Everyday Grids Answers Today: Every Day, Every Game
● Updated October 26, 20267 gamesNormal & Hard
This page is the home of every Everyday Grids answer: today’s solutions, the last few days for each game, this month’s calendar and a link to every month since November 2024. Whether you need a quick cheat for Zip, Gems, Lotus, Panda, Libra, Sums or Hashi, a hint to get unstuck, or a full solution walkthrough with a step by step explanation, start here. Updated daily, latest answers: October 26, 2026.
About Everyday Grids
Everyday Grids is a relaxing daily logic-puzzle game by Fanatee. Each day brings seven new grids, each in a Normal and a Hard level, and every single one is solvable with pure logic, with no guessing and no luck. It’s a calm daily brain workout with streaks, missions and badges, and it’s free on Android and iPhone.
Every game page shows the solved Normal and Hard grids first, followed by a step by step explanation you can follow move by move. Open today’s answers below, pick a game to see its three most recent days, or choose any date from the calendar further down.
Every day, as soon as the new Normal and Hard puzzles are released.
Which games are covered?
All seven: Zip, Gems, Lotus, Panda, Libra, Sums and Hashi, in both Normal and Hard levels.
How do I find an older puzzle?
Tap a day in this month's calendar, or open any month below it and pick the date.
Do you explain how to solve them?
Yes. Every game page has a Step by Step Explanation that shows each logical move on the grid.
💛
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.
Everyday Grids Hashi Answer Today: October 26, 2026
● Normal & Hard solved Hashi📅 Mon, Oct 26, 2026
📅This page is for October 26, 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 26, 2026 (26 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 26, 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 26, 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).
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 13 logical steps, no guessing.
1/13
The 4 at R1C2 needs 4 bridges. Its possible links (bridges allowed): R1C6 (0-2), R3C2 (0-2). The other links can carry at most 2, so R1C6 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R3C2 needs at least 2 → exactly 2 bridges.
2/13
The 4 at R1C6 needs 4 bridges. Its possible links (bridges allowed): R1C2 (2), R3C6 (0-2). The other links can carry at most 2, so R3C6 needs at least 2 → exactly 2 bridges.
3/13
The 1 at R2C1 needs 1 bridge. Its possible links (bridges allowed): R4C1 (0-1). It has no other link, so R4C1 must carry it → exactly 1 bridge.
4/13
The 3 at R3C2 needs 3 bridges. Its possible links (bridges allowed): R1C2 (2), R3C4 (0-2). The other links can carry at most 2, so R3C4 needs at least 1, and the other links already carry at least 2, so R3C4 can take at most 1 → exactly 1 bridge.
5/13
The 2 at R3C4 needs 2 bridges. Its possible links (bridges allowed): R3C2 (1), R3C6 (0-2), R6C4 (0-2). The other links already carry at least 1, so R3C6 can take at most 1; the other links already carry at least 1, so R6C4 can take at most 1.
6/13
The 3 at R3C6 needs 3 bridges. Its possible links (bridges allowed): R1C6 (2), R3C4 (0-1), R6C6 (0-2). The other links already carry at least 2, so R6C6 can take at most 1.
7/13
The 3 at R4C1 needs 3 bridges. Its possible links (bridges allowed): R2C1 (1), R6C1 (0-2). The other links can carry at most 1, so R6C1 needs at least 2 → exactly 2 bridges.
8/13
The 4 at R6C1 needs 4 bridges. Its possible links (bridges allowed): R4C1 (2), R6C4 (0-2). The other links can carry at most 2, so R6C4 needs at least 2 → exactly 2 bridges.
9/13
The 4 at R6C4 needs 4 bridges. Its possible links (bridges allowed): R3C4 (0-1), R6C1 (2), R6C6 (0-2). The other links can carry at most 3, so R6C6 needs at least 1.
10/13
Putting 1 bridge on R3C4–R3C6 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 R3C4–R3C6 gets at most no bridge.
11/13
The 2 at R3C4 needs 2 bridges. Its possible links (bridges allowed): R3C2 (1), R6C4 (0-1). The other links can carry at most 1, so R6C4 needs at least 1 → exactly 1 bridge.
12/13
The 3 at R3C6 needs 3 bridges. Its possible links (bridges allowed): R1C6 (2), R6C6 (0-1). The other links can carry at most 2, so R6C6 needs at least 1 → exactly 1 bridge.
13/13
The 4 at R6C4 needs 4 bridges. Its possible links (bridges allowed): R3C4 (1), R6C1 (2), R6C6 (1-2). The other links already carry at least 3, so R6C6 can take at most 1 → exactly 1 bridge. ✅ Solved!
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 24 logical steps, no guessing.
1/24
The 4 at R1C4 needs 4 bridges. Its possible links (bridges allowed): R1C2 (0-1), R1C7 (0-2), R3C4 (0-2). The other links can carry at most 3, so R1C7 needs at least 1; the other links can carry at most 3, so R3C4 needs at least 1.
2/24
The 2 at R1C7 needs 2 bridges. Its possible links (bridges allowed): R1C4 (1-2), R4C7 (0-2). The other links already carry at least 1, so R4C7 can take at most 1.
3/24
The 1 at R2C5 needs 1 bridge. Its possible links (bridges allowed): R4C5 (0-1). It has no other link, so R4C5 must carry it → exactly 1 bridge.
4/24
The 3 at R3C4 needs 3 bridges. Its possible links (bridges allowed): R1C4 (1-2), R3C1 (0-2). The other links can carry at most 2, so R3C1 needs at least 1.
5/24
There is already a bridge on R3C1–R3C4. Bridges can't cross, so R1C2–R4C2 is now impossible.
6/24
The 1 at R1C2 needs 1 bridge. Its possible links (bridges allowed): R1C4 (0-1). It has no other link, so R1C4 must carry it → exactly 1 bridge.
7/24
The 2 at R3C1 needs 2 bridges. Its possible links (bridges allowed): R3C4 (1-2), R5C1 (0-2). The other links already carry at least 1, so R5C1 can take at most 1.
8/24
The 4 at R4C7 needs 4 bridges. Its possible links (bridges allowed): R1C7 (0-1), R4C5 (0-2), R6C7 (0-2). The other links can carry at most 3, so R4C5 needs at least 1; the other links can carry at most 3, so R6C7 needs at least 1.
9/24
The 3 at R4C5 needs 3 bridges. Its possible links (bridges allowed): R2C5 (1), R4C2 (0-2), R4C7 (1-2), R6C5 (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 R6C5 can take at most 1.
10/24
The 2 at R4C2 needs 2 bridges. Its possible links (bridges allowed): R4C5 (0-1), R6C2 (0-2). The other links can carry at most 1, so R6C2 needs at least 1.
11/24
The 3 at R5C1 needs 3 bridges. Its possible links (bridges allowed): R3C1 (0-1), R7C1 (0-2). The other links can carry at most 2, so R3C1 needs at least 1 → exactly 1 bridge; the other links can carry at most 1, so R7C1 needs at least 2 → exactly 2 bridges.
12/24
The 2 at R3C1 needs 2 bridges. Its possible links (bridges allowed): R3C4 (1-2), R5C1 (1). The other links already carry at least 1, so R3C4 can take at most 1 → exactly 1 bridge.
13/24
The 3 at R3C4 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.
14/24
The 4 at R1C4 needs 4 bridges. Its possible links (bridges allowed): R1C2 (1), R1C7 (1-2), R3C4 (2). The other links already carry at least 3, so R1C7 can take at most 1 → exactly 1 bridge.
15/24
The 2 at R1C7 needs 2 bridges. Its possible links (bridges allowed): R1C4 (1), R4C7 (0-1). The other links can carry at most 1, so R4C7 needs at least 1 → exactly 1 bridge.
16/24
The 2 at R6C2 needs 2 bridges. Its possible links (bridges allowed): R4C2 (1-2), R6C5 (0-2). The other links already carry at least 1, so R6C5 can take at most 1.
17/24
The 3 at R6C5 needs 3 bridges. Its possible links (bridges allowed): R4C5 (0-1), R6C2 (0-1), R6C7 (0-2). The other links can carry at most 2, so R6C7 needs at least 1.
