Everyday Grids Hashi Answers October 27, 2026 [Step By Step]

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Everyday Grids Hashi Answer Today: October 27, 2026

● Normal & Hard solved Hashi📅 Tue, Oct 27, 2026
📅This page is for October 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 October 27, 2026 (27 October 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 Oct 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.

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Everyday Grids Hashi Answers and Solutions for October 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).

Normal Answer ↓Hard Answer ↓

Everyday Grids Hashi Normal Answer, Oct 27, 2026

8×8 grid
✅ Solved Normal grid
43333164213443211
43333164213443211
  1. 0/19
    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 19 logical steps, no guessing.
  2. 1/19
    The 4 at R1C1 needs 4 bridges. Its possible links (bridges allowed): R1C4 (0-2), R4C1 (0-2). The other links can carry at most 2, so R1C4 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R4C1 needs at least 2 → exactly 2 bridges.
  3. 2/19
    The 3 at R1C4 needs 3 bridges. Its possible links (bridges allowed): R1C1 (2), R1C8 (0-2), R3C4 (0-1). The other links already carry at least 2, so R1C8 can take at most 1.
  4. 3/19
    The 3 at R1C8 needs 3 bridges. Its possible links (bridges allowed): R1C4 (0-1), R6C8 (0-2). The other links can carry at most 2, so R1C4 needs at least 1 → exactly 1 bridge; the other links can carry at most 1, so R6C8 needs at least 2 → exactly 2 bridges.
  5. 4/19
    The 3 at R1C4 needs 3 bridges. Its possible links (bridges allowed): R1C1 (2), R1C8 (1), R3C4 (0-1). The other links already carry at least 3, so R3C4 can take none.
  6. 5/19
    The 3 at R2C3 needs 3 bridges. Its possible links (bridges allowed): R2C5 (0-2), R4C3 (0-2). The other links can carry at most 2, so R2C5 needs at least 1; the other links can carry at most 2, so R4C3 needs at least 1.
  7. 6/19
    The 3 at R2C5 needs 3 bridges. Its possible links (bridges allowed): R2C3 (1-2), R5C5 (0-1). The other links can carry at most 1, so R2C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R5C5 needs at least 1 → exactly 1 bridge.
  8. 7/19
    There is already a bridge on R2C5–R5C5. Bridges can't cross, so R4C3–R4C6 is now impossible.
  9. 8/19
    The 3 at R2C3 needs 3 bridges. Its possible links (bridges allowed): R2C5 (2), R4C3 (1-2). The other links already carry at least 2, so R4C3 can take at most 1 → exactly 1 bridge.
  10. 9/19
    The 1 at R3C4 needs 1 bridge. Its possible links (bridges allowed): R6C4 (0-1). It has no other link, so R6C4 must carry it → exactly 1 bridge.
