1. Divide the grid into polyominoes.
2. Each space in the grid will have a number in it.
3. Each number must be in a polyomino with that number of squares.
4. No two polyominoes with the same number of squares may touch on any of their edges.
5. A polyomino may contain any number (including zero) of the numbers originally given.
2. Each space in the grid will have a number in it.
3. Each number must be in a polyomino with that number of squares.
4. No two polyominoes with the same number of squares may touch on any of their edges.
5. A polyomino may contain any number (including zero) of the numbers originally given.
General Tips:
Think of each number (or groups of numbers touching) as its own polyomino that needs to force its way through the rest of the puzzle to get the numbers necessary. Remember rule 4, and write a line anywhere you know there is a line, even if you have not finished the polyomino it's in yet. Also remember rule 5, polyominoes often use a bajillion different givens or none at all (which I'll go into later).
Theorem 1: If an unfinished polyomino has only one cell it can go to, the polyomino's number must go in that cell.
This theorem is so obvious it's almost stupid to mention it as a theorem, but oh well. The proof is so simple I won't show it here.
Consider the following (simple) puzzle:
By the general tip mentioned earlier, we should put lines in to make this more usable:
Now, all of these are unfinished polyominoes with only one cell it can go to. According to this theorem, all of these numbers will go out one like this:
Now, again, the theorem states they will all go out again (and the threes will close off):
It would work on the fours again, and you can solve the puzzle. I think I've shown enough though.
Theorem 2: If an unfinished polyomino can not be finished by going as far as it can in all directions other than one cell, the polyomino's number must be in that one cell.
This might be hard to understand, but I'll give examples. This proof is also pretty simple, but examples show it better (Theorem 1 is actually a special case of theorem 2).
Consider this not-puzzle-but-used-as-an-example:
The three can't get finished by all directions other than down, so it must go down according to this theorem:
(note that I just put in these extra lines. I'm not going to mention them anymore)
Now, the four has the same thing. It can't get finished unless it goes down as well:
That's all you can solve of this thing.
There are other crazier examples as well:
The six, while it has a lot of space to the left, still can't get finished there. It must go right at least one:
Now that's the same thing with the sevens, right? There's only four empty spots to the left of them, so they both have to go one to the right, right? No! Remember rule 5. The two sevens could connect through the left, and then only one would have to go out to the right. You need to be careful about what you assume when solving logic puzzles.
Here's one more case that might not seem obvious:
The six in R1C1 can be completed without going right, right? It has all of the space in the world going down. Let's think about this:
(note: I'm actually doing a small amount of bifurcation. These sixes do not have to be here, and are purely hypothetical to show my point later.)
Going down, it's gotten five sixes so far. Let's extend it one more...
Wait, now there's seven sixes! That can't happen! This means that this polyomino can't be completed without going right, and must go right:
You need to be really careful about what you do when there's others of the same number around.
One more way you can use this theorem:
While we don't know exactly what the six does, it does have to go out that hole because of this theorem:
Something kind of related to these two theorems is that you can't do something that will choke off another number. Consider this partial "solution" to the image above:
The two is choked out and can't do anything. The reason this would happen is if you forgot one of these two theorems when the two could still get out while you were solving things, then you for some reason thought the six had to go where the two was supposed to.
Theorem 3: If an unfinished polyomino has exactly as much space around it as it needs to be finished, it must use all of that space.
This one is kind of a duh as well.
The four in R1C1 has only four places to go to. It can't go to R1C3 because then there would be five fours. It can only go here:
(This "puzzle" actually has no solution. See if you can figure out why.)
Implied Polyominoes:
This is special enough that it's its own section. An implied polyomino is one which uses no givens. Smaller implied polyominoes are easy to use, but larger ones have some strange things going on with them. Let's consider that first example puzzle after solving the threes and fours:
There are two open spots left. Obviously, they need to both be ones:
These ones are implied polyominoes.
Implied polyominoes can become a bit more confusing, too. Let's say our original puzzle was this instead:
This mostly solves as:
Now, we have three spaces left that must be implied polyominoes. They can't be threes, because that would touch other threes and break rule 4. The only other one that works (you can try) is a one and two twos touching each other. Which goes where? The two polyomino can't touch the other one, so the one must go there, and the two at the other area:
But, the best implied polyominoes are the really big ones.
What can go in that top left corner? It can't be a one, there's another one there. What about a two? Well, that would have to go down and hit that other two, so it doesn't work either. Three? Nope, same thing as the two. It has to be at least a four, but it could be more. We do know, however, that the two and three can't go left because of our implied polyomino (which I'll mark with light blue):
The two must extend by Theorem 1:
Now, the bottom right can't be a one. It can't be a three either, or else it would connect to the three in R3C2 and make four threes. This must be an implied polyomino of two, four, or more:
While we don't know yet what's in these implied polyominoes, we know they aren't threes. This three in R3C2 can now be finished by Theorem 3:
The implied polyominoes we have are closed off now, and can be replaced with what we now know they are:
What about the top right? It's implied too. It can't be the one and two twos this time (you can try if you want), but it can be the threes:
Implied polyominoes can be a lot bigger than this, but it's hard to make good puzzles in a 4x4 grid. That's why I like bigger puzzles.
Implied polyominoes can be crazy. Remember, it usually helps to mark where they are so that they can actually be used as the walls they can be. Also remember that even if something seems like an implied polyomino it might just be a really big one connected to a big number somewhere else.
Theorem 4: If a polyomino not going into a cell makes an implied polyomino that is the same number as an adjacent polyomino, that original polyomino must go into the cell.
