Everyday Grids Hashi Answer Today: May 19, 2026

This page has the complete Everyday Grids Hashi answers for May 19, 2026 (19 May 2026): the solved grids for both the Normal and Hard Hashi levels, plus a full solution walkthrough with a step by step explanation of every move. Need a quick cheat to check your grid, a hint to get unstuck, or a complete Hashi guide? It’s all below, and each Everyday Grids Hashi solution is verified against the puzzle rules, so it matches the May 19, 2026 puzzle in the app exactly.
About Everyday Grids Hashi
Everyday Grids is a relaxing daily logic-puzzle game by Fanatee with seven different grids: Zip, Gems, Lotus, Panda, Libra, Sums and Hashi. Everyday Grids Hashi is one of them. Join all islands with straight bridges. An island’s number is how many bridges touch it; max 2 per pair, no crossing, and everything must connect. Every level can be solved with pure logic, with no guessing and no luck, and a brand-new Normal and Hard Hashi puzzle is released every day. The game is free to download on Android and iPhone.
Everyday Grids Hashi Answers and Solutions for May 19, 2026
Each level below shows the solved Hashi grid first, so you can copy the answer straight away. If you’d rather learn how it’s done, tap Step by Step Explanation under the grid: the grid and the current step stay on screen together while you follow the solution one logical move at a time (R = row, C = column, counted from the top-left).
Everyday Grids Hashi Normal Answer, May 19, 2026
8×8 grid- 0/19Each island's number is how many bridges touch it. Bridges are straight, at most 2 between the same pair, never cross, and everything must join up. Tap Start to follow all 19 logical steps, no guessing.
- 1/19The 4 at R1C3 needs 4 bridges. Its possible links (bridges allowed): R1C6 (0-2), R4C3 (0-2). The other links can carry at most 2, so R1C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R4C3 needs at least 2 → exactly 2 bridges.
- 2/19The 4 at R3C4 needs 4 bridges. Its possible links (bridges allowed): R3C6 (0-2), R5C4 (0-2). The other links can carry at most 2, so R3C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R5C4 needs at least 2 → exactly 2 bridges.
- 3/19There is already a bridge on R3C4–R5C4. Bridges can't cross, so R4C3–R4C8 is now impossible.
- 4/19The 5 at R3C6 needs 5 bridges. Its possible links (bridges allowed): R1C6 (0-2), R3C4 (2), R8C6 (0-2). The other links can carry at most 4, so R1C6 needs at least 1; the other links can carry at most 4, so R8C6 needs at least 1.
- 5/19The 4 at R1C6 needs 4 bridges. Its possible links (bridges allowed): R1C3 (2), R1C8 (0-2), R3C6 (1-2). The other links already carry at least 3, so R1C8 can take at most 1.
- 6/19The 2 at R1C8 needs 2 bridges. Its possible links (bridges allowed): R1C6 (0-1), R4C8 (0-2). The other links can carry at most 1, so R4C8 needs at least 1.
- 7/19The 3 at R4C1 needs 3 bridges. Its possible links (bridges allowed): R4C3 (0-2), R8C1 (0-2). The other links can carry at most 2, so R4C3 needs at least 1; the other links can carry at most 2, so R8C1 needs at least 1.
- 8/19The 3 at R4C3 needs 3 bridges. Its possible links (bridges allowed): R1C3 (2), R4C1 (1-2). The other links already carry at least 2, so R4C1 can take at most 1 → exactly 1 bridge.
- 9/19The 3 at R4C1 needs 3 bridges. Its possible links (bridges allowed): R4C3 (1), R8C1 (1-2). The other links can carry at most 1, so R8C1 needs at least 2 → exactly 2 bridges.
- 10/19The 3 at R4C8 needs 3 bridges. Its possible links (bridges allowed): R1C8 (1-2), R6C8 (0-2). The other links can carry at most 2, so R6C8 needs at least 1.
