UNDIRECTED GRAPHS TODAY BBM ALGORITHMS DEPT. OF COMPUTER ENGINEERING. Undirected graphs. Graph applications

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1 BBM - ALGORIHMS ODAY DEP. O COMPUER ENGINEERING UNDIRECED GRAPHS Undirected Graphs Graph API Depth-first search Challenges Acknowledgement: he course slides are adapted from the slides prepared by R. Sedgewick and K. Wayne of Princeton Uniersity. Undirected graphs Graph. Set of ertices connected pairwise by edges. Why study graph algorithms? housands of practical applications. Hundreds of graph algorithms known. Interesting and broadly useful abstraction. Challenging branch of computer science and discrete math. Graph applications graph ertex edge communication telephone, computer fiber optic cable circuit gate, register, processor wire mechanical joint rod, beam, spring financial stock, currency transactions transportation street intersection, airport highway, airway route internet class C network connection game board position legal moe social relationship person, actor friendship, moie cast neural network neuron synapse protein network protein protein-protein interaction chemical compound molecule bond

2 Graph terminology Path. Sequence of ertices connected by edges. Cycle. Path whose first and last ertices are the same. wo ertices are connected if there is a path between them. cycle of length ertex of degree edge ertex path of length connected components Some graph-processing problems Path. Is there a path between s and t? Shortest path. What is the shortest path between s and t? Cycle. Is there a cycle in the graph? Euler tour. Is there a cycle that uses each edge exactly once? Hamilton tour. Is there a cycle that uses each ertex exactly once? Connectiity. Is there a way to connect all of the ertices? MS. What is the best way to connect all of the ertices? Biconnectiity. Is there a ertex whose remoal disconnects the graph? Planarity. Can you draw the graph in the plane with no crossing edges? Graph isomorphism. Do two adjacency lists represent the same graph? Challenge. Which of these problems are easy? difficult? intractable? Graph API Depth-first search Challenges UNDIRECED GRAPHS Graph representation Graph drawing. Proides intuition about the structure of the graph. two drawings of the same graph Caeat. Intuition can be misleading.

3 Graph representation Graph API Vertex representation. his lecture: use integers between and V. Applications: conert between names and integers with symbol table. A Anomalies. B D C E self-loop G Anomalies parallel edges symbol table public class Graph Graph(int V) Graph(In in) create an empty graph with V ertices create a graph from input stream oid addedge(int, int w) add an edge -w Iterable<Integer> adj(int ) ertices adjacent to int V() int E() String tostring() In in = new In(args[]); Graph G = new Graph(in); for (int = ; < G.V(); ++) for (int w : G.adj()) StdOut.println( + "-" + w); number of ertices number of edges string representation read graph from input stream print out each edge (twice) Graph API: sample client ypical graph-processing code. n Undirected Graphs Graph input format. V tinyg.txt E Input format for G % jaa est tinyg.txt task implementation compute the degree of compute maximum degree compute aerage degree public static int degree(graph G, int ) int degree = ; for (int w : G.adj()) degree++; return degree; public static int maxdegree(graph G) int max = ; for (int = ; < G.V(); ++) if (degree(g, ) > max) max = degree(g, ); return max; public static double aeragedegree(graph G) return. * G.E() / G.V(); In in = new In(args[]); Graph G = new Graph(in); for (int = ; < G.V(); ++) for (int w : G.adj()) StdOut.println( + "-" + w); read graph from input stream print out each edge (twice) count self-loops public static int numberofselfloops(graph G) int count = ; for (int = ; < G.V(); ++) for (int w : G.adj()) if ( == w) count++; return count/; // each edge counted twice public String tostring() String s = V + " ertices, " + E + " edges\n";

4 Set-of-edges graph representation Maintain a list of the edges (linked list or array). Adjacency-matrix graph representation Maintain a two-dimensional V-by-V boolean array; for each edge w in graph: adj[][w] = adj[w][] = true. two entries for each edge Adjacency-list graph representation Adjacency-list graph representation: Jaa implementation Maintain ertex-indexed array of lists. Bag objects public class Graph priate final int V; priate Bag<Integer>[] adj; adjacency lists ( using Bag data type ) adj[] representations of the same edge public Graph(int V) this.v = V; adj = (Bag<Integer>[]) new Bag[V]; for (int = ; < V; ++) adj[] = new Bag<Integer>(); public oid addedge(int, int w) adj[].add(w); adj[w].add(); create empty graph with V ertices add edge -w (parallel edges allowed) Adjacency-lists representation (undirected graph) public Iterable<Integer> adj(int ) return adj[]; iterator for ertices adjacent to

