Everyday Grids Hashi Answer Today: April 10, 2025

This page has the complete Everyday Grids Hashi answers for April 10, 2025 (10 April 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 Apr 10, 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 April 10, 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, Apr 10, 2025
10×10 grid- 0/28Each 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 28 logical steps, no guessing.
- 1/28The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C10 (0-2), R4C1 (0-2). The other links can carry at most 2, so R1C10 needs at least 1; the other links can carry at most 2, so R4C1 needs at least 1.
- 2/28The 3 at R1C10 needs 3 bridges. Its possible links (bridges allowed): R1C1 (1-2), R3C10 (0-2). The other links can carry at most 2, so R3C10 needs at least 1.
- 3/28The 3 at R3C2 needs 3 bridges. Its possible links (bridges allowed): R3C6 (0-2), R5C2 (0-1). The other links can carry at most 1, so R3C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R5C2 needs at least 1 → exactly 1 bridge.
- 4/28The 3 at R3C6 needs 3 bridges. Its possible links (bridges allowed): R3C2 (2), R3C10 (0-2). The other links can carry at most 2, so R3C10 needs at least 1, and the other links already carry at least 2, so R3C10 can take at most 1 → exactly 1 bridge.
- 5/28The 4 at R3C10 needs 4 bridges. Its possible links (bridges allowed): R1C10 (1-2), R3C6 (1), R8C10 (0-1). The other links can carry at most 2, so R1C10 needs at least 2 → exactly 2 bridges; the other links can carry at most 3, so R8C10 needs at least 1 → exactly 1 bridge.
- 6/28The 3 at R1C10 needs 3 bridges. Its possible links (bridges allowed): R1C1 (1-2), R3C10 (2). The other links already carry at least 2, so R1C1 can take at most 1 → exactly 1 bridge.
- 7/28The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C10 (1), R4C1 (1-2). The other links can carry at most 1, so R4C1 needs at least 2 → exactly 2 bridges.
- 8/28The 4 at R4C1 needs 4 bridges. Its possible links (bridges allowed): R1C1 (2), R6C1 (0-2). The other links can carry at most 2, so R6C1 needs at least 2 → exactly 2 bridges.
- 9/28The 1 at R5C2 needs 1 bridge. Its possible links (bridges allowed): R3C2 (1), R7C2 (0-1). The other links already carry at least 1, so R7C2 can take none.
- 10/28The 4 at R6C1 needs 4 bridges. Its possible links (bridges allowed): R4C1 (2), R6C3 (0-2), R8C1 (0-1). The other links can carry at most 3, so R6C3 needs at least 1.
- 11/28The 2 at R7C2 needs 2 bridges. Its possible links (bridges allowed): R10C2 (0-2). It has no other link, so R10C2 must carry both → exactly 2 bridges.
- 12/28There is already a bridge on R7C2–R10C2. Bridges can't cross, so R8C1–R8C3 is now impossible.
- 13/28The 1 at R8C1 needs 1 bridge. Its possible links (bridges allowed): R6C1 (0-1). It has no other link, so R6C1 must carry it → exactly 1 bridge.
- 14/28The 4 at R6C1 needs 4 bridges. Its possible links (bridges allowed): R4C1 (2), R6C3 (1-2), R8C1 (1). The other links already carry at least 3, so R6C3 can take at most 1 → exactly 1 bridge.
- 15/28The 4 at R8C3 needs 4 bridges. Its possible links (bridges allowed): R6C3 (0-2), R8C5 (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 R8C5 needs at least 2 → exactly 2 bridges.
- 16/28The 3 at R6C3 needs 3 bridges. Its possible links (bridges allowed): R6C1 (1), R6C7 (0-2), R8C3 (2). The other links already carry at least 3, so R6C7 can take none.
- 17/28The 3 at R6C7 needs 3 bridges. Its possible links (bridges allowed): R6C9 (0-2), R8C7 (0-2). The other links can carry at most 2, so R6C9 needs at least 1; the other links can carry at most 2, so R8C7 needs at least 1.
- 18/28The 2 at R6C9 needs 2 bridges. Its possible links (bridges allowed): R6C7 (1-2), R10C9 (0-2). The other links already carry at least 1, so R10C9 can take at most 1.
- 19/28The 4 at R8C5 needs 4 bridges. Its possible links (bridges allowed): R8C3 (2), R8C7 (0-2). The other links can carry at most 2, so R8C7 needs at least 2 → exactly 2 bridges.
- 20/28The 1 at R8C10 needs 1 bridge. Its possible links (bridges allowed): R3C10 (1), R8C7 (0-1). The other links already carry at least 1, so R8C7 can take none.
