3/25/10 1 15-251 Great Theoretical Ideas in Computer Science Graphs II Lecture 19, March 25, 2010 Victor Adamchik Danny Sleator Recap Theorem: Let G be a graph with n nodes and e edges The following are equivalent: 1. G is a tree (connected, acyclic) 3. G is connected and n = e + 1 4. G is acyclic and n = e + 1 5. G is acyclic and if any two non-adjacent points are joined by a line, the resulting graph has exactly one cycle 2. Every two nodes of G are joined by a unique path The number of labeled trees on n nodes is nn-2 Cayley’s Formula A graph is planar if it can be drawn in the plane without crossing edges3/25/10 2 http://www.planarity.net Planar Graphs Euler’s Formula If G is a connected planar graph with n vertices, e edges and f faces, then n – e + f = 2 A coloring of a graph is an assignment of a color to each vertex such that no neighboring vertices have the same color Graph Coloring Spanning Trees A spanning tree of a graph G is a tree that touches every node of G and uses only edges from G Every connected graph has a spanning tree Implementing Graphs Adjacency Matrix Suppose we have a graph G with n vertices. The adjacency matrix is the n x n matrix A=[aij] with: aij = 1 if (i,j) is an edge aij = 0 if (i,j) is not an edge Good for dense graphs!3/25/10 3 Example A = 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 Counting Paths The number of paths of length k from node i to node j is the entry in position (i,j) in the matrix Ak A2 = 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 3 2 2 2 2 3 2 2 2 2 3 2 2 2 2 3 = Adjacency List Suppose we have a graph G with n vertices. The adjacency list is the list that contains all the nodes that each node is adjacent to Good for sparse graphs! Example 1 2 3 4 1: 2,3 2: 1,3,4 3: 1,2,4 4: 2,3 http://www.math.ucsd.edu/~fan/hear/ “Can you hear the shape of a graph?” Graphical Muzak Finding Optimal Trees Trees have many nice properties (uniqueness of paths, no cycles, etc.) We may want to compute the “best” tree approximation to a graph If all we care about is communication, then a tree may be enough. We want a tree with smallest communication link costs3/25/10 4 Finding Optimal Trees Problem: Find a minimum spanning tree, that is, a tree that has a node for every node in the graph, such that the sum of the edge weights is minimum 4 8 7 9 6 11 9 5 8 7 Tree Approximations Kruskal’s Algorithm A simple algorithm for finding a minimum spanning tree Finding an MST: Kruskal’s Algorithm Create a forest where each node is a separate tree Make a sorted list of edges S While S is non-empty: Remove an edge with minimal weight If it connects two different trees, add the edge. Otherwise discard it. 1 8 7 9 10 3 5 4 7 9 Applying the Algorithm Analyzing the Algorithm The algorithm outputs a spanning tree T. Let M be a minimum spanning tree. Let e be the first edge chosen by the algorithm that is not in M. N = M+e-f is another spanning tree. Suppose that it’s not minimal. (For simplicity, assume all edge weights in graph are distinct) If we add e to M, it creates a cycle. Since this cycle isn’t fully contained in T, it has an edge f not in T.3/25/10 5 Analyzing the Algorithm N = M+e-f is another spanning tree. Claim: e < f, and therefore N < M Suppose not: e > f Then f would have been visited before e by the algorithm, but not added, because adding it would have formed a cycle. But all of these cycle edges are also edges of M, since e was the first edge not in M. This contradicts the assumption M is a tree. Greed is Good (In this case…) The greedy algorithm, by adding the least costly edges in each stage, succeeds in finding an MST But — in math and life — if pushed too far, the greedy approach can lead to bad results. TSP: Traveling Salesman Problem Given a number of cities and the costs of traveling from any city to any other city, what is the cheapest round-trip route that visits each city at least once and then returns to the starting city? TSP from Trees We can use an MST to derive a TSP tour that is no more expensive than twice the optimal tour. Idea: walk “around” the MST and take shortcuts if a node has already been visited. We assume that all pairs of nodes are connected, and edge weights satisfy the triangle inequality d(x,y) ≤ d(x,z) + d(z,y) Tours from Trees This is a 2-competitive algorithm Shortcuts only decrease the cost, so Cost(Greedy Tour) ≤ 2 Cost(MST) ≤ 2 Cost(Optimal Tour) Bipartite Graph A graph is bipartite if the nodes can be partitioned into two sets V1 and V2 such that all edges go only between V1 and V2 (no edges go from V1 to V1 or from V2 to V2)3/25/10 6 Dancing Partners A group of 100 boys and girls attend a dance. Every boy knows 5 girls, and every girl knows 5 boys. Can they be matched into dance partners so that each pair knows each other? Dancing Partners Perfect Matchings Regular Bipartite Matching Theorem: If every node in a bipartite graph has the same degree d ≥ 1, then the graph has a perfect matching. Note: if degrees are the same then |A| = |B|, where A is the set of nodes “on the left” and B is the set of nodes “on the right” A matching is a set of edges, no two of which share a vertex. The matching is perfect if it includes every vertex. If there are m boys, there are md edges If there are n girls, there are nd edges Proof: Claim: If degrees are the same then |A| = |B| A Matter of Degree The Regular Bipartite Matching Theorem follows from a stronger theorem, which we now come to. (Remind me to return to the proof of the RBMT later.) The Marriage Theorem Theorem: A bipartite graph has a perfect matching if and only if |A| = |B| = n and for all k ∈ [1,n]: for any subset of k nodes of A there are at least k nodes of B that are connected to at least one of them. The condition fails for this graph The Marriage Theorem For any subset of (say) k nodes of A there are at least k nodes of B that are connected to at least one of them3/25/10 7 k >n-k n-k <k The …
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