# 1 discrete structures & algorithms graphs and trees: ii eece 320

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- 1 Discrete Structures & Algorithms Graphs and Trees: II EECE 320
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- 2 Topics for the day Some examples Questions on graphs and solutions Ideas for other proofs Minimal spanning trees
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- 3 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
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- 4 Example 1 Prove: A graph G is 2-colourable iff it contains no cycle of odd length. What does 2-colorable mean? We can color the vertices of the graph with at most two colors such that two adjacent nodes do not get the same color. Adjacent nodes? Two vertices are adjacent if there is an edge connecting the two vertices. Cycle? Of odd length? If there is a path that starts at a vertex and ends at the same vertex without repeating an edge, then the graph has a cycle. The number of edges along the cycle is the length of the cycle.
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- 5 Prove: A graph G is 2-colourable iff it contains no cycle of odd length. First: Assume that the graph is 2-colourable and prove that it contains no odd cycle. Select a 2-colouring of G. Consider an arbitrary cycle v 1, v 2, , v k, v 1. The vertices with even subscript must be one colour and the vertices with odd subscripts must be a different colour. Since v 1 and v k must be coloured differently, k must be even. But, k is the length of the cycle so all cycles must be of even length. v1v1 v2v2 v3v3 v4v4
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- 6 Prove: A graph G is 2-colourable iff it contains no cycle of odd length. Second: Assume that the graph has no cycle of odd length and prove that it must be 2-colourable. If we can 2-color every connected component of G, then we can 2-color all of G. Thus, it suffices to show that we can 2-color an arbitrary component of G. Consider an arbitrary component H of G. Consider some spanning tree T of H. Choose any 2-colouring of T. Any tree can be coloured with at most 2 colours because we can think of a tree as having parents and children and we colour each generation separately. Let x-y be some edge not in T and consider the path from some vertex v to x. Let z be the last vertex on the path from v to x that also lies on the path from v to y. Of the paths from z to x and from z to y, exactly one must have odd length else the these two paths together with the edge x-y would form an odd length cycle. Thus x and y belong to different generations wrt z and will have different colours. v x y H w z
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- 7 Example 2 Problem: If a graph G has 15 edges and its complementary graph G has 13 edges, how many vertices does G have? v1v1 v2v2 v3v3 v4v4 G v2v2 v3v3 v4v4 G v1v1 Complementary graph G: An edge between vertices x and y exists in G iff x and y are not adjacent in G.
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- 8 Problem: If a graph G has 15 edges and its complementary graph G has 13 edges, how many vertices does G have? The maximum number of edges a graph with n vertices can have is n(n-1)/2. The complement of a complete graph has no edges. If E is the set of edges in G and E the set of edges in G then: |E|+|E| = n(n- 1)/2. Solve for n when n(n-1)/2 = 28. n = 8.
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- 9 Use elementary properties of graphs to construct proofs. Start with properties that you understand easily: degree, number of edges and number of vertices, paths, cycles, complements, spanning trees, dual graph,
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- 10 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
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- 11 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
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- 12 4 8 7 9 6 11 9 5 8 7 Tree Approximations
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- 13 Finding a MST: Kruskals 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.
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- 14 1 8 7 9 10 3 5 4 7 9 Applying the Algorithm
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- 15 Proving Correctness of the Algorithm The algorithm outputs a spanning tree T. Then let M be a minimum spanning tree. MT Let e be the first edge chosen by the algorithm that is in T but not in M. N = M+e-f is another spanning tree. [By contradiction] Suppose that its 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 isnt fully contained in T, it has (at least) an edge f not in T.
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- 16 Proving Correctness of the Algorithm (cont) 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. Then N < M this M is not a minimal spanning tree. Contradiction!
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- 17 Greed is Good (In this case) The greedy algorithm, by adding the least costly edges in each stage, succeeds in finding an MST Obviously greedy approaches do not work for all problems Some examples where they do work?
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- 18 TSP: Traveling Salesman Problem Given: a number of cities and the costs of traveling from any city to any other city, Q: What is the cheapest round-trip route that visits each city exactly once and then returns to the starting city? The problem is in NP
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- 19 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. Modified TSP: 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) Complexity still NP
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- 20 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)
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- 21 Shortest-paths in a graph
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- 22 Shortest path problem Input data: Directed graph: G= (V, E) Start and end node: s, t Length l e for each edge e Output: shortest path from s to t [cost of path = sum of all edges]
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- 23 Dijkstras algorithm Idea: Maintain a set of explored nodes S for which we have determined the shortest path distance d(u) from s to u. Initialize S = {s}, d(s) = 0. Repeatedly choose unexplored node v to minimize add v to S, and set d(v) = (v). Until all nodes are chosen Shortest path from a node in u to any other unexplored node
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- 24 Dijkstras algorithm Idea: Maintain a set of explored nodes S for which we have determined the shortest path distance d(u) from s to u. Initialize S = {s}, d(s) = 0. Repeatedly choose unexplored node v to minimize add v to S, and set d(v) = (v). Until all nodes are chosen Shortest path from a node in u to any other unexplored node
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- 25 Example
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- 26 Dijkstra: Complexity analysis Initialize S = {s}, d(s) = 0. Repeatedly choose unexplored node v to minimize add v to S, and set d(v) = (v).
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- 27 Bipartite Graph A graph is bipartite if the nodes can be partitioned into two sets V 1 and V 2 such that all edges go only between V 1 and V 2 (no edges go from V 1 to V 1 or from V 2 to V 2 )
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- 28 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?
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- 29 Dancing Partners
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- 30 Perfect Matchings Perfect Matching: A matching that covers all nodes in the graph with one edge going to each node. 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
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- 31 If there are m boys, there are md edges If there are n girls, there are nd edges Proof: Claim: If all nodes have the same degree then |A| = |B| A Matter of Degree
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- 32 The Marriage Theorem Theorem: A bipartite graph has a perfect matching if and only if |A| = |B| 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.
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- 33 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 them
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- 34 k At most n-k n-k At least 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 nodes in A The Feeling is Mutual
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- 35 Proof of Marriage Theorem Call a bipartite graph matchable if it has the same number of nodes on left a