Everyday Grids Hashi Answer Today: May 4, 2025

This page has the complete Everyday Grids Hashi answers for May 4, 2025 (4 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 4, 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 4, 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 4, 2025
8×8 grid- 0/23Each 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 23 logical steps, no guessing.
- 1/23The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C3 (0-2), R3C1 (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 R3C1 needs at least 1.
- 2/23The 4 at R1C5 needs 4 bridges. Its possible links (bridges allowed): R1C3 (0-2), R1C8 (0-2). The other links can carry at most 2, so R1C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R1C8 needs at least 2 → exactly 2 bridges.
- 3/23The 4 at R1C3 needs 4 bridges. Its possible links (bridges allowed): R1C1 (1-2), R1C5 (2), R4C3 (0-2). The other links already carry at least 3, so R4C3 can take at most 1.
- 4/23The 3 at R1C8 needs 3 bridges. Its possible links (bridges allowed): R1C5 (2), R6C8 (0-2). The other links can carry at most 2, so R6C8 needs at least 1, and the other links already carry at least 2, so R6C8 can take at most 1 → exactly 1 bridge.
- 5/23The 1 at R2C2 needs 1 bridge. Its possible links (bridges allowed): R2C4 (0-1). It has no other link, so R2C4 must carry it → exactly 1 bridge.
- 6/23There is already a bridge on R2C2–R2C4. Bridges can't cross, so R1C3–R4C3 is now impossible.
- 7/23The 4 at R1C3 needs 4 bridges. Its possible links (bridges allowed): R1C1 (1-2), R1C5 (2). The other links can carry at most 2, so R1C1 needs at least 2 → exactly 2 bridges.
- 8/23The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C3 (2), R3C1 (1-2). The other links already carry at least 2, so R3C1 can take at most 1 → exactly 1 bridge.
- 9/23The 2 at R2C4 needs 2 bridges. Its possible links (bridges allowed): R2C2 (1), R2C7 (0-2). The other links can carry at most 1, so R2C7 needs at least 1, and the other links already carry at least 1, so R2C7 can take at most 1 → exactly 1 bridge.
- 10/23The 3 at R2C7 needs 3 bridges. Its possible links (bridges allowed): R2C4 (1), R4C7 (0-2). The other links can carry at most 1, so R4C7 needs at least 2 → exactly 2 bridges.
- 11/23The 2 at R3C1 needs 2 bridges. Its possible links (bridges allowed): R1C1 (1), R6C1 (0-2). The other links can carry at most 1, so R6C1 needs at least 1, and the other links already carry at least 1, so R6C1 can take at most 1 → exactly 1 bridge.
- 12/23The 4 at R4C3 needs 4 bridges. Its possible links (bridges allowed): R4C7 (0-2), R6C3 (0-2). The other links can carry at most 2, so R4C7 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R6C3 needs at least 2 → exactly 2 bridges.
- 13/23The 2 at R6C1 needs 2 bridges. Its possible links (bridges allowed): R3C1 (1), R6C3 (0-2), R8C1 (0-2). The other links already carry at least 1, so R6C3 can take at most 1; the other links already carry at least 1, so R8C1 can take at most 1.
- 14/23The 4 at R6C6 needs 4 bridges. Its possible links (bridges allowed): R6C3 (0-2), R6C8 (0-2). The other links can carry at most 2, so R6C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R6C8 needs at least 2 → exactly 2 bridges.
- 15/23The 5 at R6C3 needs 5 bridges. Its possible links (bridges allowed): R4C3 (2), R6C1 (0-1), R6C6 (2), R8C3 (0-2). The other links already carry at least 4, so R8C3 can take at most 1.
- 16/23The 3 at R6C8 needs 3 bridges. Its possible links (bridges allowed): R1C8 (1), R6C6 (2), R8C8 (0-2). The other links already carry at least 3, so R8C8 can take none.
- 17/23The 2 at R8C1 needs 2 bridges. Its possible links (bridges allowed): R6C1 (0-1), R8C3 (0-2). The other links can carry at most 1, so R8C3 needs at least 1.
- 18/23The 4 at R8C3 needs 4 bridges. Its possible links (bridges allowed): R6C3 (0-1), R8C1 (1-2), R8C8 (0-2). The other links can carry at most 3, so R8C8 needs at least 1.
