HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Algorithms for Radio Networks Exercise 7 Stefan Rührup

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 2 Mini-Exam No. 2 X X X X X Task 1 –Which graphs are Weak c-Spanners?  Yao (k>6)  SparsY (k>6)  SymmY (k>6) –Which implications are true (for appropriate constants c, c’, c’’ and d ≥ 2)?  weak c’-Spanner  (c’’,d)-Power Spanner  c-Spanner  (c’’,d)-Power Spanner  c-Spanner  weak c’-Spanner –Is the Minimum Spanning Tree a c-Spanner?  Yes.  No. X

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 3 Mini-Exam No. 2 Task 1 –Which trade-offs or incompatibilities do exist?  Energy x Congestion  Congestion x Dilation  Dilation x Energy –The HL Graph... ... is a c-Spanner, if  > 2  /(  -1) ... has constant node degree ... is a weak c-Spanner, if  > 2  /(  -1) X X X X X

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 4 Yao-GraphSparsYSymmY Task 2 Draw the Yao Graph, the Sparsified Yao Graph and the Symmetric Yao Graph for the given node set.

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 5 Task 3 Given the following node set V={v 1,...,v 5 }. The nodes have the following priorities: v1v1 v2v2 v5v5 v3v3 v4v4 1+  nodepriority v1v1 3 v2v2 5 v3v3 1 v4v4 4 v5v5 2

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 6 Task 3 Calculate domination radius and publication radius for layers 0 and 1 of a HL Graph with parameters r 0 =1 and  =  =2! dom. radiuspub. radius L0L0 r 0 = 1  · r 0 = 2 L1L1 r 1 = r 0 ·  1 = 2  · r 1 = 4

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 7 Task 3 Draw the HL Graph of the following node set. dom. radiuspub. radius L0L0 r 0 = 1  · r 0 = 2 L1L1 r 1 = r 0 ·  1 = 2  · r 1 = 4 v1v1 v2v2 v5v5 v3v3 v4v4 1+  L 0 -domination L 0 -publication/ L 1 -domination

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 8 Task 3 Draw the HL Graph of the following node set. dom. radiuspub. radius L0L0 r 0 = 1  · r 0 = 2 L1L1 r 1 = r 0 ·  1 = 2  · r 1 = 4 v1v1 v2v2 v5v5 v3v3 v4v L 1 -domination V(L 0 ) = { v 1,...,v 5 } V(L 1 ) = { v 2,v 4 } V(L 2 ) = { v 2 } layer-0 edges layer-1 edges

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 9 Task 3 v1v1 v2v2 v5v5 v3v3 v4v L 1 -domination v1v1 v2v2 v5v5 v3v3 v4v L 1 -publication An alternative solution...

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 10 Task 4 Prove that the Yao Graph is a weak 2-Spanner for k>6. u v w Hint: Consider a node u and show that for every (target) node v there is either an edge (u,v) or another node w in the same sector with |w- v| < |u-v|, i.e. on a path from u to v there is always a node that reduces the distance to the target node v in the same sector. Show that such path never leaves the circle around v with radius |u-v|. To show: For each pair of nodes (u,v) there is a path from u to v (connectivity) and the path never leaves the circle C(u, 2·|u-v|) (weak spanner property).

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 11 Task 4 Prove that the Yao Graph is a weak 2-Spanner for k>6. Consider some nodes u and v. Case 1: If v is the nearest neighbor in sector S(u,v), then there is an edge (u,v) Case 2: If there is a node w the nearest neighbor in sector S(u,v), i.e. |u-w| < |u-v|, then there is an edge (u,w). We apply cases 1 and 2 again to construct a path from w to v. u v u v w

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 12 Task 4 Proof... |w-v| 6  In each step there is a progress, i.e. the distance to the target becomes smaller. So, the path P(u,v) never leaves the circle C(v, |u-v|) (progress circle) which has diameter 2·|u-v|. Therefore, it never leaves the circle C(u, 2·|u-v|) (weak spanner circle). u v w progress circle weak spanner circle

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 13 Exercise 15 For any finite point set V  R d and a layer L i of a Hierarchical Layer Graph with parameters  ≥  > 1 we have the following 1.For any point u, the number of points in layer L i with |u-v| ≤ c r i is at most (2c+1) d. 2.The degree of the sub-graph L i ist at most (2  +1) d. 3.The interference number of L i is bounded by (2  +1) 2d.

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 14 Exercise For any point u, the number of points in layer L i with |u-v| ≤ c r i is at most (2c+1) d. d = 2 Min. distance for layer i: |u-v| ≥ r i All circles with radius r i /2 and a layer-i node as center do not intersect. If |u-v| ≤ c r i then C(u, r i /2) lies inside C(v, r i /2 + c r i ) Number of small circles inside the big one: π(r i /2 + c r i ) 2 / π(r i /2 ) 2 = (2c+1) 2 u v r i /2 c r i r i /2 + c r i

HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity 15 Exercise 15 2.The degree of the sub-graph L i ist at most (2  +1) d. Within the domination radius  r i there are at most (2  + 1) d nodes (follows from 1.) 3.The interference number of L i is bounded by (2  +1) 2d. Two edges can only interfere when the distance is at most  r i. There are most (2  + 1) d nodes within this distance and their degree is at most (2  + 1) d.