18/24
The 4 at R6C7 needs 4 bridges. Its possible links (bridges allowed): R4C7 (1-2), R6C5 (1-2). The other links can carry at most 2, so R4C7 needs at least 2 → exactly 2 bridges; the other links can carry at most 2, so R6C5 needs at least 2 → exactly 2 bridges.
19/24
The 4 at R4C7 needs 4 bridges. Its possible links (bridges allowed): R1C7 (1), R4C5 (1-2), R6C7 (2). The other links already carry at least 3, so R4C5 can take at most 1 → exactly 1 bridge.
20/24
The 4 at R7C1 needs 4 bridges. Its possible links (bridges allowed): R5C1 (2), R7C6 (0-2). The other links can carry at most 2, so R7C6 needs at least 2 → exactly 2 bridges.
21/24
Putting 2 bridges on R4C2–R6C2 would use up every remaining bridge of the 2 islands in that group, cutting them off from the rest. Everything has to join up, so R4C2–R6C2 gets at most 1 bridge.
22/24
The 2 at R4C2 needs 2 bridges. Its possible links (bridges allowed): R4C5 (0-1), R6C2 (1). The other links can carry at most 1, so R4C5 needs at least 1 → exactly 1 bridge.
23/24
The 3 at R4C5 needs 3 bridges. Its possible links (bridges allowed): R2C5 (1), R4C2 (1), R4C7 (1), R6C5 (0-1). The other links already carry at least 3, so R6C5 can take none.
24/24
The 2 at R6C2 needs 2 bridges. Its possible links (bridges allowed): R4C2 (1), R6C5 (0-1). The other links can carry at most 1, so R6C5 needs at least 1 → exactly 1 bridge. ✅ Solved!
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 26, 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 26, 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.
💛
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.
Everyday Grids Sums Answer Today: October 26, 2026
● Normal & Hard solved Sums📅 Mon, Oct 26, 2026
📅This page is for October 26, 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 Sums answers for October 26, 2026 (26 October 2026): the solved grids for both the Normal and Hard Sums 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 Sums guide? It’s all below, and each Everyday Grids Sums solution is verified against the puzzle rules, so it matches the Oct 26, 2026 puzzle in the app exactly.
About Everyday Grids Sums
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 Sums is one of them. Remove numbers so the ones left match every row total, column total and colour-area total. Every level can be solved with pure logic, with no guessing and no luck, and a brand-new Normal and Hard Sums puzzle is released every day. The game is free to download on Android and iPhone.
Everyday Grids Sums Answers and Solutions for October 26, 2026
Each level below shows the solved Sums 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).
Keep the numbers that add up to every row total (left), column total (top) and colour-area total (small tag). Tap Start to follow all 9 logical steps, no guessing.
1/9
Row 5 must total 10 using 7, 7, 1, 1, 9. The ways to make 10 are: 1+9 (some using different copies of the same number). Every one uses R5C5, so keep it; none uses R5C1, R5C2, so remove them.
2/9
Column 1 must total 13 using 6, 6, 5, 7. The ways to make 13 are: 6+7 (some using different copies of the same number). Every one uses R4C1, so keep it; none uses R3C1, so remove it.
3/9
Row 3 must total 15 using 7, 6, 3, 6. The only way to make 15 is 6 + 3 + 6. So keep R3C3, R3C4, R3C5; remove R3C2.
4/9
Column 5 must total 20; the kept 6+9 = 15 so far, so the undecided numbers (1, 7, 4) must add up to 5. The only way to make 5 is 1 + 4. So keep R1C5, R4C5; remove R2C5.
5/9
Row 2 must total 18 using 6, 9, 3, 5. The only way to make 18 is 6 + 9 + 3. So keep R2C1, R2C2, R2C3; remove R2C4.
6/9
Column 3 must total 11; the kept 3+6 = 9 so far, so the undecided numbers (1, 4, 1) must add up to 2. The only way to make 2 is 1 + 1. So keep R1C3, R5C3; remove R4C3.
7/9
Row 1 must total 10; the kept 1+1 = 2 so far, so the undecided numbers (6, 9, 8) must add up to 8. The only way to make 8 is 8. So keep R1C4; remove R1C1, R1C2.
8/9
Column 4 must total 11; the kept 8+3 = 11 so far, so the undecided numbers (2, 1) must add up to 0. Nothing more is needed. Remove R4C4, R5C4.
9/9
Row 4 must total 13; the kept 7+4 = 11 so far, so the undecided numbers (2) must add up to 2. The only way to make 2 is 2. So keep R4C2. ✅ Solved!
Keep the numbers that add up to every row total (left), column total (top) and colour-area total (small tag). Tap Start to follow all 14 logical steps, no guessing.
1/14
Row 6 must total 13 using 6, 4, 3, 8, 8, 8, 4. The ways to make 13 are: 6+4+3, 6+3+4. Every one uses R6C1, R6C3, so keep them; none uses R6C4, R6C5, R6C6, so remove them.
2/14
Yellow area must total 15 using 6, 4, 8, 3. The only way to make 15 is 4 + 8 + 3. So keep R6C7, R7C6, R7C7; remove R5C6.
3/14
Row 7 must total 27; the kept 8+3 = 11 so far, so the undecided numbers (2, 6, 2, 6, 7) must add up to 16. The only way to make 16 is 2 + 6 + 2 + 6. So keep R7C1, R7C2, R7C3, R7C4; remove R7C5.
4/14
Column 3 must total 18; the kept 3+2 = 5 so far, so the undecided numbers (8, 5, 3, 6, 6) must add up to 13. The only way to make 13 is 8 + 5. So keep R1C3, R2C3; remove R3C3, R4C3, R5C3.
5/14
Row 3 must total 22 using 7, 1, 2, 8, 7, 9. The only way to make 22 is 7 + 8 + 7. So keep R3C1, R3C5, R3C6; remove R3C2, R3C4, R3C7.
6/14
Sky-blue area must total 21; the kept 8 = 8 so far, so the undecided numbers (7, 5, 6, 4) must add up to 13. The only way to make 13 is 7 + 6. So keep R1C1, R1C4; remove R1C2, R2C4.
7/14
Teal area must total 11; the kept 6+3+2 = 11 so far, so the undecided numbers (5, 2, 4) must add up to 0. Nothing more is needed. Remove R4C2, R5C2, R6C2.
8/14
Row 5 must total 11 using 4, 9, 7, 1. The only way to make 11 is 4 + 7. So keep R5C1, R5C5; remove R5C4, R5C7.
9/14
Column 5 must total 17; the kept 8+7 = 15 so far, so the undecided numbers (1, 9, 1) must add up to 2. The only way to make 2 is 1 + 1. So keep R1C5, R4C5; remove R2C5.
10/14
Column 7 must total 13; the kept 4+3 = 7 so far, so the undecided numbers (1, 4, 6) must add up to 6. The only way to make 6 is 6. So keep R4C7; remove R1C7, R2C7.
11/14
Row 4 must total 8; the kept 1+6 = 7 so far, so the undecided numbers (6, 6, 1) must add up to 1. The only way to make 1 is 1. So keep R4C6; remove R4C1, R4C4.
12/14
Green area must total 16; the kept 5 = 5 so far, so the undecided numbers (2, 9) must add up to 11. The only way to make 11 is 2 + 9. So keep R2C1, R2C2.
13/14
Row 1 must total 24; the kept 7+8+6+1 = 22 so far, so the undecided numbers (2) must add up to 2. The only way to make 2 is 2. So keep R1C6.
14/14
Row 2 must total 16; the kept 2+9+5 = 16 so far, so the undecided numbers (2) must add up to 0. Nothing more is needed. Remove R2C6. ✅ Solved!
Remove numbers so the ones left match every row total, column total and colour-area total.
Is there a Normal and a Hard Sums puzzle every day?
Yes. Everyday Grids releases a Normal and a Hard Sums level every day, and both answers and solutions are on this page.
Are these Everyday Grids Sums answers for October 26, 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 Sums puzzle looks different. What should I do?
Check the date first: this page is for October 26, 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.
💛
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.
Everyday Grids Libra Answer Today: October 26, 2026
● Normal & Hard solved Libra📅 Mon, Oct 26, 2026
📅This page is for October 26, 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 Libra answers for October 26, 2026 (26 October 2026): the solved grids for both the Normal and Hard Libra 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 Libra guide? It’s all below, and each Everyday Grids Libra solution is verified against the puzzle rules, so it matches the Oct 26, 2026 puzzle in the app exactly.
