Everyday Grids Hashi Answer Today: May 16, 2025

This page has the complete Everyday Grids Hashi answers for May 16, 2025 (16 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 16, 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 16, 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 16, 2025
8×8 grid- 0/12Each 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 12 logical steps, no guessing.
- 1/12The 3 at R1C2 needs 3 bridges. Its possible links (bridges allowed): R1C8 (0-2), R3C2 (0-2). The other links can carry at most 2, so R1C8 needs at least 1; the other links can carry at most 2, so R3C2 needs at least 1.
- 2/12The 4 at R1C8 needs 4 bridges. Its possible links (bridges allowed): R1C2 (1-2), R6C8 (0-2). The other links can carry at most 2, so R1C2 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R6C8 needs at least 2 → exactly 2 bridges.
- 3/12The 3 at R1C2 needs 3 bridges. Its possible links (bridges allowed): R1C8 (2), R3C2 (1-2). The other links already carry at least 2, so R3C2 can take at most 1 → exactly 1 bridge.
- 4/12The 2 at R2C6 needs 2 bridges. Its possible links (bridges allowed): R6C6 (0-2). It has no other link, so R6C6 must carry both → exactly 2 bridges.
- 5/12The 2 at R3C2 needs 2 bridges. Its possible links (bridges allowed): R1C2 (1), R6C2 (0-2). The other links can carry at most 1, so R6C2 needs at least 1, and the other links already carry at least 1, so R6C2 can take at most 1 → exactly 1 bridge.
- 6/12The 2 at R5C1 needs 2 bridges. Its possible links (bridges allowed): R8C1 (0-2). It has no other link, so R8C1 must carry both → exactly 2 bridges.
- 7/12The 2 at R6C2 needs 2 bridges. Its possible links (bridges allowed): R3C2 (1), R6C6 (0-2). The other links can carry at most 1, so R6C6 needs at least 1, and the other links already carry at least 1, so R6C6 can take at most 1 → exactly 1 bridge.
- 8/12The 7 at R6C6 needs 7 bridges. Its possible links (bridges allowed): R2C6 (2), R6C2 (1), R6C8 (0-2), R8C6 (0-2). The other links can carry at most 5, so R6C8 needs at least 2 → exactly 2 bridges; the other links can carry at most 5, so R8C6 needs at least 2 → exactly 2 bridges.
- 9/12The 4 at R6C8 needs 4 bridges. Its possible links (bridges allowed): R1C8 (2), R6C6 (2), R8C8 (0-1). The other links already carry at least 4, so R8C8 can take none.
- 10/12The 3 at R8C1 needs 3 bridges. Its possible links (bridges allowed): R5C1 (2), R8C3 (0-2). The other links can carry at most 2, so R8C3 needs at least 1, and the other links already carry at least 2, so R8C3 can take at most 1 → exactly 1 bridge.
- 11/12The 2 at R8C3 needs 2 bridges. Its possible links (bridges allowed): R8C1 (1), R8C6 (0-2). The other links can carry at most 1, so R8C6 needs at least 1, and the other links already carry at least 1, so R8C6 can take at most 1 → exactly 1 bridge.
- 12/12The 4 at R8C6 needs 4 bridges. Its possible links (bridges allowed): R6C6 (2), R8C3 (1), R8C8 (0-1). The other links can carry at most 3, so R8C8 needs at least 1 → exactly 1 bridge. ✅ Solved!
Hashi Normal answer as text
Everyday Grids Hashi Hard Answer, May 16, 2025
8×8 grid- 0/35Each 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 35 logical steps, no guessing.
- 1/35The 3 at R1C8 needs 3 bridges. Its possible links (bridges allowed): R1C6 (0-2), R4C8 (0-2). The other links can carry at most 2, so R1C6 needs at least 1; the other links can carry at most 2, so R4C8 needs at least 1.
- 2/35The 3 at R3C3 needs 3 bridges. Its possible links (bridges allowed): R1C3 (0-2), R5C3 (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 R5C3 needs at least 1.
- 3/35There is already a bridge on R1C3–R3C3. Bridges can't cross, so R2C2–R2C5 is now impossible.
- 4/35There is already a bridge on R3C3–R5C3. Bridges can't cross, so R4C1–R4C6 is now impossible.
- 5/35The 2 at R1C3 needs 2 bridges. Its possible links (bridges allowed): R1C1 (0-2), R1C6 (0-2), R3C3 (1-2). The other links already carry at least 1, so R1C1 can take at most 1; the other links already carry at least 1, so R1C6 can take at most 1.
- 6/35The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C3 (0-1), R4C1 (0-2). The other links can carry at most 1, so R4C1 needs at least 1.
- 7/35The 1 at R2C2 needs 1 bridge. Its possible links (bridges allowed): R7C2 (0-1). It has no other link, so R7C2 must carry it → exactly 1 bridge.
- 8/35There is already a bridge on R2C2–R7C2. Bridges can't cross, so R6C1–R6C8 is now impossible.
- 9/35The 2 at R4C1 needs 2 bridges. Its possible links (bridges allowed): R1C1 (1-2), R6C1 (0-2). The other links already carry at least 1, so R6C1 can take at most 1.
- 10/35The 4 at R5C3 needs 4 bridges. Its possible links (bridges allowed): R3C3 (1-2), R5C5 (0-2). The other links can carry at most 2, so R3C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R5C5 needs at least 2 → exactly 2 bridges.
