Everyday Grids Hashi Answers August 14, 2026 [Step By Step]

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Everyday Grids Hashi Answer Today: August 14, 2026

● Normal & Hard solved Hashi📅 Fri, Aug 14, 2026
📅This page is for August 14, 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 August 14, 2026 (14 August 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 Aug 14, 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.

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Everyday Grids Hashi Answers and Solutions for August 14, 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).

Normal Answer ↓Hard Answer ↓

Everyday Grids Hashi Normal Answer, Aug 14, 2026

7×7 grid
✅ Solved Normal grid
13334243322233
13334243322233
  1. 0/20
    Each 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 20 logical steps, no guessing.
  2. 1/20
    The 1 at R1C1 needs 1 bridge. Its possible links (bridges allowed): R1C3 (0-1). It has no other link, so R1C3 must carry it → exactly 1 bridge.
  3. 2/20
    The 3 at R1C3 needs 3 bridges. Its possible links (bridges allowed): R1C1 (1), R1C7 (0-2). The other links can carry at most 1, so R1C7 needs at least 2 → exactly 2 bridges.
  4. 3/20
    The 3 at R1C7 needs 3 bridges. Its possible links (bridges allowed): R1C3 (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. 4/20
    The 3 at R2C2 needs 3 bridges. Its possible links (bridges allowed): R2C6 (0-2), R4C2 (0-2). The other links can carry at most 2, so R2C6 needs at least 1; the other links can carry at most 2, so R4C2 needs at least 1.
  6. 5/20
    The 4 at R2C6 needs 4 bridges. Its possible links (bridges allowed): R2C2 (1-2), R4C6 (0-2). The other links can carry at most 2, so R2C2 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R4C6 needs at least 2 → exactly 2 bridges.
  7. 6/20
    The 3 at R2C2 needs 3 bridges. Its possible links (bridges allowed): R2C6 (2), R4C2 (1-2). The other links already carry at least 2, so R4C2 can take at most 1 → exactly 1 bridge.
  8. 7/20
    The 2 at R3C7 needs 2 bridges. Its possible links (bridges allowed): R1C7 (1), R5C7 (0-2). The other links can carry at most 1, so R5C7 needs at least 1, and the other links already carry at least 1, so R5C7 can take at most 1 → exactly 1 bridge.
  9. 8/20
    The 4 at R4C2 needs 4 bridges. Its possible links (bridges allowed): R2C2 (1), R4C4 (0-2), R7C2 (0-2). The other links can carry at most 3, so R4C4 needs at least 1; the other links can carry at most 3, so R7C2 needs at least 1.
  10. 9/20
    The 3 at R4C6 needs 3 bridges. Its possible links (bridges allowed): R2C6 (2), R4C4 (0-2), R6C6 (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.
  11. 10/20
    The 2 at R5C7 needs 2 bridges. Its possible links (bridges allowed): R3C7 (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. 11/20
    The 2 at R6C6 needs 2 bridges. Its possible links (bridges allowed): R4C6 (0-1), R6C4 (0-2). The other links can carry at most 1, so R6C4 needs at least 1.
  13. 12/20
    The 2 at R6C4 needs 2 bridges. Its possible links (bridges allowed): R4C4 (0-2), R6C6 (1-2). The other links already carry at least 1, so R4C4 can take at most 1.
  14. 13/20
    The 3 at R7C2 needs 3 bridges. Its possible links (bridges allowed): R4C2 (1-2), R7C7 (0-2). The other links can carry at most 2, so R7C7 needs at least 1.
  15. 14/20
    The 3 at R7C7 needs 3 bridges. Its possible links (bridges allowed): R5C7 (1), R7C2 (1-2). The other links can carry at most 1, so R7C2 needs at least 2 → exactly 2 bridges.
  16. 15/20
    The 3 at R7C2 needs 3 bridges. Its possible links (bridges allowed): R4C2 (1-2), R7C7 (2). The other links already carry at least 2, so R4C2 can take at most 1 → exactly 1 bridge.
  17. 16/20
    The 4 at R4C2 needs 4 bridges. Its possible links (bridges allowed): R2C2 (1), R4C4 (1-2), R7C2 (1). The other links can carry at most 2, so R4C4 needs at least 2 → exactly 2 bridges.
  18. 17/20
    Putting 1 bridge on R4C4–R4C6 would use up every remaining bridge of the 12 islands in that group, cutting them off from the rest. Everything has to join up, so R4C4–R4C6 gets at most no bridge.
  19. 18/20
    The 3 at R4C4 needs 3 bridges. Its possible links (bridges allowed): R4C2 (2), R6C4 (0-1). The other links can carry at most 2, so R6C4 needs at least 1 → exactly 1 bridge.
  20. 19/20
    The 3 at R4C6 needs 3 bridges. Its possible links (bridges allowed): R2C6 (2), R6C6 (0-1). The other links can carry at most 2, so R6C6 needs at least 1 → exactly 1 bridge.
  21. 20/20
    The 2 at R6C4 needs 2 bridges. Its possible links (bridges allowed): R4C4 (1), R6C6 (1-2). The other links already carry at least 1, so R6C6 can take at most 1 → exactly 1 bridge. ✅ Solved!
Hashi Normal answer as text
Bridges (═ double,, single): R1C1, R1C3, R1C3 ═ R1C7, R1C7, R3C7, R2C2 ═ R2C6, R2C2, R4C2, R2C6 ═ R4C6, R3C7, R5C7, R4C2 ═ R4C4, R4C2, R7C2, R4C4, R6C4, R4C6, R6C6, R5C7, R7C7, R6C4, R6C6, R7C2 ═ R7C7.
Playing the Hard level? Jump to the Hard Hashi answer ↓