- 11/19The 3 at R5C4 needs 3 bridges. Its possible links (bridges allowed): R3C4 (2), R7C4 (0-2). The other links can carry at most 2, so R7C4 needs at least 1, and the other links already carry at least 2, so R7C4 can take at most 1 → exactly 1 bridge.
- 12/19The 4 at R6C8 needs 4 bridges. Its possible links (bridges allowed): R4C8 (1-2), R8C8 (0-2). The other links can carry at most 2, so R4C8 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R8C8 needs at least 2 → exactly 2 bridges.
- 13/19The 3 at R4C8 needs 3 bridges. Its possible links (bridges allowed): R1C8 (1-2), R6C8 (2). The other links already carry at least 2, so R1C8 can take at most 1 → exactly 1 bridge.
- 14/19The 2 at R1C8 needs 2 bridges. Its possible links (bridges allowed): R1C6 (0-1), R4C8 (1). The other links can carry at most 1, so R1C6 needs at least 1 → exactly 1 bridge.
- 15/19The 4 at R1C6 needs 4 bridges. Its possible links (bridges allowed): R1C3 (2), R1C8 (1), R3C6 (1-2). The other links already carry at least 3, so R3C6 can take at most 1 → exactly 1 bridge.
- 16/19The 5 at R3C6 needs 5 bridges. Its possible links (bridges allowed): R1C6 (1), R3C4 (2), R8C6 (1-2). The other links can carry at most 3, so R8C6 needs at least 2 → exactly 2 bridges.
- 17/19The 1 at R7C2 needs 1 bridge. Its possible links (bridges allowed): R7C4 (0-1). It has no other link, so R7C4 must carry it → exactly 1 bridge.
- 18/19The 4 at R8C1 needs 4 bridges. Its possible links (bridges allowed): R4C1 (2), R8C6 (0-2). The other links can carry at most 2, so R8C6 needs at least 2 → exactly 2 bridges.
- 19/19The 5 at R8C6 needs 5 bridges. Its possible links (bridges allowed): R3C6 (2), R8C1 (2), R8C8 (0-2). The other links can carry at most 4, so R8C8 needs at least 1, and the other links already carry at least 4, so R8C8 can take at most 1 → exactly 1 bridge. ✅ Solved!
Hashi Normal answer as text
Everyday Grids Hashi Hard Answer, May 19, 2026
9×9 grid- 0/41Each island's number is how many bridges touch it. Bridges are straight, at most 2 between the same pair, never cross, and everything must join up. Tap Start to follow all 41 logical steps, no guessing.
- 1/41The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C4 (0-2), R3C1 (0-2). The other links can carry at most 2, so R1C4 needs at least 1; the other links can carry at most 2, so R3C1 needs at least 1.
- 2/41The 4 at R1C4 needs 4 bridges. Its possible links (bridges allowed): R1C1 (1-2), R1C7 (0-1), R4C4 (0-2). The other links can carry at most 3, so R4C4 needs at least 1.
- 3/41The 2 at R1C9 needs 2 bridges. Its possible links (bridges allowed): R1C7 (0-1), R4C9 (0-2). The other links can carry at most 1, so R4C9 needs at least 1.
- 4/41The 4 at R3C1 needs 4 bridges. Its possible links (bridges allowed): R1C1 (1-2), R3C3 (0-1), R6C1 (0-2). The other links can carry at most 3, so R6C1 needs at least 1.
- 5/41The 3 at R4C2 needs 3 bridges. Its possible links (bridges allowed): R4C4 (0-2), R8C2 (0-2). The other links can carry at most 2, so R4C4 needs at least 1; the other links can carry at most 2, so R8C2 needs at least 1.
- 6/41There is already a bridge on R4C2–R4C4. Bridges can't cross, so R3C3–R9C3 is now impossible.
- 7/41There is already a bridge on R4C2–R8C2. Bridges can't cross, so R6C1–R6C4 is now impossible.
- 8/41The 1 at R3C3 needs 1 bridge. Its possible links (bridges allowed): R3C1 (0-1). It has no other link, so R3C1 must carry it → exactly 1 bridge.
