The Complexity of Graph Pebbling

Pebbling and optimal pebbling in graphs

“…are the indegree and outdegree of v under H. When H is orderable under D (by σ), the balance of any vertex v is nonnegative, since it is the number of pebbles at v after applying σ. The No-Cycle Lemma states that if H is orderable under D, then it has an acyclic subgraph H that is orderable under D and gives balance to each vertex at least as large as does H. The lemma was proved in Crull et al [3] and in Moews [11] and has a short proof in Milans and Clark [10].…”

“…The No-Cycle Lemma states that if H is orderable under D, then it has an acyclic subgraph H ′ that is orderable under D and gives balance to each vertex at least as large as does H. The lemma was proved in [3] and in [11] and has a short proof in [10].…”

“…Moews [12] proved that (4/3) k ≤ Π OP T (Q k ) ≤ (4/3) k+O(log k) and proved a related result for Π OP T on cartesian product graphs. Milans and Clark [10] proved that computing Π OP T is NP-hard on arbitrary graphs.…”

“…For a tree, Chung [4] showed how to calculate the pebbling number from the set of decompositions into paths. For general graphs, Milans and Clark [10] showed that recognizing Π (G) ≤ k is a Π P 2 -complete problem, meaning that it is complete for the class of languages whose complements are recognizable in polynomial time by nondeterministic machines.…”

Graham’s pebbling conjecture on products of many cycles

“…We show that t pebbles cannot be moved to (x 0 , x 0 ) starting from D. Suppose that we are given a sequence of pebbling moves which starts from D and ends with t or more pebbles on (x 0 , x 0 ). Using the methods from Section 2 of [6], we may, perhaps after omitting certain moves from this sequence, reorder the sequence so that it consists of a concatenation of t subsequences, each of which starts with a sequence of moves which moves two pebbles to a vertex adjacent to (x 0 , x 0 ) and finishes by moving one pebble from this vertex to (x 0 , x 0 ), leaving no pebbles anywhere other than (x 0 , x 0 ), (x 2 , x 2 ), and (x 3 , x 3 ).…”