# Re: N -ROOM LIGHTS PROBLEM

*From*: wade@xxxxxxxxxx*Date*: 4 Apr 2006 06:11:17 -0700

Harris wrote:

However, if you just want to find any one solution (i.e. just find a way to light up all cells), just

apply my aforementioned solution once horizontally and once vertically, end then feed the rest of the

(N-2)x(N-2) new matrix to the same function in a recursive way. With a little care on initial and final

conditions, I think this can be solved in afew lines of code, and it can always be converted to a non-

recursive version by introducing a queue as container for "closed" and "open" sub-matrices.

Perhaps you could show an example for the N=3 case. You said that for

N mod 3 == 0, you would start with

010

I claim that for N=3 there is no solution that has 010 in the top row.

So how do you get from 010 to a valid solution?

Even simpler, for N=2, the only solution is

11

11

and I don't see how your '10' leads to that. Can you show your steps?

.

**Follow-Ups**:**Re: N -ROOM LIGHTS PROBLEM***From:*Harris

**References**:**N -ROOM LIGHTS PROBLEM***From:*ALIABBAS J PETIWALA

**Re: N -ROOM LIGHTS PROBLEM***From:*Harris

**Re: N -ROOM LIGHTS PROBLEM***From:*wade

**Re: N -ROOM LIGHTS PROBLEM***From:*Harris

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