The Pebble Game Geri Grolinger York University

The Pebble Game Used for studying time-space trade-off Used for studying time-space trade-off One player game, played on a DAG One player game, played on a DAG

Output nodes Nodes Input nodes Formalization: Directed acyclic graph Directed acyclic graph Bounded in-degree Bounded in-degree

Three main rules: 2. A pebble can be placed on a node v if all predecessors of the node v are marked with pebbles 3. A pebble can be removed from a node at any time Note: a pebble removed from the graph can be ‘reused’ 1. A pebble can be placed on any input node on any input node

Strategy: sequence of legal moves which ends in pebbling the distinguished node f The Goal: to place a pebble on some previously distinguished node f while minimizing the number of pebbles used A move: placing or removing one of the pebbles according to the three given rules f

moves and 7 pebbles Example 1

moves and 3 pebbles Example 2

Interpretation: 1. A pebble can be placed on any input node ~ LOAD on any input node ~ LOAD 2. A pebble can be placed on a node v if all predecessors of the node a node v if all predecessors of the node v are marked with pebbles ~ COMPUTE v are marked with pebbles ~ COMPUTE 3. A pebble can be removed form a node at any time ~ DELETE ~ # REGISERS Use as few pebbles as possible ~ # REGISERS ~ TIME Use as few moves as possible ~ TIME input nodes nodes output nodes

In general: How many pebbles are required to pebble a graph with n nodes? with n nodes?

Pyramid graph P k :

Fact 1: Every pebbling strategy for Pk (k > 1) must use AT LEAST k + 1 pebbles. That is Ω ( ) pebbles expressed in number of edges n. n √

Pyramid graph P k : k = 5 We need at least: k + 1 = 6

Pyramid graph P k : Let’s consider having k = 5 pebbles

Arbitrary graph with restricted in-degree (d =2): Fact 2: Number of pebbles needed to pebble a graph of in-degree 2 is O(n/log n) (n = # nodes in the graph).

Arbitrary graph with restricted in-degree (d =2): Proof : Recursive pebbling strategy Cases Recursions for each case Solutions: P(n) ≤ cn / log n = O(n/log n) O(n/log n)

References: 1.Gems of theoretical computer science U. Schöning, R. J. Pruim 2. Asymptotically Tight Bounds on Time-Space Trade-offs in a Pebble Game T. Lengauer, R. E. Tarjan 3. Theoretical Models 2002/03 P. van Emde Boas

Thank you for your attention Thank you for your attention Questions ?