Room and Reason Tips

1. Make each cell white or black.

2. The grid is divided into rectangular "rooms".  Rooms with numbers in their corners must have that number of black cells in them.  Rooms without numbers may have any number of black cells in them.

3. White cells may not go through more than two rooms in an uninterrupted horizontal or vertical line.

4. No two black cells may share an edge.

5. All white cells must be connected to each other through their edges.

This puzzle type is very interesting.  There are many different things that you can do with it.

If you haven't already, check out the (big breath) Japanese-Crossword-Puzzle-Like Puzzle Tips.

So first, make sure you mark all rooms with a 0 white.  I won't show an example because that's simple.

Remember rule 3.  If you have a situation like this:

R1C1 must be black, because if it was white it would make a line of white cells go through more than two rooms:

Any time you see this:

A black cell needs to be in R3C2 for the same reason:

And the same thing in this case:

Theorem 1: If a room has a number N in its corner with dimensions 1x(2*N-1), every other cell in the room is black.

Ack, a formula!  Here's a few examples to show you the idea:

Ifthen

Ifthen

Ifthen

You should get it now.

Theorem 2: If a room has a number N in its corner with dimensions 2xN, the room is filled as much as possible with diagonally adjacent black cells.

Here's some examples of this:

Ifthenor

Skipping to four, just to show what I mean better:

Ifthenor

Think about that last case.  In a bigger grid, let's say everything we know about that:

So, this can be one of these two:

There's something common to both of these:

You can always figure those white spots out on the edge of bigger 2xN rooms.  On the edge, in fact, you can figure out a bit more!

This is one of these two:

(note that I wrote more than needed for this example, I just wanted to show all of the stuff you could figure out based on what it can be)

This is all that's common to both solutions, so you can write all of this:

One more special case of these is when you have a two in a corner or a three on a wall:

These are the two possibilities for the two:

In the first image, R1C4 is a white cell not connected to the others, so that solution doesn't work.  It mus be the second one.

Here's the one for threes:

You can try the other way, but the only way it works is this:

These theorems can also be used if you have a room where a bunch of cells are already done, but the rest of it makes its own "mini-room" that follows one of these patterns.

There's a couple weird ones for 3x3 rooms as well:

Theorem 3: If a 3x3 room has a four in the corner, all of the middle-edge cells are white.

In picture form:

Ifthen

If you put a black cell in one of those, it forces the other ones to be black as well, and then the middle will be a choked white cell.

Theorem 4: If a 3x3 room has a five in the corner, all of the corner cells and the center cell will be black.

Ifthen

I'm sure there are more weird theorems for bigger rooms, but they aren't encountered enough to be worthwhile.

Room and Reason puzzles are more about actual thinking instead of applying theorems.  I made an example puzzle to show the type of thinking you usually use here:

(this puzzle is full-sized because you can't really do anything good in a small puzzle with this type)

First, I recognize that 3x3 room with the four, and will use Theorem 3:

Now, by the part of Theorem 2 about a three against a wall, we can do this:

These black cells go here because of rule 3 (and a few related white cells):

Think of the two in the top right.  It can be one of these two (I'm going to use blue to show possibilities):

These are the common cells between the two:

Now, by applying rule three again:

With this new information, think of this possibility of the two again:

This chokes the white cells in the top right.  So, it must be the other way:

The four now has only four cells that can be black, so they must be black:

Now we can use rule 3 on this part:

Now, because all of the white cells need to be touching, we can color all of these white:

Consider the room starting at R6C3.  It still needs two more black spots, and has a 2x2 area of unknowns.  Well, that's just like it was a 2x2 room with a 2 in it!  It could be these things:

The common cells between these are:

That may not have actually helped, but we still did all of that.  We can use rule 3:

Now consider one of those possibilities for the three room:

Do you see what's wrong here?  That big section of white cells in the bottom left area doesn't connect anywhere else!  This doesn't work.  It must be the other way:

This one here must go here:

We have a rule 3 thing here on the right that we've had for a while:

These two rooms on the right with a one already have their one black cell, so the rest are white:

Now what?  We're going to have to use a bit of bifurcation.  Let's see what will happen if a put a black cell here:

How did I know to do that?  Because I made this puzzle.  If you come across something like this, you just have to try several different areas before you find one that works.  Now, that 3 still needs one more black cell, and there are two possibilities for it:

The first one chokes out that five-cell area the three is in, so it doesn't work.  The second doesn't have that problem, but it choked the bottom left area again!  That means that our original assumption doesn't work, no matter what we try.  That cell I first colored blue must be white:

Now by rule three:

and, now we just fill in the last black cell:

From this example puzzle, there's a few things you should realize:

1. Remember rule 3.  It's used all over the place.

2. Remember that all white cells are connected, especially when considering different possibilities.  Some of the possibilities will stop you from connecting them all.

3. Realize that while some of the theorems were used, they weren't used that often.  That's how this puzzle is.