CE 221 Data Structures and Algorithms Chapter 9 : Graphs Part I - PowerPoint PPT Presentation

CE 221 Data Structures and Algorithms Chapter 9 : Graphs Part I (Topological Sort & Shortest Path Algorithms) Text: Read Weiss, § 9.1 – 9.3 Izmir University of Economics 1

Definitions - I • A graph G=(V, E) consists of a set of vertices , V, and a set of edges , E. • Each edge is a pair (v, w), where v, w є V. • If the pair is ordered then G is directed ( digraph ). • Vertex w is adjacent to v iff (v, w) є E. • In an undirected graph with edge (v, w), w is adjacent to v and v is adjacent to w. • Sometimes and edge has a third component, weight or cost . Izmir University of Economics 2

Definitions - II • A path in a graph is w 1 , w 2 ,...,w N such that (w i , w i+1 ) є E for 1≤i<N. The length of such a path is the number of edges on the path. If a path from a vertex to itself contains no edges, then the path length is zero. If G contains an edge (v, v), then the path v, v is called a loop . • A simple path is a path such that all vertices are distinct, except that the first and the last could be the same. Izmir University of Economics 3

Definitions - III • A cycle in a directed graph is a path of length at least 1 such that w 1 =w N . This cycle is simple if the path is simple. For undirected graphs, the edges are required to be distinct (Why?). • A directed graph is acyclic if it has no cycles ( DAG ). • An undirected graph is connected if there is a path from every vertex to every other vertex. A directed graph with this property is called strongly connected . If directed graph is not, but underlying undirected graph is, it is weakly connected . A complete graph is a graph in which there is an edge between every pair of vertices. Izmir University of Economics 4

Representation of Graphs - I • One simple way is to use a two- dimensional array ( adjacency matrix representation). If vertices are numbered starting at 1, A[u][v]=true if (u, v) є E. Space requirement is Θ (|V| 2 ). • If the graph is not dense (sparse), adjacency lists may be used. The space requirement is O(|E|+|V|). Izmir University of Economics 5

Representation of Graphs - II Izmir University of Economics 6

Topological Sort - I • A topological sort is an ordering of vertices in a DAG, such that if there is path from v i to v j , then v j appears after v i in the ordering. • A simple algorithm to find a topological ordering is first to find any vertex with no incoming edges. We can then print this vertex, and remove it, along with its edges. Then apply the same strategy to the rest of the graph. To formalize this, define the indegree of a vertex v as the number of edges (u, v). Izmir University of Economics 7

Topological Sort – Initial Attempt • running time of the algorithm is O(|V| 2 ). Izmir University of Economics 8

Topological Sort – A Better Algorithm • We can remove the inefficiency by keeping all the unassigned vertices of indegree 0 in a special data structure (queue or stack). When a new vertex with degree zero is needed, it is returned by removing one from the queue, and when the indegrees of adjacent vertices are decremented, they are inserted into the queue if the indegree falls to zero. The running time is O(|E|+|V|) Izmir University of Economics 9

Homework Assignments • 9.1, 9.2, 9.3, 9.4, 9.38 • You are requested to study and solve the exercises. Note that these are for you to practice only. You are not to deliver the results to me. Izmir University of Economics 10

Shortest-Path Algorithms • The input is a weighted graph: associated with each edge (v i , v j ) is a cost c i,j . The cost of a path v 1 v 2 ...v N is ∑c i,i+1 for i in [1..N-1]. This is weighted path length , the unweighted path length on the other hand is merely the number of edges on the path, namely, N-1. • Single-source Shortest-Path Problem : Given as input a weighted graph G=(V, E), and a distinguished vertex, s , find the shortest weighted path from s to every other vertex in G. Izmir University of Economics 11

Negative Cost Cycles • In the graph to the left , the shortest path from v 1 to v 6 has a cost of 6 and the path itself is v 1 v 4 v 7 v 6 . The shortest unweighted path has 2 edges. • In the graph to the right , we have a negative cost. The path from v 5 to v 4 has cost 1, but a shorter path exists by following the loop v 5 v 4 v 2 v 5 v 4 which has cost -5. This path is still not the shortest, because we could stay in the loop arbitrarily long. Izmir University of Economics 12

