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Everyday Grids Hashi Answer Today: July 27, 2026
📅This page is for July 27, 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 July 27, 2026 (27 July 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 Jul 27, 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 July 27, 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, Jul 27, 2026
6×6 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 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.
- 2/13The 2 at R1C3 needs 2 bridges. Its possible links (bridges allowed): R1C1 (1-2), R1C5 (0-2), R5C3 (0-2). The other links already carry at least 1, so R1C5 can take at most 1; the other links already carry at least 1, so R5C3 can take at most 1.
- 3/13The 2 at R1C5 needs 2 bridges. Its possible links (bridges allowed): R1C3 (0-1), R3C5 (0-2). The other links can carry at most 1, so R3C5 needs at least 1.
- 4/13The 1 at R2C6 needs 1 bridge. Its possible links (bridges allowed): R5C6 (0-1). It has no other link, so R5C6 must carry it → exactly 1 bridge.
- 5/13The 5 at R3C1 needs 5 bridges. Its possible links (bridges allowed): R1C1 (1-2), R3C5 (0-2), R5C1 (0-2). The other links can carry at most 4, so R3C5 needs at least 1; the other links can carry at most 4, so R5C1 needs at least 1.
- 6/13There is already a bridge on R3C1–R3C5. Bridges can't cross, so R1C3–R5C3 is now impossible.
- 7/13The 2 at R3C5 needs 2 bridges. Its possible links (bridges allowed): R1C5 (1-2), R3C1 (1-2). The other links already carry at least 1, so R1C5 can take at most 1 → exactly 1 bridge; the other links already carry at least 1, so R3C1 can take at most 1 → exactly 1 bridge.
- 8/13The 2 at R1C5 needs 2 bridges. Its possible links (bridges allowed): R1C3 (0-1), R3C5 (1). The other links can carry at most 1, so R1C3 needs at least 1 → exactly 1 bridge.
- 9/13The 2 at R1C3 needs 2 bridges. Its possible links (bridges allowed): R1C1 (1-2), R1C5 (1). The other links already carry at least 1, so R1C1 can take at most 1 → exactly 1 bridge.
- 10/13The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C3 (1), R3C1 (1-2). The other links can carry at most 1, so R3C1 needs at least 2 → exactly 2 bridges.
- 11/13The 5 at R3C1 needs 5 bridges. Its possible links (bridges allowed): R1C1 (2), R3C5 (1), R5C1 (1-2). The other links can carry at most 3, so R5C1 needs at least 2 → exactly 2 bridges.
- 12/13The 4 at R5C1 needs 4 bridges. Its possible links (bridges allowed): R3C1 (2), R5C3 (0-2). The other links can carry at most 2, so R5C3 needs at least 2 → exactly 2 bridges.
- 13/13The 4 at R5C3 needs 4 bridges. Its possible links (bridges allowed): R5C1 (2), R5C6 (0-2). The other links can carry at most 2, so R5C6 needs at least 2 → exactly 2 bridges. ✅ Solved!
Hashi Normal answer as text
Bridges (═ double,, single): R1C1, R1C3, R1C1 ═ R3C1, R1C3, R1C5, R1C5, R3C5, R2C6, R5C6, R3C1, R3C5, R3C1 ═ R5C1, R5C1 ═ R5C3, R5C3 ═ R5C6.
Everyday Grids Hashi Hard Answer, Jul 27, 2026
7×7 grid✅ Solved Hard 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 6 at R1C5 needs 6 bridges. Its possible links (bridges allowed): R1C3 (0-2), R1C7 (0-2), R3C5 (0-2). The other links can carry at most 4, so R1C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R1C7 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R3C5 needs at least 2 → exactly 2 bridges.
- 2/28The 3 at R1C3 needs 3 bridges. Its possible links (bridges allowed): R1C1 (0-2), R1C5 (2), R3C3 (0-2). The other links already carry at least 2, so R1C1 can take at most 1; the other links already carry at least 2, so R3C3 can take at most 1.
- 3/28The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C3 (0-1), R3C1 (0-2). The other links can carry at most 1, so R3C1 needs at least 1.
- 4/28The 3 at R1C7 needs 3 bridges. Its possible links (bridges allowed): R1C5 (2), R3C7 (0-2). The other links can carry at most 2, so R3C7 needs at least 1, and the other links already carry at least 2, so R3C7 can take at most 1 → exactly 1 bridge.
- 5/28The 2 at R2C2 needs 2 bridges. Its possible links (bridges allowed): R2C4 (0-2), R4C2 (0-1). The other links can carry at most 1, so R2C4 needs at least 1.
- 6/28There is already a bridge on R2C2–R2C4. Bridges can't cross, so R1C3–R3C3 is now impossible.
- 7/28The 3 at R1C3 needs 3 bridges. Its possible links (bridges allowed): R1C1 (0-1), R1C5 (2). The other links can carry at most 2, so R1C1 needs at least 1 → exactly 1 bridge.
