Linear Index Codes are Optimal Up To Five Nodes Lawrence Ong The University of Newcastle, Australia March 05 BIRS Workshop: Between Shannon and Hamming: Network Information Theory and Combinatorics
Unicast Index Coding x, x,..., x n }{{}, x i p bits Sender index code βp bits K K K n n x x x n What is the minimum β, denoted as β? (broadcast rate / coded packets) Lawrence Ong /6
Directed Side-Information Graph Example: i j Node i knows x j x, x, x }{{} Sender x x, x x x x Lawrence Ong /6
Undirected Side-Information Graph For symmetrical side information i j i j Node i knows x j & Node j knows x i Example: x, x, x }{{} Sender x x,x x x x x Lawrence Ong /6
Problem Formulation Using Graphs Index-Coding Problem Given G and p, find β (G, p). Problem open to date, except for special cases, e.g., Bar-Yossef et al. (0): Directed acyclic Undirected perfect Undirected odd hole Undirected odd anti-hole Lawrence Ong /6
Unicast: Results for Up To Five Nodes β (G, p) found for the following: G = n total non-isomorphic packet size (p). 5 6 8 9608 9 Bar-Yossef et al. 0 Arbabjolfaei et al. 0 Lawrence Ong 5/6
Our Recent Results We found β (G, p) for the following: G = n 5 total non-isomorphic 6 8 9608 9 (8) packet size (p). Bar-Yossef et al. 0 Arbabjolfaei et al. 0 Ong 0 Lawrence Ong 6/6
Up to Five Users (Ong, 0) G = n 5 total non-isomorphic 6 8 9608 9 (8) packet size (p). Bar-Yossef et al. 0 Arbabjolfaei et al. 0 Ong 0 For 988 graphs, bit-wise linear encoding is optimal For the rest (8 graphs), double-bit linear encoding is optimal Results on achieving β (G, ) using sufficiently large p omitted here (Yu & Neely 0, Arbabjolfaei & Kim 0) Lawrence Ong 7/6
Maximum Acyclic Induced Subgraph (MAIS) A Lower Bound to β (Bar-Yossef et al. 0) Example: MAIS(G) β (G, p), for any p Graph G MAIS(G) = β (G, p) Lawrence Ong 8/6
A Key Result (Ong, 0) For any G and any p, if MAIS(G) n, then MAIS(G) = β (G, p), which is achievable by bit-wise linear codes. Any graph Remove at most nodes Acyclic graph Lawrence Ong 9/6
A brief proof for the following:. If MAIS(G) n, then MAIS(G) = β (G, p).. Linear codes are optimal up to five nodes. Lawrence Ong 0/6
Proof: If MAIS(G) n then MAIS(G) = β (G, p) Case : MAIS(G) = n G is acyclic Send n symbols uncoded Example: (x, x, x ) β (G, p) = Lawrence Ong /6
Proof: If MAIS(G) n then MAIS(G) = β (G, p) Case : MAIS(G) = n G contains a cycle Use cyclic code on one cycle to save one symbol Example: (x x, x x ) β (G, p) = Lawrence Ong /6
Proof: If MAIS(G) n then MAIS(G) = β (G, p) Case a: MAIS(G) = n & there are two disjoint cycles Use cyclic code on each disjoint cycle to save two symbols Example: (x x, x x, x x 5, x 5 x 6 ) 5 6 β (G, p) = Lawrence Ong /6
Proof: If MAIS(G) n then MAIS(G) = β (G, p) Case b: MAIS(G) = n & there are NO two disjoint cycles Problem: Cannot use cyclic codes to save two symbols Lawrence Ong /6
A Solution for Case b Lemma If and MAIS(G) = n there are NO two disjoint cycles, then G contains a subgraph of the following form: Lawrence Ong 5/6
Suppose Lemma is true. Let the subgraph G contain n n nodes. Code with n symbols Send n n uncoded symbols Total code length = (n ) + (n n ) = n. Save two symbols. Lawrence Ong 6/6
A Coding Scheme on the Subgraph G 5 a b c x a x b x c 6 Send x i x j, for each vertex i except the two red nodes, j N + G (i) where N + G (i) = out-neighbourhood. Lawrence Ong 7/6
Sketch of Proof for Lemma (Existence of Subgraph) MAIS(G) = n cycles Pick a centre cycle. No two disjoint cycles All other cycles share nodes with centre cycle. one cycle sharing node with centre cycle all other cycles share nodes with centre cycle Lawrence Ong 8/6
Theorem For any G and any p, if then MAIS(G) n, MAIS(G) = β (G, p) & bit-wise linear codes are optimal. Lawrence Ong 9/6
Proof: Linear Codes Are Optimal for n 5 For any G, MAIS(G) n n =,, MAIS(G) n β (G, p) found n = If MAIS(G) =,, MAIS(G) n β (G, p) found Else (MAIS(G) = ) Any two nodes contains a cycle Each node i knows all x j, j i MAIS(G) = achievable using x x x n β (G, p) found Lawrence Ong 0/6
Proof: Linear Codes Are Optimal for n 5 For n = 5 If MAIS(G) =,, 5 MAIS(G) n β (G) found If MAIS(G) = Any two nodes contains a cycle x x x n optimal Need to solve (n = 5) & (MAIS = ) Lawrence Ong /6
(n = 5) & (MAIS = ): Classification Case : No undirected cycle Case : undirected -cycle 5 5 Case : No undirected -cycle -cycle Case : No undirected -, -cycle 5-cycle 5 5 Lawrence Ong /6
(n = 5) & (MAIS = ): Lemmas Lemma Any induced subgraph with nodes must contain a cycle. Lemma Any induced subgraph with nodes must contain an (undirected) edge. Example: 5 Lawrence Ong /6
(n = 5) & (MAIS = ): Necessary Arcs and Edges Example: No undirected cycle and edges Only non-isomorphic configurations: 5 5 5 Lawrence Ong /6
(n = 5) & (MAIS = ): Necessary Arcs and Edges Example: No undirected cycle and edges Applying Lemmas and : 5 5 {,,, 5} has no edge x x x 5 not possible x x x 5 5 four sub-cases Lawrence Ong 5/6
(n = 5) & (MAIS = ): Classified graphs according to undirected cycles Need to consider only configurations number of nodes (n) 5 total non-isomorphic 6 8 9608 8 0 β = packet size (p). β =.5 β =.5 MAIS lower bound is tight Lawrence Ong 6/6
Thank you!