About Everyday Grids Libra
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 Libra is one of them. Fill each cell with a flame or an ice cube: 3 of each per row and column, never three alike in a row, no two identical rows or columns. = means same, × means different. Every level can be solved with pure logic, with no guessing and no luck, and a brand-new Normal and Hard Libra puzzle is released every day. The game is free to download on Android and iPhone.
Everyday Grids Libra Answers and Solutions for October 26, 2026
Each level below shows the solved Libra 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).
Each row and column needs 3 🔥 and 3 🧊, never three of the same in a row, and no two rows (or columns) can be identical. = means same, × means different. Tap Start to follow all 25 logical steps, no guessing.
1/25
The × sign between R1C1 and R2C1 means they are opposite. R1C1 is 🧊, so R2C1 is 🔥.
2/25
The × sign between R2C6 and R1C6 means they are opposite. R2C6 is 🔥, so R1C6 is 🧊.
3/25
Two 🔥 sit side by side in column 4 (R5C4, R6C4). A third would make three in a row, so the cell next to them gets 🧊.
4/25
R2C6 and R4C6 are both 🔥 with a gap between them in column 6. Filling the gap with 🔥 would make three in a row, so it must be 🧊.
5/25
Two 🧊 sit side by side in row 3 (R3C5, R3C6). A third would make three in a row, so the cell next to them gets 🔥.
6/25
Column 4 already has its 3 🔥, so every empty cell left in it is 🧊.
7/25
The × sign between R2C4 and R2C3 means they are opposite. R2C4 is 🧊, so R2C3 is 🔥.
8/25
The × sign between R2C3 and R2C2 means they are opposite. R2C3 is 🔥, so R2C2 is 🧊.
9/25
R1C4 and R1C6 are both 🧊 with a gap between them in row 1. Filling the gap with 🧊 would make three in a row, so it must be 🔥.
10/25
Row 1 already has its 3 🧊, so every empty cell left in it is 🔥.
11/25
Row 2 already has its 3 🔥, so every empty cell left in it is 🧊.
12/25
Two 🔥 sit side by side in column 3 (R1C3, R2C3). A third would make three in a row, so the cell next to them gets 🧊.
13/25
The = sign between R3C3 and R4C3 means they are the same. R3C3 is 🧊, so R4C3 is 🧊.
14/25
The × sign between R4C3 and R5C3 means they are opposite. R4C3 is 🧊, so R5C3 is 🔥.
15/25
Row 3 already has its 3 🧊, so every empty cell left in it is 🔥.
16/25
The × sign between R3C1 and R4C1 means they are opposite. R3C1 is 🔥, so R4C1 is 🧊.
17/25
Two 🧊 sit side by side in row 4 (R4C3, R4C4). A third would make three in a row, so the cells on both ends get 🔥.
18/25
Two 🔥 sit side by side in row 5 (R5C3, R5C4). A third would make three in a row, so the cells on both ends get 🧊.
19/25
Column 2 already has its 3 🔥, so every empty cell left in it is 🧊.
20/25
Column 3 already has its 3 🔥, so every empty cell left in it is 🧊.
21/25
Two 🧊 sit side by side in row 6 (R6C2, R6C3). A third would make three in a row, so the cell next to them gets 🔥.
22/25
Column 1 already has its 3 🔥, so every empty cell left in it is 🧊.
23/25
Row 5 already has its 3 🧊, so every empty cell left in it is 🔥.
24/25
Column 5 already has its 3 🧊, so every empty cell left in it is 🔥.
25/25
Two 🔥 sit side by side in row 6 (R6C4, R6C5). A third would make three in a row, so the cell next to them gets 🧊. ✅ Solved!
Each row and column needs 3 🔥 and 3 🧊, never three of the same in a row, and no two rows (or columns) can be identical. = means same, × means different. Tap Start to follow all 25 logical steps, no guessing.
1/25
The × sign between R4C4 and R3C4 means they are opposite. R4C4 is 🧊, so R3C4 is 🔥.
2/25
The × sign between R6C2 and R6C3 means they are opposite. R6C2 is 🔥, so R6C3 is 🧊.
3/25
R2C1 and R2C3 are both 🔥 with a gap between them in row 2. Filling the gap with 🔥 would make three in a row, so it must be 🧊.
4/25
Try every way to fill row 2 with 3 🔥 and 3 🧊, no three in a row, obeying its = / × signs. Only 2 fit. They are: 🔥🧊🔥🧊🔥🧊 · 🔥🧊🔥🧊🧊🔥. All of them agree here: R2C4 → 🧊.
5/25
Try every way to fill row 3 with 3 🔥 and 3 🧊, no three in a row, obeying its = / × signs. Only 2 fit. They are: 🔥🔥🧊🔥🧊🧊 · 🔥🧊🔥🔥🧊🧊. All of them agree here: R3C1 → 🔥; R3C5, R3C6 → 🧊.
6/25
Two 🔥 sit side by side in column 1 (R2C1, R3C1). A third would make three in a row, so the cells on both ends get 🧊.
7/25
The × sign between R4C1 and R5C1 means they are opposite. R4C1 is 🧊, so R5C1 is 🔥.
8/25
Column 1 already has its 3 🔥, so every empty cell left in it is 🧊.
9/25
Try every way to fill row 6 with 3 🔥 and 3 🧊, no three in a row, obeying its = / × signs. Only 1 fits. They are: 🧊🔥🧊🧊🔥🔥. All of them agree here: R6C5, R6C6 → 🔥; R6C4 → 🧊.
10/25
R4C4 and R6C4 are both 🧊 with a gap between them in column 4. Filling the gap with 🧊 would make three in a row, so it must be 🔥.
11/25
Column 4 already has its 3 🧊, so every empty cell left in it is 🔥.
12/25
Try every way to fill row 5 with 3 🔥 and 3 🧊, no three in a row, obeying its = / × signs. Only 2 fit. They are: 🔥🧊🧊🔥🔥🧊 · 🔥🧊🧊🔥🧊🔥. All of them agree here: R5C2, R5C3 → 🧊.
13/25
Two 🧊 sit side by side in column 3 (R5C3, R6C3). A third would make three in a row, so the cell next to them gets 🔥.
14/25
R2C3 and R4C3 are both 🔥 with a gap between them in column 3. Filling the gap with 🔥 would make three in a row, so it must be 🧊.
15/25
Row 3 already has its 3 🧊, so every empty cell left in it is 🔥.
16/25
Column 3 already has its 3 🧊, so every empty cell left in it is 🔥.
17/25
Two 🔥 sit side by side in row 1 (R1C3, R1C4). A third would make three in a row, so the cells on both ends get 🧊.
18/25
Row 1 already has its 3 🧊, so every empty cell left in it is 🔥.
19/25
Column 2 already has its 3 🧊, so every empty cell left in it is 🔥.
20/25
R1C5 and R3C5 are both 🧊 with a gap between them in column 5. Filling the gap with 🧊 would make three in a row, so it must be 🔥.
21/25
The × sign between R2C5 and R2C6 means they are opposite. R2C5 is 🔥, so R2C6 is 🧊.
22/25
Two 🧊 sit side by side in column 6 (R2C6, R3C6). A third would make three in a row, so the cell next to them gets 🔥.
23/25
Row 4 already has its 3 🔥, so every empty cell left in it is 🧊.
24/25
Two 🧊 sit side by side in column 5 (R3C5, R4C5). A third would make three in a row, so the cell next to them gets 🔥.
25/25
Two 🔥 sit side by side in row 5 (R5C4, R5C5). A third would make three in a row, so the cell next to them gets 🧊. ✅ Solved!
Fill each cell with a flame or an ice cube: 3 of each per row and column, never three alike in a row, no two identical rows or columns. = means same, × means different.
Is there a Normal and a Hard Libra puzzle every day?
Yes. Everyday Grids releases a Normal and a Hard Libra level every day, and both answers and solutions are on this page.
Are these Everyday Grids Libra answers for October 26, 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 Libra puzzle looks different. What should I do?
Check the date first: this page is for October 26, 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.
💛
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.
