zero sum


Pebbling a cartesian product ★★★

Author(s): Graham

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

Conjecture   $ p(G_1 \Box G_2) \le p(G_1) p(G_2) $.

Keywords: pebbling; zero sum

Davenport's constant ★★★

Author(s):

For a finite (additive) abelian group $ G $, the 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.

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

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 \} $.

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 $$|S| \ge |H| \left( 1 - rk(M) + \sum_{Q \in G/H} rk(w^{-1}(Q)) \right).$$

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.

Conjecture   $ s'(G) = s(G) + |G| - 1 $ for every finite group $ G $.

Keywords: subsequence sum; zero sum

Few subsequence sums in Z_n x Z_n ★★

Author(s): Bollobas; Leader

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 $.

Keywords: subsequence sum; zero sum

Olson's Conjecture ★★

Author(s): Olson

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 $.

Keywords: zero sum

Syndicate content