  11. 10/19
    The 6 at R4C1 needs 6 bridges. Its possible links (bridges allowed): R1C1 (2), R4C3 (0-2), R7C1 (0-2). The other links can carry at most 4, so R4C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 4, so R7C1 needs at least 2 → exactly 2 bridges.
  12. 11/19
    The 4 at R4C3 needs 4 bridges. Its possible links (bridges allowed): R2C3 (1), R4C1 (2), R7C3 (0-2). The other links can carry at most 3, so R7C3 needs at least 1, and the other links already carry at least 3, so R7C3 can take at most 1 → exactly 1 bridge.
  13. 12/19
    The 2 at R4C6 needs 2 bridges. Its possible links (bridges allowed): R6C6 (0-2). It has no other link, so R6C6 must carry both → exactly 2 bridges.
  14. 13/19
    The 3 at R6C4 needs 3 bridges. Its possible links (bridges allowed): R3C4 (1), R6C6 (0-2), R8C4 (0-1). The other links can carry at most 2, so R6C6 needs at least 1.
  15. 14/19
    The 4 at R6C6 needs 4 bridges. Its possible links (bridges allowed): R4C6 (2), R6C4 (1-2), R6C8 (0-2). The other links already carry at least 3, so R6C8 can take at most 1.
  16. 15/19
    The 4 at R6C8 needs 4 bridges. Its possible links (bridges allowed): R1C8 (2), R6C6 (0-1), R8C8 (0-1). The other links can carry at most 3, so R6C6 needs at least 1 → exactly 1 bridge; the other links can carry at most 3, so R8C8 needs at least 1 → exactly 1 bridge.
  17. 16/19
    The 4 at R6C6 needs 4 bridges. Its possible links (bridges allowed): R4C6 (2), R6C4 (1-2), R6C8 (1). The other links already carry at least 3, so R6C4 can take at most 1 → exactly 1 bridge.
  18. 17/19
    The 3 at R6C4 needs 3 bridges. Its possible links (bridges allowed): R3C4 (1), R6C6 (1), R8C4 (0-1). The other links can carry at most 2, so R8C4 needs at least 1 → exactly 1 bridge.
  19. 18/19
    The 3 at R7C1 needs 3 bridges. Its possible links (bridges allowed): R4C1 (2), R7C3 (0-2). The other links can carry at most 2, so R7C3 needs at least 1, and the other links already carry at least 2, so R7C3 can take at most 1 → exactly 1 bridge.
  20. 19/19
    The 1 at R8C4 needs 1 bridge. Its possible links (bridges allowed): R6C4 (1), R8C8 (0-1). The other links already carry at least 1, so R8C8 can take none. ✅ Solved!
Hashi Normal answer as text
Bridges (═ double,, single): R1C1 ═ R1C4, R1C1 ═ R4C1, R1C4, R1C8, R1C8 ═ R6C8, R2C3 ═ R2C5, R2C3, R4C3, R2C5, R5C5, R3C4, R6C4, R4C1 ═ R4C3, R4C1 ═ R7C1, R4C3, R7C3, R4C6 ═ R6C6, R6C4, R6C6, R6C4, R8C4, R6C6, R6C8, R6C8, R8C8, R7C1, R7C3.
Playing the Hard level? Jump to the Hard Hashi answer ↓

