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?

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...


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