Technically I would say these are all logic...but it does help to use logic and guesses to help eliminate what doesn't work. I like these - but they are a bit tricky.
Technically, trial and error is a form of logical deduction.
It seems to me that all puzzles on this site are a variant of "eliminate the impossible until only 1 possibility remains," a challenge for which trial and error is particularly well-suited. All higher reasoning is simply cases where "I don't need to try this option because I already know it doesn't work."
Some higher logic I found useful for starting this puzzle: - 2 islands that must both touch an edge must not touch each other. - If an island that touches an edge touches another island, the other island must not touch an edge.
This was enough to get me through the first half or so, but I had to resort to trial and error to finish up.
Posted 28th Dec 2023 at 18:34
JoergWausW Daily subscriber Completion time: 2:26Used 'show wrong moves' Used 'check puzzle' when incorrect
Here my two cents:
1) empty corners:
If there is nothing in a 2x2 area in the corner: there has to be at least one white cell. Which white area can reach it? In this one: the bottom right corner (here: the 2x2 above the 1) can only be reached by the 9, that makes the top right corner...? And the bottom left?
2) Usually those puzzles have to main problems: a) "too few white cells" or b) "too many white cells". In this one, the right half is "too few", the left one "too many". In case of b) you have to figure out, how to keep the black cells connected.
In case of a) you have to figure out how to avoid black 2x2s - see my 1)st cent above
In case of b) there might be a logic chain of "the only escape" - in this puzzle: how to figure out where the second white cell of the 2 has to be...
* Solve the two 1s * Solve the 2 * Solve the 9, which is fully forced by its need to reach a 2x2 near the bottom-right, and to avoid any 2x2 black areas en route (actually, you can place 8 of it to the bottom-right, which forces the 3, which forces the 9th square of the 9) * Solve the 3 as above * Now the 8 is fully forced by a need to reach the top-right 2x2, while keeping all the blacks connected * So now you have the 7, 5, 6 and 4 in an area where you will need to squash them to make them all fit. You can guess the 4 runs along the bottom, which is very likely the case by uniqueness but you don't need to draw this in if you don't want - it's just an observation. * There is a 2x2 to the right of the 5 and below the 7. If the 7 comes down it forces a black square to its left which can't work - wherever it goes to connect you can't complete either the 7 or the 5. So the 5 must go right one square. * Observing very little room, place the 7 as compactly as possible and fill in the 5 below the same way, now forced by the black squares. * Place the 4 along the bottom (now forced by the need for the 6 to go to the bottom-left 2x2) * The 6 goes where it must to complete the remaining 2x2s so they aren't fully shaded. Done.
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It seems to me that all puzzles on this site are a variant of "eliminate the impossible until only 1 possibility remains," a challenge for which trial and error is particularly well-suited. All higher reasoning is simply cases where "I don't need to try this option because I already know it doesn't work."
Some higher logic I found useful for starting this puzzle:
- 2 islands that must both touch an edge must not touch each other.
- If an island that touches an edge touches another island, the other island must not touch an edge.
This was enough to get me through the first half or so, but I had to resort to trial and error to finish up.
1) empty corners:
If there is nothing in a 2x2 area in the corner: there has to be at least one white cell. Which white area can reach it? In this one: the bottom right corner (here: the 2x2 above the 1) can only be reached by the 9, that makes the top right corner...? And the bottom left?
2) Usually those puzzles have to main problems: a) "too few white cells" or b) "too many white cells". In this one, the right half is "too few", the left one "too many". In case of b) you have to figure out, how to keep the black cells connected.
In case of a) you have to figure out how to avoid black 2x2s - see my 1)st cent above
In case of b) there might be a logic chain of "the only escape" - in this puzzle: how to figure out where the second white cell of the 2 has to be...
Does this help?
* Solve the 2
* Solve the 9, which is fully forced by its need to reach a 2x2 near the bottom-right, and to avoid any 2x2 black areas en route (actually, you can place 8 of it to the bottom-right, which forces the 3, which forces the 9th square of the 9)
* Solve the 3 as above
* Now the 8 is fully forced by a need to reach the top-right 2x2, while keeping all the blacks connected
* So now you have the 7, 5, 6 and 4 in an area where you will need to squash them to make them all fit. You can guess the 4 runs along the bottom, which is very likely the case by uniqueness but you don't need to draw this in if you don't want - it's just an observation.
* There is a 2x2 to the right of the 5 and below the 7. If the 7 comes down it forces a black square to its left which can't work - wherever it goes to connect you can't complete either the 7 or the 5. So the 5 must go right one square.
* Observing very little room, place the 7 as compactly as possible and fill in the 5 below the same way, now forced by the black squares.
* Place the 4 along the bottom (now forced by the need for the 6 to go to the bottom-left 2x2)
* The 6 goes where it must to complete the remaining 2x2s so they aren't fully shaded.
Done.
You can however view other players' statistics and comments in the tables above.