In Bridges, the closest-looking connection is not always the useful one. We will count what each island can still accept, rule out crossings, and finally check that the entire board forms one network.
The fixed practice board is bridges-2027-05-11, an existing bank puzzle scheduled for that date. This lesson makes it available for practice now; it is not today’s puzzle. The letters A–J label the islands in this lesson so that every connection is easy to identify. Their numbered clues still specify the required bridge totals.
Start with the clues
An island can connect to the nearest visible island horizontally or vertically. A connection carries one or two bridges. It cannot run diagonally, pass through another island or cross another connection.
An island’s clue counts individual bridges. A double connection contributes two to each endpoint. Each potential neighbour supplies at most two; a neighbour with less remaining capacity may supply less.
Every island must belong to one connected network. Matching all the numbered clues is necessary, but a collection of separate finished groups is not a solution. In the diagrams, an undecided connection is a possibility, not an already placed bridge.
A letter identifies an island; the number beside it is its clue. Dotted lines are possible routes, not placed bridges. A thin route with × is ruled out. Solid single or double lines are placed bridges; red lines changed in this step.
Coordinates count from the top left: R2C3 means row 2, column 3.
| Island | Position | Clue |
|---|---|---|
| A | R7C3 | 4 |
| B | R2C3 | 3 |
| C | R7C6 | 4 |
| D | R1C6 | 3 |
| E | R1C2 | 3 |
| F | R4C2 | 4 |
| G | R6C2 | 2 |
| H | R2C5 | 2 |
| I | R4C5 | 3 |
| J | R6C5 | 2 |
The game opens its normal saved board for this ID. This lesson does not change or clear your progress. Read the steps here or try the board first.
Follow the deductions
Every before-and-after board is included below. The optional step view shows one deduction at a time; it does not make game moves.
DEDUCTION 1 OF 9
Island A must use both neighbours twice
A at R7C3 needs four bridges. Its only visible neighbours are B above and C to the right. Each connection can carry at most two, so the total capacity is exactly four. Place double connections A–B and A–C. Using only one bridge on either connection would leave A short.
Mark: A–C: 2 bridges; B–A: 2 bridges.
DEDUCTION 2 OF 9
Subtract A–B from B’s total
B at R2C3 needs three bridges. The double connection to A already supplies two. B’s only other visible neighbour is H to the right, so B–H must be a single connection. A double there would raise B’s total to four.
Mark: B–H: 1 bridge.
DEDUCTION 3 OF 9
C needs two more from D
C at R7C6 needs four bridges and already has two from A. Its only other visible neighbour is D above at R1C6. Place a double C–D connection to supply the remaining two. It does not pass through any intervening island.
Mark: D–C: 2 bridges.
DEDUCTION 4 OF 9
D needs a single bridge to E
D needs three bridges. Two now arrive from C, leaving one. E at R1C2 is D’s only other visible neighbour, so place a single D–E connection. Count the individual bridges, not just the number of neighbouring islands.
Mark: E–D: 1 bridge.
DEDUCTION 5 OF 9
E’s remaining capacity points down to F
E needs three bridges and has one from D. Its other visible neighbour is F at R4C2, so E–F must carry two bridges. G farther down column 2 is not a direct neighbour of E: F blocks that line of sight.
Mark: E–F: 2 bridges.
DEDUCTION 6 OF 9
Complete H with a single connection
H at R2C5 needs two bridges. It already receives one from B, and its only other visible neighbour is I at R4C5. Place one bridge H–I. This satisfies H while leaving I with further capacity.
Mark: H–I: 1 bridge.
DEDUCTION 7 OF 9
Use A–B to rule out two crossings
A–B runs vertically along column 3 between rows 2 and 7. The possible horizontal connections F–I on row 4 and G–J on row 6 would both cross it. Neither connection can be used, so rule both out. The same exclusion would apply even if A–B carried just one bridge.
Mark: F–I: no bridge; G–J: no bridge.
DEDUCTION 8 OF 9
F now has only G for its remaining two
F needs four bridges. E–F already supplies two, and the crossing connection F–I is unavailable. The only remaining route is F–G, so make it a double. That also supplies all two bridges required by G.
Mark: F–G: 2 bridges.
DEDUCTION 9 OF 9
Finish I and J with a double connection
I needs three bridges. One arrives from H; F–I is ruled out, so the remaining two must go to J at R6C5. Place a double I–J connection. J now has its required two bridges, and its alternative connection to G was already excluded by the crossing rule.
Mark: I–J: 2 bridges.
Check the whole solution
Check the totals at A through J: 4, 3, 4, 3, 3, 4, 2, 2, 3 and 2. A connection contributes its count at both ends. Then check that no used connections cross and that every used connection joins visible neighbours.
The network check is concrete here: follow G–F–E–D–C–A–B–H–I–J. That continuous route visits every island. This establishes connectedness separately from the number checks. There was no need to guess a connection during this particular solve.
The crossing exclusions could have been recorded immediately after step 1. We delayed them to show how several simple remaining-count deductions can be made first. The point is to preserve valid constraints, not to copy one mandatory move order.
In Playfield’s game, selecting the same island pair cycles one bridge, two bridges, then none. Check compares placed connections with the answer, and Hint reveals a connection from it. Those functions do not provide the deduction sequence shown here. Use the fixed-board link to practise the reasoning; changing or resetting that board does not require changing today’s puzzle.
Check yourself
Try each question before opening its explanation. These examples separate what is forced from what merely looks possible.
Practice: use the remaining capacity
At this point C has a double connection to A and a clue of 4. What must happen between C and D, and why is a single connection insufficient?
Show explanation
C–D must be double. C still needs two bridges, and D is its only other visible neighbour. A single would leave C with a total of three and no other route for the fourth bridge.
Practice: every number matches, but is it solved?
| Island | Position | Clue |
|---|---|---|
| A | R1C1 | 2 |
| B | R1C5 | 2 |
| C | R5C1 | 2 |
| D | R5C5 | 2 |
This separate four-island example has a double A–B connection across the top and a double C–D connection across the bottom. Every island has its required two bridges and nothing crosses. Why is the board unfinished, and how can you repair it?
Show explanation
A and B form one group, while C and D form another; there is no route between the groups. Replace each horizontal double with a single, then add single connections A–C and B–D down the sides. Each island still has two bridges, and all four now belong to one connected rectangle.
Take the method to another puzzle
Return to a line or island when a crossing constraint changes. Keep undecided choices open; a plausible placement is not a proof. These selected teaching boards have the deduction sequences shown here. Other daily boards may need different techniques.
Use your browser’s Print command for all diagrams and explanations. Practice answers print with the lesson.