UNDIRECTED GRAPHS BBM ALGORITHMS DEPT. OF COMPUTER ENGINEERING

Transcription

1 BBM - ALGORIHMS DEP. O COMPUER ENGINEERING UNDIRECED GRAPHS Acknowledgement: he course slides are adapted from the slides prepared by R. Sedgewick and K. Wayne of Princeton Uniersity.

2 Undirected Graphs Graph API Depth-first search Breadth-first search Connected components Challenges ODAY

3 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.

4 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

5 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 edge ertex path of length ertex of degree connected components

6 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?

7 Graph API Depth-first search Breadth-first search Connected components Challenges UNDIRECED GRAPHS

8 Graph representation Graph drawing. Proides intuition about the structure of the graph. two drawings of the same graph Caeat. Intuition can be misleading.

9 Graph representation 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

10 Graph API 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() number of ertices number of edges string representation 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)

11 Graph API: sample client Graph input format. V tinyg.txt E Input format for G % jaa est tinyg.txt 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)

12 ypical graph-processing code compute the degree of public static int degree(graph G, int ) { int degree = ; for (int w : G.adj()) degree++; return degree; } compute maximum degree public static int maxdegree(graph G) { int max = ; for (int = ; < G.V(); ++) if (degree(g, ) > max) max = degree(g, ); return max; } compute aerage degree public static double aeragedegree(graph G) { return. * G.E() / G.V(); } 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 }

13 Set-of-edges graph representation Maintain a list of the edges (linked list or array).

14 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

15 Adjacency-list graph representation Maintain ertex-indexed array of lists. Bag objects adj[] representations of the same edge

16 Adjacency-list graph representation: Jaa implementation public class Graph { priate final int V; priate Bag<Integer>[] adj; public Graph(int V) { this.v = V; adj = (Bag<Integer>[]) new Bag[V]; for (int = ; < V; ++) adj[] = new Bag<Integer>(); } adjacency lists ( using Bag data type ) create empty graph with V ertices public oid addedge(int, int w) { adj[].add(w); adj[w].add(); } add edge -w (parallel edges allowed) } public Iterable<Integer> adj(int ) { return adj[]; } iterator for ertices adjacent to

17 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 = ) wo graphs (V = )

18 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 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() * disallows parallel edges

19 Graph API Depth-first search Breadth-first search Connected components Challenges UNDIRECED GRAPHS

20 Maze exploration Maze graphs. Vertex = intersection. Edge = passage. intersection passage Goal. Explore eery intersection in the maze.

21 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.

22 Depth-first search Goal. Systematically search through a graph. Idea. Mimic maze exploration. DS (to isit a ertex ) Mark as isited. Recursiely isit all unmarked ertices w adjacent to. ypical applications. ind all ertices connected to a gien source ertex. ind a path between two ertices. Design challenge. How to implement?

23 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. public class Paths Paths(Graph G, int s) 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 Paths paths = new Paths(G, s); for (int = ; < G.V(); ++) if (paths.haspatho()) StdOut.println(); print all ertices connected to s

24 Depth-first search o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. V tinyg.txt E graph G

25 Depth-first search o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] edgeo[] ertices reachable from

26 Depth-first search 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

27 Depth-first search public class DepthirstPaths { priate boolean[] marked; priate int[] edgeo; priate int s; marked[] = true if connected to s edgeo[] = preious ertex on path from s to public DepthirstSearch(Graph G, int s) {... dfs(g, s); } initialize data structures find ertices connected to s } priate oid dfs(graph G, int ) { marked[] = true; for (int w : G.adj()) if (!marked[w]) { dfs(g, w); edgeo[w] = ; } } recursie DS does the work

28 Depth-first search properties 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) set of unmarked ertices source s x set of marked ertices no such edge can exist Running time: Each ertex connected to s is isited once. w

29 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. public boolean haspatho(int ) { return marked[]; } 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[]

