Presentation on theme: "Minimum Spanning Trees Definition Two properties of MST’s Prim and Kruskal’s Algorithm –Proofs of correctness Boruvka’s algorithm Verifying an MST Randomized."— Presentation transcript:
Minimum Spanning Trees Definition Two properties of MST’s Prim and Kruskal’s Algorithm –Proofs of correctness Boruvka’s algorithm Verifying an MST Randomized algorithm in linear time
Problem Definition Input –Weighted, connected undirected graph G=(V,E) Weight (length) function w on each edge e in E Task –Compute a spanning tree of G of minimum total weight Spanning tree –If there are n nodes in G, a spanning tree consists of n-1 edges such that no cycles are formed
Example ABC D EFG Input: ABC D EFG Output:
Two Properties of MST’s Cycle Property: For any cycle C in a graph, the heaviest edge in C does not appear in the minimum spanning tree –Used to rule edges out Cut Property: For any proper non-empty subset X of the vertices, the lightest edge with exactly one endpoint in X belongs to the minimum spanning forest –Used to rule edges in
Cycle Property Illustration/Proof Proof by contradiction: –Suppose T is an MST with such an edge e. –Derive a contradiction showing that T is not an MST ABC D EFG ABC D EFG
Cut Property Illustration/Proof Proof by contradiction: –Suppose T is an MST without such an edge e. –Derive a contradiction showing that T is not an MST ABC D EFG ABC D EFG
Three Classic Greedy Algorithms Kruskal’s approach –Select the minimum weight edge that does not form a cycle Prim’s approach –Choose an arbitrary start node v –At any point in time, we have connected component N containing v and other nodes V-N –Choose the minimum weight edge from N to V-N Boruvka’s approach –Prim “in parallel”
Illustrate the execution of Kruskal and Prim’s algorithms on the following input graph. Let D be the arbitrary start node for Prim’s algorithm. Example ABC D EFG
Prim Implementation Use a priority queue to organize nodes in V-N to facilitate finding closest node. –Extract-Min operation –Decrease-key operation Describe how we could implement Prim using a priority queue. ABC D EFG
Running Time of Prim How many extract-min operations will we perform? How many decrease-key operations will we perform? How much time to build initial priority queue? Given binary heap implementation, what is the running time? Fibonacci heap: –Decrease-key drops to O(1) amortized time
Kruskal Implementation Kruskal’s Algorithm –Adding edge (u,v) to the current set of edges T forms a cycle if and only if u and v are in the same connected component. ABC D EFG
Disjoint Set Data Stucture (Ch 21) Given a universe U of objects –Maintain a collection of sets S i such that Union i S i = U S i intersect S j is empty –Find-set(x): Returns set S i that contains x –Merge(S i, S j ): Returns new set S k = S i union S j Describe how we can implement cycle detection with this data structure. ABC D EFG
Running Time of Kruskal How many merges will we perform? How many Find-set operations will we perform? –Each can be implemented in amortized (V) time where is a very slow growing function. What other operations do we need to implement? Overall running time?
Proofs of Correctness Why do we know each edge that is added in Kruskal’s algorithm is part of an MST? Why do we know that each edge added in Prim’s algorithm is part of an MST?
Boruvka’s Algorithm Prim “in parallel” Boruvka Step: –We have a graph of vertices –For each v in V, select the minimum weight edge connected to v –Update Contract all selected edges, replacing each connected component by a single vertex Delete loops, and keep only the lowest weight edge in a multi- edge Run Boruvka steps until we have a single node
Boruvka Step Illustration ABC D EFG ABC D EFG A’B’ 7
Boruvka’s Algorithm Analysis Correctness: –How can we verify that each edge we add is part of an MST? Running time: –What is the running time of a Boruvka Step? –How many Boruvka steps must we implement in the worst case?
Verification Problem Input –Weighted, connected undirected graph G=(V,E) Weight (length) function w on each edge e in E –A spanning tree T of G Task –Answer yes/no if T is a minimum spanning tree for G Two key concepts –Decision problem: problem with yes/no answer –Verification: Is it easier to verify an answer as correct as compared to generating an answer?
Key idea: T(u,v) Suppose we have a tree T For each edge (u,v) not in T, let T(u,v) be the heaviest edge on the (u,v) path in T If w(u,v) > w(T(u,v)), then (u,v) should not be in T. ABC D EFG Consider edge (B,F) with weight 7. T(B,F) = (C,G) which has weight 6. (B,F) is appropriately not in T as w(B,F) > w(C,G).
Verification Algorithm For each edge (u,v) not in T, find T(u,v). Compare w (u,v) to w(T(u,v)) If all are ok, then return yes, else return no. Running time? –O(E) time to perform comparisons –Sophisticated techniques to find all the T(u,v) in O(E) time.
Randomized Algorithm 1.Run Baruvka’s Step 2 times a)If there were originally n nodes, how many in reduced problem 2.Random Sampling a)Make a smaller subgraph by choosing each edge with probability ½ b)Recursively compute minimum spanning tree (or perhaps forest) F on this reduced graph c)Use verification algorithm to eliminate any edges (u,v) not in F whose weight is more than weight of F(u,v). 3.Apply algorithm recursively to the remaining graph to compute a spanning tree T’ a)Knit together tree from step 1 with F’ to form spanning tree
Graph G o Step 1: Run 2 Baruvka Steps: Contracted Graph G 1 List of edges E 1 Step 2: Construct G 2 by sampling ½ edges Step 2: Recursively Construct T 2 for graph G 2 Step 2: Use T 2 to identify edges E 2 that cannot be in MST for G 1 Graph G 3 = G 1 – E 2 Step 3: Recursively Construct T 3 for graph G 3 Step 3: Return T 3 melded with edges E 1 as final tree T 4 Tree T 4 = T 3 union E 1 Visualization
Graph G o Step 1: Run 2 Baruvka Steps: Contracted Graph G 1 List of edges E 1 Step 2: Construct G 2 by sampling ½ edges Step 2: Recursively Construct T 2 for graph G 2 Step 2: Use T 2 to identify edges E 2 that cannot be in MST for G 1 Graph G 3 = G 1 – E 2 Step 3: Recursively Construct T 3 for graph G 3 Step 3: Return T 3 melded with edges E 1 as final tree T 4 Tree T 4 = T 3 union E 1 Analysis Intuition G 2 has at most ½ the edges of G 0 G 3 edges bounded by ½ nodes of G 0