1. 1 Graph Theory: Solved and Unsolved Problems Roger House Scientific Café Coffee Catz Sebastopol, CA 2010 April 22 Copyright © 2010 Roger House

2. 2 What we're going to talk about Graphs –What's a graph? –What's a graph good for? Graph problems –The Königsberg bridge problem –The planarity problem –The Traveling Salesman Problem (TSP)

3. 3 What's a graph?

4. 4 Remember these graphs?

5. 5 And these?

6. 6 And graphs you can eat

7. 7 None of these graphs are the kinds of graphs we're interested in at the moment

8. 8 A different kind of graph We're going to deal with much different kinds of graphs, namely, mathematical graphs.

9. 9 Definition of a graph Definition: A graph G consists of a finite nonempty set V of vertices together with a set E of unordered pairs of distinct vertices of V. The pair e = {u,v} of vertices in E is called an edge of G.

10. 10 There will be a quiz on the definition

11. 11 There will be a quiz on the definition But not on the one above

12. 12 There will be a quiz on the definition But not on the one above On the one below

13. 13 A friendlier definition A graph is dots connected by lines.

14. 14 Fancy terminology The dots are called vertices. (Or points, or nodes, or...)‏

15. 15 Fancy terminology The dots are called vertices. (Or points, or nodes, or...)‏ The lines are called edges. (The graph above has 7 vertices and 9 edges.)‏

16. 16 Fancy terminology The dots are called vertices. (Or points, or nodes, or...)‏ The lines are called edges. (The graph above has 7 vertices and 9 edges.)‏ Two vertices connected by an edge are adjacent.

17. 17 Fancy terminology The dots are called vertices. (Or points, or nodes, or...)‏ The lines are called edges. (The graph above has 7 vertices and 9 edges.)‏ Two vertices connected by an edge are adjacent. An edge is incident with its two vertices.

18. 18 Fancy terminology The dots are called vertices. (Or points, or nodes, or...)‏ The lines are called edges. (The graph above has 7 vertices and 9 edges.)‏ Two vertices connected by an edge are adjacent. An edge is incident with its two vertices. The degree of a vertex is the number of edges incident with the vertex.

19. 19 Complete graphs

20. 20 Bipartite graphs

21. 21 More graphs

22. 22 Yet more graphs

23. 23 What's a graph good for?

24. Road network graph 24 The vertices in G are intersections, protection points, and entrance points.

25. Modeling an electrical circuit

26. 26

27. 27 #include structure

28. 28 Function call graph

29. 29 Unix family tree

30. 30 Are there enough graphs? # vertices # graphs 1 1 2 2 3 4

31. 31 Are there enough graphs? # vertices #graphs 1 1 2 2 3 4 4 11

32. 32 Are there enough graphs? # vertices #graphs 1 1 2 2 3 4 4 11 5 34

33. 33 Are there enough graphs? # vertices #graphs 1 1 2 2 3 4 4 11 5 34 6 156

34. 34 Are there enough graphs? # vertices #graphs 1 1 2 2 3 4 4 11 5 34 6 156 7 1044

35. 35 Are there enough graphs? # vertices #graphs 1 1 2 2 3 4 4 11 5 34 6 156 7 1044 8 12,346

36. 36 Are there enough graphs? # vertices #graphs 1 1 2 2 3 4 4 11 5 34 6 156 7 1044 8 12,346 9 274,668

37. 37 Are there enough graphs? # vertices #graphs 1 1 2 2 3 4 4 11 5 34 6 156 7 1044 8 12,346 9 274,668 10 12,005,168

38. 38 The Königsberg bridge problem

39. 39 What's the problem? Where: Königsberg, Prussia (now Kaliningrad, Russia) When: Early 1700's What: How to start at one place, cross each bridge exactly once, and end up back where you started from

40. 40 Abstracting a bit

41. 41 More abstraction

42. 42 The essence of the problem

43. 43 A graph problem Pick any vertex V as a start vertex.

44. 44 A graph problem Pick any vertex V as a start vertex. Pick any edge incident to V which has not yet been traveled, and travel along it to another vertex, marking the edge as traveled.

45. 45 A graph problem Pick any vertex V as a start vertex. Pick any edge incident to V which has not yet been traveled, and travel along it to another vertex, marking the edge as traveled. Keep traveling from vertex to vertex along untraveled edges, marking the edges as traveled, until you return to V and all edges have been traveled exactly once.

46. 46 Euler to the rescue Leonhard Euler (1707-1783), a Swiss mathematician discovered everything.

47. 47 Euler to the rescue Leonhard Euler (1707-1783), a Swiss mathematician discovered everything. Well, almost everything.

48. 48 Euler to the rescue In 1735 Euler proved that the problem has no solution.

49. 49 Euler to the rescue In 1735 Euler proved that the problem has no solution. He proved more: If a connected graph has a closed eulerian circuit, then the degree of every vertex is even.

50. 50 Are the degrees all even?

51. 51 Say we add a couple of bridges

52. 52 Now is there an eulerian circuit? The degree of each vertex is even

53. 53 Now is there an eulerian circuit? The degree of each vertex is even So Euler's theorem tells us there is an eulerian circuit, right?

54. 54 Now is there an eulerian circuit? The degree of each vertex is even So Euler's theorem tells us there is an eulerian circuit, right? NO! NO! NO!

