Thanks for the reply. However, I'm not so sure about the intention
part of the comment
I got the solve(x+y==3, x,y), i.e. asking solve for more than one
variables strict from the current documentation (solve? )
except I dropped the redundant equation 2x+2y==6.
Why don't solve just call the main solve() routine? but calling
Expression.solve() instead.
I think solve(x+y==3,x,y) should give
[[x == -r1 + 3, y == r1]]
while solve(x+y==3,y,x) should give
[[y == -r2 + 3, x == r2]]
Currently these can be achieved by add a redundant equation, e.g. 1==1
to the list.
On Sep 13, 9:34 am, kcrisman <kcris...@gmail.com> wrote:
> On Sep 13, 12:08 pm, Pong <wypon...@gmail.com> wrote:
>
> > y,z=var('y,z'); solve(6*x + 10*y + 15*z ==1,x,y,z) gives
> > ([{x: -5/3*y - 5/2*z + 1/6}], [1])
>
> So wacky. Definitely a bug, needless to say.
>
> > ([x == -y + 3], [1])
>
> > My questions are:
> > 1) Why the notation are different in the 2 and 3-variable case? One
> > gives x: and the other x==
>
> Well, you are asking for solutions to both x *and* y, but the solve is
> really designed to solve one in terms of the other(s). I just think
> no one has pointed out this weirdness before. So you are using it in
> a way it wasn't intended. I'm not sure whether we should add the
> use, or have a warning, or both, or neither, or documentation, or...
>
> > 2) What the [1] in both cases stand for ?
>
> Compare:
>
> sage: a = x+y==3
> sage: type(a)
> <type 'sage.symbolic.expression.Expression'>
> sage: a.solve(x,y)
> ([x == -y + 3], [1])
> sage: a.solve(x,False)
> [x == -y + 3]
> sage: a.solve(x,True)
> ([x == -y + 3], [1])
>
> So we are giving a non-False input to the second argument (whether we
> should get multiplicities) and so we get a multiplicity of 1. Which
> doesn't really make sense anyway.
>
> Maybe we should put in a check for the expression having more than one
> variable before we send it to a.solve(), or maybe it just needs that
> long-awaited rewrite... seehttp://trac.sagemath.org/sage_trac/wiki/symbolics
> for some issues with solve.
>
> > 3) In order to get the parameters appear in the solution, one need to
> > add a redundant equation. Shall we improve on that?
> > e.g.
> > solve([x+y==3,2*x+2*y==6],x,y)
> > [[x == -r1 + 3, y == r1]]
>
> I think this is because when you do that, now the input is multiple,
> and instead of doing Expression.solve(), it does the main solve()
> routine, which deals with this.
>
> Thoughts from others who care about "solve"? I'm not quite sure what
> the best solution is. The easiest one is making really big letters in
> the documentation that say "don't do this" and trying to catch these
> cases.
>
> - kcrisman
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