[seqfan] Re: Unique Products Regarding Binary Matrices
Rick Shepherd
rlshepherd2 at gmail.com
Fri Jun 11 16:43:22 CEST 2010
Hi Leroy,
I also highly suspect that solutions exist: It only took me a few tries by
hand to find a 9-by-9 with 16 unique products from the 18 possible {1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 24, 27}. If I have time
later, I'll give this more thought.
I also agree with Hugo's statement: Exchanging all 0s and 1s in any given
solution would also result in a solution (and of course there would be
symmetries of the square by rotating and flipping). Different sequences
could be generated by including or ignoring these.
If solutions do exist and they're not too numerous, I could imagine this
being a good somewhat Sudoku-like puzzle (with some entries given upfront).
Rick
On Thu, Jun 10, 2010 at 11:16 AM, Leroy Quet <q1qq2qqq3qqqq at yahoo.com>wrote:
> 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
>
>
>
> [ ( [ ([( [ ( ([[o0Oo0Ooo0Oo(0)oO0ooO0oO0o]]) ) ] )]) ] ) ]
>
>
> --- 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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> >
>
>
>
>
>
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