thnx.
I see that the problem can be also formulated as marix problem. but
the way i did it is in this case the more natural one for me.
is there any reason why it only works this way and solve does not lead
to any result?
greatz Johannes
On 18 Jul., 14:45, luisfe <lftab...@yahoo.es> wrote:
> On Jul 16, 1:33 am, Johannes <dajo.m...@web.de> wrote:
>
>
>
>
>
>
>
>
>
> > a very easy example would be this:
>
> > sage: p1 = vector([-3,1,1])
> > sage: p2 = vector([1,-3,1])
> > sage: p = vector([0,-2,1])
>
> > #now i'm looking for some x,y such that
> > #x * p1 + y * p1 == p
>
> > x,y = var('x,y')
> > sage: assume(x > 0)
> > sage: assume(y > 0)
> > sage: solve([x * p1 + y * p2 == p],x,y)
> > []
>
> > #but: x = 1/4 and y = 3/4 is a solution of this equation.
>
> > in the end i need this kind of calculation for every latticepoint on the
> > border of a lattice-simplex. like the example above shows how it should
> > work for a line, it also should be extendable to n+1 points on each n
> > dimensional facet of the simplex, where the points p0,....,pn are given
> > as the vertices of the facet.
>
> > greatz Johannes
>
> This is a linear algebra problem. You have a vector p that is a linear
> combinations of others p1,p2 and you want the coordinates of p in
> terms of B. This is only a change of basis problem. Assuming that the
> points of B form a basis of the linear span of B you can do:
>
> sage: m = matrix([p1,p2])
> sage: m.solve_left(p)
> (1/4, 3/4)
>
> The rows of m are the vectors p1 and p2. You want to express p as
> combination of these rows.
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