May be you want to use ` WeightedIntegerVectors(15,[1,1])` or restricted
partitions inside gap. This, for a single linear Diophantine equation like
yours might be a fast approach.
Pedro
On Friday, April 17, 2020 at 7:17:12 PM UTC+2, Bert Henry wrote:
>
> I have the equation
> x + y = 15
> an
On Sun, Apr 19, 2020 at 7:41 AM Bert Henry wrote:
>
>
> wow, I didn‘t expect, that may „simple“ problem needs such deep math. I will
> look for the math of polyhedrons to understand, what you wrote, because in
> some number-crosswords (I don‘t know the correct english word) you search for
>
wow, I didn‘t expect, that may „simple“ problem needs such deep math. I
will look for the math of polyhedrons to understand, what you wrote,
because in some number-crosswords (I don‘t know the correct english word)
you search for solutions of the m entioned type. Also you need it in some
Matthias is hinting at a possible reformulation
of the problem as finding integral points in a
polyhedron. Let me expand.
In RR^2, consider the set S of all (x, y) satisfying:
x >= 1
x <= 9
y >= 1
y <= 9
x + y = 15
or if one prefers,
-1 + x >= 0
@Matthias,
thanks for your answer, but I don‘t underrstand it. In your link, I can‘t
fInd the solution for my problem. Would you give me a hunt, where to search?
Am Freitag, 17. April 2020 19:17:12 UTC+2 schrieb Bert Henry:
>
> I have the equation
> x + y = 15
> an I'm looking for solution only
http://doc.sagemath.org/html/en/reference/discrete_geometry/sage/geometry/polyhedron/base.html#sage.geometry.polyhedron.base.Polyhedron_base.integral_points
On Friday, April 17, 2020 at 11:16:30 AM UTC-7, Bert Henry wrote:
>
> I tried it with
> var('x, y')
> assume(x,"integer")
> assume(x>0)
>
I tried it with
var('x, y')
assume(x,"integer")
assume(x>0)
assume(y, "integer")
assume(y>0)
solve(x+y==15,x,y)
The result was
(t_0, -t_0 + 15)
obviously right, but not 6,9 7,8 8,7 and 9,6
Am Freitag, 17. April 2020 19:17:12 UTC+2 schrieb Bert Henry:
>
> I have the equation
> x + y = 15
> an
