EDIT: I added step-by step pictures of the first quarter (first 5 pictures) of the puzzle. note: The flawed logic of the fourth deduction is improved here
The more condense original solution:
EDIT: I added step-by step pictures of the first quarter (first 5 pictures) of the puzzle. note: The flawed logic of the fourth deduction is improved here
The more condense original solution:
My solution, withoutnow with comment (al least for nowand adapted to fix some oversights)
Picture 1: Plan: Exclude possibilities one by one, using situations that may lead quickly to contradiction
- very wet or dry rows/columns
- rows with few options (top row: 3, bottom row: 4)
- wet rows directly above dry rows (7 above 4, 6 above 4)
- and adjacent dry + wet columns (4-8, 4-7)
Obvious first candidate : tile 13: If the indicated tiles are wet, the rest is dry, including 11,12,14 on the row above. Only 4 tiles (the red) can thus be wet there - contradiction -> 13 is dry on those 2 rows.
Picture 2: On the top row , only blue, only orange or only yellow is wet. Assume blue. Then the 4-8 columns are now 2-7 and all 5 (dark blue) tiles that give the 7 more water must be used. This also makes light blue wet, and the rest of column 3 and 4 must then be dry. Especially, in the bottom row 28 and 29 will be dry, allowing only the 5 red tiles to be wet. - contradiction -> 1 is dry on the top row.
Picture 3: Assume 25 is fully filled with water. Then 21 is fully dry, and also 18 or 20. T get to 7 on the row above 16 mus be wet there. Thus the bottom of 16 is wet, and 18 and 20 are completely dry. All remaining tiles of column 2 and 5 must then be wet, leading to contradiction on the bottom row.
Picture 4: Assume 9 is wet on row 4 (part 1) The 4-8 columns are now 3-8, and 5 of the 6 possibilities to reduce this (dark blue) must be used. If 21 is used, only 1 of the 4 yellow tiles can be used , and column 2 cannot reach 6. So the rest is used.
Picture 5: Assume 9 is wet on row 4 (part 2) Thus 3 and 11 must be fully wet, as well as the bottom of 22. This defines row 1,4 and 5 fully, which forces water due to varies column constraints , and 30 to be dry. Row 10 now forces 23 and 26 to be dry there, and then column 7 the bottom of 16 to be wet and the rest of row 8 to be dry. Now both orange tiles need to be wet. - contradiction -> 9 is dry on row 4. In addition (column 2), the bottom half of 18 must be wet.
Picture 6: Assume 11 is mostly dry (part 1)
Then 14 and the bottom of 10 must be wet due to row 4+5. This leaves 1 remaining wet tile (a pink) for column 11, i.e. not at the top row; i.e in the top row only 4 is wet. All 4 now remaining spots in column 5 to compensate the difference in column 4 and 5 (dark blue) must be used, and 15 must be dry.
Picture 7: Assume 11 is mostly dry (part 2)
The rest of row 9 must be dry. On row 8, block 16 and 20 are then dry, and column 6 has 6 dry tiles. - contradiction -> 11 is wet on row 5.
Picture 8: Assume 29 is not completely dry (part 1)
Then (row 12) 28 and 23 are dry. With that (1 dry missing in column 2 and 3) we must add some water to 1, 18 and 26.
Now assume 21 is dry, then (column 3) 2 and 26 cannot both be fully dry and thus (column 4) 15 cannot be wet on row 7.
If (row 10) 26 is wet, the rest is dry and row 9 cannot have 4 wet tiles, so 26 is partly dry, and the rest of column 5 wet.
Then we can fill in row 1, with a contradiction on row 2. Thus 21 must be partly blue.
Picture 9: Assume 29 is not completely dry (part 2)
Since 21 is partly wet, the rest of row 9 is dry.
16 dry makes 18,19 and 14 wet on row 7 (1 dry spot left); 21 wet on row 8 1 dry spot left and 12 wet in column 8, which makes the rest of column 5 dry.
column 6 (1 dry left) and 7 (1 wet left) define 3 and 7 further.
Picture 10: Assume 29 is not completely dry (part 3)
At row 4 10 must be wet, which makes the top of 24 dry, which makes the rest of row 10 wet, which makes the rest of column 11 dry.
Now (row 1 and 2) 4 , and the rest of 2 must be wet, thus bot the red tiles blue, and row11 cannot have 7 wet tiles. - contradiction -> 29 is dry. In addition 28 and 23 are wet on row 12.
Picture 11: Assume 16 is wet on row 8
Now (column 8) 23 is dry on row 10, and (row 9) 21 is dry, which makes at least one yellow area wet, which makes the top of 26 dry.
The rest of column 5 must be wet, which leads to contradiction on row 1 or 2. -> 16 is dry up to row 8.
Picture 12: assume 12 is wet on row 5
Then the rest of row 5 (esp. 8) is dry, which (row 4) makes the bottom of 10 wet.