- 21/28The 5 at R8C7 needs 5 bridges. Its possible links (bridges allowed): R6C7 (1-2), R8C5 (2), R10C7 (0-2). The other links can carry at most 4, so R10C7 needs at least 1.
- 22/28The 4 at R10C2 needs 4 bridges. Its possible links (bridges allowed): R7C2 (2), R10C4 (0-2). The other links can carry at most 2, so R10C4 needs at least 2 → exactly 2 bridges.
- 23/28The 3 at R10C4 needs 3 bridges. Its possible links (bridges allowed): R10C2 (2), R10C7 (0-2). The other links can carry at most 2, so R10C7 needs at least 1, and the other links already carry at least 2, so R10C7 can take at most 1 → exactly 1 bridge.
- 24/28The 3 at R10C7 needs 3 bridges. Its possible links (bridges allowed): R8C7 (1-2), R10C4 (1), R10C9 (0-2). The other links already carry at least 2, so R10C9 can take at most 1.
- 25/28The 2 at R10C9 needs 2 bridges. Its possible links (bridges allowed): R6C9 (0-1), R10C7 (0-1). The other links can carry at most 1, so R6C9 needs at least 1 → exactly 1 bridge; the other links can carry at most 1, so R10C7 needs at least 1 → exactly 1 bridge.
- 26/28The 2 at R6C9 needs 2 bridges. Its possible links (bridges allowed): R6C7 (1-2), R10C9 (1). The other links already carry at least 1, so R6C7 can take at most 1 → exactly 1 bridge.
- 27/28The 3 at R6C7 needs 3 bridges. Its possible links (bridges allowed): R6C9 (1), R8C7 (1-2). The other links can carry at most 1, so R8C7 needs at least 2 → exactly 2 bridges.
- 28/28The 5 at R8C7 needs 5 bridges. Its possible links (bridges allowed): R6C7 (2), R8C5 (2), R10C7 (1-2). The other links already carry at least 4, so R10C7 can take at most 1 → exactly 1 bridge. ✅ Solved!
Hashi Normal answer as text
Everyday Grids Hashi Hard Answer, Apr 10, 2025
9×9 grid- 0/29Each 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 29 logical steps, no guessing.
- 1/29The 3 at R1C9 needs 3 bridges. Its possible links (bridges allowed): R1C7 (0-1), R3C9 (0-2). The other links can carry at most 2, so R1C7 needs at least 1 → exactly 1 bridge; the other links can carry at most 1, so R3C9 needs at least 2 → exactly 2 bridges.
- 2/29The 1 at R1C7 needs 1 bridge. Its possible links (bridges allowed): R1C5 (0-1), R1C9 (1). The other links already carry at least 1, so R1C5 can take none.
- 3/29The 3 at R1C5 needs 3 bridges. Its possible links (bridges allowed): R1C1 (0-2), R3C5 (0-2). The other links can carry at most 2, so R1C1 needs at least 1; the other links can carry at most 2, so R3C5 needs at least 1.
- 4/29The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C5 (1-2), R9C1 (0-2). The other links already carry at least 1, so R9C1 can take at most 1.
- 5/29The 3 at R2C2 needs 3 bridges. Its possible links (bridges allowed): R2C4 (0-2), R6C2 (0-2). The other links can carry at most 2, so R2C4 needs at least 1; the other links can carry at most 2, so R6C2 needs at least 1.
- 6/29The 3 at R2C4 needs 3 bridges. Its possible links (bridges allowed): R2C2 (1-2), R4C4 (0-2). The other links can carry at most 2, so R4C4 needs at least 1.
- 7/29The 4 at R3C5 needs 4 bridges. Its possible links (bridges allowed): R1C5 (1-2), R3C9 (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 R3C9 needs at least 2 → exactly 2 bridges.
- 8/29The 3 at R1C5 needs 3 bridges. Its possible links (bridges allowed): R1C1 (1-2), R3C5 (2). The other links already carry at least 2, so R1C1 can take at most 1 → exactly 1 bridge.
- 9/29The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C5 (1), R9C1 (0-1). The other links can carry at most 1, so R9C1 needs at least 1 → exactly 1 bridge.
- 10/29The 4 at R3C9 needs 4 bridges. Its possible links (bridges allowed): R1C9 (2), R3C5 (2), R7C9 (0-2). The other links already carry at least 4, so R7C9 can take none.
- 11/29The 5 at R6C2 needs 5 bridges. Its possible links (bridges allowed): R2C2 (1-2), R6C4 (0-2), R8C2 (0-1). The other links can carry at most 3, so R2C2 needs at least 2 → exactly 2 bridges; the other links can carry at most 3, so R6C4 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R8C2 needs at least 1 → exactly 1 bridge.
