 0 or 1 or 2 bridgesA2B3C2D2E3F2
A, B, C label the top row left to right; D, E, F label the bottom row. The middle islands B and E need 3 bridges; the four corner islands need 2.

After enforcing both island counts and connectivity, is this ladder uniquely solved?

Make a prediction before opening the reasoning.

Compare the predictions
  1. Yes: the all-single layout is forced.

    Not supported. It is valid, but two other connected layouts satisfy exactly the same clues.

  2. No: three connected solutions remain.

    Supported. The connected choices are (x,y) = (1,1), (1,2) and (2,1).

  3. No connected solution exists.

    Not supported. The all-single layout already connects every island and meets each clue.

Predictions stay on this page; they are not saved or sent.

The position and the question

The seven possible routes are A–B, B–C, D–E, E–F, A–D, B–E and C–F. There are no diagonal routes and no route may pass through an intervening island. Each route can carry zero, one or two bridges. All six islands must form one network.

This is an intentionally ambiguous construction. It asks a question that checking the rules alone cannot settle: if a bridge layout is connected and every island total is right, must it be the only valid layout? We will enumerate every possibility on this exact ladder.

Show the reasoning

  1. Reduce seven route counts to two choices

    Call the top-left route A–B x and the top-right route B–C y. Island A forces A–D = 2 − x, then island D forces D–E = x. Similarly, C–F = 2 − y and E–F = y. The middle islands both require B–E = 3 − x − y. Every route count is therefore determined by x and y.

  2. List every arithmetic possibility

    Both x and y belong to {0,1,2}. Keeping B–E between zero and two requires 1 ≤ x + y ≤ 3. Exactly seven pairs survive: (0,1), (0,2), (1,0), (1,1), (1,2), (2,0) and (2,1). This is a complete enumeration of the count-valid layouts, not a selection of examples.

  3. Reject four disconnected arrangements

    If x = 0, A–D is a double bridge and the left pair has no connection to the other four islands. If y = 0, C–F is a double bridge and the right pair is isolated. That removes (0,1), (0,2), (1,0) and (2,0). The remaining three pairs—(1,1), (1,2) and (2,1)—each produce one connected network.

    Bridge networkA–B: 1 bridge; B–C: 1 bridge; D–E: 1 bridge; E–F: 1 bridge; A–D: 1 bridge; B–E: 1 bridge; C–F: 1 bridgeA2B3C2D2E3F2
    One valid solution: x = y = 1, so all seven routes are single bridges.

  4. Show why connected does not mean unique

    For (x,y) = (1,2), A–B and D–E are single, B–C and E–F are double, A–D is single, and B–E and C–F are absent. All six islands still connect through the left side. The reflected (2,1) layout is also valid. Thus the all-single layout is a solution, but the clues do not force it. Connectivity removes four candidates without choosing between the other three.

    Bridge networkA–B: 1 bridge; B–C: 2 bridges; D–E: 1 bridge; E–F: 2 bridges; A–D: 1 bridge; B–E: 0 bridges; C–F: 0 bridges××A2B3C2D2E3F2
    A second valid solution: x = 1, y = 2. Connection through A–D keeps the whole ladder in one network.

The tempting shortcut

Every island total is right and all islands connect, so the all-single solution must be forced.

Where the shortcut breaks

Those checks prove that the displayed layout is valid. They do not exclude the two other connected layouts with the same clues.

A better decision

To establish uniqueness, compare every valid layout or prove that each route has the same value in all of them.

Changed conditions

Lower both middle clues B and E from 3 to 2, leaving the four corner clues unchanged. How many connected solutions remain?

Bridge networkA–B: 1 bridge; B–C: 1 bridge; D–E: 1 bridge; E–F: 1 bridge; A–D: 1 bridge; B–E: 0 bridges; C–F: 1 bridge×A2B2C2D2E2F2
Transfer solution: all clues 2; the six outer routes form one cycle.
Check the changed-condition answer

Now B–E = 2 − x − y, so x + y ≤ 2. Six count-valid pairs remain: (0,0), (0,1), (0,2), (1,0), (1,1) and (2,0). Any pair with x = 0 or y = 0 isolates a side pair. Only (1,1) connects all six islands: the six outer routes are single and the middle B–E route is absent. The changed puzzle has exactly one solution.

Take this back to your next game

Correct counts and connectivity establish validity. Uniqueness is a separate claim about every other possible layout.

How the claim was checked

Enumerate all 3⁷ assignments on the seven routes; filter by all six island totals, then graph connectivity. Repeat with both middle clues lowered to 2.

  • 7 count-valid layouts for clues 2,3,2 / 2,3,2
  • 3 connected layouts, so the original clues are ambiguous
  • Changed clues all 2: 6 count-valid layouts, exactly 1 connected

Read the fixed positions and supporting proof data. Computational checks establish only the claim and position stated here. They do not establish a player difficulty rating or make an unproved strategy optimal.

Playfield Arcade · Updated 4 October 2026 · Original case design, AI-assisted writing and code, with reproducible position checks. Our method · Report a correction