- 8/28The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C3 (1), R3C1 (1-2). The other links already carry at least 1, so R3C1 can take at most 1 → exactly 1 bridge.
- 9/28The 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.
- 10/28There is already a bridge on R2C4–R4C4. Bridges can't cross, so R3C3–R3C5 is now impossible.
- 11/28The 2 at R3C1 needs 2 bridges. Its possible links (bridges allowed): R1C1 (1), R3C3 (0-2), R5C1 (0-2). The other links already carry at least 1, so R3C3 can take at most 1; the other links already carry at least 1, so R5C1 can take at most 1.
- 12/28The 2 at R3C3 needs 2 bridges. Its possible links (bridges allowed): R3C1 (0-1), R5C3 (0-2). The other links can carry at most 1, so R5C3 needs at least 1.
- 13/28There is already a bridge on R3C3–R5C3. Bridges can't cross, so R4C2–R4C4 is now impossible.
- 14/28The 3 at R3C5 needs 3 bridges. Its possible links (bridges allowed): R1C5 (2), R3C7 (0-2), R7C5 (0-2). The other links already carry at least 2, so R3C7 can take at most 1; the other links already carry at least 2, so R7C5 can take at most 1.
- 15/28The 4 at R3C7 needs 4 bridges. Its possible links (bridges allowed): R1C7 (1), R3C5 (0-1), R6C7 (0-2). The other links can carry at most 3, so R3C5 needs at least 1 → exactly 1 bridge; the other links can carry at most 2, so R6C7 needs at least 2 → exactly 2 bridges.
- 16/28The 3 at R3C5 needs 3 bridges. Its possible links (bridges allowed): R1C5 (2), R3C7 (1), R7C5 (0-1). The other links already carry at least 3, so R7C5 can take none.
- 17/28The 1 at R4C2 needs 1 bridge. Its possible links (bridges allowed): R2C2 (0-1). It has no other link, so R2C2 must carry it → exactly 1 bridge.
- 18/28There is already a bridge on R2C2–R4C2. Bridges can't cross, so R3C1–R3C3 is now impossible.
- 19/28The 2 at R2C2 needs 2 bridges. Its possible links (bridges allowed): R2C4 (1-2), R4C2 (1). The other links already carry at least 1, so R2C4 can take at most 1 → exactly 1 bridge.
- 20/28The 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.
- 21/28The 2 at R3C1 needs 2 bridges. Its possible links (bridges allowed): R1C1 (1), R5C1 (0-1). The other links can carry at most 1, so R5C1 needs at least 1 → exactly 1 bridge.
- 22/28The 2 at R3C3 needs 2 bridges. Its possible links (bridges allowed): R5C3 (1-2). It has no other link, so R5C3 must carry both → exactly 2 bridges.
- 23/28The 4 at R4C4 needs 4 bridges. Its possible links (bridges allowed): R2C4 (2), R6C4 (0-2). The other links can carry at most 2, so R6C4 needs at least 2 → exactly 2 bridges.
- 24/28The 4 at R6C4 needs 4 bridges. Its possible links (bridges allowed): R4C4 (2), R6C7 (0-2). The other links can carry at most 2, so R6C7 needs at least 2 → exactly 2 bridges.
- 25/28The 4 at R7C1 needs 4 bridges. Its possible links (bridges allowed): R5C1 (0-2), R7C3 (0-2). The other links can carry at most 2, so R5C1 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R7C3 needs at least 2 → exactly 2 bridges.
- 26/28The 3 at R5C1 needs 3 bridges. Its possible links (bridges allowed): R3C1 (1), R5C3 (0-2), R7C1 (2). The other links already carry at least 3, so R5C3 can take none.
- 27/28The 4 at R5C3 needs 4 bridges. Its possible links (bridges allowed): R3C3 (2), R7C3 (0-2). The other links can carry at most 2, so R7C3 needs at least 2 → exactly 2 bridges.
- 28/28The 6 at R7C3 needs 6 bridges. Its possible links (bridges allowed): R5C3 (2), R7C1 (2), R7C5 (0-2). The other links can carry at most 4, so R7C5 needs at least 2 → exactly 2 bridges. ✅ Solved!
Hashi Hard answer as text
Bridges (═ double,, single): R1C1, R1C3, R1C1, R3C1, R1C3 ═ R1C5, R1C5 ═ R1C7, R1C5 ═ R3C5, R1C7, R3C7, R2C2, R2C4, R2C2, R4C2, R2C4 ═ R4C4, R3C1, R5C1, R3C3 ═ R5C3, R3C5, R3C7, R3C7 ═ R6C7, R4C4 ═ R6C4, R5C1 ═ R7C1, R5C3 ═ R7C3, R6C4 ═ R6C7, R7C1 ═ R7C3, R7C3 ═ R7C5.
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 July 27, 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 July 27, 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 July 27, 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!
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