- 9/41The 8 at R4C4 needs 8 bridges. Its possible links (bridges allowed): R1C4 (1-2), R4C2 (1-2), R4C6 (0-2), R6C4 (0-2). The other links can carry at most 6, so R1C4 needs at least 2 → exactly 2 bridges; the other links can carry at most 6, so R4C2 needs at least 2 → exactly 2 bridges; the other links can carry at most 6, so R4C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 6, so R6C4 needs at least 2 → exactly 2 bridges.
- 10/41The 3 at R4C2 needs 3 bridges. Its possible links (bridges allowed): R4C4 (2), R8C2 (1-2). The other links already carry at least 2, so R8C2 can take at most 1 → exactly 1 bridge.
- 11/41The 2 at R4C9 needs 2 bridges. Its possible links (bridges allowed): R1C9 (1-2), R4C6 (0-2), R6C9 (0-2). The other links already carry at least 1, so R4C6 can take at most 1; the other links already carry at least 1, so R6C9 can take at most 1.
- 12/41The 3 at R5C8 needs 3 bridges. Its possible links (bridges allowed): R2C8 (0-2), R8C8 (0-2). The other links can carry at most 2, so R2C8 needs at least 1; the other links can carry at most 2, so R8C8 needs at least 1.
- 13/41There is already a bridge on R2C8–R5C8. Bridges can't cross, so R4C6–R4C9 is now impossible.
- 14/41There is already a bridge on R5C8–R8C8. Bridges can't cross, so R6C6–R6C9 is now impossible.
- 15/41The 2 at R2C8 needs 2 bridges. Its possible links (bridges allowed): R2C6 (0-2), R5C8 (1-2). The other links already carry at least 1, so R2C6 can take at most 1.
- 16/41The 2 at R2C6 needs 2 bridges. Its possible links (bridges allowed): R2C8 (0-1), R4C6 (0-2). The other links can carry at most 1, so R4C6 needs at least 1.
- 17/41The 4 at R4C6 needs 4 bridges. Its possible links (bridges allowed): R2C6 (1-2), R4C4 (2), R6C6 (0-2). The other links already carry at least 3, so R6C6 can take at most 1.
- 18/41The 3 at R6C1 needs 3 bridges. Its possible links (bridges allowed): R3C1 (1-2), R9C1 (0-2). The other links can carry at most 2, so R9C1 needs at least 1.
- 19/41The 2 at R6C4 needs 2 bridges. Its possible links (bridges allowed): R4C4 (2), R6C6 (0-2). The other links already carry at least 2, so R6C6 can take none.
- 20/41The 2 at R6C6 needs 2 bridges. Its possible links (bridges allowed): R4C6 (0-1), R8C6 (0-2). The other links can carry at most 1, so R8C6 needs at least 1.
- 21/41The 2 at R6C9 needs 2 bridges. Its possible links (bridges allowed): R4C9 (0-1), R9C9 (0-2). The other links can carry at most 1, so R9C9 needs at least 1.
- 22/41The 3 at R8C2 needs 3 bridges. Its possible links (bridges allowed): R4C2 (1), R8C6 (0-2). The other links can carry at most 1, so R8C6 needs at least 2 → exactly 2 bridges.
- 23/41The 3 at R8C6 needs 3 bridges. Its possible links (bridges allowed): R6C6 (1-2), R8C2 (2), R8C8 (0-2). The other links already carry at least 2, so R6C6 can take at most 1 → exactly 1 bridge; the other links already carry at least 3, so R8C8 can take none.
- 24/41The 2 at R6C6 needs 2 bridges. Its possible links (bridges allowed): R4C6 (0-1), R8C6 (1). The other links can carry at most 1, so R4C6 needs at least 1 → exactly 1 bridge.
- 25/41The 4 at R4C6 needs 4 bridges. Its possible links (bridges allowed): R2C6 (1-2), R4C4 (2), R6C6 (1). The other links already carry at least 3, so R2C6 can take at most 1 → exactly 1 bridge.