- 19/23The 2 at R8C8 needs 2 bridges. Its possible links (bridges allowed): R8C3 (1-2). It has no other link, so R8C3 must carry both → exactly 2 bridges.
- 20/23Putting 1 bridge on R6C1–R6C3 would use up every remaining bridge of the 14 islands in that group, cutting them off from the rest. Everything has to join up, so R6C1–R6C3 gets at most no bridge.
- 21/23The 2 at R6C1 needs 2 bridges. Its possible links (bridges allowed): R3C1 (1), R8C1 (0-1). The other links can carry at most 1, so R8C1 needs at least 1 → exactly 1 bridge.
- 22/23The 5 at R6C3 needs 5 bridges. Its possible links (bridges allowed): R4C3 (2), R6C6 (2), R8C3 (0-1). The other links can carry at most 4, so R8C3 needs at least 1 → exactly 1 bridge.
- 23/23The 2 at R8C1 needs 2 bridges. Its possible links (bridges allowed): R6C1 (1), R8C3 (1-2). The other links already carry at least 1, so R8C3 can take at most 1 → exactly 1 bridge. ✅ Solved!
Hashi Normal answer as text
Everyday Grids Hashi Hard Answer, May 4, 2025
9×9 grid- 0/31Each 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 31 logical steps, no guessing.
- 1/31The 4 at R1C1 needs 4 bridges. Its possible links (bridges allowed): R1C6 (0-2), R5C1 (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 R5C1 needs at least 2 → exactly 2 bridges.
- 2/31The 6 at R1C6 needs 6 bridges. Its possible links (bridges allowed): R1C1 (2), R1C9 (0-2), R3C6 (0-2). The other links can carry at most 4, so R1C9 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R3C6 needs at least 2 → exactly 2 bridges.
- 3/31The 4 at R1C9 needs 4 bridges. Its possible links (bridges allowed): R1C6 (2), R9C9 (0-2). The other links can carry at most 2, so R9C9 needs at least 2 → exactly 2 bridges.
- 4/31The 2 at R3C2 needs 2 bridges. Its possible links (bridges allowed): R3C6 (0-2), R6C2 (0-1). The other links can carry at most 1, so R3C6 needs at least 1.
- 5/31The 7 at R3C6 needs 7 bridges. Its possible links (bridges allowed): R1C6 (2), R3C2 (1-2), R3C8 (0-2), R5C6 (0-2). The other links can carry at most 6, so R3C8 needs at least 1; the other links can carry at most 6, so R5C6 needs at least 1.
- 6/31The 3 at R3C8 needs 3 bridges. Its possible links (bridges allowed): R3C6 (1-2), R6C8 (0-2). The other links can carry at most 2, so R6C8 needs at least 1.
- 7/31The 3 at R5C4 needs 3 bridges. Its possible links (bridges allowed): R5C1 (0-2), R5C6 (0-2). The other links can carry at most 2, so R5C1 needs at least 1; the other links can carry at most 2, so R5C6 needs at least 1.
- 8/31There is already a bridge on R5C1–R5C4. Bridges can't cross, so R3C2–R6C2 is now impossible.
- 9/31The 2 at R3C2 needs 2 bridges. Its possible links (bridges allowed): R3C6 (1-2). It has no other link, so R3C6 must carry both → exactly 2 bridges.
- 10/31The 4 at R5C1 needs 4 bridges. Its possible links (bridges allowed): R1C1 (2), R5C4 (1-2), R7C1 (0-2). The other links already carry at least 3, so R7C1 can take at most 1.
- 11/31The 2 at R5C6 needs 2 bridges. Its possible links (bridges allowed): R3C6 (1-2), R5C4 (1-2). The other links already carry at least 1, so R3C6 can take at most 1 → exactly 1 bridge; the other links already carry at least 1, so R5C4 can take at most 1 → exactly 1 bridge.
- 12/31The 7 at R3C6 needs 7 bridges. Its possible links (bridges allowed): R1C6 (2), R3C2 (2), R3C8 (1-2), R5C6 (1). The other links can carry at most 5, so R3C8 needs at least 2 → exactly 2 bridges.