Everyday Grids Hashi Hard Answer, Oct 27, 2026

9×9 grid
✅ Solved Hard grid
223322434344221132441323
223322434344221132441323
  1. 0/38
    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 38 logical steps, no guessing.
  2. 1/38
    The 3 at R2C6 needs 3 bridges. Its possible links (bridges allowed): R2C8 (0-2), R4C6 (0-2). The other links can carry at most 2, so R2C8 needs at least 1; the other links can carry at most 2, so R4C6 needs at least 1.
  3. 2/38
    The 3 at R2C8 needs 3 bridges. Its possible links (bridges allowed): R2C6 (1-2), R4C8 (0-2). The other links can carry at most 2, so R4C8 needs at least 1.
  4. 3/38
    There is already a bridge on R2C8–R4C8. Bridges can't cross, so R3C7–R3C9 is now impossible.
  5. 4/38
    The 2 at R3C7 needs 2 bridges. Its possible links (bridges allowed): R5C7 (0-2). It has no other link, so R5C7 must carry both → exactly 2 bridges.
  6. 5/38
    There is already a bridge on R3C7–R5C7. Bridges can't cross, so R4C6–R4C8 is now impossible.
  7. 6/38
    The 4 at R4C8 needs 4 bridges. Its possible links (bridges allowed): R2C8 (1-2), R8C8 (0-2). The other links can carry at most 2, so R2C8 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R8C8 needs at least 2 → exactly 2 bridges.
  8. 7/38
    There is already a bridge on R4C8–R8C8. Bridges can't cross, so R7C7–R7C9 is now impossible.
  9. 8/38
    The 3 at R2C8 needs 3 bridges. Its possible links (bridges allowed): R2C6 (1-2), R4C8 (2). The other links already carry at least 2, so R2C6 can take at most 1 → exactly 1 bridge.
  10. 9/38
    The 3 at R2C6 needs 3 bridges. Its possible links (bridges allowed): R2C8 (1), R4C6 (1-2). The other links can carry at most 1, so R4C6 needs at least 2 → exactly 2 bridges.
  11. 10/38
    The 3 at R4C6 needs 3 bridges. Its possible links (bridges allowed): R2C6 (2), R4C2 (0-2), R6C6 (0-2). The other links already carry at least 2, so R4C2 can take at most 1; the other links already carry at least 2, so R6C6 can take at most 1.
  12. 11/38
    The 4 at R4C2 needs 4 bridges. Its possible links (bridges allowed): R1C2 (0-2), R4C6 (0-1), R8C2 (0-2). The other links can carry at most 3, so R1C2 needs at least 1; the other links can carry at most 3, so R8C2 needs at least 1.
  13. 12/38
    The 2 at R1C2 needs 2 bridges. Its possible links (bridges allowed): R1C9 (0-2), R4C2 (1-2). The other links already carry at least 1, so R1C9 can take at most 1.
  14. 13/38
    The 2 at R1C9 needs 2 bridges. Its possible links (bridges allowed): R1C2 (0-1), R3C9 (0-2). The other links can carry at most 1, so R3C9 needs at least 1.
  15. 14/38
    The 2 at R3C9 needs 2 bridges. Its possible links (bridges allowed): R1C9 (1-2), R7C9 (0-2). The other links already carry at least 1, so R7C9 can take at most 1.
  16. 15/38
    The 3 at R5C3 needs 3 bridges. Its possible links (bridges allowed): R5C5 (0-2), R7C3 (0-2). The other links can carry at most 2, so R5C5 needs at least 1; the other links can carry at most 2, so R7C3 needs at least 1.
  17. 16/38
    The 4 at R5C5 needs 4 bridges. Its possible links (bridges allowed): R5C3 (1-2), R5C7 (0-2), R7C5 (0-1). The other links can carry at most 3, so R5C7 needs at least 1.
  18. 17/38
    There is already a bridge on R5C5–R5C7. Bridges can't cross, so R4C6–R6C6 is now impossible.
  19. 18/38
    The 3 at R4C6 needs 3 bridges. Its possible links (bridges allowed): R2C6 (2), R4C2 (0-1). The other links can carry at most 2, so R4C2 needs at least 1 → exactly 1 bridge.
  20. 19/38
    The 2 at R6C6 needs 2 bridges. Its possible links (bridges allowed): R8C6 (0-2). It has no other link, so R8C6 must carry both → exactly 2 bridges.
  21. 20/38
    There is already a bridge on R6C6–R8C6. Bridges can't cross, so R7C5–R7C7 is now impossible.
  22. 21/38
    The 2 at R7C3 needs 2 bridges. Its possible links (bridges allowed): R5C3 (1-2), R7C5 (0-1), R9C3 (0-2). The other links already carry at least 1, so R9C3 can take at most 1.
  23. 22/38
    The 3 at R7C9 needs 3 bridges. Its possible links (bridges allowed): R3C9 (0-1), R9C9 (0-2). The other links can carry at most 2, so R3C9 needs at least 1 → exactly 1 bridge; the other links can carry at most 1, so R9C9 needs at least 2 → exactly 2 bridges.
