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Suppose you have N numbers, A1 .. An, you calculate all sums of 2, e.g A1 + A2, A1 + A3 etc., and name them S12, S13 etc., so you end up with n * (n - 1) / 2 sums,

My question is, given all these sub sums, how can I recover those N numbers,

If I have only 3 numbers, that's easy,

A1 = (S12 + S13 + S23) / 2 - S23
A3 = (S12 + S13 + S23) / 2 - S12
A2 = (S12 + S13 + S23) / 2 - S13

But when it comes more than 3 numbers, is there a more general formula that would work ? It looks like solving some complexed equation with linear algebra, but I'm not sure

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2 Answers 2

Compute $A_1$ using your method : $A_1 = \frac{S_{12} + S_{13} - S_{23}}2$

After that use : $\ A_{n+1}=S_{n,n+1}-A_n$

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Edit: Original version of this answer contained several mistakes. Hopefully I've cleared those out.

First you can compute the sum of all $n$ of the $a_i$; let's call this sum $S$. If you add up all the $s_{ij}$, you'll have added up each number $n-1$ times, so the result will be $(n-1)S$, and you can divide to find $S$.

Now if you want a particular $a_k$ -- well, what distinguishes it from the others? The $s_{kj}$ it appears in. So we'll add up all the $s_{kj}$; we add up all of them so we're not distinguishing any particular $a_i$ other than $a_k$. But if the others appear in an even mixture, this means we can use our computation of $S$ (which tells us what happens when each element appears in an even mixture). In this new sum we've just computed, each $a_i$ for $i\ne k$ will appear only once in the sum, but $a_k$ will appear $n-1$ times. Thus this sum is $S+(n-2)a_k$. Since you know $S$, you can solve for $a_k$.

If we put this all together, we find that $$a_k=\frac{\sum_j s_{kj}-\frac{\sum_{i<j} s_{ij}}{n-1}}{n-2}=\frac{1}{n-1}\sum_j s_{kj}-\frac{1}{(n-2)(n-1)}\sum_{\substack{i<j \\ i,j\ne k}}s_{ij}.$$

(Note that as per Raymond Manzoni's answer, this is way more information than needed to find each $a_i$; you've given us $\binom{n}{2}$ equations, when only $n$ are necessary. But this is the symmetric way of doing it.)

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