15 251 Great Theoretical Ideas in Computer Science Upcoming Events Quiz this Thursday Test in 2 weeks Graphs II Lecture 21 November 6 2007 Recap Theorem Let G be a graph with n nodes and e edges The following are equivalent 1 G is a tree connected acyclic 2 Every two nodes of G are joined by a unique path 3 G is connected and n e 1 4 G is acyclic and n e 1 5 G is acyclic and if any two nonadjacent points are joined by a line the resulting graph has exactly one cycle Cayley s Formula The number of labeled trees on n nodes is nn 2 A graph is planar if it can be drawn in the plane without crossing edges Euler s Formula If G is a connected planar graph with n vertices e edges and f faces then n e f 2 Graph Coloring A coloring of a graph is an assignment of a color to each vertex such that no neighboring vertices have the same color 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 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 costs 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 Tree Approximations 7 8 4 5 9 7 8 6 11 9 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 Applying the Algorithm 7 4 1 5 9 9 10 8 7 3 Analyzing the Algorithm The algorithm outputs a spanning tree T Suppose that it s not minimal For simplicity assume all edge weights in graph are distinct Let M be a minimum spanning tree Let e be the first edge chosen by the algorithm that is not in M 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 N M e f is another spanning tree 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 exactly 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 Shortcuts only decrease the cost so Cost Greedy Tour 2 Cost MST 2 Cost Optimal Tour This is a 2 competitive algorithm 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 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 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 Matter of Degree Claim If degrees are the same then A B Proof If there are m boys there are md edges If there are n girls there are nd edges 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 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 them The condition fails for this graph The Feeling is Mutual At least k At most n k k n k The condition of the theorem still holds if we swap the roles of A and B If we pick any k nodes in B they are connected to at least k Proof of Marriage Theorem Call a bipartite graph matchable if it has the same number of nodes on left and right and any k nodes on the left are connected to at least k on the right Strategy Break up the graph into two matchable parts and recursively partition each of these into two matchable parts etc until each part has only two nodes Proof of Marriage Theorem Select two nodes a A and b B connected by an edge Idea Take G1 a b and G2 everything else Problem G2 need not be matchable There could be a set of k nodes that has only k 1 neighbors Proof of Marriage Theorem a k The only way this b could fail is if one of k 1 the missing nodes is b Add this in to form G1 and take G2 to be everything else This is a matchable partition Generalized Marriage Hall s Theorem Let S S1 S2 be a set of finite subsets that satisfies For any subset T Ti of S UTi T Thus any k subsets contain at least k elements Then we can choose an element xi Si from each Si so that x1 x2 are all distinct Example Suppose that a standard deck of cards is dealt into 13 piles of 4 cards each Then it is possible to …
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