Everyday Grids Hashi Hard Answer, Aug 14, 2026

8×8 grid
✅ Solved Hard grid
233434525322252322
233434525322252322
  1. 0/26
    Each 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 26 logical steps, no guessing.
  2. 1/26
    The 3 at R1C8 needs 3 bridges. Its possible links (bridges allowed): R1C3 (0-2), R4C8 (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 R4C8 needs at least 1.
  3. 2/26
    The 4 at R2C2 needs 4 bridges. Its possible links (bridges allowed): R2C6 (0-2), R4C2 (0-2). The other links can carry at most 2, so R2C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R4C2 needs at least 2 → exactly 2 bridges.
  4. 3/26
    There is already a bridge on R2C2–R2C6. Bridges can't cross, so R1C3–R8C3 is now impossible.
  5. 4/26
    The 3 at R1C3 needs 3 bridges. Its possible links (bridges allowed): R1C1 (0-2), R1C8 (1-2). The other links can carry at most 2, so R1C1 needs at least 1.
  6. 5/26
    The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C3 (1-2), R5C1 (0-2). The other links already carry at least 1, so R5C1 can take at most 1.
  7. 6/26
    The 3 at R2C6 needs 3 bridges. Its possible links (bridges allowed): R2C2 (2), R4C6 (0-2). 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. 7/26
    The 2 at R4C8 needs 2 bridges. Its possible links (bridges allowed): R1C8 (1-2), R4C6 (0-2), R6C8 (0-2). The other links already carry at least 1, so R4C6 can take at most 1; the other links already carry at least 1, so R6C8 can take at most 1.
  9. 8/26
    The 5 at R4C6 needs 5 bridges. Its possible links (bridges allowed): R2C6 (1), R4C2 (0-2), R4C8 (0-1), R6C6 (0-2). The other links can carry at most 4, so R4C2 needs at least 1; the other links can carry at most 4, so R6C6 needs at least 1.
  10. 9/26
    The 4 at R4C2 needs 4 bridges. Its possible links (bridges allowed): R2C2 (2), R4C6 (1-2), R7C2 (0-2). The other links already carry at least 3, so R7C2 can take at most 1.
  11. 10/26
    The 5 at R5C1 needs 5 bridges. Its possible links (bridges allowed): R1C1 (0-1), R5C4 (0-2), R8C1 (0-2). The other links can carry at most 4, so R1C1 needs at least 1 → exactly 1 bridge; the other links can carry at most 3, so R5C4 needs at least 2 → exactly 2 bridges; the other links can carry at most 3, so R8C1 needs at least 2 → exactly 2 bridges.
  12. 11/26
    There is already a bridge on R5C1–R5C4. Bridges can't cross, so R4C2–R7C2 is now impossible.
  13. 12/26
    The 2 at R1C1 needs 2 bridges. Its possible links (bridges allowed): R1C3 (1-2), R5C1 (1). The other links already carry at least 1, so R1C3 can take at most 1 → exactly 1 bridge.
  14. 13/26
    The 3 at R1C3 needs 3 bridges. Its possible links (bridges allowed): R1C1 (1), R1C8 (1-2). The other links can carry at most 1, so R1C8 needs at least 2 → exactly 2 bridges.
  15. 14/26
    The 3 at R1C8 needs 3 bridges. Its possible links (bridges allowed): R1C3 (2), R4C8 (1-2). The other links already carry at least 2, so R4C8 can take at most 1 → exactly 1 bridge.
  16. 15/26