- 12/29The 3 at R2C2 needs 3 bridges. Its possible links (bridges allowed): R2C4 (1-2), R6C2 (2). The other links already carry at least 2, so R2C4 can take at most 1 → exactly 1 bridge.
- 13/29The 3 at R2C4 needs 3 bridges. Its possible links (bridges allowed): R2C2 (1), R4C4 (1-2). The other links can carry at most 1, so R4C4 needs at least 2 → exactly 2 bridges.
- 14/29The 3 at R6C4 needs 3 bridges. Its possible links (bridges allowed): R4C4 (0-2), R6C2 (2), R6C6 (0-2), R9C4 (0-2). The other links already carry at least 2, so R4C4 can take at most 1; the other links already carry at least 2, so R6C6 can take at most 1; the other links already carry at least 2, so R9C4 can take at most 1.
- 15/29The 4 at R4C4 needs 4 bridges. Its possible links (bridges allowed): R2C4 (2), R4C6 (0-2), R6C4 (0-1). The other links can carry at most 3, so R4C6 needs at least 1.
- 16/29The 2 at R7C9 needs 2 bridges. Its possible links (bridges allowed): R9C9 (0-2). It has no other link, so R9C9 must carry both → exactly 2 bridges.
- 17/29The 2 at R9C1 needs 2 bridges. Its possible links (bridges allowed): R1C1 (1), R9C4 (0-2). The other links can carry at most 1, so R9C4 needs at least 1, and the other links already carry at least 1, so R9C4 can take at most 1 → exactly 1 bridge.
- 18/29The 4 at R9C4 needs 4 bridges. Its possible links (bridges allowed): R6C4 (0-1), R9C1 (1), R9C6 (0-2). The other links can carry at most 3, so R6C4 needs at least 1 → exactly 1 bridge; the other links can carry at most 2, so R9C6 needs at least 2 → exactly 2 bridges.
- 19/29The 3 at R6C4 needs 3 bridges. Its possible links (bridges allowed): R4C4 (0-1), R6C2 (2), R6C6 (0-1), R9C4 (1). The other links already carry at least 3, so R4C4 can take none; the other links already carry at least 3, so R6C6 can take none.
- 20/29The 4 at R4C4 needs 4 bridges. Its possible links (bridges allowed): R2C4 (2), R4C6 (1-2). The other links can carry at most 2, so R4C6 needs at least 2 → exactly 2 bridges.
- 21/29The 3 at R4C6 needs 3 bridges. Its possible links (bridges allowed): R4C4 (2), R4C8 (0-2), R6C6 (0-2). The other links already carry at least 2, so R4C8 can take at most 1; the other links already carry at least 2, so R6C6 can take at most 1.
- 22/29The 2 at R4C8 needs 2 bridges. Its possible links (bridges allowed): R4C6 (0-1), R6C8 (0-2). The other links can carry at most 1, so R6C8 needs at least 1.
- 23/29The 2 at R6C8 needs 2 bridges. Its possible links (bridges allowed): R4C8 (1-2), R6C6 (0-2). The other links already carry at least 1, so R6C6 can take at most 1.
- 24/29The 3 at R6C6 needs 3 bridges. Its possible links (bridges allowed): R4C6 (0-1), R6C8 (0-1), R9C6 (0-2). The other links can carry at most 2, so R9C6 needs at least 1.
- 25/29The 6 at R9C6 needs 6 bridges. Its possible links (bridges allowed): R6C6 (1-2), R9C4 (2), R9C9 (0-2). The other links can carry at most 4, so R6C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R9C9 needs at least 2 → exactly 2 bridges.
- 26/29Putting 1 bridge on R4C6–R6C6 would use up every remaining bridge of the 19 islands in that group, cutting them off from the rest. Everything has to join up, so R4C6–R6C6 gets at most no bridge.
- 27/29The 3 at R4C6 needs 3 bridges. Its possible links (bridges allowed): R4C4 (2), R4C8 (0-1). The other links can carry at most 2, so R4C8 needs at least 1 → exactly 1 bridge.
- 28/29The 2 at R4C8 needs 2 bridges. Its possible links (bridges allowed): R4C6 (1), R6C8 (1-2). The other links already carry at least 1, so R6C8 can take at most 1 → exactly 1 bridge.
- 29/29The 3 at R6C6 needs 3 bridges. Its possible links (bridges allowed): R6C8 (0-1), R9C6 (2). The other links can carry at most 2, so R6C8 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.