If 7 is completely dry: (column 7) 23 must have water in column 7 and 12 must be completely wet, leading to contradiction in column 8. This forces row 4.
If 21 is fully dry: (column 5) 2,15 and 26 must be wet, leading to contradiction in column 4, so its base is wet, and the rest of row 9 dry. In row 8 this makes 20 dry, and thus 21 and 19 wet.
In column 5: 2 and 26 cannot both be wet because of column 3, and 15 cannot be wet together with either (column4), so only one of the red tiles is wet, which makes 8 in column 5 impossible - contradiction -> 12 is dry at row 5
Picture 13: assume 15 is dry
Then 14,18 and 19 must be wet in that row, and (column 6) the bottom of 3 and 22 must be wet, thus (row 10) the top of 26 dry, and thus (column 5) the bottom of 21 wet.
The rest of row 9 must be dry, then the rest of column 6 wet, then the rest of row 4 dry. Now (row1) 5 has to be wet, and the red tile must be dry according to its row, but wet according to its column. - contradiction -> 15 is wet from row 7.
Picture 14: Assume 21 is dry
Then the rest of row 8 is wet, and since (column 4) 2 and 26 can not both be wet, so are all other tiles of column 5. Then row 1 forces 5 to be wet, making 2 dry, making 26 wet.
This makes the top of 22,23 and 24 dry, giving row 9 9 dry tiles. - contradiction -> the bottom of 21 is wet.
Picture 15: 2,6 and the rest of row 9 are dry. Row 8 then adds water to 21 and 19 and column 2 to 1 and 18. Due to row 11, the bottom of 22 is wet.
Picture 16: assume 7 is dry (part 1)
Column 7 now makes 4 and the rest of 23 wet, row 2 makes 5 wet, and row 4 its remaining tiles.
Picture 17: assume 7 is dry (part 2)
The rest of ro1 1 is dry, and then the rest of column 5 and 6 wet. This leaves 1 wet tile in row 3 and 5.
Now the top of 18 must be other than the the green tiles, and thus the same as the other two purple tiles. Purple must be dry because of row 1, the top 19 must aslo be dry (column 8), making the rest of row 7 (yellow) wet, and thus too much of row 8.
Picture 18: With the bottom of 7 wet , row 4 can be finished, then column5, then row 1 and column 6
Picture 19: We can now add some dry tile-pairs to row 2 and 3, and a wet tile-pair to row 6. After that we can finish column 8 and (column 12) add water to 17.
Picture 20: Row 8 makes 14 dry, which allows us to finish Column 11, column 10, then row 7 and 7 , and then the rest.
My solution, without comment (al least for now)
My solution, now with comment (and adapted to fix some oversights)
Picture 1: Plan: Exclude possibilities one by one, using situations that may lead quickly to contradiction
- very wet or dry rows/columns
- rows with few options (top row: 3, bottom row: 4)
- wet rows directly above dry rows (7 above 4, 6 above 4)
- and adjacent dry + wet columns (4-8, 4-7)
Obvious first candidate : tile 13: If the indicated tiles are wet, the rest is dry, including 11,12,14 on the row above. Only 4 tiles (the red) can thus be wet there - contradiction -> 13 is dry on those 2 rows.
Picture 2: On the top row , only blue, only orange or only yellow is wet. Assume blue. Then the 4-8 columns are now 2-7 and all 5 (dark blue) tiles that give the 7 more water must be used. This also makes light blue wet, and the rest of column 3 and 4 must then be dry. Especially, in the bottom row 28 and 29 will be dry, allowing only the 5 red tiles to be wet. - contradiction -> 1 is dry on the top row.
Picture 3: Assume 25 is fully filled with water. Then 21 is fully dry, and also 18 or 20. T get to 7 on the row above 16 mus be wet there. Thus the bottom of 16 is wet, and 18 and 20 are completely dry. All remaining tiles of column 2 and 5 must then be wet, leading to contradiction on the bottom row.
Picture 4: Assume 9 is wet on row 4 (part 1) The 4-8 columns are now 3-8, and 5 of the 6 possibilities to reduce this (dark blue) must be used. If 21 is used, only 1 of the 4 yellow tiles can be used , and column 2 cannot reach 6. So the rest is used.
Picture 5: Assume 9 is wet on row 4 (part 2) Thus 3 and 11 must be fully wet, as well as the bottom of 22. This defines row 1,4 and 5 fully, which forces water due to varies column constraints , and 30 to be dry. Row 10 now forces 23 and 26 to be dry there, and then column 7 the bottom of 16 to be wet and the rest of row 8 to be dry. Now both orange tiles need to be wet. - contradiction -> 9 is dry on row 4. In addition (column 2), the bottom half of 18 must be wet.
