# How many possible combinations are there in Hua Rong Dao?

How many possible combinations are there in Hua Rong Dao?

Hua Rong Dao is a Chinese sliding puzzle game, also called Daughter in a Box in Japan. You can see a picture here:

http://en.wikipedia.org/wiki/File:Hakoiri3.jpg

And an explanation here:

http://en.wikipedia.org/wiki/Klotski#Hua_Rong_Dao

The puzzle is a 4x5 grid with these pieces

• 2x2 square (1 piece)
• 1x2 vertical (4 pieces)
• 2x1 horizontal (1 piece)
• 1x1 square (4 pieces)

Though traditionally each type of piece will have different pictures, you can treat each of the 1x2's as identical and each of the 1x1's as identical.

The goal is to slide around the pieces (not removing them) until the 2x2 "general" goes from the middle top to the middle bottom (where it may slide out of the border).

I'm not concerned in this question with the solution, but more curious about the number of combinations. Naively, I can come up with an upper bound like this

Place each piece on the board, ignoring overlaps.

The 2x2 can go in any of 3x4 = 12 squares
The 2x1 can go in any of 4x4 = 16 squares
The 1x2 can go in any of 3x5 = 15 squares
The 1x1 can go in any of 4x5 = 20 squares

If you place the pieces one at a time, subtracting out the used squares

place the 2x2 = 12 options
place each of the 2x1 = ((16 - 4) * (16 - 6) * (16 - 8) * (16 - 10)) / 4!
place the 1x2 = 15 - 6
place the 1x1 = (20-14) choose 4 = 15

multiplied together this works out to 388,800.

Is there any way I might be able to narrow this down further? The two obvious things not taken into account are blocked pieces (a 2x1 pieces will not fit into two separated squares) and the fact that not all possibilities might be accessible when sliding from the starting position.

Update: I realized that the puzzle is bilaterally symmetrical, so if you just care about meaningful differences between positions, you can divide by two.

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Questions of this general form are actually very hard and an active topic of research in combinatorics. I don't know any good references, though. – Qiaochu Yuan Jan 12 '11 at 15:45
You need to decide if you just want the number of ways to put the pieces in the box or the number of ways accessible from start. For the number of ways in the box, you could write a program that just puts the 2x2 in each location, then finds the possibilities for the 1x2s, etc. As Qiaochu says, I don't know any easy way. For accessible from start, maybe you can prove that they all are accessible, or maybe there is a parity argument (like the 14-15 puzzle). – Ross Millikan Jan 12 '11 at 17:01
Both are interesting questions. Part of the genesis of the question is that we were speculating about random arrangements of the pieces and wondering how many were legitimate states (e.g. accessible from the start). I quickly realized that it was not simple to answer either! – Will Glass Jan 12 '11 at 23:00
If you're counting total positions, it's not important to distinguish between the four small squares or between the four vertical dominoes, since the number of positions when they are distinguishable is just a constant factor (4!^2) greater than when they are indistinguishable. But if you're counting accessible positions, this may no longer be the case. – mjqxxxx Jan 13 '11 at 1:34

A straightforward search yields the figure of $4392$ for all but the $1 \times 1$ stones. The former fill $14$ out of $20$ squares, so there are $\binom{6}{4} = 15$ possibilities to place the latter. In total, we get $$4392 \times 15 = 65880.$$ These can all be generated, and one can in principle calculate the number of connected components in the resulting graph, where the edges correspond to movements of pieces.