5 Graph representations In practice. Use adjacency-lists representation. Algorithms based on iterating oer ertices adjacent to. Real-world graphs tend to be sparse. huge number of ertices, small aerage ertex degree Graph representations In practice. Use adjacency-lists representation. Algorithms based on iterating oer ertices adjacent to. Real-world graphs tend to be sparse. huge number of ertices, small aerage ertex degree sparse (E = ) dense (E = ) representation space add edge edge between and w? iterate oer ertices adjacent to? list of edges E E E adjacency matrix V * V adjacency lists E + V degree() degree() wo graphs (V = ) * disallows parallel edges Graph API Depth-first search Challenges UNDIRECED GRAPHS Maze exploration Maze graphs. Vertex = intersection. Edge = passage. intersection passage Goal. Explore eery intersection in the maze.

6 rémaux maze exploration Algorithm. Unroll a ball of string behind you. Mark each isited intersection and each isited passage. Retrace steps when no unisited options. Depth-first search Goal. Systematically search through a graph. Idea. Mimic maze exploration. ypical applications. DS (to isit a ertex ) Mark as isited. Recursiely isit all unmarked ertices w adjacent to. ind all ertices connected to a gien source ertex. ind a path between two ertices. Design challenge. How to implement? Design pattern for graph processing Design pattern. Decouple graph data type from graph processing. Create a Graph object. Pass the Graph to a graph-processing routine, e.g., Paths. Query the graph-processing routine for information. Depth-first search o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. public class Paths Paths(Graph G, int s) Paths paths = new Paths(G, s); for (int = ; < G.V(); ++) if (paths.haspatho()) StdOut.println(); find paths in G from source s boolean haspatho(int ) is there a path from s to? Iterable<Integer> patho(int ) path from s to ; null if no such path print all ertices connected to s graph G V tinyg.txt E

7 Depth-first search Depth-first search o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. edgeo[] Goal. ind all ertices connected to s (and a path). Idea. Mimic maze exploration. Algorithm. Use recursion (ball of string). Mark each isited ertex (and keep track of edge taken to isit it). Return (retrace steps) when no unisited options. Data structures. boolean[] marked to mark isited ertices. int[] edgeo to keep tree of paths. (edgeo[w] == ) means that edge -w taken to isit w for first time ertices reachable from Depth-first search Depth-first search properties public class DepthirstPaths priate boolean[] marked; priate int[] edgeo; priate int s; public DepthirstSearch(Graph G, int s)... dfs(g, s); priate oid dfs(graph G, int ) = true; for (int w : G.adj()) if (!marked[w]) dfs(g, w); edgeo[w] = ; = true if connected to s edgeo[] = preious ertex on path from s to initialize data structures find ertices connected to s recursie DS does the work Proposition. DS marks all ertices connected to s in time proportional to the sum of their degrees. Pf. Correctness: - if w marked, then w connected to s (why?) - if w connected to s, then w marked (if w unmarked, then consider last edge on a path from s to w that goes from a marked ertex to an unmarked one) Running time: Each ertex connected to s is isited once. set of unmarked ertices source s x w set of marked ertices no such edge can exist

8 Depth-first search properties Proposition. After DS, can find ertices connected to s in constant time and can find a path to s (if one exists) in time proportional to its length. Pf. edgeo[] is a parent-link representation of a tree rooted at s. Graph API Depth-first search Challenges UNDIRECED GRAPHS public boolean haspatho(int ) return ; public Iterable<Integer> patho(int ) if (!haspatho()) return null; Stack<Integer> path = new Stack<Integer>(); for (int x = ; x!= s; x = edgeo[x]) path.push(x); path.push(s); return path; x path edgeo[] Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. tinycg.txt V E edgeo[] graph G add to

9 Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de de Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de de

10 Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] done de Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de de

11 Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de de Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] done de

12 Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de de Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] done de

13 Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de de Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] done de

14 Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de de Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de done

15 Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de de Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. edgeo[] edgeo[] de done

16 Remoe ertex from. Add to all unmarked ertices adjacent to and mark them. Depth-first search. Put unisited ertices on a stack.. Put unisited ertices on a. Shortest path. ind path from s to t that uses fewest number of edges. edgeo[] BS (from source ertex s) Put s onto a IO, and mark s as isited. Repeat until the is empty: - remoe the least recently added ertex - add each of 's unisited neighbors to the, and mark them as isited. done Intuition. BS examines ertices in increasing distance from s. properties Proposition. BS computes shortest path (number of edges) from s in a connected graph in time proportional to E + V. Pf. [correctness] Queue always consists of zero or more ertices of distance k from s, followed by zero or more ertices of distance k +. Pf. [running time] Each ertex connected to s is isited once. standard drawing dist = dist = dist = public class BreadthirstPaths priate boolean[] marked; priate boolean[] edgeo[]; priate final int s; priate oid bfs(graph G, int s) Queue<Integer> q = new Queue<Integer>(); q.en(s); marked[s] = true; while (!q.isempty()) int = q.de(); for (int w : G.adj()) if (!marked[w]) q.en(w); marked[w] = true; edgeo[w] = ;