30 Graph API Depth-first search Breadth-first search Connected components Challenges UNDIRECED GRAPHS

31 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. tinycg.txt V E graph G

32 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] add to queue

33 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

34 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

35 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

36 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

37 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] done

38 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

39 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

40 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

41 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

42 Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. Breadth-first search dequeue edgeo[] queue

43 Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. Breadth-first search done edgeo[] queue

44 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

45 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

46 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

47 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] done

48 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

49 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

50 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

51 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] done

52 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

53 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

54 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

55 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

56 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] done

57 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

58 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

59 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] dequeue

60 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. queue edgeo[] done

61 Breadth-first search Repeat until queue is empty: Remoe ertex from queue. Add to queue all unmarked ertices adjacent to and mark them. edgeo[] done

62 Breadth-first search Depth-first search. Put unisited ertices on a stack. Breadth-first search. Put unisited ertices on a queue. Shortest path. ind path from s to t that uses fewest number of edges. BS (from source ertex s) Put s onto a IO queue, and mark s as isited. Repeat until the queue is empty: - remoe the least recently added ertex - add each of 's unisited neighbors to the queue, and mark them as isited. Intuition. BS examines ertices in increasing distance from s.

63 Breadth-first search 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 =

64 Breadth-first search 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.enqueue(s); marked[s] = true; while (!q.isempty()) { int = q.dequeue(); for (int w : G.adj()) { if (!marked[w]) { q.enqueue(w); marked[w] = true; edgeo[w] = ; } } } }

65 Graph API Depth-first search Breadth-first search Connected components Challenges UNDIRECED GRAPHS

66 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

67 Connected components 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.

68 Connected components Def. A connected component is a maximal set of connected ertices. connected components

69 Connected components Goal. Partition ertices into connected components. Connected components 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

70 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] graph G

71 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

72 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

73 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

74 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

75 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

76 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

77 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

78 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

79 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

80 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

81 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

82 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

83 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

84 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

85 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

86 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

87 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

88 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

89 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

90 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

91 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

92 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

93 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] connected component connected component:

94 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] check

95 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

96 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

97 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

98 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

99 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] connected component:

100 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] check

101 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

102 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

103 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

104 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

105 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

106 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

107 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

108 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

109 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] isit

110 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

111 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

112 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] connected component:

113 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] check

114 Connected components o isit a ertex : Mark ertex as isited. Recursiely isit all unmarked ertices adjacent to. marked[] cc[] done

115 inding connected components with DS public class CC { priate boolean marked[]; priate int[] id; priate int count; public CC(Graph G) { marked = new boolean[g.v()]; id = new int[g.v()]; for (int = ; < G.V(); ++) { if (!marked[]) { dfs(g, ); count++; } } } public int count() public int id(int ) priate oid dfs(graph G, int ) id[] = id of component containing number of components run DS from one ertex in each component see next slide }

116 inding connected components with DS (continued) public int count() { return count; } public int id(int ) { return id[]; } priate oid dfs(graph G, int ) { marked[] = true; id[] = count; for (int w : G.adj()) if (!marked[w]) dfs(g, w); } number of components id of component containing all ertices discoered in same call of dfs hae same id

117 Graph API Depth-first search Breadth-first search Connected components Challenges UNDIRECED GRAPHS

118 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

119 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)

120 Graph-processing challenge 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

121 Graph-processing challenge 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. (see textbook) Impossible. simple DS-based solution

122 Bridges of Königsberg he Seen Bridges of Königsberg. [Leonhard Euler ] 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).

123 Graph-processing challenge 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

124 Graph-processing challenge 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. Eulerian tour No one knows. Impossible. (classic graph-processing problem)

125 Graph-processing challenge Problem. ind a cycle that isits eery 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

126 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. Hamiltonian cycle No one knows. (classical NP-complete problem) Impossible

127 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 ,,,,,,

128 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. graph isomorphism is longstanding open problem ,,,,,,

129 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

130 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. linear-time DS-based planarity algorithm discoered by arjan in s (too complicated for practitioners)