Everyday Grids Hashi Answer Today: October 19, 2026

This page has the complete Everyday Grids Hashi answers for October 19, 2026 (19 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 19, 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 October 19, 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, Oct 19, 2026
6×6 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): R1C4 (0-2), R3C1 (0-2). The other links can carry at most 2, so R1C4 needs at least 1; the other links can carry at most 2, so R3C1 needs at least 1.
- 2/13The 2 at R5C2 needs 2 bridges. Its possible links (bridges allowed): R5C4 (0-2). It has no other link, so R5C4 must carry both → exactly 2 bridges.
- 3/13The 3 at R5C4 needs 3 bridges. Its possible links (bridges allowed): R1C4 (0-2), R5C2 (2), R5C6 (0-2). The other links already carry at least 2, so R1C4 can take at most 1; the other links already carry at least 2, so R5C6 can take at most 1.
- 4/13The 3 at R5C6 needs 3 bridges. Its possible links (bridges allowed): R3C6 (0-2), R5C4 (0-1). The other links can carry at most 1, so R3C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R5C4 needs at least 1 → exactly 1 bridge.
- 5/13The 3 at R5C4 needs 3 bridges. Its possible links (bridges allowed): R1C4 (0-1), R5C2 (2), R5C6 (1). The other links already carry at least 3, so R1C4 can take none.
- 6/13The 3 at R1C4 needs 3 bridges. Its possible links (bridges allowed): R1C1 (1-2), R1C6 (0-2). The other links can carry at most 2, so R1C6 needs at least 1.
- 7/13The 2 at R1C6 needs 2 bridges. Its possible links (bridges allowed): R1C4 (1-2), R3C6 (0-2). The other links already carry at least 1, so R3C6 can take at most 1.
- 8/13The 4 at R3C6 needs 4 bridges. Its possible links (bridges allowed): R1C6 (0-1), R3C1 (0-2), R5C6 (2). The other links can carry at most 3, so R3C1 needs at least 1.
- 9/13The 2 at R6C1 needs 2 bridges. Its possible links (bridges allowed): R3C1 (0-2). It has no other link, so R3C1 must carry both → exactly 2 bridges.
- 10/13The 4 at R3C1 needs 4 bridges. Its possible links (bridges allowed): R1C1 (1-2), R3C6 (1-2), R6C1 (2). The other links already carry at least 3, so R1C1 can take at most 1 → exactly 1 bridge; the other links already carry at least 3, so R3C6 can take at most 1 → exactly 1 bridge.
- 11/13The 3 at R1C1 needs 3 bridges. Its possible links (bridges allowed): R1C4 (1-2), R3C1 (1). The other links can carry at most 1, so R1C4 needs at least 2 → exactly 2 bridges.
- 12/13The 3 at R1C4 needs 3 bridges. Its possible links (bridges allowed): R1C1 (2), R1C6 (1-2). The other links already carry at least 2, so R1C6 can take at most 1 → exactly 1 bridge.
- 13/13The 2 at R1C6 needs 2 bridges. Its possible links (bridges allowed): R1C4 (1), R3C6 (0-1). The other links can carry at most 1, so R3C6 needs at least 1 → exactly 1 bridge. ✅ Solved!
Hashi Normal answer as text
Everyday Grids Hashi Hard Answer, Oct 19, 2026
7×7 grid- 0/20Each 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.
- 1/20The 4 at R1C7 needs 4 bridges. Its possible links (bridges allowed): R1C4 (0-2), R5C7 (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 R5C7 needs at least 2 → exactly 2 bridges.
- 2/20The 3 at R3C6 needs 3 bridges. Its possible links (bridges allowed): R3C4 (0-2), R6C6 (0-2). The other links can carry at most 2, so R3C4 needs at least 1; the other links can carry at most 2, so R6C6 needs at least 1.
- 3/20The 2 at R3C4 needs 2 bridges. Its possible links (bridges allowed): R1C4 (0-2), R3C2 (0-2), R3C6 (1-2), R6C4 (0-2). The other links already carry at least 1, so R1C4 can take at most 1; the other links already carry at least 1, so R3C2 can take at most 1; the other links already carry at least 1, so R6C4 can take at most 1.
