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Everyday Grids Hashi Answer Today: April 25, 2025
📅This page is for April 25, 2025. Playing a different day? Scroll down to More Days and pick your date, or open the all-days calendar at the bottom. New puzzles unlock at midnight in your own time zone, so if it is still the previous day where you live, use the previous day link.
This page has the complete Everyday Grids Hashi answers for April 25, 2025 (25 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 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 April 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, Apr 25, 2025
7×7 grid✅ Solved Normal 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): R1C5 (0-2), R6C2 (0-2). The other links can carry at most 2, so R1C5 needs at least 1; the other links can carry at most 2, so R6C2 needs at least 1.
- 2/12There is already a bridge on R1C2–R6C2. Bridges can't cross, so R2C1–R2C4 is now impossible.
- 3/12The 2 at R1C5 needs 2 bridges. Its possible links (bridges allowed): R1C2 (1-2). It has no other link, so R1C2 must carry both → exactly 2 bridges.
- 4/12The 3 at R1C2 needs 3 bridges. Its possible links (bridges allowed): R1C5 (2), R6C2 (1-2). The other links already carry at least 2, so R6C2 can take at most 1 → exactly 1 bridge.
- 5/12The 2 at R2C1 needs 2 bridges. Its possible links (bridges allowed): R7C1 (0-2). It has no other link, so R7C1 must carry both → exactly 2 bridges.
- 6/12The 3 at R6C2 needs 3 bridges. Its possible links (bridges allowed): R1C2 (1), R6C6 (0-2). The other links can carry at most 1, so R6C6 needs at least 2 → exactly 2 bridges.
- 7/12The 3 at R6C6 needs 3 bridges. Its possible links (bridges allowed): R4C6 (0-2), R6C2 (2). The other links can carry at most 2, so R4C6 needs at least 1, and the other links already carry at least 2, so R4C6 can take at most 1 → exactly 1 bridge.
- 8/12The 2 at R4C6 needs 2 bridges. Its possible links (bridges allowed): R4C4 (0-2), R6C6 (1). The other links can carry at most 1, so R4C4 needs at least 1, and the other links already carry at least 1, so R4C4 can take at most 1 → exactly 1 bridge.
- 9/12The 2 at R4C4 needs 2 bridges. Its possible links (bridges allowed): R2C4 (0-2), R4C6 (1). The other links can carry at most 1, so R2C4 needs at least 1, and the other links already carry at least 1, so R2C4 can take at most 1 → exactly 1 bridge.
- 10/12The 2 at R2C4 needs 2 bridges. Its possible links (bridges allowed): R2C7 (0-2), R4C4 (1). 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.
- 11/12The 2 at R2C7 needs 2 bridges. Its possible links (bridges allowed): R2C4 (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.
- 12/12The 3 at R7C1 needs 3 bridges. Its possible links (bridges allowed): R2C1 (2), 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. ✅ Solved!
Hashi Normal answer as text
Bridges (═ double,, single): R1C2 ═ R1C5, R1C2, R6C2, R2C1 ═ R7C1, R2C4, R2C7, R2C4, R4C4, R2C7, R7C7, R4C4, R4C6, R4C6, R6C6, R6C2 ═ R6C6, R7C1, R7C7.
Everyday Grids Hashi Hard Answer, Apr 25, 2025
8×8 grid✅ Solved Hard 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 4 at R1C1 needs 4 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 2 → exactly 2 bridges; the other links can carry at most 2, so R3C1 needs at least 2 → exactly 2 bridges.
- 2/29The 5 at R1C3 needs 5 bridges. Its possible links (bridges allowed): R1C1 (2), R1C7 (0-2), R4C3 (0-1). The other links can carry at most 3, so R1C7 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R4C3 needs at least 1 → exactly 1 bridge.
- 3/29There is already a bridge on R1C3–R4C3. Bridges can't cross, so R3C1–R3C4 is now impossible.
- 4/29The 2 at R1C7 needs 2 bridges. Its possible links (bridges allowed): R1C3 (2), R4C7 (0-2). The other links already carry at least 2, so R4C7 can take none.
- 5/29The 4 at R3C1 needs 4 bridges. Its possible links (bridges allowed): R1C1 (2), R8C1 (0-2). The other links can carry at most 2, so R8C1 needs at least 2 → exactly 2 bridges.
- 6/29The 4 at R3C4 needs 4 bridges. Its possible links (bridges allowed): R3C8 (0-2), R5C4 (0-2). The other links can carry at most 2, so R3C8 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R5C4 needs at least 2 → exactly 2 bridges.
- 7/29There is already a bridge on R3C4–R5C4. Bridges can't cross, so R4C3–R4C5 is now impossible.
- 8/29The 3 at R3C8 needs 3 bridges. Its possible links (bridges allowed): R3C4 (2), R5C8 (0-2). The other links can carry at most 2, so R5C8 needs at least 1, and the other links already carry at least 2, so R5C8 can take at most 1 → exactly 1 bridge.
- 9/29The 2 at R4C7 needs 2 bridges. Its possible links (bridges allowed): R4C5 (0-1), R7C7 (0-2). The other links can carry at most 1, so R7C7 needs at least 1.
- 10/29There is already a bridge on R4C7–R7C7. Bridges can't cross, so R5C6–R5C8 is now impossible.
- 11/29The 3 at R5C2 needs 3 bridges. Its possible links (bridges allowed): R5C4 (0-2), R7C2 (0-2). The other links can carry at most 2, so R5C4 needs at least 1; the other links can carry at most 2, so R7C2 needs at least 1.