Everyday Grids Panda Answer Today: October 26, 2026
● Normal & Hard solved Panda📅 Mon, Oct 26, 2026
📅This page is for October 26, 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 Panda answers for October 26, 2026 (26 October 2026): the solved grids for both the Normal and Hard Panda 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 Panda guide? It’s all below, and each Everyday Grids Panda solution is verified against the puzzle rules, so it matches the Oct 26, 2026 puzzle in the app exactly.
About Everyday Grids Panda
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 Panda is one of them. Give every bamboo its own panda directly above, below, left or right of it. Pandas never touch, and the edge numbers count pandas per row and column. Every level can be solved with pure logic, with no guessing and no luck, and a brand-new Normal and Hard Panda puzzle is released every day. The game is free to download on Android and iPhone.
Everyday Grids Panda Answers and Solutions for October 26, 2026
Each level below shows the solved Panda 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).
Each bamboo needs its own panda right next to it (up, down, left or right). Pandas never touch, and the edge numbers count pandas per row and column. Tap Start to follow all 9 logical steps, no guessing.
1/9
Every panda must sit directly above, below, left or right of a bamboo. These 11 cells have no bamboo next to them, so they are grass.
2/9
Row 1 still needs 2 more pandas. Pandas in a line can't sit side by side, giving 3 ways to place them, but 2 of those would break a rule elsewhere (a bamboo left with no spot, a row/column overfilled or short, or pandas touching), leaving only 1 way. Every one of them uses R1C1, R1C3, so those are pandas. Pandas never touch, not even diagonally, so the 2 cells around them become grass.
3/9
Row 1 already has its 2 pandas, so the remaining cells (R1C5) are grass.
4/9
Row 2 needs 1 panda in total and has exactly 1 open cell left for the missing one, so R2C5 must be a panda. Pandas never touch, not even diagonally, so the 1 cell around it becomes grass.
5/9
Column 4 needs 1 panda in total and has exactly 1 open cell left for the missing one, so R5C4 must be a panda. Pandas never touch, not even diagonally, so the 3 cells around it become grass.
6/9
Row 4 needs 1 panda in total and has exactly 1 open cell left for the missing one, so R4C1 must be a panda. Pandas never touch, not even diagonally, so the 2 cells around it become grass.
7/9
Row 3 needs 1 panda in total and has exactly 1 open cell left for the missing one, so R3C3 must be a panda.
8/9
Row 5 needs 2 pandas in total and has exactly 1 open cell left for the missing one, so R5C6 must be a panda.
9/9
Row 6 needs 1 panda in total and has exactly 1 open cell left for the missing one, so R6C2 must be a panda. ✅ Solved!
Each bamboo needs its own panda right next to it (up, down, left or right). Pandas never touch, and the edge numbers count pandas per row and column. Tap Start to follow all 12 logical steps, no guessing.
1/12
Every panda must sit directly above, below, left or right of a bamboo. These 10 cells have no bamboo next to them, so they are grass.
2/12
Row 1 still needs 2 more pandas. Pandas in a line can't sit side by side, giving 5 ways to place them, but 3 of those would break a rule elsewhere (a bamboo left with no spot, a row/column overfilled or short, or pandas touching), leaving 2 ways. Every one of them uses R1C4, so that's a panda. Pandas never touch, not even diagonally, so the 3 cells around it become grass.
3/12
Row 6 still needs 2 more pandas. Pandas in a line can't sit side by side, giving 8 ways to place them, but 6 of those would break a rule elsewhere (a bamboo left with no spot, a row/column overfilled or short, or pandas touching), leaving 2 ways. Every one of them uses R6C7, so that's a panda. Pandas never touch, not even diagonally, so the 2 cells around it become grass.
4/12
Column 6 needs 2 pandas in total and has exactly 2 open cells left for the missing ones, so R1C6, R4C6 must all be pandas. Pandas never touch, not even diagonally, so the 4 cells around them become grass.
5/12
Row 1 already has its 2 pandas, so the remaining cells (R1C1) are grass.
6/12
Row 2 needs 1 panda in total and has exactly 1 open cell left for the missing one, so R2C2 must be a panda. Pandas never touch, not even diagonally, so the 1 cell around it becomes grass.
7/12
Row 3 needs 1 panda in total and has exactly 1 open cell left for the missing one, so R3C4 must be a panda. Pandas never touch, not even diagonally, so the 2 cells around it become grass.
8/12
Row 4 already has its 1 panda, so the remaining cells (R4C1) are grass.
9/12
Column 4 already has its 2 pandas, so the remaining cells (R6C4, R7C4) are grass.
10/12
Row 7 needs 2 pandas in total and has exactly 2 open cells left for the missing ones, so R7C1, R7C3 must all be pandas. Pandas never touch, not even diagonally, so the 2 cells around them become grass.
11/12
Row 6 needs 2 pandas in total and has exactly 1 open cell left for the missing one, so R6C5 must be a panda.
12/12
Column 2 needs 2 pandas in total and has exactly 1 open cell left for the missing one, so R5C2 must be a panda. Pandas never touch, not even diagonally, so the 1 cell around it becomes grass. ✅ Solved!
Give every bamboo its own panda directly above, below, left or right of it. Pandas never touch, and the edge numbers count pandas per row and column.
Is there a Normal and a Hard Panda puzzle every day?
Yes. Everyday Grids releases a Normal and a Hard Panda level every day, and both answers and solutions are on this page.
Are these Everyday Grids Panda answers for October 26, 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 Panda puzzle looks different. What should I do?
Check the date first: this page is for October 26, 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.
💛
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.
Everyday Grids Lotus Answer Today: October 26, 2026
● Normal & Hard solved Lotus📅 Mon, Oct 26, 2026
📅This page is for October 26, 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 Lotus answers for October 26, 2026 (26 October 2026): the solved grids for both the Normal and Hard Lotus 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 Lotus guide? It’s all below, and each Everyday Grids Lotus solution is verified against the puzzle rules, so it matches the Oct 26, 2026 puzzle in the app exactly.
About Everyday Grids Lotus
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 Lotus is one of them. Place one lotus in every row, every column and every colour zone. Lotuses can never touch, not even diagonally. Every level can be solved with pure logic, with no guessing and no luck, and a brand-new Normal and Hard Lotus puzzle is released every day. The game is free to download on Android and iPhone.
Everyday Grids Lotus Answers and Solutions for October 26, 2026
Each level below shows the solved Lotus 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).
Each row, column and colour zone needs exactly one lotus, and lotuses can't touch, not even diagonally. Tap Start to follow all 19 logical steps, no guessing.
1/19
The purple zone can only put its lotus in R5C1, R6C1, R7C1. Whichever of those it uses, the lotus would share a row or column with or touch each of R1C1, R2C1, R3C1, R4C1, R6C2. So none of them can hold a lotus.
2/19
Row 2 can only put its lotus in R2C2, R2C3, R2C4, R2C5, R2C6, R2C7. Whichever of those it uses, the lotus would share a column or zone with or touch R3C6. So that cell can't hold a lotus.
3/19
The green zone can only put its lotus in R3C3, R4C2, R4C3, R5C2. Whichever of those it uses, the lotus would share a row or column with or touch each of R3C2, R5C3. So none of them can hold a lotus.
4/19
The sky-blue zone can only put its lotus in R7C2, R7C3. Whichever of those it uses, the lotus would share a row or column with or touch each of R6C3, R7C1, R7C4, R7C5, R7C6, R7C7. So none of them can hold a lotus.
5/19
Column 1 can only put its lotus in R5C1, R6C1. Whichever of those it uses, the lotus would share a row with or touch R5C2. So that cell can't hold a lotus.
6/19
Column 5 can only put its lotus in R1C5, R2C5, R3C5, R4C5, R5C5, R6C5. Whichever of those it uses, the lotus would share a row or zone with or touch R5C6. So that cell can't hold a lotus.
7/19
The green zone can only put its lotus in R3C3, R4C2, R4C3. Whichever of those it uses, the lotus would share a row with or touch R4C4. So that cell can't hold a lotus.
8/19
The brown zone can only put its lotus in R4C7, R5C7, R6C7. Whichever of those it uses, the lotus would share a column with each of R1C7, R2C7, R3C7. So none of them can hold a lotus.
