This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CME 305: Discrete Mathematics and Algorithms Instructor: Professor Amin Saberi ([email protected]) HW#2 Solutions 1. Let T be a spanning tree of a graph G with an edge cost function c . We say that T has the cycle property if for any edge e / ∈ T , c ( e ) ≥ c ( e ) for all e in the cycle generated by adding e to T . Also, T has the cut property if for any edge e ∈ T , c ( e ) ≤ c ( e ) for all e in the cut defined by e . Show that the following three statements are equivalent: (a) T has the cycle property. (b) T has the cut property. (c) T is a minimum cost spanning tree. Remark 1 : Note that removing e ∈ T creates two trees with vertex sets V 1 and V 2 . A cut defined by e ∈ T is the set of edges of G with one endpoint in V 1 and the other in V 2 (with the exception of e itself). Remark 2 : The cycle and cut properties we have defined in this problem are slightly different from what they usually mean. Solution: In order to show that (a), (b), and (c) are equivalent, it is enough to show: (a) ⇔ (c), and (b) ⇔ (c). (c) ⇒ (a) : By contradiction suppose T does not have the cycle property: there exists e / ∈ T such that T ∪ { e } has a cycle C in which there exists e ∈ T and e ∈ C where c ( e ) > c ( e ). Let tree T be the tree obtained by adding e to T and removing e ; T is a tree with cost strictly less than cost of T which is contradicting with T being an MST. (a) ⇒ (c) : By contradiction suppose T is not an MST: let e be the first edge that was picked by Kruskal’s algorithm but des not belong to T . Adding e to T would create cycle C . Since T has cycle property, c ( e ) ≤ c ( e ), e ∈ C . Therefore, all e ∈ C , e 6 = e have been visited by the Kruskal’s algorithm. We have two cases: case 1 : All e ∈ C \{ e } were picked by the algorithm. In this case the algorithm would not pick e because it creates a cycle with the existing edges. case 2 : There exists e * ∈ C , e * 6 = e such that it was not picked by the algorithm. The reason for not picking an edge is that it would create a cycle with the existing edges. However, since e was the first edge picked by the algorithm that does to belong to T , this would mean that T has a cycle, which is a contradiction. (a) ⇒ (b) : By contradiction suppose T does not have the cut property: there exists e ∈ T such that there exits edge e = ( v 1 ,v 2 ) such that v 1 ∈ V 1 and v 2 ∈ V 2 ( V 1 and V 2 are the set of vertices of the two connected components after removing e , see Remark 2.), and c ( e ) < c ( e ). Since T is connected graph there exist path P T between v 1 and v 2 such that all the edges of P T belong to T . Adding e to P T creates a cycle in which there exist e ∈ T where c ( e ) > c ( e ) which is contradicting with T having the cycle property. (b) ⇒ (a) : By contradiction suppose T does not have the cycle property: there exists edge e such that when adding it to T and creating cycle C , there exists e ∈ C , where c ( e ) > c ( e ). In T , if we remove...
View
Full Document
 Winter '09
 Graph Theory, edges, Bipartite graph

Click to edit the document details