- 11/35The 3 at R3C3 needs 3 bridges. Its possible links (bridges allowed): R1C3 (1-2), R5C3 (2). The other links already carry at least 2, so R1C3 can take at most 1 → exactly 1 bridge.
- 12/35The 4 at R5C5 needs 4 bridges. Its possible links (bridges allowed): R2C5 (0-1), R5C3 (2), R5C7 (0-2). The other links can carry at most 3, so R5C7 needs at least 1.
- 13/35There is already a bridge on R5C5–R5C7. Bridges can't cross, so R4C6–R8C6 is now impossible.
- 14/35The 2 at R6C1 needs 2 bridges. Its possible links (bridges allowed): R4C1 (0-1), R8C1 (0-2). The other links can carry at most 1, so R8C1 needs at least 1.
- 15/35The 3 at R6C8 needs 3 bridges. Its possible links (bridges allowed): R4C8 (0-2), R8C8 (0-2). The other links can carry at most 2, so R4C8 needs at least 1; the other links can carry at most 2, so R8C8 needs at least 1.
- 16/35The 3 at R4C8 needs 3 bridges. Its possible links (bridges allowed): R1C8 (1-2), R4C6 (0-2), R6C8 (1-2). The other links already carry at least 2, so R4C6 can take at most 1.
- 17/35The 2 at R4C6 needs 2 bridges. Its possible links (bridges allowed): R1C6 (0-2), R4C8 (0-1). The other links can carry at most 1, so R1C6 needs at least 1.
- 18/35There is already a bridge on R1C6–R4C6. Bridges can't cross, so R2C5–R2C7 is now impossible.
- 19/35The 1 at R2C5 needs 1 bridge. Its possible links (bridges allowed): R5C5 (0-1). It has no other link, so R5C5 must carry it → exactly 1 bridge.
- 20/35The 1 at R2C7 needs 1 bridge. Its possible links (bridges allowed): R5C7 (0-1). It has no other link, so R5C7 must carry it → exactly 1 bridge.
- 21/35There is already a bridge on R2C7–R5C7. Bridges can't cross, so R4C6–R4C8 is now impossible.
- 22/35The 2 at R4C6 needs 2 bridges. Its possible links (bridges allowed): R1C6 (1-2). It has no other link, so R1C6 must carry both → exactly 2 bridges.
- 23/35The 3 at R1C6 needs 3 bridges. Its possible links (bridges allowed): R1C3 (0-1), R1C8 (1-2), R4C6 (2). The other links already carry at least 3, so R1C3 can take none; the other links already carry at least 2, so R1C8 can take at most 1 → exactly 1 bridge.
- 24/35The 2 at R1C3 needs 2 bridges. Its possible links (bridges allowed): R1C1 (0-1), R3C3 (1). The other links can carry at most 1, so R1C1 needs at least 1 → exactly 1 bridge.
- 25/35The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C3 (1), R4C1 (1-2). The other links already carry at least 1, so R4C1 can take at most 1 → exactly 1 bridge.
- 26/35The 3 at R1C8 needs 3 bridges. Its possible links (bridges allowed): R1C6 (1), R4C8 (1-2). The other links can carry at most 1, so R4C8 needs at least 2 → exactly 2 bridges.
- 27/35The 2 at R4C1 needs 2 bridges. Its possible links (bridges allowed): R1C1 (1), R6C1 (0-1). The other links can carry at most 1, so R6C1 needs at least 1 → exactly 1 bridge.
- 28/35The 3 at R4C8 needs 3 bridges. Its possible links (bridges allowed): R1C8 (2), R6C8 (1-2). The other links already carry at least 2, so R6C8 can take at most 1 → exactly 1 bridge.
- 29/35The 4 at R5C5 needs 4 bridges. Its possible links (bridges allowed): R2C5 (1), R5C3 (2), R5C7 (1-2). The other links already carry at least 3, so R5C7 can take at most 1 → exactly 1 bridge.
- 30/35The 3 at R5C7 needs 3 bridges. Its possible links (bridges allowed): R2C7 (1), R5C5 (1), R7C7 (0-2). The other links can carry at most 2, so R7C7 needs at least 1, and the other links already carry at least 2, so R7C7 can take at most 1 → exactly 1 bridge.
- 31/35The 2 at R6C1 needs 2 bridges. Its possible links (bridges allowed): R4C1 (1), R8C1 (1-2). The other links already carry at least 1, so R8C1 can take at most 1 → exactly 1 bridge.
- 32/35The 3 at R6C8 needs 3 bridges. Its possible links (bridges allowed): R4C8 (1), R8C8 (1-2). The other links can carry at most 1, so R8C8 needs at least 2 → exactly 2 bridges.
- 33/35The 2 at R7C2 needs 2 bridges. Its possible links (bridges allowed): R2C2 (1), R7C7 (0-2). The other links can carry at most 1, so R7C7 needs at least 1, and the other links already carry at least 1, so R7C7 can take at most 1 → exactly 1 bridge.
- 34/35The 2 at R8C1 needs 2 bridges. Its possible links (bridges allowed): R6C1 (1), R8C6 (0-2). The other links can carry at most 1, so R8C6 needs at least 1, and the other links already carry at least 1, so R8C6 can take at most 1 → exactly 1 bridge.
- 35/35The 2 at R8C6 needs 2 bridges. Its possible links (bridges allowed): R8C1 (1), R8C8 (0-2). The other links can carry at most 1, so R8C8 needs at least 1, and the other links already carry at least 1, so R8C8 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.