Graphs - II. Announcements. Where did I leave that book? Where did I leave that book? Where did I leave that book? CS 2110, Fall 2016

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1 Graphs - II CS, Fall Announcements A will be available tonight or tomorrow. Gries lunch canceled today Gries office hours today from to only Q. Why do programmers confuse Halloween and Christmas? Answer. Because oct = dec! Decimal 8 9 Octal Decimal 8 9 Where did I leave that book? Octal hop:// Where did I leave that book? Where did I leave that book? Go as far down a path as possible before backtracking hop:// hop://

2 Search Depth-first search Breadth-first search Graph Algorithms Shortest paths Dijkstra's algorithm Minimum spanning trees Prim's algorithm Kruskal's algorithm Representa[ons of Graphs Adjacency List Adjacency Matrix Adjacency Matrix or Adjacency List? Defini[ons: n = number of ver[ces m = number of edges d(u) = degree of u = number of edges leaving u Adjacency Matrix Uses space O(n ) Can iterate over all edges in [me O(n ) Can answer Is there an edge from u to v? in O() [me BeOer for dense graphs (lots of edges) Adjacency List Uses space O(m + n) Can iterate over all edges in [me O(m + n) Can answer Is there an edge from u to v? in O(d(u)) [me BeOer for sparse graphs (fewer edges) Given a graph and one of its nodes u (say node below) Given a graph and one of its nodes u (say node below) We want to visit each node reachable from u (nodes,,,, ) There are many paths to some nodes. How do we visit all nodes efficiently, without doing extra work? Suppose all nodes

3 Suppose all nodes from node : {,,,, Suppose all nodes from node : {,,,, from : {,, Green: visited Blue: unvisited Green: visited Blue: unvisited from node : {,, Green: visited Blue: unvisited from node : {,, from : none Not even itself, because it s already been visited! /** Visit all nodes that are REACHABLE from u. Precondi[on: u is not visited*/ Let u be The nodes REACHABLE from are,,,, Start End

4 Let u be The nodes REACHABLE from are,,,, Let u be The nodes REACHABLE from are,,,, Let u be (visited) The nodes to be visited are,,, Let u be (visited) The nodes to be visited are,,, Have to do DFS on all unvisited neighbors of u! Suppose the for loop visits neighbors in numerical order. Then dfs() visits the nodes in this order: Suppose the for loop visits neighbors in numerical order. Then dfs() visits the nodes in this order:,

5 Suppose the for loop visits neighbors in numerical order. Then dfs() visits the nodes in this order:,, Suppose the for loop visits neighbors in numerical order. Then dfs() visits the nodes in this order:,,, Suppose the for loop visits neighbors in numerical order. Then dfs() visits the nodes in this order:,,,, Suppose n nodes are REACHABLE along e edges (in total). What is Worst-case execu[on? Worst-case space? Example: Use different way (other than array visited) to know whether a node has been visited Example: We really haven t said what data structures are used to implement the graph That s all there is to basic DFS. You may have to change it to fit a par[cular situa[on. If you don t have this spec and you do something different, it s probably wrong. in OO fashion public class Node { boolean visited; List<Node> neighbors; Each node of the graph is an object of type Node /** This node is unvisited. Visit all nodes REACHABLE from this node */ public void dfs() { No need for a visited= true; parameter. The for (Node n: neighbors) { object is the node. if (!n.visited) n.dfs();

6 wrioen itera[vely = (u); // Not Java! // inv: all nodes that have to be visited are // REACHABLE from some node in s while ( s is not empty ) { // Remove top stack node, put in u wrioen itera[vely = (u); Call dfs() wrioen itera[vely = (u); Call dfs() Itera[on wrioen itera[vely = (u); Call dfs() Itera[on wrioen itera[vely = (u); Call dfs() Itera[on wrioen itera[vely = (u); Call dfs() Itera[on

7 wrioen itera[vely = (u); Call dfs() Itera[on wrioen itera[vely = (u); Call dfs() Itera[on wrioen itera[vely = (u); Call dfs() Itera[on wrioen itera[vely = (u); Call dfs() Itera[on wrioen itera[vely = (u); Call dfs() Itera[on wrioen itera[vely = (u); Call dfs() Itera[on

8 wrioen itera[vely = (u); Call dfs() Itera[on Yes, is put on the stack twice, once for each edge to it. It will be visited only once. // Not Java! // inv: all nodes that have to be visited are // REACHABLE from some node in s while ( q is not empty ) { // Remove first node in queue, put in u // Add to end of queue Call bfs() Call bfs() Itera[on Call bfs() Itera[on Call bfs() Itera[on 8

9 Call bfs() Itera[on Call bfs() Itera[on Call bfs() Itera[on Call bfs() Itera[on Call bfs() Itera[on Call bfs() Itera[on 9

10 Call bfs() Itera[on Call bfs() Itera[on Call bfs() Itera[on Breadth first: () Node u () All nodes edge from u () All nodes edges from u () All nodes edges from u