- 26/41The 2 at R2C6 needs 2 bridges. Its possible links (bridges allowed): R2C8 (0-1), R4C6 (1). The other links can carry at most 1, so R2C8 needs at least 1 → exactly 1 bridge.
- 27/41There is already a bridge on R2C6–R2C8. Bridges can't cross, so R1C7–R9C7 is now impossible.
- 28/41The 2 at R2C8 needs 2 bridges. Its possible links (bridges allowed): R2C6 (1), R5C8 (1-2). The other links already carry at least 1, so R5C8 can take at most 1 → exactly 1 bridge.
- 29/41The 3 at R5C8 needs 3 bridges. Its possible links (bridges allowed): R2C8 (1), R8C8 (1-2). The other links can carry at most 1, so R8C8 needs at least 2 → exactly 2 bridges.
- 30/41The 2 at R9C1 needs 2 bridges. Its possible links (bridges allowed): R6C1 (1-2), R9C3 (0-2). The other links already carry at least 1, so R9C3 can take at most 1.
- 31/41The 3 at R9C3 needs 3 bridges. Its possible links (bridges allowed): R9C1 (0-1), R9C7 (0-2). The other links can carry at most 2, so R9C1 needs at least 1 → exactly 1 bridge; the other links can carry at most 1, so R9C7 needs at least 2 → exactly 2 bridges.
- 32/41The 2 at R9C1 needs 2 bridges. Its possible links (bridges allowed): R6C1 (1-2), R9C3 (1). The other links already carry at least 1, so R6C1 can take at most 1 → exactly 1 bridge.
- 33/41The 3 at R6C1 needs 3 bridges. Its possible links (bridges allowed): R3C1 (1-2), R9C1 (1). The other links can carry at most 1, so R3C1 needs at least 2 → exactly 2 bridges.
- 34/41The 4 at R3C1 needs 4 bridges. Its possible links (bridges allowed): R1C1 (1-2), R3C3 (1), R6C1 (2). The other links already carry at least 3, so R1C1 can take at most 1 → exactly 1 bridge.
- 35/41The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C4 (1-2), R3C1 (1). The other links can carry at most 1, so R1C4 needs at least 2 → exactly 2 bridges.
- 36/41The 4 at R1C4 needs 4 bridges. Its possible links (bridges allowed): R1C1 (2), R1C7 (0-1), R4C4 (2). The other links already carry at least 4, so R1C7 can take none.
- 37/41The 1 at R1C7 needs 1 bridge. Its possible links (bridges allowed): R1C9 (0-1). It has no other link, so R1C9 must carry it → exactly 1 bridge.
- 38/41The 2 at R1C9 needs 2 bridges. Its possible links (bridges allowed): R1C7 (1), R4C9 (1-2). The other links already carry at least 1, so R4C9 can take at most 1 → exactly 1 bridge.
- 39/41The 2 at R4C9 needs 2 bridges. Its possible links (bridges allowed): R1C9 (1), R6C9 (0-1). The other links can carry at most 1, so R6C9 needs at least 1 → exactly 1 bridge.
- 40/41The 2 at R6C9 needs 2 bridges. Its possible links (bridges allowed): R4C9 (1), R9C9 (1-2). The other links already carry at least 1, so R9C9 can take at most 1 → exactly 1 bridge.
- 41/41The 3 at R9C7 needs 3 bridges. Its possible links (bridges allowed): R9C3 (2), R9C9 (0-2). The other links can carry at most 2, so R9C9 needs at least 1, and the other links already carry at least 2, so R9C9 can take at most 1 → exactly 1 bridge. ✅ Solved!
Hashi Hard answer as text
How to Solve Everyday Grids Hashi: Quick Tips
- Islands with few neighbours are easiest: a 4 with two neighbours needs two bridges to each.
- Two 1-islands can't connect to each other, they would be cut off from the rest.
- Bridges never cross, so a placed bridge blocks every crossing link.
- Everything must join into one group, so never close off a small set of islands.