This sounds confusing, so here's an example:
Think about the top left corner. If the two goes to the right, that would force an implied 1-omino to be in the top left. But there's already a 1-omino next to it. So, the two must go left:
Now, a similar thing is happening in the upper right corner. If those threes go down, there must be an implied 2-omino in the corner, and that can't happen. So:
A similar thing (not explicitly said by the theorem, but still basically the same thing) is something like this:
If the fours extend into R1C2, there must be an implied 1-omino in R1C1. That can't happen, so we can draw lines around the fours there:
Counting:
This is a technique that I rarely use (and others use a lot more). Consider the following puzzle:
You might be able to find the solution by some type of trial and error, but what's a good way to find it? Well, think about this. The number of cells in the puzzle is 16. We have two 3-ominoes, 1 1-omino, one 4-omino, and one 5-omino. Add up all of these, and we get 3*2 + 1 + 4 + 5 = 16. There are no implied polyominoes at all. Now, if we can just find somewhere that only one of our polyominoes can get to, that might help us. Well, consider R1C2. The three can't get there (there's other threes around) and the four definitely can't. That means, because there are no implied polyominoes, the five MUST go there.
By Theorem 2, the five must also go here:
Now, by Theorem 3, the threes are solved:
The rest of the puzzle is solved with Theorem 1.
Counting can also be used when implied polyominoes ARE around. Consider this puzzle:
We have 1*3 + 3 + 4 + 5 = 15 cells accounted for here. That means there will be an implied 1-omino somewhere. It can't be close to any of these other ones, however. Think of R1C2. It can't be our 1-omino, and the three and four can't reach it. It must be a five.
By Theorem 1:
By Theorem 4:
We still have the same situation as we started in. We have 8 cells left, with 3 + 4 = 7 cells accounted for. R3C2 can't be a 1 or a 3, so it must be the four:
Theorem 1:
Theorem 4 (first on the fours):
And now we have our 1:
This is special enough that it's its own section. An implied polyomino is one which uses no givens. Smaller implied polyominoes are easy to use, but larger ones have some strange things going on with them. Let's consider that first example puzzle after solving the threes and fours:
There are two open spots left. Obviously, they need to both be ones:
These ones are implied polyominoes.
Implied polyominoes can become a bit more confusing, too. Let's say our original puzzle was this instead:
This mostly solves as:
Now, we have three spaces left that must be implied polyominoes. They can't be threes, because that would touch other threes and break rule 4. The only other one that works (you can try) is a one and two twos touching each other. Which goes where? The two polyomino can't touch the other one, so the one must go there, and the two at the other area:
But, the best implied polyominoes are the really big ones.
What can go in that top left corner? It can't be a one, there's another one there. What about a two? Well, that would have to go down and hit that other two, so it doesn't work either. Three? Nope, same thing as the two. It has to be at least a four, but it could be more. We do know, however, that the two and three can't go left because of our implied polyomino (which I'll mark with light blue):
The two must extend by Theorem 1:
Now, the bottom right can't be a one. It can't be a three either, or else it would connect to the three in R3C2 and make four threes. This must be an implied polyomino of two, four, or more:
While we don't know yet what's in these implied polyominoes, we know they aren't threes. This three in R3C2 can now be finished by Theorem 3:
The implied polyominoes we have are closed off now, and can be replaced with what we now know they are:
What about the top right? It's implied too. It can't be the one and two twos this time (you can try if you want), but it can be the threes:
Implied polyominoes can be a lot bigger than this, but it's hard to make good puzzles in a 4x4 grid. That's why I like bigger puzzles.
Implied polyominoes can be crazy. Remember, it usually helps to mark where they are so that they can actually be used as the walls they can be. Also remember that even if something seems like an implied polyomino it might just be a really big one connected to a big number somewhere else.
Theorem 4: If a polyomino not going into a cell makes an implied polyomino that is the same number as an adjacent polyomino, that original polyomino must go into the cell.
This sounds confusing, so here's an example:
Think about the top left corner. If the two goes to the right, that would force an implied 1-omino to be in the top left. But there's already a 1-omino next to it. So, the two must go left:
Now, a similar thing is happening in the upper right corner. If those threes go down, there must be an implied 2-omino in the corner, and that can't happen. So:
A similar thing (not explicitly said by the theorem, but still basically the same thing) is something like this:
If the fours extend into R1C2, there must be an implied 1-omino in R1C1. That can't happen, so we can draw lines around the fours there:
Counting:
This is a technique that I rarely use (and others use a lot more). Consider the following puzzle:
You might be able to find the solution by some type of trial and error, but what's a good way to find it? Well, think about this. The number of cells in the puzzle is 16. We have two 3-ominoes, 1 1-omino, one 4-omino, and one 5-omino. Add up all of these, and we get 3*2 + 1 + 4 + 5 = 16. There are no implied polyominoes at all. Now, if we can just find somewhere that only one of our polyominoes can get to, that might help us. Well, consider R1C2. The three can't get there (there's other threes around) and the four definitely can't. That means, because there are no implied polyominoes, the five MUST go there.
By Theorem 2, the five must also go here:
Now, by Theorem 3, the threes are solved:
The rest of the puzzle is solved with Theorem 1.
Counting can also be used when implied polyominoes ARE around. Consider this puzzle:
We have 1*3 + 3 + 4 + 5 = 15 cells accounted for here. That means there will be an implied 1-omino somewhere. It can't be close to any of these other ones, however. Think of R1C2. It can't be our 1-omino, and the three and four can't reach it. It must be a five.
By Theorem 1:
By Theorem 4:
We still have the same situation as we started in. We have 8 cells left, with 3 + 4 = 7 cells accounted for. R3C2 can't be a 1 or a 3, so it must be the four:
Theorem 1:
Theorem 4 (first on the fours):
And now we have our 1:
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