Polycubeous Tips

1. Make a three-dimensional grid out of the two-dimensional ones by stacking them on top of each other, with the grid on the top being the top section of the 3D grid.  (The example puzzle is a dome when stacked this way.)

2. Divide the grid into polycubes.
3. Each space in the grid will have a number in it.
4. Each number must be in a polycube with that number of cubes.
5. No two polycubes with the same number of cubes may touch on any of their (square) edges.
6. A polycube may contain any number (including zero) of the numbers originally given.


Polycubeous should almost be a variant of Polyominous.  They work almost exactly the same, other than that Polycubeous is a bit more confusing.  Thus, I'm just going to link to the Polyominous Tips and solve an example puzzle.  If you for some reason haven't read the Polyominous tips before this, read those first, because I'm not going to show anything there here again.

Even though it might cause some confusion, I'm going to say that going from a lower "level" of the puzzle to a higher one is going "above", while going lower is going "below".  Up, down, left and right will still be normal on a particular level.  Also, I'm going to call the bottom level level 1, and the highest one level 4.  So, level 1 is the one on the bottom of the picture.  To extend the RC designations of cells, I'll say L after R and C.  So, R2C3 on level 2 would be R2C3L2.

There won't be any lines for boundaries between levels.  That would just be too confusing.  While you're solving these you might want to write something.

So, on to the example:

First, I'm just going to write lines around all of the ones:

Now, R2C2L1 and R2C3L1, both extend above with Theorem 1:

R2C4L1 extends up by Theorem 1 as well:

Oh yeah, R4C2L1 and R4C2L2 are connected, so that two is done:

The five in R4C3L1 extends a whole bunch with Theorem 1, first to the right then above a bunch (and then gets finished):

The four in R3C3L1 must extend above by Theorem 1, and then gets connected to the four in R4C3L2 and is finished:

Now the five in R2C3L1 goes above a couple as well:

The three in R2C4L2 goes up then above:

Oh whoops, that five earlier actually connected to the top five as well and got finished:

The twos in R1C3L3 and R2C4L3 now go above:

There's an implied 1 in the very top right:

The three in R3C4L3 needs to go left because of Theorem 2 (it can only go one above) and then connects with the other three:

This four extends here through Theorem 1:

This six needs to go here because of Theorem 4:

Now this five could actually have gone above for a while now:

There are four fives in that polycube now, R3C2L1, R3C2L2, R3C2L3, and R4C2L3.  A five can't go in R2C2L3 because then there would be six fives, so there are lines here:

Now that five in R2C1L3 goes above and down, to R4C2L4 - wait, that connects to the other five!  What happened?

Well, I made a mistake.  I decided to keep it because it's good to know how to fix mistakes like this (and I didn't feel like rewriting all of this stuff)  Now we have to look back at what we did and figure out what went wrong.  You should look back and see if you can find it.



Anyway, the mistake was when I extended that four in R1C2L2.  I didn't realize that it could have gone below to R1C1L1.  Thankfully, the four still mostly goes there with Theorem 2, and that highest four didn't change any of our logic.  That six down there didn't have to go there either.  This is actually how much we know:

As I said before, this thankfully doesn't mess with the other stuff we did after that.  Now, the five does go above one with Theorem 1:

It can only go two places going up and right on the top level, so it must go down according to Theorem 2:

Now, consider if the five went down one more:

Is there anything wrong with this?  Well, think about the five polyomino that includes the five in R4C2L3.  That's the only open spot for it, and it must go either left or above.  If this blue five is here, both of those ways are gone.  This blue five can't go here:

Now the five finishes by Theorem 1 or Theorem 3:

NOW the four goes down:

This three goes here:

Now this two goes down:

Now this six is done with theorem 3:

Now, that five earlier will finally go above one more:

And the four is solved with Theorem 3, and there's an implied 1 up there as well:

Moral of the story: don't mess up.