9/19
Row 1 can only put its lotus in R1C2, R1C3, R1C4, R1C5, R1C6. Whichever of those it uses, the lotus would share a column or zone with or touch each of R2C5, R4C6. So none of them can hold a lotus.
10/19
Row 2 can only put its lotus in R2C2, R2C3, R2C4, R2C6. Whichever of those it uses, the lotus would share a zone with or touch each of R1C5, R3C5. So none of them can hold a lotus.
11/19
Row 3 can only put its lotus in R3C3, R3C4. Whichever of those it uses, the lotus would share a column with or touch each of R2C3, R2C4, R4C3. So none of them can hold a lotus.
12/19
Column 5 can only put its lotus in R4C5, R5C5, R6C5. Whichever of those it uses, the lotus would share a row or zone with or touch each of R3C4, R5C4. So none of them can hold a lotus.
13/19
Only one spot is left in row 3, so its lotus goes at R3C3. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (4 cells marked ✕).
14/19
Only one spot is left in row 2, so its lotus goes at R2C6. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
15/19
Only one spot is left in row 7, so its lotus goes at R7C2. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
16/19
Only one spot is left in row 1, so its lotus goes at R1C4. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
17/19
Only one spot is left in row 4, so its lotus goes at R4C7. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
18/19
Only one spot is left in row 6, so its lotus goes at R6C5. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (1 cells marked ✕).
19/19
Only one spot is left in row 5, so its lotus goes at R5C1. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (0 cells marked ✕). ✅ Solved!
Each row, column and colour zone needs exactly one lotus, and lotuses can't touch, not even diagonally. Tap Start to follow all 13 logical steps, no guessing.
1/13
Only one spot is left in the yellow zone, so its lotus goes at R5C3. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (18 cells marked ✕).
2/13
Only one spot is left in the teal zone, so its lotus goes at R8C2. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (11 cells marked ✕).
3/13
Column 1 can only put its lotus in R1C1, R2C1, R3C1, R4C1, R6C1. Whichever of those it uses, the lotus would share a zone with each of R7C4, R7C5. So none of them can hold a lotus.
4/13
Row 7 can only put its lotus in R7C6, R7C7, R7C8. Whichever of those it uses, the lotus would share a column with or touch R6C7. So that cell can't hold a lotus.
5/13
Column 4 can only put its lotus in R1C4, R2C4, R3C4. Whichever of those it uses, the lotus would share a row or zone with each of R1C5, R1C6, R2C5, R2C6, R2C7, R4C5. So none of them can hold a lotus.
6/13
Column 5 can only put its lotus in R3C5, R6C5. Whichever of those it uses, the lotus would share a row or zone with each of R3C6, R4C6. So none of them can hold a lotus.
7/13
Column 6 can only put its lotus in R6C6, R7C6. Whichever of those it uses, the lotus would share a row with or touch each of R6C5, R7C7. So none of them can hold a lotus.
8/13
Only one spot is left in column 5, so its lotus goes at R3C5. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (5 cells marked ✕).
9/13
Only one spot is left in column 4, so its lotus goes at R1C4. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (3 cells marked ✕).
10/13
Only one spot is left in column 7, so its lotus goes at R4C7. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (4 cells marked ✕).
11/13
Only one spot is left in row 6, so its lotus goes at R6C1. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (1 cells marked ✕).
12/13
Only one spot is left in row 2, so its lotus goes at R2C8. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (1 cells marked ✕).
13/13
Only one spot is left in row 7, so its lotus goes at R7C6. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (0 cells marked ✕). ✅ Solved!
Place one lotus in every row, every column and every colour zone. Lotuses can never touch, not even diagonally.
Is there a Normal and a Hard Lotus puzzle every day?
Yes. Everyday Grids releases a Normal and a Hard Lotus level every day, and both answers and solutions are on this page.
Are these Everyday Grids Lotus answers for October 26, 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 Lotus puzzle looks different. What should I do?
Check the date first: this page is for October 26, 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.
💛
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.
Everyday Grids Gems Answer Today: October 26, 2026
● Normal & Hard solved Gems📅 Mon, Oct 26, 2026
📅This page is for October 26, 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 Gems answers for October 26, 2026 (26 October 2026): the solved grids for both the Normal and Hard Gems 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 Gems guide? It’s all below, and each Everyday Grids Gems solution is verified against the puzzle rules, so it matches the Oct 26, 2026 puzzle in the app exactly.
About Everyday Grids Gems
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 Gems is one of them. Split the grid into rectangles (gems). Each gem holds exactly one number, and that number is its size in cells. Every level can be solved with pure logic, with no guessing and no luck, and a brand-new Normal and Hard Gems puzzle is released every day. The game is free to download on Android and iPhone.
Everyday Grids Gems Answers and Solutions for October 26, 2026
Each level below shows the solved Gems 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).
Every number needs one rectangle of exactly that many cells, and rectangles can't overlap or contain two numbers. Tap Start to follow all 17 logical steps, no guessing.
1/17
Draw a 6 wide × 2 tall rectangle for the 12 at R7C5 . Of the 24 rectangles of area 12 that could hold it, only this one still works, of the others, 23 contain another number.
2/17
No other number has a rectangle that can cover R1C6, R2C6, so they all must belong to the 12 at R2C6 . That narrows it to 2 possible shapes.
3/17
No other number has a rectangle that can cover R1C7, R2C7, so they all must belong to the 4 at R2C7 . That narrows it to 2 possible shapes.
4/17
No other number has a rectangle that can cover R3C1, R4C1, R5C1, so they all must belong to the 6 at R4C1 . That narrows it to 2 possible shapes.
5/17
No other number has a rectangle that can cover R3C8, R4C8, R6C8, R7C8, so they all must belong to the 6 at R4C8 . That narrows it to 2 possible shapes.
6/17
No other number has a rectangle that can cover R1C8, so it must belong to the 4 at R2C7 . That leaves it only one possible shape.
7/17
Draw a 2 wide × 2 tall rectangle for the 4 at R2C7 . Of the 8 rectangles of area 4 that could hold it, only this one still works, of the others, 4 contain another number, 2 overlap cells claimed by another number, 1 misses cells this number must cover.
8/17
Draw a 1 wide × 6 tall rectangle for the 6 at R4C8 . Of the 9 rectangles of area 6 that could hold it, only this one still works, of the others, 5 contain another number, 1 misses cells this number must cover, 2 overlap a finished rectangle.
9/17
Draw a 1 wide × 2 tall rectangle for the 2 at R8C7 . Of the 3 rectangles of area 2 that could hold it, only this one still works, of the others, 2 overlap a finished rectangle.
10/17
No other number has a rectangle that can cover R4C3, R4C4, R4C6, R5C3, R5C4, so they all must belong to the 12 at R4C4 . That leaves it only one possible shape.
11/17
Draw a 4 wide × 3 tall rectangle for the 12 at R4C4 . Of the 36 rectangles of area 12 that could hold it, only this one still works, of the others, 4 overlap cells claimed by another number, 8 miss cells this number must cover, 3 overlap a finished rectangle, 20 contain another number.
12/17
Draw a 2 wide × 1 tall rectangle for the 2 at R6C6 . Of the 4 rectangles of area 2 that could hold it, only this one still works, of the others, 1 contains another number, 2 overlap a finished rectangle.
13/17
Draw a 1 wide × 3 tall rectangle for the 3 at R5C7 . Of the 5 rectangles of area 3 that could hold it, only this one still works, of the others, 4 overlap a finished rectangle.
14/17
Draw a 3 wide × 1 tall rectangle for the 3 at R6C5 . Of the 6 rectangles of area 3 that could hold it, only this one still works, of the others, 3 contain another number, 2 overlap a finished rectangle.
15/17
Draw a 6 wide × 2 tall rectangle for the 12 at R2C6 . Of the 22 rectangles of area 12 that could hold it, only this one still works, of the others, 13 overlap a finished rectangle, 1 overlaps cells already claimed by another number, 7 contain another number.
16/17
Draw a 2 wide × 3 tall rectangle for the 6 at R4C1 . Of the 9 rectangles of area 6 that could hold it, only this one still works, of the others, 7 overlap a finished rectangle, 1 contains another number.
17/17
Draw a 2 wide × 1 tall rectangle for the 2 at R6C2 . Of the 4 rectangles of area 2 that could hold it, only this one still works, of the others, 3 overlap a finished rectangle. ✅ Solved!
