Spanning tree

Spanning tree Lecture 4

Minimum Spanning Tree Problem • Instance: An undirected graph G, weights c: E(G) → R . • Task: Find a spanning tree inG of minimum weight or decide that G is not connected.

Maximum Weight Forest Problem • Instance: An undirected graph G, weights c: E(G) → R . • Task: Find a forest inG of maximum weight.

Equivalent Problems • We say that a problem Plinearly reduces to a problem Qif there are functions f and g, each computable in linear time, such that f transforms an instance x of Pto an instance y of Q,and g transforms a solution f (x) to a solution of x. • If Plinearly reducesto Qand Q linearly reducesto P , then both problem are called equivalent.

MVF ⇔ MST Proposition 4.1. The Maximum Weight Forest Problem and Minimum Spanning Tree Problem the are equivalent.

Proof (1) • Given an instance (G, c) of the Maximum Weight Forest Problem. • Delete all edges of negative weight, set . • Add a minimum set F of edges (with arbitrary weight) to make the graph connected. • Let us call the resulting graph G’. • Then instance (G',c') of the Minimum Spanning Tree Problem is equivalent in the following sence: Deleting the edges of F froma minimum weight spanning tree in (G',c') yields a maximum forest in (G, c).

Proof (2) • Given an instance (G, c) of the Minimum Spanning Tree Problem. • Set for all e E(G), where • All edges in a new instance have a positive weight. • Then the instance (G, c) of the Maximum Weight Forest Problem is equivalent, since all spanning trees have the same number of edges.

Optimality conditions of MST Theorem 4.2. Let (G ,c) be an instance of the Minimum Spanning Tree Problem, and letT b a spanning tree in G. Then the following statements are equivalent: • T is optimum. • For every e={x, y} ∊ E(G)\ E(T), noedge on the x-y-path in T has higher cost than e. • For every e ∊ E(T), e is a minimum cost edge of (V(C)), where C is a connected component of T– e.

Exercise 4.1 • Proof (a)(b) . • T is optimum. • For every e={x, y} ∊ E(G)\ E(T), noedge on the x-y-path in T has higher cost than e.

Proof (a)(b) Suppose (b) is violated. • Lete = {x,y} E(G)\ E(T) and let f be an edge on the x-y-path in T with c(f) > c(e). • Then (T – f ) +e is a spanning tree with lower cost.

Exercise 4.2 • Proof (b)(c) . b) For every e={x, y} ∊ E(G)\ E(T), noedge on the x-y-path in T has higher cost than e. c) For every e ∊ E(T), e is a minimum cost edge of (V(C)), where C is a connected component of T– e.

Proof (b)(c) Suppose (c) is violated. • Lete  E(T), C is a connected component of T – e and f ={x,y} ∊(V(C)) with c(f) < c(e). • Observe that the x-y-path in T must contain an edge of (V(C)), but the only such edge is e. • So (b) is violated.

Proof (c)(a) Suppose T satisfies (c), and let T*be an optimum spanning tree with E(T)∩ E(T*) as large as possible. We show thatT = T*. • Suppose there is an edgee = {x,y} E(T)\ E(T*). • Let C be a connected component of T– e. • T* + e contains a circuit D. • Since e E(D)∩ δ(V(C)), at least one more edge f (f≠ e) of D must belong to δ(V(C)). • Observe that (T* + e)– f is a spanning tree. • Since T* is optimum, c(e) ≥ c(f). But since (c) holds for T , we also have c(f) ≥ c(e). So c(f) = c(e) and (T* + e)– f is another optimum spanning tree. • This is a contradiction.

Kruskal’s Algorithm Input: A connected undirected graph G, weights c: E(G) → R . Output: A spanning tree T of minimum weight. • Sort the edges such that c(e1) ≤ c(e2) ≤…≤ c(em). • Set T  (V(G), ). • For i  1 to m do: If T+ei contains no circuit then set T  T +ei.

Kruskal’s Algorithm (2) Theorem 4.3. Kruskal’s Algorithm works correctly.

Prim’s Algorithm Input: A connected undirected graph G, weights c: E(G) → R . Output: A spanning tree T of minimum weight. • Choose v ∊ V(G). Set T  ({v}, ). • While V(T) ≠V(G) do: Choose an edge e ∊ G(V(T)) of minimum cost. Set T  T +e.

Prim’s Algorithm (2) Theorem 4.5. Prim’s Algorithm works correctly. Its running time is O(n2).

Maximum Weight Branching Problem • Instance: A digraph G, weights c: E(G) → R . • Task: Find a maximum weight branching in G.

Minimum Weight Arborescence Problem • Instance: A digraph G, weights c: E(G) → R . • Task: Find a minimum weight spanning arborescence in G or decide that none exists.

Minimum Weight Rooted Arborescence Problem • Instance: A digraph G, a vertex r ∊V(G), weights c: E(G) → R . • Task: Find a minimum weight spanning arborescence rooted at r in G or decide that none exists.

MVB ⇔ MWA ⇔ MWRA Proposition 4.6. The Maximum Weight Branching Problem, the Minimum Weight Arborescence Problem, and the Minimum Weight Rooted Arborescence Problem are all equivalent.

Homework Prove the following proposition. Proposition 4.6. The Maximum Weight Branching Problem, the Minimum Weight Arborescence Problem, and the Minimum Weight Rooted Arborescence Problem are all equivalent.

Maximum Weight Branching Problem • Instance: A digraph G, weights c: E(G) → R . • Task: Find a maximum weight branching in G.