Shortest Path Length: Problems We will examine 4 algorithms to solve four versions of the problem 1.Unweighted shortest path  O(|E|+|V|) 2.Weighted shortest path without negative edges  O(|E|log|V|) using queues 3.Weighted shortest path with negative edges  O(|E| . |V|) 4.Weighted shortest path of acyclic graphs  linear time Izmir University of Economics 13

Unweighted Shortest Paths • Using some vertex, s , which is an input parameter, find the shortest path from s to all other vertices in an unweighted graph. Assume s=v 3 . Izmir University of Economics 14

Unweighted Shortest Paths • Algorithm : find vertices that are at distance 1, 2, ... N-1 by processing vertices in layers ( breadth-first search ) Izmir University of Economics 15

Unweighted Shortest Paths Izmir University of Economics 16

Unweighted Shortest Paths • Complexity O(|V| 2 ) Izmir University of Economics 17

Unweighted Shortest Paths - Improvement • At any point in time there are only two types of unknown vertices that have d v ≠∞ . Some have d v = currDist and the rest have d v = currDist +1. • We can make use of a queue data structure. • O(|E|+|V|) Izmir University of Economics 18

Weighted Shortest Path Dijkstra’s Algorithm • With weighted shortest path,distance d v is tentative. It turns out to be the shortest path length from s to v using only known vertices as intermediates. • Greedy algorithm: proceeds in stages doing the best at each stage. Dijkstra’s algorithm selects a vertex v with smallest d v among all unknown vertices and declares it known. Remainder of the stage consists of updating the values d w for all edges (v, w). Izmir University of Economics 19

Dijkstra’s Algorithm - Example ► ► ► Izmir University of Economics 20

Dijkstra’s Algorithm - Example ► ► ► ► • A proof by contradiction will show that this algorithm always works as long as no edge has a negative cost. Izmir University of Economics 21

Dijkstra’s Algorithm - Pseudocode • If the vertices are sequentially scanned to find minimum d v , each phase will take O(|V|) to find the minimum, thus O(|V| 2 ) over the course of the algorithm. • The time for updates is constant and at most one update per edge for a total of O(|E|). • Therefore the total time spent is O(|V| 2 +|E|). • If the graph is dense, OPTIMAL. Izmir University of Economics 22

Dijkstra’s Algorithm -What if the graph is sparse? • If the graph is sparse |E|= θ (|V|), algorithm is too slow. The distances of vertices need to be kept in a priority queue . • Selection of vertex with minimum distance via deleteMin , and updates via decreaseKey operation. Hence; O( |E|log|V| +|V|log|V|) • find operations are not supported, so you need to be able to maintain locations of d i in the heap and update them as they change. • Alternative: insert w and d w with every update. Izmir University of Economics 23

Graphs with negative edge costs • Dijkstra’s algorithm does not work with negative edge costs. Once a vertex u is known, it is possible that from some other unknown vertex v , there is a path back to u that is very negative. • Algorithm : A combination of weighted and unweighted algorithms. Forget about the concept of known vertices. Izmir University of Economics 24

Graphs with negative edge costs - I • O(|E|*|V|). Each vertex can dequeue at most O(|V|) times. ( Why? Algorithm computes shortest paths with at most 0, 1, ..., |V|-1 edges in this order ). Hence, the result! • If negative cost cycles, then each vertex should be checked to have been dequeued at most |V| times. Izmir University of Economics 25

Acyclic Graphs • If the graph is known to be acyclic, the order in which vertices are declared known, can be set to be the topological order. • Running time = O(|V|+|E|) • This selection rule works because when a vertex is selected, its distance can no longer be lowered, since by topological ordering rule it has no incoming edges emanating from unknown nodes. Izmir University of Economics 26

Homework Assignments • 9.5, 9.7, 9.10, 9.40, 9.42, 9.44, 9.46, 9.47, 9.52 • You are requested to study and solve the exercises. Note that these are for you to practice only. You are not to deliver the results to me. Izmir University of Economics 27