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Topic: Tiling the plane with checkerboard patterns
Replies: 21   Last Post: Jul 14, 2010 10:33 PM

 Messages: [ Previous | Next ]
 Rouben Rostamian Posts: 193 Registered: 12/6/04
Re: Tiling the plane with checkerboard patterns
Posted: Jul 11, 2010 12:39 PM

On Sat, Jul 10, 2010 at 05:11:58PM +0000, Avni Pllana wrote:
> On Fri, Jul 9, 2010 Rouben Rostamian wrote:
> > Here is the Maple's version of your formula:
> >
> > a := proc(n::posint)
> > local N, b;
> > N := n^2;
> > b := binomial(N,i);
> > return sum( iquo(b,N) + irem(b, N), i=0..N);
> > end proc:

>
> the term ? quo(b,N) + rem(b,N) ? gives enough insight into the
> nature of the problem. It is exactly the number of equivalence classes
> of nxn binary matrices that correspond to binomial(N,i) combinations
> with respect to horizontal and vertical shifts. Since there are (n
> horizontal shifts)x(n vertical shifts), the binomial(N,i) should be
> divided by N.
>
> I wrote a Matlab program implementing the above ideas and according to
> the numerical results for n=1,2,3, I was able to derive the general
> formula that fits perfectly to your numerical results for a(n),
> n=1,?,5. The general formula is to nice not to be true, and a formal
> proof of it ?is left to the reader?.

Avni, I agree with you that your formula for a(n) is too nice
not to be true and chances are good that it is true. I tried to
convince myself that it is correct but I couldn't quite get it.
The "irem(b,N)" terms in the formula accounts for patterns
which are invariant under certain shifts. It certainly does
the right thing for the cases n=1,2,3,4,5 because the formula
agrees with independently computed results. I would love to
prove that it is true in general but I am unable to.

Rouben

Date Subject Author
7/6/10 Rouben Rostamian
7/7/10 Avni Pllana
7/8/10 mark
7/9/10 Rouben Rostamian
7/9/10 Avni Pllana
7/9/10 Rouben Rostamian
7/10/10 Avni Pllana
7/10/10 mark
7/11/10 Rouben Rostamian
7/11/10 mark
7/11/10 Rouben Rostamian
7/11/10 mark
7/12/10 Rouben Rostamian
7/13/10 mark
7/14/10 Rouben Rostamian
7/14/10 mark
7/11/10 Rouben Rostamian
7/14/10 Avni Pllana
7/14/10 Rouben Rostamian
7/13/10 Mary Krimmel
7/14/10 Rouben Rostamian
7/14/10 mark