    The 4 at R4C2 needs 4 bridges. Its possible links (bridges allowed): R2C2 (2), R4C6 (1-2). The other links can carry at most 2, so R4C6 needs at least 2 → exactly 2 bridges.
  17. 16/26
    The 3 at R5C4 needs 3 bridges. Its possible links (bridges allowed): R5C1 (2), R7C4 (0-2). The other links can carry at most 2, so R7C4 needs at least 1, and the other links already carry at least 2, so R7C4 can take at most 1 → exactly 1 bridge.
  18. 17/26
    The 2 at R6C6 needs 2 bridges. Its possible links (bridges allowed): R4C6 (1-2), R6C8 (0-2). The other links already carry at least 1, so R6C8 can take at most 1.
  19. 18/26
    The 2 at R7C2 needs 2 bridges. Its possible links (bridges allowed): R7C4 (0-2). It has no other link, so R7C4 must carry both → exactly 2 bridges.
  20. 19/26
    The 5 at R7C4 needs 5 bridges. Its possible links (bridges allowed): R5C4 (1), R7C2 (2), R7C7 (0-2). The other links can carry at most 3, so R7C7 needs at least 2 → exactly 2 bridges.
  21. 20/26
    The 3 at R8C1 needs 3 bridges. Its possible links (bridges allowed): R5C1 (2), R8C3 (0-2). The other links can carry at most 2, so R8C3 needs at least 1, and the other links already carry at least 2, so R8C3 can take at most 1 → exactly 1 bridge.
  22. 21/26
    The 2 at R8C3 needs 2 bridges. Its possible links (bridges allowed): R8C1 (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.
  23. 22/26
    The 2 at R8C8 needs 2 bridges. Its possible links (bridges allowed): R6C8 (0-2), R8C3 (1). 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.
  24. 23/26
    Putting 2 bridges on R4C6–R6C6 would use up every remaining bridge of the 5 islands in that group, cutting them off from the rest. Everything has to join up, so R4C6–R6C6 gets at most 1 bridge.
  25. 24/26
    The 5 at R4C6 needs 5 bridges. Its possible links (bridges allowed): R2C6 (1), R4C2 (2), R4C8 (0-1), R6C6 (1). The other links can carry at most 4, so R4C8 needs at least 1 → exactly 1 bridge.
  26. 25/26
    The 2 at R4C8 needs 2 bridges. Its possible links (bridges allowed): R1C8 (1), R4C6 (1), R6C8 (0-1). The other links already carry at least 2, so R6C8 can take none.
  27. 26/26
    The 2 at R6C6 needs 2 bridges. Its possible links (bridges allowed): R4C6 (1), R6C8 (0-1). The other links can carry at most 1, so R6C8 needs at least 1 → exactly 1 bridge. ✅ Solved!
Hashi Hard answer as text
Bridges (═ double,, single): R1C1, R1C3, R1C1, R5C1, R1C3 ═ R1C8, R1C8, R4C8, R2C2 ═ R2C6, R2C2 ═ R4C2, R2C6, R4C6, R4C2 ═ R4C6, R4C6, R4C8, R4C6, R6C6, R5C1 ═ R5C4, R5C1 ═ R8C1, R5C4, R7C4, R6C6, R6C8, R6C8, R8C8, R7C2 ═ R7C4, R7C4 ═ R7C7, R8C1, R8C3, R8C3, R8C8.
↑ Back to the Normal Hashi answer

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 August 14, 2026

ZipGemsLotusPandaLibraSumsHashi

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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 August 14, 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 August 14, 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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