[Magictour] Re: pandiagonal

Sir,
         Thanks for the mail.

> Since no closed magic tour on 12x12 is known to this date, I suggest that
we
> put our efforts on enumerating all semi-regular closed 12x12 magic tours.

It is to inform you that hundreds of closed magic tours on 12x12 is known.
Just go through 'The Games and Puzzles Journal' # 24.

> However, my first tests show that enumerating them is pretty slow, even
> though the same program solved the 136 start-end semi-regular magic tour
in
> 3 minutes (I can post the program if someone is interested).

I am interesred in the programme. Kindly send it to me.


With regards,

Yours truly,

Awani Kumar

awanieva@eth.net

12th August 2003

----- Original Message -----
From: "Jean-Charles Meyrignac" <euler@free.fr>
To: <magictour@ml.free.fr>
Sent: Sunday, August 10, 2003 2:31 PM
Subject: [Magictour] Re: pandiagonal


> >that's McKay,BTW., not to be confused with our Mackay.
> >And he only counts the closed tours, so that's just one of our
> >136 ranges. He gets 1.3e13 tours , with our 200 cycles
> >per knight-move that would be 2.6e15 cycles or 30 days.
> >But that's only for the maximum nodes in the tree.
> >Fei Lu gets 4e4 tours per second with 300MHz , so that's
> >1.3e5/s*GHz or 1 year with a 3GHz Computer, so it's doable.
> >Hmm, maybe Fei Lu is counting open tours too, I'll have to check.
> >Anyway... JC, Hugues , is that a project for us ?!?
>
> It may be.
> Fei Lu used a trick to reduce the number of nodes, because of some
> properties of closed knight tours.
> I already implemented it, and could possibly add it to my ASM generator.
> Since no closed magic tour on 12x12 is known to this date, I suggest that
we
> put our efforts on enumerating all semi-regular closed 12x12 magic tours.
> However, my first tests show that enumerating them is pretty slow, even
> though the same program solved the 136 start-end semi-regular magic tour
in
> 3 minutes (I can post the program if someone is interested).
> Having a long running time leads to the problem of saving the computation
> state, since when you shut down your computer, all computation is lost.
This
> is why we need to find a way to either add other constraints, or split the
> computation ranges into smaller computation ranges. I'm open to any idea.
>
> One day, I'll start the writing of a database which will enumerate the
first
> jumps, and every client will connect to this database, get one or more
> starting tours, compute them and return them to the server.
> This way, we can really finetune the program so that a range takes a given
> amount of time !
> (This is a lot of work !)
> Perhaps will I write this server for the non-crossing knight tour problem
> first, then use it for the tours enumeration.
> This way, we could enumerate all possible 8x8 tours.
>
> > John Beasley, in an afterthought to a recent e-mail writes:
> >
> > Here's a thought. A pan-diagonal nxn square must satisfy 4n
constraints,
> > which goes up linearly with n. The number of knight's tours on an nxn
> board
> > goes up (I would think) by something not too far short of factorial-n.
> Does
> > there exist a suitably huge N such that a knight's tour forming a
> > pan-diagonal magic square is possible?
> >
> I think that a pandiagonal magic square may exist on the 12x12 board, and
> DOES exist on the 16x16 board.
> Enumerating regular 12x12 magic squares will surely lead to interesting
> discoveries in this area.
> 16x16 is currently too big to compute in reasonable time, except if we
find
> good constraints.
>
> JC
>
>
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Received on Thu Dec 04 14:24:18 2001