Gems Normal answer as text
12: R1C1 to R2C6 (6×2); 4: R1C7 to R2C8 (2×2); 6: R3C1 to R5C2 (2×3); 12: R3C3 to R5C6 (4×3); 3: R3C7 to R5C7 (1×3); 6: R3C8 to R8C8 (1×6); 2: R6C1 to R6C2 (2×1); 3: R6C3 to R6C5 (3×1); 2: R6C6 to R6C7 (2×1); 12: R7C1 to R8C6 (6×2); 2: R7C7 to R8C7 (1×2).
Every number needs one rectangle of exactly that many cells, and rectangles can't overlap or contain two numbers. Tap Start to follow all 19 logical steps, no guessing.
1/19
Draw a 1 wide × 8 tall rectangle for the 8 at R5C8 . Of the 8 rectangles of area 8 that could hold it, only this one still works, of the others, 7 contain another number.
2/19
Draw a 5 wide × 1 tall rectangle for the 5 at R8C5 . Of the 5 rectangles of area 5 that could hold it, only this one still works, of the others, 3 contain another number, 1 overlaps a finished rectangle.
3/19
Draw a 2 wide × 1 tall rectangle for the 2 at R8C2 . Of the 3 rectangles of area 2 that could hold it, only this one still works, of the others, 1 overlaps a finished rectangle, 1 contains another number.
4/19
No other number has a rectangle that can cover R1C1, R4C1, R4C2, so they all must belong to the 12 at R4C2 . That leaves it only one possible shape.
5/19
Draw a 2 wide × 6 tall rectangle for the 12 at R4C2 . Of the 24 rectangles of area 12 that could hold it, only this one still works, of the others, 20 contain another number, 3 miss cells this number must cover.
6/19
No other number has a rectangle that can cover R1C3, so it must belong to the 2 at R1C3 . That narrows it to 2 possible shapes.
7/19
No other number has a rectangle that can cover R1C5, R1C6, so they all must belong to the 2 at R1C5 . That leaves it only one possible shape.
8/19
Draw a 2 wide × 1 tall rectangle for the 2 at R1C5 . Of the 3 rectangles of area 2 that could hold it, only this one still works, of the others, 2 miss cells this number must cover.
9/19
No other number has a rectangle that can cover R1C4, so it must belong to the 2 at R1C3 . That leaves it only one possible shape.
10/19
Draw a 2 wide × 1 tall rectangle for the 2 at R1C3 . Of the 3 rectangles of area 2 that could hold it, only this one still works, of the others, 1 overlaps a finished rectangle, 1 misses cells this number must cover.
11/19
No other number has a rectangle that can cover R1C7, R3C7, so they all must belong to the 3 at R3C7 . That leaves it only one possible shape.
12/19
Draw a 1 wide × 3 tall rectangle for the 3 at R3C7 . Of the 5 rectangles of area 3 that could hold it, only this one still works, of the others, 1 misses cells this number must cover, 1 overlaps a finished rectangle, 2 contain another number.
13/19
Draw a 4 wide × 1 tall rectangle for the 4 at R2C4 . Of the 10 rectangles of area 4 that could hold it, only this one still works, of the others, 5 overlap a finished rectangle, 4 contain another number.
14/19
Draw a 4 wide × 3 tall rectangle for the 12 at R5C5 . Of the 36 rectangles of area 12 that could hold it, only this one still works, of the others, 9 overlap a finished rectangle, 26 contain another number.
15/19
Draw a 1 wide × 2 tall rectangle for the 2 at R4C7 . Of the 4 rectangles of area 2 that could hold it, only this one still works, of the others, 2 overlap a finished rectangle, 1 contains another number.
16/19
Draw a 1 wide × 2 tall rectangle for the 2 at R6C7 . Of the 4 rectangles of area 2 that could hold it, only this one still works, of the others, 3 overlap a finished rectangle.
17/19
Draw a 4 wide × 1 tall rectangle for the 4 at R7C4 . Of the 10 rectangles of area 4 that could hold it, only this one still works, of the others, 2 contain another number, 7 overlap a finished rectangle.
18/19
Draw a 2 wide × 1 tall rectangle for the 2 at R7C2 . Of the 4 rectangles of area 2 that could hold it, only this one still works, of the others, 2 overlap a finished rectangle, 1 contains another number.
19/19
Draw a 4 wide × 1 tall rectangle for the 4 at R3C4 . Of the 11 rectangles of area 4 that could hold it, only this one still works, of the others, 7 overlap a finished rectangle, 3 contain another number. ✅ Solved!
Gems Hard answer as text
12: R1C1 to R6C2 (2×6); 2: R1C3 to R1C4 (2×1); 2: R1C5 to R1C6 (2×1); 3: R1C7 to R3C7 (1×3); 8: R1C8 to R8C8 (1×8); 4: R2C3 to R2C6 (4×1); 4: R3C3 to R3C6 (4×1); 12: R4C3 to R6C6 (4×3); 2: R4C7 to R5C7 (1×2); 2: R6C7 to R7C7 (1×2); 2: R7C1 to R7C2 (2×1); 4: R7C3 to R7C6 (4×1); 2: R8C1 to R8C2 (2×1); 5: R8C3 to R8C7 (5×1).
Split the grid into rectangles (gems). Each gem holds exactly one number, and that number is its size in cells.
Is there a Normal and a Hard Gems puzzle every day?
Yes. Everyday Grids releases a Normal and a Hard Gems level every day, and both answers and solutions are on this page.
Are these Everyday Grids Gems answers for October 26, 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 Gems puzzle looks different. What should I do?
Check the date first: this page is for October 26, 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.
💛
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.
📅This page is for October 26, 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 Zip answers for October 26, 2026 (26 October 2026): the solved grids for both the Normal and Hard Zip 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 Zip guide? It’s all below, and each Everyday Grids Zip solution is verified against the puzzle rules, so it matches the Oct 26, 2026 puzzle in the app exactly.
About Everyday Grids Zip
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 Zip is one of them. Draw one continuous path through every cell exactly once, starting at 1 and visiting the numbered seeds in order. The path can’t cross walls. Every level can be solved with pure logic, with no guessing and no luck, and a brand-new Normal and Hard Zip puzzle is released every day. The game is free to download on Android and iPhone.
Everyday Grids Zip Answers and Solutions for October 26, 2026
Each level below shows the solved Zip 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).
Start at 🌻 1. A cell in the middle of the path needs exactly two connections (in and out); the start and finish need one. Walls and the board edge close sides off. Tap Start to follow all 30 logical steps, no guessing.
1/30
R1C1 needs two connections (in and out). Closed sides, up/left: edge of the board. So the path must go right and down from here.
2/30
R1C5 needs two connections (in and out). Closed sides, up/right: edge of the board. So the path must go left and down from here.
3/30
R3C5 needs two connections (in and out). Closed sides, right: edge of the board; left: wall. So the path must go up and down from here.
4/30
Dot 5 at R2C5 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
5/30
R5C1 needs two connections (in and out). Closed sides, down/left: edge of the board. So the path must go up and right from here.
6/30
R5C3 needs two connections (in and out). Closed sides, down: edge of the board; right: wall. So the path must go up and left from here.
7/30
R5C2 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
8/30
Seed 6 (R5C4) needs two connections (in and out). Closed sides, down: edge of the board; left: wall. So the path must go up and right from here.
9/30
R5C5 needs two connections (in and out). Closed sides, down/right: edge of the board. So the path must go up and left from here.
10/30
R4C5 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
11/30
Linking R1C4 and R2C4 is impossible: it would put the numbers in the order 6→5→7 on one stretch of path, but they must be visited in sequence. Mark that side closed.
12/30
R1C4 needs two connections (in and out). Closed sides, up: edge of the board; down: it would put the numbers in the order 6→5→7 on one stretch of path, but they must be visited in sequence. So the path must go left and right from here.
13/30
Linking R3C2 and R4C2 is impossible: it would connect seed 4 straight to seed 1. Mark that side closed.
14/30
Seed 4 (R3C2) needs two connections (in and out). Closed sides, right: wall, down: it would connect seed 4 straight to seed 1. So the path must go up and left from here.
15/30
Linking R3C1 and R4C1 is impossible: it would put the numbers in the order 4→2→3 on one stretch of path, but they must be visited in sequence. Mark that side closed.
