tony . reix on Fri, 17 Aug 2018 17:56:07 +0200


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Re: Equation solver


My mistake.
That should have been:  N=6, C(i,j)=0 or 1, with i=0..N-1, j=0..N-1).

Sorry, the output is produced by an automatic tool. I cannot give variables a name based on letters.
Here are equations for N=2, 3, 4, with a solution for each.

N=2 :
C(0,0) + C(0,1) + C(1,0) + C(1,1) != 0 (mod 4)

0  0 
0  1

N=3 :
C(0,0) + C(0,1) + C(1,0) + C(1,1) != 0 (mod 4)
C(0,1) + C(0,2) + C(1,1) + C(1,2) != 0 (mod 4)
C(1,0) + C(1,1) + C(2,0) + C(2,1) != 0 (mod 4)
C(0,0) + C(0,2) + C(2,0) + C(2,2) != 0 (mod 4)
C(1,1) + C(1,2) + C(2,1) + C(2,2) != 0 (mod 4)

0  0  0
0  1  0
0  0  1

N=4:
C(0,0) + C(0,1) + C(1,0) + C(1,1) != 0 (mod 4)
C(0,1) + C(0,2) + C(1,1) + C(1,2) != 0 (mod 4)
C(0,2) + C(0,3) + C(1,2) + C(1,3) != 0 (mod 4)
C(1,0) + C(1,1) + C(2,0) + C(2,1) != 0 (mod 4)
C(0,0) + C(0,2) + C(2,0) + C(2,2) != 0 (mod 4)
C(1,1) + C(1,2) + C(2,1) + C(2,2) != 0 (mod 4)
C(0,1) + C(0,3) + C(2,1) + C(2,3) != 0 (mod 4)
C(1,2) + C(1,3) + C(2,2) + C(2,3) != 0 (mod 4)
C(2,0) + C(2,1) + C(3,0) + C(3,1) != 0 (mod 4)
C(1,0) + C(1,2) + C(3,0) + C(3,2) != 0 (mod 4)
C(2,1) + C(2,2) + C(3,1) + C(3,2) != 0 (mod 4)
C(0,0) + C(0,3) + C(3,0) + C(3,3) != 0 (mod 4)
C(1,1) + C(1,3) + C(3,1) + C(3,3) != 0 (mod 4)
C(2,2) + C(2,3) + C(3,2) + C(3,3) != 0 (mod 4)

 0  0  0  0
 0  1  0  1
 0  0  1  1
 1  0  0  0

Regards,

Tony


De: "Jacques Gélinas" <jacquesg00@hotmail.com>
À: "tony reix" <tony.reix@laposte.net>, pari-users@pari.math.u-bordeaux.fr
Envoyé: Vendredi 17 Août 2018 15:04:34
Objet: RE: Equation solver

> I'm looking for an "Equation Solver" that would be able to solve equations like below
> (where: N=6, C(i,j)=0 or 1, with i=0..N-1, j=0..N1).

What is N1 ?

Please give the system of equations for N=1, N=2, and N=3 using variables such as x,y,z,...

In the example you gave, the (mod 4) could be deleted by rejecting the (0,0,0,...) and (1,1,1,...)
solutions later. We thus get a system of equations for variables restricted to the vertices of
the unit hypercube.

And yes GP/PARI has sophisticated functions to explore this kind of space.
Try
                  ?? forvec

to start (new functions are added at each release, it seems).

Jacques Gélinas