On 11/7/07, anvera <[EMAIL PROTECTED]> wrote:
>
>
> Why not? Does order really matters here? Look at the symmetry of the
> problem. Put 3,4,5,5 and then 4,5,6,7 at the right side. Look at the
> solutions. How they differ? Is this natural?


Though the order is not imp you cant tell which 2 variables for a particular
equation.


On Nov 7, 6:55 pm, Andrey <[EMAIL PROTECTED]> wrote:
> > Won't work.
> > You just can't build such a system, because you don't know in which
> > order values should appear in right part of system. Say, we have the
> > follwing input of 6 numbers: 3, 4, 5, 5, 6, 7 and we are supposed to
> > find 4 values (4 * (4 - 1)  /  2 = 6) x1, x2, x3, x4
> >
> > What the system will look like?
> >
> > x1 + x2                 = ?
> > x1         + x3         = ?
> > x1                 + x4 = ?
> >         x2 + x3         = ?
> >         x2         + x4 = ?
> >                 x3 + x4 = ?
> >
> > On 7 нояб, 20:16, anvera <[EMAIL PROTECTED]> wrote:
> >
> > > I have not developed entirely the idea, but I am sure it works.
> > > Just write the corresponding linear system. You will have n unknowns
> > > and n(n-1)/2 equations. Provided that the system is consistent you can
> > > find a solution by Gaussian elimination.
> > > For the complexity, you can do it in less than n^3/3 +O(n^2)
> > > operations, because it is simpler than just inverting a nxn matrix.
> > > Inverting the matrix can let you solve the problem for many instances.
> > > It would be interesting to exploit the special structure of this
> > > matrix in order to speed up the computation. Please post if you find
> > > such an improvement.
> >
> > > Best Regards,
> >
> > > Antonio
> >
> > > On Nov 7, 5:16 pm, Andrey <[EMAIL PROTECTED]> wrote:
> >
> > > > Any set of n integers form n(n - 1)/2 sums by adding every possible
> > > > pair.
> > > > The task is to find the n integers given the set of sums.
> >
> > > > Any ideas?
> >
> > > > I've found out the solution but I doubt it the best one...
>
>
> >
>


-- 
Ciao,
Ajinkya

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