16/30
R3C1 needs two connections (in and out). Closed sides, left: edge of the board; down: it would put the numbers in the order 4→2→3 on one stretch of path, but they must be visited in sequence. So the path must go up and right from here.
17/30
R2C1 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
18/30
Seed 2 (R4C1) needs two connections (in and out). Closed sides, left: edge of the board; up: it would put the numbers in the order 4→2→3 on one stretch of path, but they must be visited in sequence. So the path must go right and down from here.
19/30
Dot 1 at R4C2 already has its one connection, it is the start of the path, so the path cannot use its other side(s). Mark them as closed.
20/30
Linking R1C2 and R2C2 is impossible: it would close a loop. Mark that side closed.
21/30
R1C2 needs two connections (in and out). Closed sides, up: edge of the board; down: it would close a loop. So the path must go left and right from here.
22/30
R1C3 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
23/30
R2C2 needs two connections (in and out). Closed sides, up: it would close a loop, left: R2C1 already has its two connections. So the path must go right and down from here.
24/30
Linking R2C3 and R2C4 is impossible: it would put the numbers in the order 6→5→4→7 on one stretch of path, but they must be visited in sequence. Mark that side closed.
25/30
R2C3 needs two connections (in and out). Closed sides, up: R1C3 already has its two connections, right: it would put the numbers in the order 6→5→4→7 on one stretch of path, but they must be visited in sequence. So the path must go left and down from here.
26/30
As the finish, seed 7 (R2C4) needs exactly one connection. Closed sides, up: it would put the numbers in the order 6→5→7 on one stretch of path, but they must be visited in sequence, left: it would put the numbers in the order 6→5→4→7 on one stretch of path, but they must be visited in sequence, right: R2C5 already has its two connections. So the path must go down from here.
27/30
Linking R3C3 and R3C4 is impossible: it would put the numbers in the order 6→5→4→7 on one stretch of path, but they must be visited in sequence. Mark that side closed.
28/30
R3C3 needs two connections (in and out). Closed sides, left: wall, right: it would put the numbers in the order 6→5→4→7 on one stretch of path, but they must be visited in sequence. So the path must go up and down from here.
29/30
Dot 3 at R4C3 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
30/30
R3C4 needs two connections (in and out). Closed sides, left: it would put the numbers in the order 6→5→4→7 on one stretch of path, but they must be visited in sequence, right: wall. So the path must go up and down from here. ✅ Solved!
Start at 🌻 1. A cell in the middle of the path needs exactly two connections (in and out); the start and finish need one. Walls and the board edge close sides off. Tap Start to follow all 38 logical steps, no guessing.
1/38
R1C1 needs two connections (in and out). Closed sides, up/left: edge of the board. So the path must go right and down from here.
2/38
R2C1 needs two connections (in and out). Closed sides, left: edge of the board; right: wall. So the path must go up and down from here.
3/38
R2C3 needs two connections (in and out). Closed sides, right: wall, down: wall. So the path must go up and left from here.
4/38
R4C2 needs two connections (in and out). Closed sides, up: wall, right: wall. So the path must go left and down from here.
5/38
R4C3 needs two connections (in and out). Closed sides, left: wall, right: wall. So the path must go up and down from here.
6/38
R4C4 needs two connections (in and out). Closed sides, left: wall, right: wall. So the path must go up and down from here.
7/38
Seed 2 (R4C5) needs two connections (in and out). Closed sides, up: wall, left: wall. So the path must go right and down from here.
8/38
R6C1 needs two connections (in and out). Closed sides, down/left: edge of the board. So the path must go up and right from here.
9/38
Seed 4 (R6C3) needs two connections (in and out). Closed sides, down: edge of the board; up: wall. So the path must go left and right from here.
10/38
R6C2 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
11/38
R6C4 needs two connections (in and out). Closed sides, down: edge of the board; up: wall. So the path must go left and right from here.
12/38
R6C6 needs two connections (in and out). Closed sides, down/right: edge of the board. So the path must go up and left from here.
13/38
Dot 3 at R6C5 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
14/38
Linking R1C6 and R2C6 is impossible: it would connect seed 8 straight to seed 1. Mark that side closed.
15/38
As the finish, seed 8 (R1C6) needs exactly one connection. Closed sides, up/right: edge of the board; down: it would connect seed 8 straight to seed 1. So the path must go left from here.
16/38
Linking R1C4 and R1C5 is impossible: it would put the numbers in the order 6→8 on one stretch of path, but they must be visited in sequence. Mark that side closed.
17/38
Seed 6 (R1C4) needs two connections (in and out). Closed sides, up: edge of the board; right: it would put the numbers in the order 6→8 on one stretch of path, but they must be visited in sequence. So the path must go left and down from here.
18/38
R1C3 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
19/38
Seed 5 (R1C2) needs two connections (in and out). Closed sides, up: edge of the board; right: R1C3 already has its two connections. So the path must go left and down from here.
20/38
R2C2 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
21/38
R1C5 needs two connections (in and out). Closed sides, up: edge of the board; left: it would put the numbers in the order 6→8 on one stretch of path, but they must be visited in sequence. So the path must go right and down from here.
22/38
R3C2 needs two connections (in and out). Closed sides, up: R2C2 already has its two connections, down: wall. So the path must go left and right from here.
23/38
R3C1 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
24/38
R3C3 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
25/38
R4C1 needs two connections (in and out). Closed sides, left: edge of the board; up: R3C1 already has its two connections. So the path must go right and down from here.
26/38
R5C1 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
27/38
R5C2 needs two connections (in and out). Closed sides, left: R5C1 already has its two connections, down: R6C2 already has its two connections. So the path must go up and right from here.
28/38
R5C3 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
29/38
R5C4 needs two connections (in and out). Closed sides, left: R5C3 already has its two connections, down: wall. So the path must go up and right from here.
30/38
R5C5 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
31/38
R5C6 needs two connections (in and out). Closed sides, right: edge of the board; left: R5C5 already has its two connections. So the path must go up and down from here.
32/38
R4C6 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
33/38
R3C6 needs two connections (in and out). Closed sides, right: edge of the board; down: R4C6 already has its two connections. So the path must go up and left from here.
34/38
Dot 1 at R2C6 already has its one connection, it is the start of the path, so the path cannot use its other side(s). Mark them as closed.
35/38
Linking R2C4 and R3C4 is impossible: it would close a loop. Mark that side closed.
36/38
R2C4 needs two connections (in and out). Closed sides, left: wall, down: it would close a loop. So the path must go up and right from here.
37/38
Dot 7 at R2C5 already has both of its connections, so the path cannot use its other side(s). Mark them as closed.
38/38
R3C4 needs two connections (in and out). Closed sides, up: it would close a loop, left: R3C3 already has its two connections. So the path must go right and down from here. ✅ Solved!
Draw one continuous path through every cell exactly once, starting at 1 and visiting the numbered seeds in order. The path can't cross walls.
Is there a Normal and a Hard Zip puzzle every day?
Yes. Everyday Grids releases a Normal and a Hard Zip level every day, and both answers and solutions are on this page.
Are these Everyday Grids Zip answers for October 26, 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 Zip puzzle looks different. What should I do?
Check the date first: this page is for October 26, 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.
💛
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.
📅This page is for October 26, 2026. Playing a different day? Scroll down to More Days and pick your date, or open the calendar on the all-answers page. 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 every Everyday Grids answer for October 26, 2026 (26 October 2026) in one place: the solved grids for all seven puzzles, Zip, Gems, Lotus, Panda, Libra, Sums and Hashi, in both Normal and Hard mode. Use it as a quick cheat sheet to check the Oct 26, 2026 grids, or open any game for a full solution walkthrough with hints and a step by step explanation of every move.
About Everyday Grids
Everyday Grids is a calm daily logic-puzzle game by Fanatee. Each day it releases seven fresh grids, each with a Normal and a Hard level, so there are 14 puzzles in total, and every one can be cracked with pure logic and no guessing. Short sessions, streaks and a relaxing style make it a perfect daily brain workout, and it’s free on Android and iPhone.
Everyday Grids Answers and Solutions for October 26, 2026
Each game below shows its Normal answer and its Hard answer one after the other. Under every solved grid there is a Step by Step Explanation button that opens the full Everyday Grids solution for that level, one logical move at a time.
