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Everyday Grids Hashi Answer Today: May 5, 2026
📅This page is for May 5, 2026. 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 May 5, 2026 (5 May 2026): 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 5, 2026 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 5, 2026
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 5, 2026
8×8 grid✅ Solved Normal grid
- 0/13Each 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 13 logical steps, no guessing.
- 1/13The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C3 (0-2). It has no other link, so R1C3 must carry both → exactly 2 bridges.
- 2/13The 6 at R1C3 needs 6 bridges. Its possible links (bridges allowed): R1C1 (2), R1C5 (0-2), R7C3 (0-2). The other links can carry at most 4, so R1C5 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R7C3 needs at least 2 → exactly 2 bridges.
- 3/13There is already a bridge on R1C3–R7C3. Bridges can't cross, so R6C2–R6C8 is now impossible.
- 4/13The 3 at R1C5 needs 3 bridges. Its possible links (bridges allowed): R1C3 (2), R1C8 (0-2). The other links can carry at most 2, so R1C8 needs at least 1, and the other links already carry at least 2, so R1C8 can take at most 1 → exactly 1 bridge.
- 5/13The 2 at R1C8 needs 2 bridges. Its possible links (bridges allowed): R1C5 (1), R6C8 (0-2). The other links can carry at most 1, so R6C8 needs at least 1, and the other links already carry at least 1, so R6C8 can take at most 1 → exactly 1 bridge.
- 6/13The 1 at R3C6 needs 1 bridge. Its possible links (bridges allowed): R5C6 (0-1). It has no other link, so R5C6 must carry it → exactly 1 bridge.
- 7/13The 1 at R4C2 needs 1 bridge. Its possible links (bridges allowed): R6C2 (0-1). It has no other link, so R6C2 must carry it → exactly 1 bridge.
- 8/13The 3 at R5C6 needs 3 bridges. Its possible links (bridges allowed): R3C6 (1), R7C6 (0-2). The other links can carry at most 1, so R7C6 needs at least 2 → exactly 2 bridges.
- 9/13The 2 at R6C2 needs 2 bridges. Its possible links (bridges allowed): R4C2 (1), R8C2 (0-2). The other links can carry at most 1, so R8C2 needs at least 1, and the other links already carry at least 1, so R8C2 can take at most 1 → exactly 1 bridge.
- 10/13The 2 at R6C8 needs 2 bridges. Its possible links (bridges allowed): R1C8 (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.
- 11/13The 3 at R7C3 needs 3 bridges. Its possible links (bridges allowed): R1C3 (2), R7C6 (0-2). The other links can carry at most 2, so R7C6 needs at least 1, and the other links already carry at least 2, so R7C6 can take at most 1 → exactly 1 bridge.
- 12/13The 3 at R8C2 needs 3 bridges. Its possible links (bridges allowed): R6C2 (1), R8C4 (0-2). The other links can carry at most 1, so R8C4 needs at least 2 → exactly 2 bridges.
- 13/13The 3 at R8C4 needs 3 bridges. Its possible links (bridges allowed): R8C2 (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 Normal answer as text
Bridges (═ double,, single): R1C1 ═ R1C3, R1C3 ═ R1C5, R1C3 ═ R7C3, R1C5, R1C8, R1C8, R6C8, R3C6, R5C6, R4C2, R6C2, R5C6 ═ R7C6, R6C2, R8C2, R6C8, R8C8, R7C3, R7C6, R8C2 ═ R8C4, R8C4, R8C8.
Everyday Grids Hashi Hard Answer, May 5, 2026
9×9 grid✅ Solved Hard grid
- 0/36Each 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 36 logical steps, no guessing.
- 1/36The 4 at R1C1 needs 4 bridges. Its possible links (bridges allowed): R1C5 (0-2), R3C1 (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 R3C1 needs at least 2 → exactly 2 bridges.
- 2/36The 3 at R2C2 needs 3 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 1; the other links can carry at most 2, so R5C2 needs at least 1.
- 3/36There is already a bridge on R2C2–R5C2. Bridges can't cross, so R3C1–R3C5 is now impossible.