- 12/29The 2 at R5C8 needs 2 bridges. Its possible links (bridges allowed): R3C8 (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.
- 13/29The 3 at R7C2 needs 3 bridges. Its possible links (bridges allowed): R5C2 (1-2), R7C5 (0-2). The other links can carry at most 2, so R7C5 needs at least 1.
- 14/29There is already a bridge on R7C2–R7C5. Bridges can't cross, so R5C4–R8C4 is now impossible.
- 15/29The 5 at R5C4 needs 5 bridges. Its possible links (bridges allowed): R3C4 (2), R5C2 (1-2), R5C6 (0-2). The other links can carry at most 4, so R5C6 needs at least 1.
- 16/29There is already a bridge on R5C4–R5C6. Bridges can't cross, so R4C5–R7C5 is now impossible.
- 17/29The 1 at R4C5 needs 1 bridge. Its possible links (bridges allowed): R4C7 (0-1). It has no other link, so R4C7 must carry it → exactly 1 bridge.
- 18/29The 2 at R4C7 needs 2 bridges. Its possible links (bridges allowed): R4C5 (1), R7C7 (1-2). The other links already carry at least 1, so R7C7 can take at most 1 → exactly 1 bridge.
- 19/29The 2 at R5C6 needs 2 bridges. Its possible links (bridges allowed): R5C4 (1-2), R8C6 (0-2). The other links already carry at least 1, so R8C6 can take at most 1.
- 20/29The 2 at R7C5 needs 2 bridges. Its possible links (bridges allowed): R7C2 (1-2), R7C7 (0-2). The other links already carry at least 1, so R7C7 can take at most 1.
- 21/29The 2 at R7C7 needs 2 bridges. Its possible links (bridges allowed): R4C7 (1), R7C5 (0-1). The other links can carry at most 1, so R7C5 needs at least 1 → exactly 1 bridge.
- 22/29There is already a bridge on R7C5–R7C7. Bridges can't cross, so R5C6–R8C6 is now impossible.
- 23/29The 2 at R5C6 needs 2 bridges. Its possible links (bridges allowed): R5C4 (1-2). It has no other link, so R5C4 must carry both → exactly 2 bridges.
- 24/29The 5 at R5C4 needs 5 bridges. Its possible links (bridges allowed): R3C4 (2), R5C2 (1-2), R5C6 (2). The other links already carry at least 4, so R5C2 can take at most 1 → exactly 1 bridge.
- 25/29The 3 at R5C2 needs 3 bridges. Its possible links (bridges allowed): R5C4 (1), R7C2 (1-2). The other links can carry at most 1, so R7C2 needs at least 2 → exactly 2 bridges.
- 26/29The 3 at R7C2 needs 3 bridges. Its possible links (bridges allowed): R5C2 (2), R7C5 (1-2). The other links already carry at least 2, so R7C5 can take at most 1 → exactly 1 bridge.
- 27/29The 4 at R8C1 needs 4 bridges. Its possible links (bridges allowed): R3C1 (2), R8C4 (0-2). The other links can carry at most 2, so R8C4 needs at least 2 → exactly 2 bridges.
- 28/29The 4 at R8C4 needs 4 bridges. Its possible links (bridges allowed): R8C1 (2), R8C6 (0-2). The other links can carry at most 2, so R8C6 needs at least 2 → exactly 2 bridges.
- 29/29The 3 at R8C6 needs 3 bridges. Its possible links (bridges allowed): R8C4 (2), R8C8 (0-2). The other links can carry at most 2, so R8C8 needs at least 1, and the other links already carry at least 2, so R8C8 can take at most 1 → exactly 1 bridge. ✅ Solved!
Hashi Hard answer as text
Bridges (═ double,, single): R1C1 ═ R1C3, R1C1 ═ R3C1, R1C3 ═ R1C7, R1C3, R4C3, R3C1 ═ R8C1, R3C4 ═ R3C8, R3C4 ═ R5C4, R3C8, R5C8, R4C5, R4C7, R4C7, R7C7, R5C2, R5C4, R5C2 ═ R7C2, R5C4 ═ R5C6, R5C8, R8C8, R7C2, R7C5, R7C5, R7C7, R8C1 ═ R8C4, R8C4 ═ R8C6, R8C6, R8C8.
How to Solve Everyday Grids Hashi: Quick Tips
Want to beat tomorrow’s Everyday Grids Hashi without help? These are the logical moves our step by step solutions use most often:
- 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.
Other Everyday Grids Answers for April 25, 2025
More Days: Everyday Grids Hashi Answers by Date
Everyday Grids Hashi FAQ
How do you play Everyday Grids Hashi?
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.
Is there a Normal and a Hard Hashi puzzle every day?
Yes. Everyday Grids releases a Normal and a Hard Hashi level every day, and both answers and solutions are on this page.
Are these Everyday Grids Hashi answers for April 25, 2025 correct?
Yes. Each grid is solved with pure logic step by step and checked against the puzzle rules, so it matches the one valid solution.
My Hashi puzzle looks different. What should I do?
Check the date first: this page is for April 25, 2025. Use the date links under More Days to find your day, and if it still looks different, leave a comment so we can check it.
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Thanks for solving with us!
Come back tomorrow for a fresh set of grids. Is your puzzle different from the one on this page, are you stuck on a step, or did you spot a mistake? Tell us in the comments with the game, the level (Normal or Hard) and your date, and share any suggestions too. We read every comment and use it to improve these answers.
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