- 4/20The 4 at R1C4 needs 4 bridges. Its possible links (bridges allowed): R1C2 (0-2), R1C7 (2), R3C4 (0-1). The other links can carry at most 3, so R1C2 needs at least 1.
- 5/20The 2 at R1C2 needs 2 bridges. Its possible links (bridges allowed): R1C4 (1-2), R3C2 (0-2). The other links already carry at least 1, so R3C2 can take at most 1.
- 6/20The 3 at R3C2 needs 3 bridges. Its possible links (bridges allowed): R1C2 (0-1), R3C4 (0-1), R6C2 (0-2). The other links can carry at most 2, so R6C2 needs at least 1.
- 7/20The 1 at R4C1 needs 1 bridge. Its possible links (bridges allowed): R7C1 (0-1). It has no other link, so R7C1 must carry it → exactly 1 bridge.
- 8/20The 4 at R5C7 needs 4 bridges. Its possible links (bridges allowed): R1C7 (2), R7C7 (0-2). The other links can carry at most 2, so R7C7 needs at least 2 → exactly 2 bridges.
- 9/20The 3 at R6C2 needs 3 bridges. Its possible links (bridges allowed): R3C2 (1-2), R6C4 (0-2). The other links can carry at most 2, so R6C4 needs at least 1.
- 10/20The 4 at R6C6 needs 4 bridges. Its possible links (bridges allowed): R3C6 (1-2), R6C4 (0-2). The other links can carry at most 2, so R3C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R6C4 needs at least 2 → exactly 2 bridges.
- 11/20The 3 at R3C6 needs 3 bridges. Its possible links (bridges allowed): R3C4 (1-2), R6C6 (2). The other links already carry at least 2, so R3C4 can take at most 1 → exactly 1 bridge.
- 12/20The 3 at R6C4 needs 3 bridges. Its possible links (bridges allowed): R3C4 (0-1), R6C2 (1-2), R6C6 (2). The other links already carry at least 3, so R3C4 can take none; the other links already carry at least 2, so R6C2 can take at most 1 → exactly 1 bridge.
- 13/20The 3 at R6C2 needs 3 bridges. Its possible links (bridges allowed): R3C2 (1-2), R6C4 (1). The other links can carry at most 1, so R3C2 needs at least 2 → exactly 2 bridges.
- 14/20The 2 at R7C1 needs 2 bridges. Its possible links (bridges allowed): R4C1 (1), R7C3 (0-2). The other links can carry at most 1, so R7C3 needs at least 1, and the other links already carry at least 1, so R7C3 can take at most 1 → exactly 1 bridge.
- 15/20The 2 at R7C3 needs 2 bridges. Its possible links (bridges allowed): R7C1 (1), R7C5 (0-2). The other links can carry at most 1, so R7C5 needs at least 1, and the other links already carry at least 1, so R7C5 can take at most 1 → exactly 1 bridge.
- 16/20The 3 at R7C5 needs 3 bridges. Its possible links (bridges allowed): R7C3 (1), R7C7 (0-2). The other links can carry at most 1, so R7C7 needs at least 2 → exactly 2 bridges.
- 17/20Putting 2 bridges on R1C2–R1C4 would use up every remaining bridge of the 9 islands in that group, cutting them off from the rest. Everything has to join up, so R1C2–R1C4 gets at most 1 bridge.
- 18/20The 2 at R1C2 needs 2 bridges. Its possible links (bridges allowed): R1C4 (1), R3C2 (0-1). The other links can carry at most 1, so R3C2 needs at least 1 → exactly 1 bridge.
- 19/20The 4 at R1C4 needs 4 bridges. Its possible links (bridges allowed): R1C2 (1), R1C7 (2), R3C4 (0-1). The other links can carry at most 3, so R3C4 needs at least 1 → exactly 1 bridge.
- 20/20The 3 at R3C2 needs 3 bridges. Its possible links (bridges allowed): R1C2 (1), R3C4 (0-1), R6C2 (2). The other links already carry at least 3, so R3C4 can take none. ✅ Solved!
Hashi Hard answer as text
How to Solve Everyday Grids Hashi: Quick Tips
- 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.