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Algorithms for Radio Networks Exercise 6. Stefan Rührup sr@upb.de. The Yao-Family. Yao -Graph nearest neighbor in each sector Spanner. ⊇ SparsY Sparsified Yao-Graph use only the shortest ingoing edges weak- & power-Spanner , constant in-degree. ⊇ SymmY
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Algorithms for Radio NetworksExercise 6 Stefan Rührup sr@upb.de
The Yao-Family Yao-Graph nearest neighborin each sector Spanner ⊇ SparsY Sparsified Yao-Graph use only the shortestingoing edges weak- & power-Spanner, constant in-degree ⊇SymmY Symmetric Yao-Graph only symmetric edges not a spanner, nor weak spanner, nor power-spanner
Exercise 12 • Prove that the Symmetric Yao Graph is connected for k>6 sectors! SymmY is not a Spanner: vm um only symmetric edges v2 u2 v1 u1
u v Exercise 12 • Prove that the Symmetric Yao Graph is connected for k>6 sectors! Proof: Induction over the distance of vertices • Edge connecting the closest pair is part of SymmY • Consider two nodes u,v: • Either u and v are connected orthere is a path from u to w and a path from w to v Case 1: Case 2: w V: |u-w| < |u-v| k>6 |v-w| < |u-v| w u v
Exercise 13 • Which implications are true (for appropriate constants c,c’,c’’ and d 2)? • Spanner weak c’-Spanner • weak c’-Spanner c-Spanner • weak c’-Spanner (c’’,d)-Power Spanner • (c’’,d)-Power Spanner weak c’-Spanner • c-Spanner (c’’,d)-Power Spanner • (c’’,d)-Power Spanner c-Spanner If an implication is not true, then give a counterexample!
v3 v2 v4 1/2 1/3 1 v1 1 vn Exercise 13 Weak Spanner, but not a Spanner: Spanner X Koch Curve Weak Spanner X Power Spanner, but not a Weak Spanner: Power Spanner Circular node chain with |vi - vi+1| = 1/i
Construction of the HL Graph • Every node on layer i is dominated by some node in layer i+1 • No nodes may dominate each other (the priority decides) • Edges are inserted in the publication radius of each node L1 publication radius L1 domination radius L1 node L0 node L1 edge
Radii and Edges of the HL Graph • Li-1 publication radius ≥ Li domination radius: ≥ > 1 • Layer-i edges are established in between r1 layer-1 domination radius L1 node L0/L1 edge r0 layer-0 domination radius L0 node L0 edge L1 edge L1 node · r0 L1 node
Exercise 14 • Draw the HL Graph of v1,...,v6 for ==2! v1 v2 v6 1+ 2+2 v5 v3 v4
Exercise 14 layer-0-edges of v2 v1 v6 v2 L0-domination v5 v3 v4 L0-publication/L1-domination
Exercise 14 No other node with higher rank within the domination radius: 6 v1 v6 v2 3 4 L0-domination v5 v3 2 1 v4 5 L0-publication/L1-domination
Exercise 14 node v1 has highest priority and becomes layer-1 node 6 v1 L0-domination v6 v2 3 4 L0-publication/L1-domination v5 v3 2 1 v4 5
Exercise 14 V(L0) = {v1,...,v6} V(L1) = {v1,v4} V(L2) = {v1} 6 v1 layer-0 edges layer-1 edges v6 v2 3 4 v5 v3 2 1 v4 5
Exercise 14 The same node set with other priorities: v5 is dominated by v3, so it will remain on layer 0 and v6 can become layer-1 node. 1 v1 V(L0) = {v1,...,v6} V(L1) = {v3,v6} V(L2) = {v3} v6 v2 4 3 layer-0 edges layer-1 edges v5 v3 5 6 v4 2