zero sum


Pebbling a cartesian product ★★★

Author(s): Graham

We let $p(G)$ denote the pebbling number of a graph $G$.

\begin{conjecture} $p(G_1 \Box G_2) \le p(G_1) p(G_2)$. \end{conjecture}

Keywords: pebbling; zero sum

Davenport's constant ★★★

Author(s):

For a finite (additive) abelian group $G$, the \emph{Davenport constant} of $G$, denoted $s(G)$, is the smallest integer $t$ so that every sequence of elements of $G$ with length $\ge t$ has a nontrivial subsequence which sums to zero.

\begin{conjecture} $s( {\mathbb Z}_n^d) = d(n-1) + 1$ \end{conjecture}

Keywords: Davenport constant; subsequence sum; zero sum

Bases of many weights ★★★

Author(s): Schrijver; Seymour

Let $G$ be an (additive) abelian group, and for every $S \subseteq G$ let ${\mathit stab}(S) = \{ g \in G : g + S = S \}$.

\begin{conjecture} Let $M$ be a matroid on $E$, let $w : E \rightarrow G$ be a map, put $S = \{ \sum_{b \in B} w(b) : B \mbox{ is a base} \}$ and $H = {\mathit stab}(S)$. Then $$|S| \ge |H| \left( 1 - rk(M) + \sum_{Q \in G/H} rk(w^{-1}(Q)) \right).$$ \end{conjecture}

Keywords: matroid; sumset; zero sum

Gao's theorem for nonabelian groups ★★

Author(s): DeVos

For every finite multiplicative group $G$, let $s(G)$ ($s'(G)$) denote the smallest integer $m$ so that every sequence of $m$ elements of $G$ has a subsequence of length $>0$ (length $|G|$) which has product equal to 1 in some order.

\begin{conjecture} $s'(G) = s(G) + |G| - 1$ for every finite group $G$. \end{conjecture}

Keywords: subsequence sum; zero sum

Few subsequence sums in Z_n x Z_n ★★

Author(s): Bollobas; Leader

\begin{conjecture} For every $0 \le t \le n-1$, the sequence in ${\mathbb Z}_n^2$ consisting of $n-1$ copes of $(1,0)$ and $t$ copies of $(0,1)$ has the fewest number of distinct subsequence sums over all zero-free sequences from ${\mathbb Z}_n^2$ of length $n-1+t$. \end{conjecture}

Keywords: subsequence sum; zero sum

Olson's Conjecture ★★

Author(s): Olson

\begin{conjecture} If $a_1,a_2,\ldots,a_{2n-1}$ is a sequence of elements from a multiplicative group of order $n$, then there exist $1 \le j_1 < j_2 \ldots < j_n \le 2n-1$ so that $\prod_{i=1}^n a_{j_i} = 1$. \end{conjecture}

Keywords: zero sum

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