[seqfan] Re: Unique Products Regarding Binary Matrices

[Leroy Quet]

I'm actually most interested in the 9-by-9 case.

I am wondering if it would make a good puzzle, trying to come up with a solution by hand.

But I don't even know if it is possible, although I highly suspect that it is.

Thanks,
Leroy Quet



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--- On Thu, 6/10/10, Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:

> From: Leroy Quet <q1qq2qqq3qqqq at yahoo.com>
> Subject: [seqfan]  Unique Products Regarding Binary Matrices
> To: seqfan at seqfan.eu
> Date: Thursday, June 10, 2010, 11:24 AM
> Say we have an n-by-n binary matrix
> (all elements either 0 or 1).
> 
> For a given row of the matrix, take the lengths of the runs
> of 0's and 1's and multiply these lengths. 
> (By "run", it is meant a string of consecutive elements in
> the row (or column) all of the same value b, bounded by the
> value 1-b or by the edge of the row (or column).)
> 
> Do this for all rows and all columns to get 2n products.
> 
> Let a(n) = the number of such n-by-n binary matrices such
> that the 2n products are all unique.
> 
> I know that a(n) = 0 for n <= 8, since A034891(n) <
> 2n for n <= 8.
> 
> Is {a(n)} in the EIS already? It would seem a bad idea to
> compute this sequence via brute-force search of all the
> 2^(n^2) matrices for a given n, since the number of
> potential matrices grows so quickly as n grows.
> 
> Thanks,
> Leroy Quet
> 
> 
> [ ( [ ([( [ ( ([[o0Oo0Ooo0Oo(0)oO0ooO0oO0o]]) ) ] )]) ] )
> ]
> 
> 
>       
> 
> 
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