Everyday Grids (also searched as Everyday Grid) is a daily logic-puzzle app by Fanatee with seven grid games: Zip, Gems, Lotus, Panda, Libra, Sums and Hashi.
When do new puzzles come out?
A new Normal and Hard level of every game is released each day, and each day has its own answers page.
Are these the correct Everyday Grids answers for October 26, 2026?
Yes. Every grid is solved logically step by step and checked against the rules.
How do I find a different day?
Use the date links under More Days, the month page, or the calendar on the all-answers page.
💛
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.
Everyday Grids Lotus Answer Today: February 9, 2025
● Normal & Hard solved Lotus📅 Sun, Feb 9, 2025
📅This page is for February 9, 2025. 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 Lotus answers for February 9, 2025 (9 February 2025): the solved grids for both the Normal and Hard Lotus 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 Lotus guide? It’s all below, and each Everyday Grids Lotus solution is verified against the puzzle rules, so it matches the Feb 9, 2025 puzzle in the app exactly.
About Everyday Grids Lotus
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 Lotus is one of them. Place one lotus in every row, every column and every colour zone. Lotuses can never touch, not even diagonally. Every level can be solved with pure logic, with no guessing and no luck, and a brand-new Normal and Hard Lotus puzzle is released every day. The game is free to download on Android and iPhone.
Everyday Grids Lotus Answers and Solutions for February 9, 2025
Each level below shows the solved Lotus 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).
Each row, column and colour zone needs exactly one lotus, and lotuses can't touch, not even diagonally. Tap Start to follow all 19 logical steps, no guessing.
1/19
Only one spot is left in the sky-blue zone, so its lotus goes at R7C9. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (18 cells marked ✕).
2/19
The purple zone can only put its lotus in R8C5, R9C2, R9C3, R9C4, R9C5. Whichever of those it uses, the lotus would share a row with or touch R9C6. So that cell can't hold a lotus.
3/19
The brown zone can only put its lotus in R5C7, R5C8. Whichever of those it uses, the lotus would share a row or column with or touch each of R4C7, R4C8, R5C1, R5C2, R5C3, R5C4, R5C5, R5C6, R6C7. So none of them can hold a lotus.
4/19
Only one spot is left in the pink zone, so its lotus goes at R6C2. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (11 cells marked ✕).
5/19
Column 1 can only put its lotus in R1C1, R2C1, R3C1, R4C1, R8C1, R9C1. Whichever of those it uses, the lotus would share a row or zone with each of R2C3, R3C3, R4C3. So none of them can hold a lotus.
6/19
Column 3 can only put its lotus in R1C3, R8C3, R9C3. Whichever of those it uses, the lotus would share a row or zone with or touch R8C4. So that cell can't hold a lotus.
7/19
Column 4 can only put its lotus in R1C4, R2C4, R3C4, R4C4, R9C4. Whichever of those it uses, the lotus would share a zone with or touch R8C3. So that cell can't hold a lotus.
8/19
Column 6 can only put its lotus in R1C6, R2C6, R3C6, R4C6, R8C6. Whichever of those it uses, the lotus would share a row or zone with or touch R2C5. So that cell can't hold a lotus.
9/19
The yellow zone can only put its lotus in R2C6, R3C6, R3C7, R3C8. Whichever of those it uses, the lotus would share a row or column with or touch each of R2C7, R3C5. So none of them can hold a lotus.
10/19
All remaining spots for the purple zone, the teal zone lie inside row 8 and row 9. Those 2 lotuses will fill those 2 rows, so no other cell there can have one: clear R8C1, R8C6, R9C1.
11/19
All remaining spots for the teal zone, the brown zone lie inside column 7 and column 8. Those 2 lotuses will fill those 2 columns, so no other cell there can have one: clear R1C7, R1C8, R2C8, R3C7, R3C8.
12/19
The yellow zone can only put its lotus in R2C6, R3C6. Whichever of those it uses, the lotus would share a column with each of R1C6, R4C6. So none of them can hold a lotus.
13/19
Only one spot is left in the cream zone, so its lotus goes at R1C5. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (8 cells marked ✕).
14/19
Only one spot is left in row 2, so its lotus goes at R2C1. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
15/19
Only one spot is left in row 4, so its lotus goes at R4C4. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
16/19
Only one spot is left in row 3, so its lotus goes at R3C6. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (0 cells marked ✕).
17/19
Only one spot is left in row 8, so its lotus goes at R8C7. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (3 cells marked ✕).
18/19
Only one spot is left in row 5, so its lotus goes at R5C8. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (0 cells marked ✕).
19/19
Only one spot is left in row 9, so its lotus goes at R9C3. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (0 cells marked ✕). ✅ Solved!
Each row, column and colour zone needs exactly one lotus, and lotuses can't touch, not even diagonally. Tap Start to follow all 18 logical steps, no guessing.
1/18
Only one spot is left in the purple zone, so its lotus goes at R1C5. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (18 cells marked ✕).
2/18
The yellow zone can only put its lotus in R2C7, R3C7. Whichever of those it uses, the lotus would share a row or column with or touch each of R2C8, R3C6, R3C8, R4C7, R5C7, R6C7, R7C7, R8C7, R9C7. So none of them can hold a lotus.
3/18
The teal zone can only put its lotus in R2C9, R3C9, R4C9. Whichever of those it uses, the lotus would share a column with each of R5C9, R6C9, R7C9, R8C9, R9C9. So none of them can hold a lotus.
4/18
The brown zone can only put its lotus in R3C4, R4C6. Whichever of those it uses, the lotus would share a row or column with or touch each of R3C7, R4C3, R4C4. So none of them can hold a lotus.
5/18
Only one spot is left in column 7, so its lotus goes at R2C7. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (4 cells marked ✕).
6/18
Column 9 can only put its lotus in R3C9, R4C9. Whichever of those it uses, the lotus would share a row with or touch R4C8. So that cell can't hold a lotus.
7/18
The green zone can only put its lotus in R5C6, R5C8. Whichever of those it uses, the lotus would share a row with each of R5C1, R5C2, R5C3, R5C4. So none of them can hold a lotus.
8/18
All remaining spots for the teal zone, the brown zone lie inside row 3 and row 4. Those 2 lotuses will fill those 2 rows, so no other cell there can have one: clear R3C1, R3C2, R3C3, R4C1, R4C2.
9/18
All remaining spots for the peach zone, the cream zone lie inside row 8 and row 9. Those 2 lotuses will fill those 2 rows, so no other cell there can have one: clear R8C3.
10/18
All remaining spots for the green zone, the brown zone, the cream zone lie inside column 6 and column 8 and column 4. Those 3 lotuses will fill those 3 columns, so no other cell there can have one: clear R6C4, R6C6, R6C8, R7C4, R7C6, R7C8, R9C4.
11/18
Row 6 can only put its lotus in R6C1, R6C2, R6C3. Whichever of those it uses, the lotus would share a column or zone with or touch each of R7C1, R7C2. So none of them can hold a lotus.
12/18
Only one spot is left in row 7, so its lotus goes at R7C3. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (5 cells marked ✕).
13/18
Only one spot is left in row 6, so its lotus goes at R6C1. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
14/18
Only one spot is left in column 2, so its lotus goes at R9C2. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
15/18
Only one spot is left in column 4, so its lotus goes at R3C4. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (2 cells marked ✕).
16/18
Only one spot is left in row 4, so its lotus goes at R4C9. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (1 cells marked ✕).
17/18
Only one spot is left in row 5, so its lotus goes at R5C6. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (1 cells marked ✕).
18/18
Only one spot is left in row 8, so its lotus goes at R8C8. Now clear everything it rules out: the rest of its row, column and zone, and the 8 cells touching it (0 cells marked ✕). ✅ Solved!
Place one lotus in every row, every column and every colour zone. Lotuses can never touch, not even diagonally.
Is there a Normal and a Hard Lotus puzzle every day?
Yes. Everyday Grids releases a Normal and a Hard Lotus level every day, and both answers and solutions are on this page.
Are these Everyday Grids Lotus answers for February 9, 2025 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 Lotus puzzle looks different. What should I do?
Check the date first: this page is for February 9, 2025. 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.
💛
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.