Branching • A digraph B is a branching if the underlying graph is a forest and each vertex v has at most one entering edge. • Equivalently, a branching is an acyclic digraph B with for all x  V(B). Proposition 4.7. Let B be a graph with for all x ∊ V(B). ThenB contains a circuit if and only if the underlying undirected graph contains a circuit.

How to solve MWB • Let G be a directed graph and c: E(G) → R. • We can ignore negative weights since such edges will never appear in optimum branching. • A first idea is to take the best entering edge for each vertex. • Of course the returning graph may contain circuits. • We must delete at least one edge of each circuit. • The following lemma says that one is enough.

Branching and circuits Lemma 4.8. (Karp [1972]) Let B0 be a maximum weight subgraph of G with for all v ∊ V(B0). Then there exists an optimum branching B of G such that for each circuit C in B0, |E(C)\ E(B)| = 1.

Proof of Lemma LetB be anoptimum branching of G containing as many edges ofB0 as possible. Let C be some circuit in B0. b1 a1 С  B0 a2 b3 Let E(C)\ E(B)={(a1, b1),…, (ak, bk)} and k greater than 1. b2 a3 We claim thatBcontainsbi-bi-1-path for eachi=1,…,k, (b0=bk).

Show thatBcontainsbi-bi-1-path for eachi. bi-1 ai-1 С  B0 [bi-1 ai]B ai PB bi+1 eE(B) bi ai+1 V(B′):=V(G)andE(B′):={(x,y)E(B)}\{e}U{(ai ,bi)} B′ contains more edges ofB0 thanBB′ is not branching. So B′ contains a circuit. B contains a bi-ai-path P.

Main Idea • To find B0,as above, and than contract every circuit of B0 in G. If we choose the weights of the resulting graph G1 correctly,any optimum branching in G1will correspond to an optimum branching inG.

Edmonds’ Branching Algorithm Input: A digraph G, weights c: E(G) → R+. Output: A maximum weight branching B of G. • Set i 0, G0  G, and c0  c. • Let Bi bea maximum weight subgraph of Gi with for all v ∊ Bi . • If Bi contains no circuits then set B  Bi and go to (5). • Construct (Gi+1,ci+1) from (Gi,ci) by doing the following for each circuit C of Bi . Contract C to a single vertex vC in Gi+1. For each edge e = ( z, y) E(Gi) with zV(C), yV(C) do: Set ci+1 (e′)  ci(e) – ci((e,C)) + ci(eC) and (e′) e, where e′ ( z, vC), (e,C)=(x,y) E(C), and eC is some cheapest edge of C. Set i:=i+1 and go to (2). • If i = 0 then stop. • For each circuit C of Bi-1 do: If there is an edge e′ ( z, vC)  E(B) thenset E(B)  (E(B)\{e′ }) ∪(e′)∪(E(C)\{((e′),C)}) else set E(B)  E(B) ∪(E(C)\{eC}). Set V(B)  V(Gi-1), i  i–1 and go to (5).

Step 4 z z e x α(e,C) vC e′ y С  Bi eC For each edge e = ( z, y) E(Gi) with zV(C), yV(C) do: Set ci+1 (e′)  ci(e) – ci((e,C)) + ci(eC) and (e′) e, where e′ ( z, vC), (e,C)=(x,y) E(C),and eC is some cheapest edge of C.

Step 6 z e z x α(e,C) y vC e′ E(B) С  Bi eC x α(e,C) y vC С  Bi eC

Edmonds’ Branching Algorithm(2) Theorem 4.9. Edmonds’ Branching Algorithm works correctly.

Proof • Applying step 4 of the algorithm, we obtain a sequence (Gi,ci), i = 0,…, k. • We show that each time just before execution of step 5, B is an optimum branching of Gi. • This is trivial, for the first time we reach step 5. • So we have to show that step 6 transforms an optimum branching B of Giinto an optimum branching B*forGi–1.

Proof (2) • LetB'i–1 be any branching of Gi–1, such that|E(C)\ E(B'i–1)| = 1for each circuit CofBi–1. • LetB'i results from B'i–1bycontracting the circuits ofBi–1. • ThenB'i is a branching of Gi.

Step 4 z z e x α(e,C) vC e′ y С  Bi eC Setci+1 (e′)  ci (e) – ci((e,C)) + ci(eC)

Induction • By the induction hypothesis, B is an optimumbranching of Gi . • So we have ci(B) ≥ ci(B'i)

Step 6 z e z x α(e,C) y vC e′ E(B) С  Bi eC x α(e,C) y vC С  Bi eC

Induction • By the induction hypothesis, B is an optimumbranching of Gi . • So we have ci(B) ≥ ci(B'i). • This, together with Lemma, implies that B* is an optimum branching of Gi–1.

Exercise4.3 • Given an undirected graphG with arbitraryweightsc:E(G) → R. We ask for a minimum weight connected spanning subgraph. Can you solve this problem efficiently.

Exercise4.4 • Given an undirected graphG withweightsc:E(G) → R and a vertex v∊ V(G). We ask for a minimum weight spanning tree in G where v is not a leaf. Can you solve this problem in polynomial time.

Spanning tree

Spanning tree

Presentation Transcript

Minimum Spanning Tree

Minimum spanning tree

Minimum Spanning Tree

Minimum Spanning Tree

Spanning Tree Algorithm

Minimum Spanning Tree

Spanning Tree exercise

Spanning Tree

Minimum spanning tree

Minimal Spanning Tree

Spanning-Tree

Spanning Tree Understanding

Minimum Spanning Tree

Minimum Spanning Tree

Minimum Spanning Tree

Implementing Spanning Tree

Minimum Spanning Tree

Spanning Tree Protocol

Minimum Spanning Tree