When does a set of sums uniquely determine a set of values?

This is a follow up to a question about determining the set of values from a set of sums.

We consider the following setup:

Consider a vector $$A$$ (which you will not see) of $$n$$ positive integers. You are given the set of sums of the (contiguously indexed) subvectors of $$A$$. For example, say

$$A = (3,2,1,2)$$

The subvectors are $$(3),(2),(1),(2), (3,2), (2,1), (1,2),(3,2,1), (2,1,2),(3,2,1,2)$$. We would be given the sums $$\{1, 2, 3, 5, 6, 8\}$$. Let us call this set of sums $$f(A)$$.

It turns out it is not always possible to uniquely determine the set of integers in $$A$$ from the pair $$(f(A), n)$$? For example:

$$A = (1,1,3)$$ and $$B = (1,2,2)$$

both give the same set of sums $$f(A) = f(B) = \{1,2,3,4,5\}$$.

However in some cases it is possible. For example, if $$n=3$$ and $$f(C) = \{1,2,3,5,6\}$$ then we know that $$\operatorname{set}(C) = \{1, 2, 3\}$$.

This raises the following question:

Is it possible to characterise which pairs $$(f(X), n)$$ uniquely determine the set of values in the array $$X$$?

• Maybe it suffices to know that in the sum, there exist a one or a zero??? – user7427029 Oct 15 '18 at 22:50
• @user7427029 I don’t quite understand your comment. Zero isn’t allowed as all the values in the arrays are positive. Also there is a 1 in the set of sums for both A and B in my example so that idea doesn’t work. – felipa Oct 16 '18 at 6:45
• Observations: $a := \min_{c\in f(X)} c \in set(X)$ and $S := \sum_{i=1}^{n} X_i = \max_{c\in f(X)} c$. Maybe one can conclude $a = X_1$ or $a = X_n$, if $S - a \in set(X)$. And then this can be applied recursively? – Frozn Oct 16 '18 at 8:48
• I should not write comments nightly. I thought of something like what @Frozn proposes. Two conjectures / ideas (I'm in a hurry): The question asks for something like the fundamental theorem of arithmetic concerning addition. Zeckendorf's theorem may help. – user7427029 Oct 16 '18 at 17:01
• You may find useful results in links from a search for finite radon transform : google.com/… – Ethan Bolker Oct 17 '18 at 19:50