One bridge. Two happy endpoints.
Each circle is an island. Its number tells you exactly how many bridges must touch it. A single bridge counts once at each end — these islands each have one of the two they need.
Count the bridges, not the neighbouring islands.
Make it a double.
The same pair of islands can share two parallel bridges. Now each 2 has exactly two bridges. That is the limit: never three bridges between the same pair.
In Hop to Play: tap an island, then its neighbour. Repeat the pair to cycle one → two → none.
Take the straight route.
Bridges run horizontally or vertically, to the next island in that direction. From A, you can go right to B or down to D. You cannot jump over B to reach C, bend around a corner, or travel diagonally.
Empty space is fine. An island in the way is not.
No crossroads at sea.
Can Hashi bridges cross? No. These two routes intersect in open space, so they cannot both be used. A double bridge has the same restriction. Bridges may meet only at an island.
This is a deliberately invalid placement example, not a puzzle to solve.
Every number matches. Something is missing.
Each island has two bridges. But the top pair and the bottom pair are separate. You cannot travel from one group to the other. Matching the numbers is only half the finish line.
All islands must belong to one connected network.
Bring everyone into the same world.
Keep one bridge across the top and one across the bottom. Use the other two bridges to join the sides. Every number still matches — and now every island can reach every other island.
This loop is allowed. Hashi asks for one connected network; it does not forbid loops.
Six islands. One connected network.
Let’s solve a whole puzzle together. Every number is a bridge count. Start with an island whose available routes leave no room for doubt.
Deduction 1 / 5
The 4 has nowhere else to go.
A can see only B to its right and D below. Each route can hold at most two bridges. To reach four, both routes must be double.
Deduction 2 / 5
Three needs just one more.
D needs three bridges and already has two from A. Its only other neighbour is E, so one bridge to the right finishes D.
Deduction 3 / 5
Three more. Two possible routes.
E has one bridge and needs three more, through B above and F to the right. Neither route can hold all three, so E needs at least one on each. B already has two of its three bridges: B–E must be exactly one.
Deduction 4 / 5
The double falls into place.
E now has two bridges. B and D have reached their numbers, leaving F as E’s only unfinished route. Add two bridges to the right.
Deduction 5 / 5
Bring the last island home.
F needs three bridges and already has two from E. Its final bridge goes up to C. Every number now matches, and you can follow bridges from any island to every other island.