- 4/36The 2 at R2C4 needs 2 bridges. Its possible links (bridges allowed): R2C2 (1-2), R8C4 (0-2). The other links already carry at least 1, so R8C4 can take at most 1.
- 5/36The 4 at R3C1 needs 4 bridges. Its possible links (bridges allowed): R1C1 (2), R9C1 (0-2). The other links can carry at most 2, so R9C1 needs at least 2 → exactly 2 bridges.
- 6/36The 4 at R3C5 needs 4 bridges. Its possible links (bridges allowed): R1C5 (0-2), R3C8 (0-2), R7C5 (0-1). The other links can carry at most 3, so R1C5 needs at least 1; the other links can carry at most 3, so R3C8 needs at least 1.
- 7/36The 4 at R1C5 needs 4 bridges. Its possible links (bridges allowed): R1C1 (2), R1C9 (0-2), R3C5 (1-2). The other links already carry at least 3, so R1C9 can take at most 1.
- 8/36The 2 at R1C9 needs 2 bridges. Its possible links (bridges allowed): R1C5 (0-1), R4C9 (0-2). The other links can carry at most 1, so R4C9 needs at least 1.
- 9/36The 3 at R3C8 needs 3 bridges. Its possible links (bridges allowed): R3C5 (1-2), R5C8 (0-2). The other links can carry at most 2, so R5C8 needs at least 1.
- 10/36The 4 at R4C9 needs 4 bridges. Its possible links (bridges allowed): R1C9 (1-2), R9C9 (0-2). The other links can carry at most 2, so R1C9 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R9C9 needs at least 2 → exactly 2 bridges.
- 11/36The 2 at R1C9 needs 2 bridges. Its possible links (bridges allowed): R1C5 (0-1), R4C9 (2). The other links already carry at least 2, so R1C5 can take none.
- 12/36The 4 at R1C5 needs 4 bridges. Its possible links (bridges allowed): R1C1 (2), R3C5 (1-2). The other links can carry at most 2, so R3C5 needs at least 2 → exactly 2 bridges.
- 13/36The 5 at R5C8 needs 5 bridges. Its possible links (bridges allowed): R3C8 (1-2), R5C6 (0-2), R8C8 (0-2). The other links can carry at most 4, so R5C6 needs at least 1; the other links can carry at most 4, so R8C8 needs at least 1.
- 14/36The 2 at R5C6 needs 2 bridges. Its possible links (bridges allowed): R5C2 (0-2), R5C8 (1-2), R8C6 (0-2). The other links already carry at least 1, so R5C2 can take at most 1; the other links already carry at least 1, so R8C6 can take at most 1.
- 15/36The 1 at R6C7 needs 1 bridge. Its possible links (bridges allowed): R9C7 (0-1). It has no other link, so R9C7 must carry it → exactly 1 bridge.
- 16/36There is already a bridge on R6C7–R9C7. Bridges can't cross, so R8C6–R8C8 is now impossible.
- 17/36The 2 at R8C6 needs 2 bridges. Its possible links (bridges allowed): R5C6 (0-1), R8C4 (0-2). The other links can carry at most 1, so R8C4 needs at least 1.
- 18/36There is already a bridge on R8C4–R8C6. Bridges can't cross, so R7C5–R9C5 is now impossible.
- 19/36The 1 at R7C5 needs 1 bridge. Its possible links (bridges allowed): R3C5 (0-1). It has no other link, so R3C5 must carry it → exactly 1 bridge.
- 20/36There is already a bridge on R3C5–R7C5. Bridges can't cross, so R5C2–R5C6 is now impossible.
- 21/36The 4 at R3C5 needs 4 bridges. Its possible links (bridges allowed): R1C5 (2), R3C8 (1-2), R7C5 (1). The other links already carry at least 3, so R3C8 can take at most 1 → exactly 1 bridge.
- 22/36The 3 at R3C8 needs 3 bridges. Its possible links (bridges allowed): R3C5 (1), R5C8 (1-2). The other links can carry at most 1, so R5C8 needs at least 2 → exactly 2 bridges.
