Number of Arrangements in a Grid

Arrangements, sorting, packing, partitions, critical path analysis, networks, graphs, ...
Post Reply
User avatar
GenePeer
Posts: 112
Joined: Sat Apr 03, 2010 12:14 pm
Contact:

Number of Arrangements in a Grid

Post by GenePeer » Sun Apr 24, 2011 5:22 pm

There's a 5x5 grid and each cell (i,j) is either black or white. How many arrangements are such that all black cells are connected* with each other and same for white cells?

*Two cells are connected if I can reach one cell from the other cell moving through cells of the same colour, plus I can only move horizontally or vertically.

Valid arrangements:
WWWWW   BBBBB   BBWWW
WWWWW   BBBBB   WBWBW
WWWWW   BBBBB   WBBBW
WWWWW   BBBBB   WWBWW
WWWWW   BBBBB   WWWWW
Invalid arrangement:
BBBWW
WBWBW
WBBBW
WWBWW
WBWWW
PS: Can this be solved for a 7x7 grid?
Image

User avatar
jaap
Posts: 540
Joined: Tue Mar 25, 2008 3:57 pm
Contact:

Re: Number of Arrangements in a Grid

Post by jaap » Sun Apr 24, 2011 7:05 pm

Instead of looking at connected regions, it may be easier to look at the boundary between them. This must be a single non-intersecting path along the edges between the squares. It is either closed loop, or a path that starts and ends on the side of the large square.
It is still tricky to find out how many there are. There likely won't be a formula for it, and you would have to write a computer program to enumerate them all.

User avatar
GenePeer
Posts: 112
Joined: Sat Apr 03, 2010 12:14 pm
Contact:

Re: Number of Arrangements in a Grid

Post by GenePeer » Sun Apr 24, 2011 11:01 pm

jaap wrote:you would have to write a computer program to enumerate them all.
Even though I haven't had the time to try it, bruteforcing the 5x5 grid may be possible as there are only $2^{25}$ arrangements to check for connectivity using a DFS. I was hoping there would be something more elegant for the 7x7 grid. Will looking at the boundary between them speed up things? I don't know how I would implement the "path enumeration".

Btw, I was brainstorming on Problem 161 (View Problem) when I thought of this problem.
Image

User avatar
jaap
Posts: 540
Joined: Tue Mar 25, 2008 3:57 pm
Contact:

Re: Number of Arrangements in a Grid

Post by jaap » Mon Apr 25, 2011 7:38 am

GenePeer wrote:Will looking at the boundary between them speed up things?
I think so. It is easier to check that the boundary path doesn't intersect itself than it is to check that both regions are connected.

User avatar
elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

Re: Number of Arrangements in a Grid

Post by elendiastarman » Tue Apr 26, 2011 11:44 pm

Perhaps it would be easier still to just construct the boundary? Implementing such an enumeration tactic for that method would probably involve recursive functions...

Wait a minute...this is (kinda, maybe) easier than I thought. There can only be one boundary, which can be visualized as walls between cells. At each point where four cells meet (pad the outside with a ring of blank cells), there can either be no walls or two walls meeting there. Any other number means that there is either a superfluous wall or indicates an intersection. Surely it is a simpler matter to toggle those points as to whether or not there are walls meeting there and do so in such a way as to avoid having more than one boundary.
Want some
3.14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679...?
Image

TripleM
Posts: 382
Joined: Fri Sep 12, 2008 2:31 am

Re: Number of Arrangements in a Grid

Post by TripleM » Wed Apr 27, 2011 9:28 am

You can do this with a DP-based method. Loop through columns one by one, at each stage keeping track of the groups of connected squares of each colour.

An example, where we can label whites with numbers and blacks with letters:

Code: Select all

WW...
WB...
WW...
BB...
WW...
The first column corresponds to state (0,0,0,A,1). When we add on the next column, we reach state (0,A,0,B,1) - the first and third columns are already joined on the left.

Thus for each possible state, loop through all 2^5 options for the next column, work out what state you get when you combine them, make sure we don't 'cut off' any groups and by the time you reach the end you're done. There aren't that many possible states to work with assuming we label them in increasing order from top to bottom.

If that makes sense.

(Will easily work for 7x7 too).

Post Reply