A Minimum problem for finite sets of real numbers with non-negative sum

Details A Minimum problem for finite sets of real numbers with non-negative sum A Minimum problem for finite sets of real numbers with non-negative sum

2011-02-23

arxiv.org/abs/1102.4761v1

Mathematics

Combinatorics

15 pages

Scientific paper

Let $n$ and $r$ be two integers such that $0 < r \le n$; we denote by $\gamma(n,r)$ [$\eta(n,r)$] the minimum [maximum] number of the non-negative partial sums of a sum $\sum_{1=1}^n a_i \ge 0$, where $a_1, \cdots, a_n$ are $n$ real numbers arbitrarily chosen in such a way that $r$ of them are non-negative and the remaining $n-r$ are negative. Inspired by some interesting extremal combinatorial sum problems raised by Manickam, Mikl\"os and Singhi in 1987 \cite{ManMik87} and 1988 \cite{ManSin88} we study the following two problems: \noindent$(P1)$ {\it which are the values of $\gamma(n,r)$ and $\eta(n,r)$ for each $n$ and $r$, $0 < r \le n$?} \noindent$(P2)$ {\it if $q$ is an integer such that $\gamma(n,r) \le q \le \eta(n,r)$, can we find $n$ real numbers $a_1, \cdots, a_n$, such that $r$ of them are non-negative and the remaining $n-r$ are negative with $\sum_{1=1}^n a_i \ge 0$, such that the number of the non-negative sums formed from these numbers is exactly $q$?} \noindent We prove that the solution of the problem $(P1)$ is given by $\gamma(n,r) = 2^{n-1}$ and $\eta(n,r) = 2^n - 2^{n-r}$. We provide a partial result of the latter problem showing that the answer is affirmative for the weighted boolean maps. With respect to the problem $(P2)$ such maps (that we will introduce in the present paper) can be considered a generalization of the multisets $a_1, \cdots, a_n$ with $\sum_{1=1}^n a_i \ge 0$. More precisely we prove that for each $q$ such that $\gamma(n,r) \le q \le \eta(n,r)$ there exists a weighted boolean map having exactly $q$ positive boolean values.

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