Bridge Links strategy: count the capacity, then keep an exit.
Solve a new six-island board by bounding each neighbour’s capacity and rejecting sealed pairs. Compare all three number-perfect arrangements, then replay the one connected answer.
A bridge can fit the numbers at both ends and still destroy the puzzle. The useful question is not just “Can I add a line here?” Ask how much each neighbour can still accept, and whether a completed group would retain a route to the rest of the board.
This lesson uses a different six-island layout from the eight-island beginner board. Its seven possible lanes have three number-perfect arrangements. Only one connects all six islands. We will separate a capacity deduction from a connectivity deduction, so you can see precisely which rule justifies each bridge.
Load the six-island capacity board
Use Practice this example beside the article, or Load strategy board in the game controls. Start with no bridges and Edits 0. The letters identify islands; the digits are their required bridge totals. The drawing shows positions, not connections.
A1 B2 C2 D3 E4 F2
The top row is row 0 and the bottom row is row 6. The three columns are 0, 3 and 6. The only lanes are AB, AD, BC, BE, CF, DE and EF. AC is not a lane because B lies between its endpoints; DF is blocked by E. No pair of these seven lanes crosses. Crossing deductions are therefore unnecessary for this example, although the normal no-crossing rule still applies.
Use a neighbour’s remaining capacity, not just the two-line limit
For an unfinished island, subtract the lines already touching it from its target. Call the result its remaining need. An empty lane can take at most two lines, but the neighbour may be able to accept fewer. Use the smaller of those two quantities as an upper bound. An occupied lane has fewer than two unused places; count only those additional places when bounding further lines.
On this empty board D needs 3. Its only neighbours are A and E. A needs just 1, so AD can carry at most 1. DE can carry at most 2. Those upper bounds sum to exactly D’s need: 1 + 2 = 3. Both bounds must be reached. AD is single and DE is double. Looking only at “two per lane” would miss the forced double.
The general calculation is: a lane must supply at least the remaining need minus the combined upper bounds of all the other lanes, or zero if that difference is negative. For DE here, 3 − 1 = 2; for AD, 3 − 2 = 1. This is a necessary lower bound, not a promise that any arrangement satisfying the bounds will be a connected solution.
- Fill D’s forced capacity. Add AD once and DE twice. D now has 3 and A has 1. Both are complete, so AB must stay empty. E has 2 of its required 4. Edits: 3.
Reject a double that would seal a proper group
B still needs 2. Its AB lane is now unavailable, leaving BC and BE. It would be locally legal to put two lines on BC: B and C would both reach 2 without exceeding a number. But then both would be full. AB, BE and CF would all have to be zero. The group containing B and C would have no exit to A, D, E or F.
A group smaller than the whole board must have some bridge to an outside island in the final answer. A completed pair with every possible exit forced to zero violates that requirement. Therefore BC cannot be double. Apply the same argument to C and F: both have target 2, so doubling CF would fill and isolate that pair. CF cannot be double either.
This is not a rule against every double or every loop. DE is already a justified double, and a closed route can be part of a larger connected answer. The forbidden condition is a separate completed group with no possible exit while other islands remain outside it.
- Use both of C’s remaining lanes. C needs 2 from BC and CF. Neither lane can exceed 1 after the sealed-pair checks, so each must carry exactly 1. C is complete; B and F each have 1. Edits: 5.
- Finish B and F. B’s remaining line must go to E because A and C are full. F’s remaining line must also go to E because C is full. E finishes with DE 2 + BE 1 + EF 1 = 4. Edits: 7.
The completed route is AD, DE, DE, BC, CF, BE, EF in Cycle bridge count mode. Select the two letters of each pair in that order; the selection clears after each edit. This is a verified seven-edit construction from empty, not a claim about a scoring competition or arbitrary starting positions. Undo restores the previous edit; Retry this board returns these same six clues to Edits 0.
Compare every number-perfect completion
After the first checkpoint, let x be the number of lines on BE. The remaining equations force BC = 2 − x, CF = x and EF = 2 − x. BE can have 0, 1 or 2 lines. The following ledger gives every completion of the island totals; the earlier capacity deduction already fixed AB 0, AD 1 and DE 2.
lane x=0 x=1 x=2 AB 0 0 0 AD 1 1 1 BC 2 1 0 BE 0 1 2 CF 0 1 2 DE 2 2 2 EF 2 1 0 groups 2 1 2
At x = 0 the groups are B–C and A–D–E–F. At x = 2 they are C–F and A–D–E–B. Each x column satisfies every number and the two-line limit; there is no crossing in this geometry. Only the middle x column is a win. A route through all six islands is A–D–E–B–C–F, with E also connected directly to F.
Try the tempting branch before opening the answer
Retry the strategy board. Make the first checkpoint AD, DE, DE, then add BC twice. Before changing anything else, predict three things: the edit count, the number of matching island totals, and whether any legal continuation can connect B and C to the other four islands. Use Check network to inspect the current groups; it reports the current drawing, not whether all future completions are possible.
Check the sealed-pair prediction
The attempt has 5 edits. A, B, C and D match, so 4 of 6 totals are complete. There are three current groups: A–D–E, B–C and the still separate F. No continuation that only adds lines can rescue B–C, because both endpoints are full. Undo the second BC line to reach Edits 4, then use CF, BE and EF to finish at Edits 7. Alternatively choose Clear connection and select B then C to remove the double entirely; that accepted removal adds one edit. Return to Cycle bridge count before rebuilding. A refusal to add an overfull line changes nothing.
Check the scope of the method
Capacity bounds are local and safe as upper bounds even when several lanes compete for the same neighbour. Adding those bounds does not prove they can all be used simultaneously. If a remaining need exceeds their sum, something in the current branch is inconsistent; do not invent an extra lane. Recheck a previous assumption or undo it.
The sealed-group argument also needs evidence. Two islands connected now are not necessarily trapped: an unfinished endpoint can still have capacity for an exit. For example, after only AD is drawn, A is complete but D has 1 of 3 and still reaches E. That pair is not sealed. Likewise a group containing every island needs no outside exit. These checks prune choices; on other boards you may still need crossing analysis or a carefully recorded branch.
For an independent self-check, stop after AD and the double DE. What is E’s remaining need, and which lanes can supply it?
Check the remaining-capacity answer
E has 2 of 4 and needs 2 more. Its only unfinished lanes are BE and EF, each initially bounded above by 2 at this checkpoint. Capacity alone permits BE 0 / EF 2, BE 1 / EF 1, or BE 2 / EF 0. The network rule rejects the first and third after the other island equations are considered. This is why the capacity and connectivity arguments do different jobs.