- 13/31The 3 at R3C8 needs 3 bridges. Its possible links (bridges allowed): R3C6 (2), R6C8 (1-2). The other links already carry at least 2, so R6C8 can take at most 1 → exactly 1 bridge.
- 14/31The 3 at R5C4 needs 3 bridges. Its possible links (bridges allowed): R5C1 (1-2), R5C6 (1). The other links can carry at most 1, so R5C1 needs at least 2 → exactly 2 bridges.
- 15/31The 4 at R5C1 needs 4 bridges. Its possible links (bridges allowed): R1C1 (2), R5C4 (2), R7C1 (0-1). The other links already carry at least 4, so R7C1 can take none.
- 16/31The 1 at R6C2 needs 1 bridge. Its possible links (bridges allowed): R6C8 (0-1). It has no other link, so R6C8 must carry it → exactly 1 bridge.
- 17/31The 3 at R6C8 needs 3 bridges. Its possible links (bridges allowed): R3C8 (1), R6C2 (1), R8C8 (0-1). The other links can carry at most 2, so R8C8 needs at least 1 → exactly 1 bridge.
- 18/31The 6 at R7C5 needs 6 bridges. Its possible links (bridges allowed): R7C3 (0-2), R7C7 (0-2), R9C5 (0-2). The other links can carry at most 4, so R7C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R7C7 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R9C5 needs at least 2 → exactly 2 bridges.
- 19/31The 3 at R7C3 needs 3 bridges. Its possible links (bridges allowed): R7C1 (0-2), R7C5 (2), R9C3 (0-2). The other links already carry at least 2, so R7C1 can take at most 1; the other links already carry at least 2, so R9C3 can take at most 1.
- 20/31The 2 at R7C1 needs 2 bridges. Its possible links (bridges allowed): R7C3 (0-1), R9C1 (0-2). The other links can carry at most 1, so R9C1 needs at least 1.
- 21/31The 4 at R7C7 needs 4 bridges. Its possible links (bridges allowed): R7C5 (2), R9C7 (0-2). The other links can carry at most 2, so R9C7 needs at least 2 → exactly 2 bridges.
- 22/31The 2 at R9C1 needs 2 bridges. Its possible links (bridges allowed): R7C1 (1-2), R9C3 (0-2). The other links already carry at least 1, so R9C3 can take at most 1.
- 23/31The 3 at R9C5 needs 3 bridges. Its possible links (bridges allowed): R7C5 (2), R9C3 (0-2), R9C7 (0-2). The other links already carry at least 2, so R9C3 can take at most 1; the other links already carry at least 2, so R9C7 can take at most 1.
- 24/31The 3 at R9C7 needs 3 bridges. Its possible links (bridges allowed): R7C7 (2), R9C5 (0-1), R9C9 (0-2). The other links already carry at least 2, so R9C9 can take at most 1.
- 25/31The 3 at R9C9 needs 3 bridges. Its possible links (bridges allowed): R1C9 (2), R9C7 (0-1). The other links can carry at most 2, so R9C7 needs at least 1 → exactly 1 bridge.
- 26/31The 3 at R9C7 needs 3 bridges. Its possible links (bridges allowed): R7C7 (2), R9C5 (0-1), R9C9 (1). The other links already carry at least 3, so R9C5 can take none.
- 27/31The 3 at R9C5 needs 3 bridges. Its possible links (bridges allowed): R7C5 (2), R9C3 (0-1). The other links can carry at most 2, so R9C3 needs at least 1 → exactly 1 bridge.
- 28/31Putting 2 bridges on R7C1–R9C1 would use up every remaining bridge of the 2 islands in that group, cutting them off from the rest. Everything has to join up, so R7C1–R9C1 gets at most 1 bridge.
- 29/31The 2 at R7C1 needs 2 bridges. Its possible links (bridges allowed): R7C3 (0-1), R9C1 (1). The other links can carry at most 1, so R7C3 needs at least 1 → exactly 1 bridge.
- 30/31The 3 at R7C3 needs 3 bridges. Its possible links (bridges allowed): R7C1 (1), R7C5 (2), R9C3 (0-1). The other links already carry at least 3, so R9C3 can take none.
- 31/31The 2 at R9C1 needs 2 bridges. Its possible links (bridges allowed): R7C1 (1), R9C3 (0-1). The other links can carry at most 1, so R9C3 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.