17 Graph API Depth-first search Challenges UNDIRECED GRAPHS Connectiity queries Def. Vertices and w are connected if there is a path between them. Goal. Preprocess graph to answer queries: is connected to w? in constant time. public class CC CC(Graph G) Depth-first search. [next few slides] find connected components in G boolean connected(int, int w) are and w connected? int count() number of connected components int id(int ) component identifier for he relation "is connected to" is an equialence relation: Reflexie: is connected to. Symmetric: if is connected to w, then w is connected to. ransitie: if connected to w and w connected to x, then connected to x. Def. A connected component is a maximal set of connected ertices. connected components id[] Remark. Gien connected components, can answer queries in constant time. Def. A connected component is a maximal set of connected ertices. connected components

18 Goal. Partition ertices into connected components. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. Initialize all ertices as unmarked. or each unmarked ertex, run DS to identify all ertices discoered as part of the same component. V tinyg.txt E V mediumg.txt E... ( additional lines) Input format for Graph constructor graph (two G examples) o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit isit

19 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit isit o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit isit

20 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit done o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit isit

21 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. done isit o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit done

22 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. done isit o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit done

23 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit isit o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. done done

24 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. connected component connected component: check o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit isit

25 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. done done o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. connected component: check

26 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit isit o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit isit

27 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit done o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. done isit

28 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. isit done o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. done connected component:

29 o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. check done inding connected components with DS inding connected components with DS (continued) public class CC priate boolean ; priate int[] id; priate int count; id[] = id of component containing number of components public int count() return count; number of components public CC(Graph G) marked = new boolean[g.v()]; id = new int[g.v()]; for (int = ; < G.V(); ++) if (!) dfs(g, ); count++; run DS from one ertex in each component public int id(int ) return id[]; priate oid dfs(graph G, int ) = true; id[] = count; for (int w : G.adj()) if (!marked[w]) dfs(g, w); id of component containing all ertices discoered in same call of dfs hae same id public int count() public int id(int ) priate oid dfs(graph G, int ) see next slide

30 UNDIRECED GRAPHS Graph-processing challenge Graph API Depth-first search Challenges Problem. Is a graph bipartite? How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible Graph-processing challenge Graph-processing challenge Problem. Is a graph bipartite? How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible. simple DS-based solution (see textbook) Problem. ind a cycle. How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible

31 Graph-processing challenge Bridges of Königsberg Problem. ind a cycle. he Seen Bridges of Königsberg. [Leonhard Euler ] How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. (see textbook) No one knows. Impossible. simple DS-based solution in Königsberg in Prussia, there is an island A, called the Kneiphof; the rier which surrounds it is diided into two branches... and these branches are crossed by seen bridges. Concerning these bridges, it was asked whether anyone could arrange a route in such a way that he could cross each bridge once and only once. Euler tour. Is there a (general) cycle that uses each edge exactly once? Answer. Yes iff connected and all ertices hae een degree. o find path. DS-based algorithm (see textbook). Graph-processing challenge Graph-processing challenge Problem. ind a cycle that uses eery edge. Assumption. Need to use each edge exactly once. Problem. ind a cycle that uses eery edge. Assumption. Need to use each edge exactly once. How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. Eulerian tour No one knows. Impossible. (classic graph-processing problem)

32 Graph-processing challenge Problem. ind a cycle that isits eery ertex exactly once. Graph-processing challenge Problem. ind a cycle that isits eery ertex. Assumption. Need to isit each ertex exactly once. How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. Hamiltonian cycle No one knows. (classical NP-complete problem) Impossible Graph-processing challenge Graph-processing challenge Problem. Are two graphs identical except for ertex names? How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible Problem. Are two graphs identical except for ertex names? How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible. graph isomorphism is longstanding open problem ,,,,,,,,,,,,

33 Graph-processing challenge Graph-processing challenge Problem. Lay out a graph in the plane without crossing edges? How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible Problem. Lay out a graph in the plane without crossing edges? How difficult? Any programmer could do it. ypical diligent algorithms student could do it. Hire an expert. Intractable. No one knows. Impossible. linear-time DS-based planarity algorithm discoered by arjan in s (too complicated for practitioners)