Optimal Index Codes for a Class of Multicast Networks with Receiver Side Information

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5 jointly as follows: (i) execute Step 1 in the SCGC to get G ; (ii) execute Step 2 in the SCGC to get a trail T ; (iii) execute Step 3 in the SCGC but instead of appending the trail T, we append T = T \({k 1 }, {(k 1, k 2 )}); (iv) remove an arc from the cycle chosen in (i). It is clear that this joint algorithm produces G. Now, after step (i), G is a cycle and hence od max = 1. Next, we consider each iteration (ii) (iii). We show by induction that if G has od max = 1, after appending T, (G T ) has od max = 1. As T can only assume a path or a cycle, T has od max = 1, and so does T. Since G T = ({k K }, ), only the vertex k K can change its outdegree when we append T to G. But the last arc (k K 1, k K ) on trail T meets the vertex k K as an incoming arc, so its od is in fact not changed. Hence, (G T ) has od max = 1. Lemma 5: Given a strongly connected graph G, after Algorithm 1 the number of arcs in the resulting graph G equals the number of vertices in G less one, i.e., A(G ) = V (G ) 1. Proof: From SCGC, after Step 1, the number of vertices equals the number of arcs. For each trail with K arcs appended to G in Step 3 of SCGC, we add K new arcs and K 1 new vertices to G. However, because the first arc will be removed in Algorithm 1, effectively, the number of vertices added to G equals the number of arcs added to G. In the last step in Algorithm 1, an arc is removed from the cycle added in Step 1 of SCGC. Hence, the number of arcs in the resulting graph G equals the number of vertices in G less one. 3) Getting the Lower Bound: Now, for any single-uniprior problem represented by a strongly connected graph G = (V, A), we execute Algorithm 1 to obtain G = (V, A ). We have l (G) (a) l (G ) (b) = A(G ) (c) = V (G ) 1 (d) = V (G) 1, where (a) is because A A, (b) follows from Theorem 1 as G is acyclic with od max = 1, (c) follows from Lemma 5, and (d) is because we only prune arcs but not vertices in Algorithm 1. Combining this lower bound with the upper bound in Sec. III-A, we have Theorem 2. Remark 5: Note that the SCGC and Algorithm 1 need not be executed for actual coding. They are only used to show that we are able to prune some arcs off a strongly connected graph until the number of arcs equals the number of nodes less one, and the resultant graph is acyclic with od max = 1. IV. GENERAL GRAPHS We now generalize the results in the previous sections to any single-uniprior problem represented by a graph G. A. The Pruning Algorithm To build on the earlier results for specific classes of graphs, we introduce Algorithm 2 that first prunes G to get G. It can be shown that Algorithm 2 runs in polynomial time with respect to the number of vertices. After Algorithm 2 terminates, we get G = G sub (G \ G sub ) where G sub N sub G sub,i is a graph consisting of strongly connected components {G sub,i }, and G \G sub (V \V sub, A \A sub ). Note that G \ G sub might not be a graph as there might exist an arc (i, j) G \ G sub where i G \ G sub and j G sub. We Initialization: G = (V, A ) G = (V, A) ; 1) Iteration: while there exists a vertex i V with (i) more than one outgoing arc, and (ii) an outgoing arc that does not belong to any cycle [denote any such arc by (i, j)] do remove, from G, all outgoing arcs of vertex i except for the arc (i, j); end 2) label each (non-trivial) strongly connected component in G as G sub,i, i {1, 2,..., N sub}; Algorithm 2: The Pruning Algorithm make the following observations. Observation 1: The sets {G sub,1, G sub,2,..., G sub,n sub, G \ G sub } are vertex and arc disjoint. Observation 2: After Step 1 in Algorithm 2, all vertices with od > 1 in G have all outgoing arcs belonging to some cycles. So all these vertices and arcs will eventually be in G sub. We now state the following main result of this paper: Theorem 3: For any single-uniprior problem, which can be represented by a graph G, after executing Algorithm 2 we have N sub l (G) = (V (G sub,i) 1) + A(G \ G sub). (5) Here, A(G \ G sub ) is the number of arcs in A \ A sub. Remark 6: If G is acyclic with od max = 1, then N sub = 0 and G = G \ G sub. Thus, we recover Theorem 1. If G is strongly connected, then N sub = 1, G sub,1 = G sub = G = G, and G \ G sub = (, ). Thus, we recover Theorem 2. We prove Theorem 3 in the following two subsections. B. Lower Bound Now, for any graph G, we first execute Algorithm 2 to get G = N sub G sub,i (G \ G sub ). For each strongly connected components G sub,i, we execute Algorithm 1 to get G sub,i. Let the final graph be G = N sub G sub,i (G \ G sub ). Then N sub A(G ) = (A(G sub,i)) + A(G \ G sub) N sub = (V (G sub,i) 1) + A(G \ G sub) N sub (6a) (6b) = (V (G sub,i) 1) + A(G \ G sub), (6c) where (6a) follows from Observation 1, (6b) follows from Lemma 5, and (6c) follows because we prune arcs but not vertices in Algorithm 1, so V (G sub,i ) = V (G sub,i ), for all i {1,..., N sub }. Following from Observation 2, each vertex in G \ G sub has od 1. After executing Algorithm 1, from Lemma 4, all G sub,i has od max = 1. Hence, G has od max = 1.

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