  24. 23/38
    The 2 at R3C9 needs 2 bridges. Its possible links (bridges allowed): R1C9 (1-2), R7C9 (1). The other links already carry at least 1, so R1C9 can take at most 1 → exactly 1 bridge.
  25. 24/38
    The 2 at R1C9 needs 2 bridges. Its possible links (bridges allowed): R1C2 (0-1), R3C9 (1). The other links can carry at most 1, so R1C2 needs at least 1 → exactly 1 bridge.
  26. 25/38
    The 2 at R1C2 needs 2 bridges. Its possible links (bridges allowed): R1C9 (1), R4C2 (1-2). The other links already carry at least 1, so R4C2 can take at most 1 → exactly 1 bridge.
  27. 26/38
    The 4 at R4C2 needs 4 bridges. Its possible links (bridges allowed): R1C2 (1), R4C6 (1), R8C2 (1-2). The other links can carry at most 2, so R8C2 needs at least 2 → exactly 2 bridges.
  28. 27/38
    The 2 at R8C2 needs 2 bridges. Its possible links (bridges allowed): R4C2 (2), R8C6 (0-2). The other links already carry at least 2, so R8C6 can take none.
  29. 28/38
    The 4 at R8C6 needs 4 bridges. Its possible links (bridges allowed): R6C6 (2), R8C8 (0-2). The other links can carry at most 2, so R8C8 needs at least 2 → exactly 2 bridges.
  30. 29/38
    There is already a bridge on R8C6–R8C8. Bridges can't cross, so R7C7–R9C7 is now impossible.
  31. 30/38
    The 1 at R7C7 needs 1 bridge. Its possible links (bridges allowed): R5C7 (0-1). It has no other link, so R5C7 must carry it → exactly 1 bridge.
  32. 31/38
    The 4 at R5C7 needs 4 bridges. Its possible links (bridges allowed): R3C7 (2), R5C5 (1-2), R7C7 (1). The other links already carry at least 3, so R5C5 can take at most 1 → exactly 1 bridge.
  33. 32/38
    The 4 at R5C5 needs 4 bridges. Its possible links (bridges allowed): R5C3 (1-2), R5C7 (1), R7C5 (0-1). The other links can carry at most 2, so R5C3 needs at least 2 → exactly 2 bridges; the other links can carry at most 3, so R7C5 needs at least 1 → exactly 1 bridge.
  34. 33/38
    The 3 at R5C3 needs 3 bridges. Its possible links (bridges allowed): R5C5 (2), R7C3 (1-2). The other links already carry at least 2, so R7C3 can take at most 1 → exactly 1 bridge.
  35. 34/38
    The 1 at R7C5 needs 1 bridge. Its possible links (bridges allowed): R5C5 (1), R7C3 (0-1). The other links already carry at least 1, so R7C3 can take none.
  36. 35/38
    The 2 at R7C3 needs 2 bridges. Its possible links (bridges allowed): R5C3 (1), R9C3 (0-1). The other links can carry at most 1, so R9C3 needs at least 1 → exactly 1 bridge.
  37. 36/38
    The 1 at R9C1 needs 1 bridge. Its possible links (bridges allowed): R9C3 (0-1). It has no other link, so R9C3 must carry it → exactly 1 bridge.
  38. 37/38
    The 3 at R9C3 needs 3 bridges. Its possible links (bridges allowed): R7C3 (1), R9C1 (1), 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.
  39. 38/38
    The 2 at R9C7 needs 2 bridges. Its possible links (bridges allowed): R9C3 (1), R9C9 (0-2). The other links can carry at most 1, so R9C9 needs at least 1, and the other links already carry at least 1, so R9C9 can take at most 1 → exactly 1 bridge. ✅ Solved!
Hashi Hard answer as text
Bridges (═ double,, single): R1C2, R1C9, R1C2, R4C2, R1C9, R3C9, R2C6, R2C8, R2C6 ═ R4C6, R2C8 ═ R4C8, R3C7 ═ R5C7, R3C9, R7C9, R4C2, R4C6, R4C2 ═ R8C2, R4C8 ═ R8C8, R5C3 ═ R5C5, R5C3, R7C3, R5C5, R5C7, R5C5, R7C5, R5C7, R7C7, R6C6 ═ R8C6, R7C3, R9C3, R7C9 ═ R9C9, R8C6 ═ R8C8, R9C1, R9C3, R9C3, R9C7, R9C7, R9C9.
↑ 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 October 27, 2026

ZipGemsLotusPandaLibraSumsHashi

More Days: Everyday Grids Hashi Answers by Date

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← Previous dayEveryday Grids Hashi Oct 26, 2026Next day →Oct 28, 2026 coming soon
📅 Everyday Grids All Game Type Answers Oct 27, 2026🗓️ Everyday Grids Answers October 2026 (All Days)🗂️ Everyday Grids Answers All Days [Choose from Calendar]

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 October 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 October 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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