Lokale Netzstrukturen Exercise 5
Juli 19, 2017
Ex 1 – a)
Definition
The undirected degree 8Yao graph over a node set V ⊂R2, denotedYK8(V), is defined as follows. For any node v ∈V partition the plane into 8 equal cones centered at v and s.t. the boundaries of two cones are parallel to the x -axis (see Fig. 1). For each such cone which contains at least one other node x 6=v , add the undirected edge vu toYK8(V), where u is the nearest neighbor of v in that cone w.r.t. Euclidean distance (ties can be broken arbitrarily).
Figure:Illustration ofYK8
Ex 1 – a) (cont’d)
Ex
Prove thatEMST(V)⊆YK8(V).
Preliminary observation:
I Ifq andq0 are two nodes in Cl(p), then
||qq0||<max{||pq||,||pq0||}.
I Consider4pqq0: Since
∠qpq0 =π/4< π/3, its opposite sideqq0 cannot be longest side of the triangle.
Ex 1 – a) (cont’d)
Ex
Prove thatEMST(V)⊆YK8(V).
Proof:
I Let EMST be the set of edges forming EMST(V) and YK the set of edges in YK8(V).
I Show: Any edgevw ∈EMST not in YK can be replaced by an edge in YK s.t.
MST property is maintained
I This proves the claim since all edges from EMST can be replaced by edges from YK
I Let vw ∈EMST \YK and assume w ∈Cl(v).
Ex 1 – a) (cont’d)
Proof (cont’d):
I Nl(v)6=∅
I There is a nearest neighboru∈Nl(v), s.t.
uv∈YK.
I It hols thatu6=w and||vu|| ≤ ||vw||, becausevw ∈EMST \YK.
I Deletevw fromEMST and obtain two subtrees one containingv and the other containing w
I Note thatuandw must be in the same component. If not, thenuw would be a shorter connecting edge for the two subtrees thanvw (by preliminary observation), contradicting the fact thatEMST is a MST.
I Hence,u andw are in same subtree and adding edgevutoEMST \ {vw}results in a spanning tree with total weight no greater than that ofEMST.
Ex 1 – b)
Definition (Minimal Dominating Set)
A dominating set for a graph G = (V,E) is a subset D of V , such that every vertex not in D is adjacent to at least one member in D. An Independent Set is minimal if no vertex can be removed from D without violating the independent set property.
Definition (Maximal Independent Set)
A Independent Set for a graph G = (V,E) is a subset I of V such that for every two vertices in I there is no edge in E connecting them. A Maximal Independent Set is an Independent Set that is not a subset of any other Independent Set.
Ex
Prove that any Maximal Independent Set for a given graph G = (V,E) is also a Minimal Dominating Set for G.
Ex 1 – c)
Ex
Proof that in an Independent set defined over a given Unit Disk GraphUDG(V), any node contains at most 5 Independent Set nodes in its Unit Disk.
Ex 1 – c) (cont’d)
Ex
Proof that in an Independent set defined over a given Unit Disk GraphUDG(V), any node contains at most 5 Independent Set nodes in its Unit Disk.
Proof:
I Considerx,u,v, s.t. u,v ∈N1(x) andu,v are I.S. nodes
I Must hold that ∠uxv > π/3, otherwise u,v would share an edge in UDG(V) (Law of Cosines)
I For k > π/3, 2π/k <6 holds
I Hence, there are at most 5 I.S.
nodes in a node’s neighborhood
Greedy Forwarding
Ex 2 – a)
Ex
Prove that progress- and distance-based Greedy Forwarding algorithms which only use edges in forwarding direction, cannot guarantee packet delivery.
Ex 2 – a)
Ex
Prove that progress- and distance-based Greedy Forwarding algorithms which only use edges in forwarding direction, cannot guarantee packet delivery.
Ex 2 – b)
Ex
Explain the differences of the forwarding areasSector,Releaux triangle, andCircle in Beaconless routing. Why is the forwarding area limited to one of these shapes, instead of using the complete positive progress area? Give an example that explains the problem.
Ex 3
Prepare an example that illustrates the different routing paths obtained when executing the algorithmsGreedy-Face-Greedy, GOAFR,FACE using before crossing, andFACE using after crossing. The example should be designed such that the routing paths differ in at least a single edge.
Ex 3 (cont’d)
Greedy-Face-Greedy
Please note!
This is the correct solution because it takes into account both Greedy and Face phase of the algorithm.
The solution presented in the exercises takes into account only Face algorithm
Ex 3 (cont’d)
Greedy-Face-Greedy
Please note!
This is the correct solution because it takes into account both Greedy and Face phase of the algorithm.
The solution presented in the exercises takes into account only Face algorithm
Ex 3 (cont’d)
Greedy Other Adaptive Face Routing (GOAFR)
Ex 3 (cont’d)
Greedy Other Adaptive Face Routing (GOAFR)
Ex 3 (cont’d)
Face using before crossing
Ex 3 (cont’d)
Face using before crossing
Ex 3 (cont’d)
Face using after crossing
Ex 3 (cont’d)
Face using after crossing
