Lecture 10 Graph algorithms: testing graph properties COMP 523: Advanced Algorithmic Techniques Lecturer: Dariusz Kowalski Lecture 10: Testing Graph Properties 1 Overview Previous lectures: Representation of graphs (adjacency matrix or list) Breadth-First Search (BFS) Depth-First Search (DFS) This lecture: Testing bipartiteness Testing for (directed) cycles Lecture 10: Testing Graph Properties 2 How to represent graphs? Given graph G = (V,E) , how to represent it? 1. Adjacency matrix: ith node is represented as ith row and ith column, edge between ith and jth nodes is represented as 1 in row i and column j, and vice versa (0 if there is no edge between i and j) 2. Adjacency list: nodes are arranged as array/list, each node record has the list of its neighbours attached 1 2 3 4 1 2 3 4 1 0 1 1 1 2 1 0 1 0 3 1 1 0 1 4 1 0 1 0 Adjacency matrix Lecture 10: Testing Graph Properties 1 2 3 4 2 1 1 1 3 4 3 2 4 3 3 Adjacency list Adjacency list Advantages: Representation takes memory O(m+n) - versus O(n2) Examining all neighbours of a given node requires time O(m/n) in average - versus O(n) Check if an edge belongs to the graph requires time O(m/n) in average - versus O(1) 1 2 3 4 1 2 3 4 1 0 1 1 1 2 1 0 1 0 3 1 1 0 1 4 1 0 1 0 Adjacency matrix Lecture 10: Testing Graph Properties Disadvantages: Advantages especially for sparse graphs! 1 2 3 4 2 1 1 1 3 4 3 2 4 3 4 Adjacency list Bipartiteness Graph G = (V,E) is bipartite iff it can be partitioned into two sets of nodes A and B such that each edge has one end in A and the other end in B. Alternatively: Graph G = (V,E) is bipartite iff all its cycles have even length. Graph G = (V,E) is bipartite iff nodes can be coloured using two colours. Question: given a graph represented as the adjacency list, how to test if graph is bipartite? Note: graphs without cycles (trees) are bipartite. bipartite: not bipartite Lecture 10: Testing Graph Properties 5 Testing bipartiteness We use following definition: Graph G = (V,E) is bipartite iff all its cycles have even length, or Graph G = (V,E) is bipartite iff it has no odd cycle. Method: use BFS search tree. Recall: BFS is a rooted spanning tree having the same layers as the original graph G (each node has the same distance from the root in BFS tree and in graph G) Algorithm: Perform BFS search and colour all nodes in odd layers red, others blue Go through all edges in adjacency list and check if each of them has two different colours at its ends - if so then G is bipartite not bipartite bipartite Lecture 10: Testing Graph Properties 6 Testing bipartiteness - why it works Property of layers: Every edge is either between two consecutive layers or within one layer (two ends are in the same layer) - it follows from BFS property 1. Suppose that graph G is not bipartite. Hence there is an odd cycle. This cycle must have an edge in one layer, and so the ends of this edge have the same colour, so the algorithm answers correctly. 2. Suppose that graph G is bipartite. Hence all its cycles have even length. Suppose, to the contrary, that there is uni-coloured edge. This edge must be in one layer. Consider one of such edges which is in the smallest possible layer. Consider a cycle containing this edge and the BFS path between its two ends. The length of this path is odd, which is a contradiction. Thus all edges are coloured properly (in sense of their ends) and algorithm is correct. bipartite not bipartite Lecture 10: Testing Graph Properties 7 Testing cycles Directed graph G (edges have direction - one end is the beginning, the other one is the end of the edge). Reversed graph Gr has the same nodes but each edge is a reversed edge from graph G. Representing directed graph: adjacency list - each node has two lists: of in-coming and out-coming edges/neighbours Problem: does the graph has a directed cycle? Lecture 10: Testing Graph Properties 8 Testing cycles Technique: use directed DFS tree for graph G. Algorithm: Find a node with no incoming edges - if not found answer cycled Build a DFS directed tree - consider only edges in proper direction. If during building DFS two nodes which are already in the stack are considered then answer cycled, otherwise answer acyclic. General remark: if graph G is not connected then do the same until no free node remains. Lecture 10: Testing Graph Properties 9 Testing cycles - analysis Property of acyclic graph: There is a node with no incoming edges 1. Suppose that graph G is acyclic. Hence there is no directed cycle and then while building DFS directed tree we never try to explore the already visited node. Answer acyclic is then proper. 2. Suppose that graph G is not acyclic. It follows that there is a cycle in it. Let v be the first node in a cycle visited by DFS search. By the property of DFS, ...