Picture 6: Assume 11 is mostly dry (part 1)
Then 14 and the bottom of 10 must be wet due to row 4+5. This leaves 1 remaining wet tile (a pink) for column 11, i.e. not at the top row; i.e in the top row only 4 is wet. All 4 now remaining spots in column 5 to compensate the difference in column 4 and 5 (dark blue) must be used, and 15 must be dry.
Picture 7: Assume 11 is mostly dry (part 2)
The rest of row 9 must be dry. On row 8, block 16 and 20 are then dry, and column 6 has 6 dry tiles. - contradiction -> 11 is wet on row 5.
Picture 8: Assume 29 is not completely dry (part 1)
Then (row 12) 28 and 23 are dry. With that (1 dry missing in column 2 and 3) we must add some water to 1, 18 and 26.
Now assume 21 is dry, then (column 3) 2 and 26 cannot both be fully dry and thus (column 4) 15 cannot be wet on row 7.
If (row 10) 26 is wet, the rest is dry and row 9 cannot have 4 wet tiles, so 26 is partly dry, and the rest of column 5 wet.
Then we can fill in row 1, with a contradiction on row 2. Thus 21 must be partly blue.
Picture 9: Assume 29 is not completely dry (part 2)
Since 21 is partly wet, the rest of row 9 is dry.
16 dry makes 18,19 and 14 wet on row 7 (1 dry spot left); 21 wet on row 8 1 dry spot left and 12 wet in column 8, which makes the rest of column 5 dry.
column 6 (1 dry left) and 7 (1 wet left) define 3 and 7 further.
Picture 10: Assume 29 is not completely dry (part 3)
At row 4 10 must be wet, which makes the top of 24 dry, which makes the rest of row 10 wet, which makes the rest of column 11 dry.
Now (row 1 and 2) 4 , and the rest of 2 must be wet, thus bot the red tiles blue, and row11 cannot have 7 wet tiles. - contradiction -> 29 is dry. In addition 28 and 23 are wet on row 12.
Picture 11: Assume 16 is wet on row 8
Now (column 8) 23 is dry on row 10, and (row 9) 21 is dry, which makes at least one yellow area wet, which makes the top of 26 dry.
The rest of column 5 must be wet, which leads to contradiction on row 1 or 2. -> 16 is dry up to row 8.
Picture 12: assume 12 is wet on row 5
Then the rest of row 5 (esp. 8) is dry, which (row 4) makes the bottom of 10 wet.
If 7 is completely dry: (column 7) 23 must have water in column 7 and 12 must be completely wet, leading to contradiction in column 8. This forces row 4.
If 21 is fully dry: (column 5) 2,15 and 26 must be wet, leading to contradiction in column 4, so its base is wet, and the rest of row 9 dry. In row 8 this makes 20 dry, and thus 21 and 19 wet.
In column 5: 2 and 26 cannot both be wet because of column 3, and 15 cannot be wet together with either (column4), so only one of the red tiles is wet, which makes 8 in column 5 impossible - contradiction -> 12 is dry at row 5
Picture 13: assume 15 is dry
Then 14,18 and 19 must be wet in that row, and (column 6) the bottom of 3 and 22 must be wet, thus (row 10) the top of 26 dry, and thus (column 5) the bottom of 21 wet.
The rest of row 9 must be dry, then the rest of column 6 wet, then the rest of row 4 dry. Now (row1) 5 has to be wet, and the red tile must be dry according to its row, but wet according to its column. - contradiction -> 15 is wet from row 7.
Picture 14: Assume 21 is dry
Then the rest of row 8 is wet, and since (column 4) 2 and 26 can not both be wet, so are all other tiles of column 5. Then row 1 forces 5 to be wet, making 2 dry, making 26 wet.
This makes the top of 22,23 and 24 dry, giving row 9 9 dry tiles. - contradiction -> the bottom of 21 is wet.
Picture 15: 2,6 and the rest of row 9 are dry. Row 8 then adds water to 21 and 19 and column 2 to 1 and 18. Due to row 11, the bottom of 22 is wet.
Picture 16: assume 7 is dry (part 1)
Column 7 now makes 4 and the rest of 23 wet, row 2 makes 5 wet, and row 4 its remaining tiles.
Picture 17: assume 7 is dry (part 2)
The rest of ro1 1 is dry, and then the rest of column 5 and 6 wet. This leaves 1 wet tile in row 3 and 5.
Now the top of 18 must be other than the the green tiles, and thus the same as the other two purple tiles. Purple must be dry because of row 1, the top 19 must aslo be dry (column 8), making the rest of row 7 (yellow) wet, and thus too much of row 8.
Picture 18: With the bottom of 7 wet , row 4 can be finished, then column5, then row 1 and column 6
Picture 19: We can now add some dry tile-pairs to row 2 and 3, and a wet tile-pair to row 6. After that we can finish column 8 and (column 12) add water to 17.
Picture 20: Row 8 makes 14 dry, which allows us to finish Column 11, column 10, then row 7 and 7 , and then the rest.