Everyday Grids Hashi Answer Today: May 25, 2025

This page has the complete Everyday Grids Hashi answers for May 25, 2025 (25 May 2025): 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 25, 2025 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 25, 2025
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 25, 2025
7×7 grid- 0/17Each 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 17 logical steps, no guessing.
- 1/17The 3 at R1C3 needs 3 bridges. Its possible links (bridges allowed): R1C7 (0-2), R4C3 (0-2). The other links can carry at most 2, so R1C7 needs at least 1; the other links can carry at most 2, so R4C3 needs at least 1.
- 2/17The 3 at R1C7 needs 3 bridges. Its possible links (bridges allowed): R1C3 (1-2), R3C7 (0-2). The other links can carry at most 2, so R3C7 needs at least 1.
- 3/17The 2 at R2C1 needs 2 bridges. Its possible links (bridges allowed): R4C1 (0-2). It has no other link, so R4C1 must carry both → exactly 2 bridges.
- 4/17The 3 at R3C7 needs 3 bridges. Its possible links (bridges allowed): R1C7 (1-2), R5C7 (0-2). The other links can carry at most 2, so R5C7 needs at least 1.
- 5/17The 5 at R4C1 needs 5 bridges. Its possible links (bridges allowed): R2C1 (2), R4C3 (0-2), R6C1 (0-2). The other links can carry at most 4, so R4C3 needs at least 1; the other links can carry at most 4, so R6C1 needs at least 1.
- 6/17The 3 at R5C7 needs 3 bridges. Its possible links (bridges allowed): R3C7 (1-2), R5C2 (0-2). The other links can carry at most 2, so R5C2 needs at least 1.
- 7/17There is already a bridge on R5C2–R5C7. Bridges can't cross, so R4C5–R7C5 is now impossible.
- 8/17The 1 at R4C5 needs 1 bridge. Its possible links (bridges allowed): R4C3 (0-1). It has no other link, so R4C3 must carry it → exactly 1 bridge.
- 9/17The 3 at R4C3 needs 3 bridges. Its possible links (bridges allowed): R1C3 (1-2), R4C1 (1-2), R4C5 (1). The other links already carry at least 2, so R1C3 can take at most 1 → exactly 1 bridge; the other links already carry at least 2, so R4C1 can take at most 1 → exactly 1 bridge.
- 10/17The 3 at R1C3 needs 3 bridges. Its possible links (bridges allowed): R1C7 (1-2), R4C3 (1). The other links can carry at most 1, so R1C7 needs at least 2 → exactly 2 bridges.
- 11/17The 3 at R1C7 needs 3 bridges. Its possible links (bridges allowed): R1C3 (2), R3C7 (1-2). The other links already carry at least 2, so R3C7 can take at most 1 → exactly 1 bridge.
- 12/17The 3 at R3C7 needs 3 bridges. Its possible links (bridges allowed): R1C7 (1), R5C7 (1-2). The other links can carry at most 1, so R5C7 needs at least 2 → exactly 2 bridges.
- 13/17The 5 at R4C1 needs 5 bridges. Its possible links (bridges allowed): R2C1 (2), R4C3 (1), R6C1 (1-2). The other links can carry at most 3, so R6C1 needs at least 2 → exactly 2 bridges.
- 14/17The 2 at R5C2 needs 2 bridges. Its possible links (bridges allowed): R5C7 (1-2), R7C2 (0-2). The other links already carry at least 1, so R7C2 can take at most 1.
- 15/17The 3 at R5C7 needs 3 bridges. Its possible links (bridges allowed): R3C7 (2), R5C2 (1-2). The other links already carry at least 2, so R5C2 can take at most 1 → exactly 1 bridge.
- 16/17The 2 at R5C2 needs 2 bridges. Its possible links (bridges allowed): R5C7 (1), R7C2 (0-1). The other links can carry at most 1, so R7C2 needs at least 1 → exactly 1 bridge.
- 17/17The 3 at R7C2 needs 3 bridges. Its possible links (bridges allowed): R5C2 (1), R7C5 (0-2). The other links can carry at most 1, so R7C5 needs at least 2 → exactly 2 bridges. ✅ Solved!
Hashi Normal answer as text
Everyday Grids Hashi Hard Answer, May 25, 2025
9×9 grid- 0/25Each 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 25 logical steps, no guessing.
- 1/25The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C3 (0-2), R4C1 (0-2). The other links can carry at most 2, so R1C3 needs at least 1; the other links can carry at most 2, so R4C1 needs at least 1.
- 2/25The 4 at R1C9 needs 4 bridges. Its possible links (bridges allowed): R1C7 (0-2), R3C9 (0-2). The other links can carry at most 2, so R1C7 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R3C9 needs at least 2 → exactly 2 bridges.
- 3/25The 4 at R2C2 needs 4 bridges. Its possible links (bridges allowed): R2C4 (0-2), R5C2 (0-2). The other links can carry at most 2, so R2C4 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R5C2 needs at least 2 → exactly 2 bridges.