- 23/36The 3 at R5C2 needs 3 bridges. Its possible links (bridges allowed): R2C2 (1-2), R8C2 (0-2). The other links can carry at most 2, so R8C2 needs at least 1.
- 24/36The 2 at R8C2 needs 2 bridges. Its possible links (bridges allowed): R5C2 (1-2), R8C4 (0-2). The other links already carry at least 1, so R8C4 can take at most 1.
- 25/36The 2 at R8C8 needs 2 bridges. Its possible links (bridges allowed): R5C8 (1-2). It has no other link, so R5C8 must carry both → exactly 2 bridges.
- 26/36The 5 at R5C8 needs 5 bridges. Its possible links (bridges allowed): R3C8 (2), R5C6 (1-2), R8C8 (2). The other links already carry at least 4, so R5C6 can take at most 1 → exactly 1 bridge.
- 27/36The 2 at R5C6 needs 2 bridges. Its possible links (bridges allowed): R5C8 (1), R8C6 (0-1). The other links can carry at most 1, so R8C6 needs at least 1 → exactly 1 bridge.
- 28/36The 2 at R8C6 needs 2 bridges. Its possible links (bridges allowed): R5C6 (1), R8C4 (1-2). The other links already carry at least 1, so R8C4 can take at most 1 → exactly 1 bridge.
- 29/36The 3 at R8C4 needs 3 bridges. Its possible links (bridges allowed): R2C4 (0-1), R8C2 (0-1), R8C6 (1). The other links can carry at most 2, so R2C4 needs at least 1 → exactly 1 bridge; the other links can carry at most 2, so R8C2 needs at least 1 → exactly 1 bridge.
- 30/36The 2 at R2C4 needs 2 bridges. Its possible links (bridges allowed): R2C2 (1-2), R8C4 (1). The other links already carry at least 1, so R2C2 can take at most 1 → exactly 1 bridge.
- 31/36The 3 at R2C2 needs 3 bridges. Its possible links (bridges allowed): R2C4 (1), R5C2 (1-2). The other links can carry at most 1, so R5C2 needs at least 2 → exactly 2 bridges.
- 32/36The 3 at R5C2 needs 3 bridges. Its possible links (bridges allowed): R2C2 (2), R8C2 (1-2). The other links already carry at least 2, so R8C2 can take at most 1 → exactly 1 bridge.
- 33/36The 4 at R9C1 needs 4 bridges. Its possible links (bridges allowed): R3C1 (2), R9C3 (0-2). The other links can carry at most 2, so R9C3 needs at least 2 → exactly 2 bridges.
- 34/36The 4 at R9C3 needs 4 bridges. Its possible links (bridges allowed): R9C1 (2), R9C5 (0-2). The other links can carry at most 2, so R9C5 needs at least 2 → exactly 2 bridges.
- 35/36The 3 at R9C5 needs 3 bridges. Its possible links (bridges allowed): R9C3 (2), R9C7 (0-2). The other links can carry at most 2, so R9C7 needs at least 1, and the other links already carry at least 2, so R9C7 can take at most 1 → exactly 1 bridge.
- 36/36The 3 at R9C7 needs 3 bridges. Its possible links (bridges allowed): R6C7 (1), R9C5 (1), 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. ✅ Solved!
Hashi Hard answer as text
Bridges (═ double,, single): R1C1 ═ R1C5, R1C1 ═ R3C1, R1C5 ═ R3C5, R1C9 ═ R4C9, R2C2, R2C4, R2C2 ═ R5C2, R2C4, R8C4, R3C1 ═ R9C1, R3C5, R3C8, R3C5, R7C5, R3C8 ═ R5C8, R4C9 ═ R9C9, R5C2, R8C2, R5C6, R5C8, R5C6, R8C6, R5C8 ═ R8C8, R6C7, R9C7, R8C2, R8C4, R8C4, R8C6, R9C1 ═ R9C3, R9C3 ═ R9C5, R9C5, R9C7, R9C7, R9C9.
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 May 5, 2026
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 May 5, 2026 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 May 5, 2026. 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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