- 4/25There is already a bridge on R2C2–R2C4. Bridges can't cross, so R1C3–R6C3 is now impossible.
- 5/25There is already a bridge on R2C2–R5C2. Bridges can't cross, so R4C1–R4C5 is now impossible.
- 6/25The 4 at R1C3 needs 4 bridges. Its possible links (bridges allowed): R1C1 (1-2), R1C5 (0-2). The other links can carry at most 2, so R1C1 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R1C5 needs at least 2 → exactly 2 bridges.
- 7/25The 3 at R1C1 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.
- 8/25The 4 at R2C4 needs 4 bridges. Its possible links (bridges allowed): R2C2 (2), R5C4 (0-2). The other links can carry at most 2, so R5C4 needs at least 2 → exactly 2 bridges.
- 9/25The 3 at R3C9 needs 3 bridges. Its possible links (bridges allowed): R1C9 (2), R5C9 (0-2). The other links can carry at most 2, so R5C9 needs at least 1, and the other links already carry at least 2, so R5C9 can take at most 1 → exactly 1 bridge.
- 10/25The 3 at R4C1 needs 3 bridges. Its possible links (bridges allowed): R1C1 (1), R6C1 (0-2). The other links can carry at most 1, so R6C1 needs at least 2 → exactly 2 bridges.
- 11/25The 4 at R4C5 needs 4 bridges. Its possible links (bridges allowed): R1C5 (0-2), R4C7 (0-2). The other links can carry at most 2, so R1C5 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R4C7 needs at least 2 → exactly 2 bridges.
- 12/25The 4 at R1C5 needs 4 bridges. Its possible links (bridges allowed): R1C3 (2), R1C7 (0-2), R4C5 (2). The other links already carry at least 4, so R1C7 can take none.
- 13/25The 4 at R1C7 needs 4 bridges. Its possible links (bridges allowed): R1C9 (2), R4C7 (0-2). The other links can carry at most 2, so R4C7 needs at least 2 → exactly 2 bridges.
- 14/25The 3 at R5C2 needs 3 bridges. Its possible links (bridges allowed): R2C2 (2), R5C4 (0-2). The other links can carry at most 2, so R5C4 needs at least 1, and the other links already carry at least 2, so R5C4 can take at most 1 → exactly 1 bridge.
- 15/25The 2 at R5C9 needs 2 bridges. Its possible links (bridges allowed): R3C9 (1), R5C4 (0-2), R7C9 (0-2). The other links already carry at least 1, so R5C4 can take at most 1; the other links already carry at least 1, so R7C9 can take at most 1.
- 16/25The 5 at R5C4 needs 5 bridges. Its possible links (bridges allowed): R2C4 (2), R5C2 (1), R5C9 (0-1), R7C4 (0-2). The other links can carry at most 4, so R7C4 needs at least 1.
- 17/25The 6 at R6C1 needs 6 bridges. Its possible links (bridges allowed): R4C1 (2), R6C3 (0-2), R9C1 (0-2). The other links can carry at most 4, so R6C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R9C1 needs at least 2 → exactly 2 bridges.
- 18/25The 2 at R7C4 needs 2 bridges. Its possible links (bridges allowed): R5C4 (1-2), R7C9 (0-2). The other links already carry at least 1, so R7C9 can take at most 1.
- 19/25The 4 at R9C1 needs 4 bridges. Its possible links (bridges allowed): R6C1 (2), R9C6 (0-2). The other links can carry at most 2, so R9C6 needs at least 2 → exactly 2 bridges.
- 20/25The 3 at R9C6 needs 3 bridges. Its possible links (bridges allowed): R9C1 (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.
- 21/25The 2 at R9C9 needs 2 bridges. Its possible links (bridges allowed): R7C9 (0-2), R9C6 (1). The other links can carry at most 1, so R7C9 needs at least 1, and the other links already carry at least 1, so R7C9 can take at most 1 → exactly 1 bridge.
- 22/25Putting 2 bridges on R5C4–R7C4 would use up every remaining bridge of the 5 islands in that group, cutting them off from the rest. Everything has to join up, so R5C4–R7C4 gets at most 1 bridge.
- 23/25The 5 at R5C4 needs 5 bridges. Its possible links (bridges allowed): R2C4 (2), R5C2 (1), R5C9 (0-1), R7C4 (1). The other links can carry at most 4, so R5C9 needs at least 1 → exactly 1 bridge.
- 24/25The 2 at R5C9 needs 2 bridges. Its possible links (bridges allowed): R3C9 (1), R5C4 (1), R7C9 (0-1). The other links already carry at least 2, so R7C9 can take none.
- 25/25The 2 at R7C4 needs 2 bridges. Its possible links (bridges allowed): R5C4 (1), R7C9 (0-1). The other links can carry at most 1, so R7C9 needs at least 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.