55. 55 Avoid logical pitfalls “A implies B” does NOT mean that “B implies A” So the fact that all the vertex degrees are even does not tell us that an eulerian circuit exists. But one might exist anyway.

56. 56 Can you find an eulerian circuit?

57. 57 Here's one 1 2 3 4 5 6 7 8 9

58. 58 Hierholzer to the rescue Carl Hierholzer (1840-1871), a German mathematician, published a paper in 1873 proving the converse of Euler's theorem:

59. 59 Hierholzer to the rescue Carl Hierholzer (1840-1871), a German mathematician, published a paper in 1873 proving the converse of Euler's theorem: If the degree of every vertex of a connected graph is even, then the graph has a closed eulerian circuit.

60. 60 Hierholzer to the rescue So now it's really easy to tell if a graph has a closed eulerian circuit or not.

61. 61 Hierholzer to the rescue So now it's really easy to tell if a graph has a closed eulerian circuit or not. Just look at the degree of every vertex.

62. 62 Hierholzer to the rescue So now it's really easy to tell if a graph has a closed eulerian circuit or not. Just look at the degree of every vertex. If all the degrees are even then a closed eulerain circuit exists. If one or more vertices has odd degree, then no eulerian circuit exist.

63. 63 You call that “rescue”? So, once we know a closed eulerian circuit exists, how do we find it?

64. 64 You call that “rescue”? So, once we know a closed eulerian circuit exists, how do we find it? Hierholzer's theorem tells us nothing about how to find a closed eulerian circuit.

65. 65 You call that “rescue”? So, once we know a closed eulerian circuit exists, how do we find it? Hierholzer's theorem tells us nothing about how to find a closed eulerian circuit. We can be assured that at least one exists, but we don't have a clue how to find one.

66. 66 Some terminology If more than one edge connects distinct vertices u and v, then we have a multigraph.

67. 67 Some terminology If more than one edge connects distinct vertices u and v, then we have a multigraph. The Königsberg Bridges graph is a multigraph.

68. 68 Some terminology If there is at most one edge between any two distinct vertices, we have a simple graph.

69. 69 Some terminology If there is at most one edge between any two distinct vertices, we have a simple graph. From now we consider only simple graphs.

70. 70 The planarity problem Some graphs are planar, which means they can be drawn in a “nice” way

71. 71 Some planar graphs

72. 72 Definition of a planar graph A graph is planar if it can be drawn in the plane in such a way that

73. 73 Definition of a planar graph A graph is planar if it can be drawn in the plane in such a way that – Each vertex is a distinct point in the plane

74. 74 Definition of a planar graph A graph is planar if it can be drawn in the plane in such a way that – Each vertex is a distinct point in the plane – Edges do not cross – in fact, two edges touch if and only if they have a common vertex, in which case they touch only at that vertex

75. 75 Definition of a planar graph A graph is planar if it can be drawn in the plane in such a way that – Each vertex is a distinct point in the plane – Edges do not cross – in fact, two edges touch if and only if they have a common vertex, in which case they touch only at that vertex – An edge does not touch any vertices except the two at its endpoints

76. 76 Some planar graphs again

77. 77 Is this graph (K 4 ) planar?

78. 78 Yes, K 4 is planar

79. 79 A bit nicer drawing

80. 80 Is this graph (K 5 -e) planar? A B C D E

81. 81 Redraw K 5 -e, swapping C and D A B C D E A B D C E

82. 82 Yes, K 5 -e is planar A B D C E A B D C E

83. 83 Is this graph (K 5 ) planar? A B C D E

84. 84 No, K 5 is not planar A B C D E A B C D E

85. 85 Not so fast How do we know K 5 is nonplanar?

86. 86 Not so fast How do we know K 5 is nonplanar? Just because we have nonplanar drawings of it does not mean that no planar drawing exists.

87. 87 Not so fast Maybe we just haven't been clever enough to find a planar drawing.

88. 88 Not so fast Maybe we just haven't been clever enough to find a planar drawing. Trust me, K 5 is nonplanar

89. 89 Is this graph (K 3,3 ) planar? This graph is also called the Utility Graph. Think of connecting three utilities, gas, water, and electric, to three houses without any of the lines crossing. Can it be done?

90. 90 No, K 3,3 is not planar

91. 91 Not so fast (again) Again, you should be asking, “How do we know K 3,3 is nonplanar?”

92. 92 Not so fast (again) Again, you should be asking, “How do we know K 3,3 is nonplanar?” Again, you'll just have to trust me on this.

93. 93 Not so fast (again) Again, you should be asking, “How do we know K 3,3 is nonplanar?” Again, you'll just have to trust me on this. Or, look at Dots and Lines by Richard J. Trudeau.

94. 94 How to tell if a graph is planar? At this point we know that K 5 and K 3,3 are nonplanar (that is, if you trust me) But, in general, given any old graph, how to tell if it is planar or nonplanar? For example...

95. 95 Which of these is planar?

96. 96 Which of these is planar? None of these graphs is planar. How do we know? How can we be sure? Let's carefully examine the first one...

97. 97 Petersen's graph After Julius Petersen (1839-1910), a Danish mathematician, who pre- sented it in 1898 (although it had already been published in 1886). Donald Knuth says the Petersen graph is "a remarkable config- uration that serves as a counter- example to many optimistic predictions about what might be true for graphs in general."

98. 98 K 3,3 is hiding in Petersen's graph

99. 99 K 3,3 is hiding in Petersen's graph

100. 100 K 3,3 is hiding in Petersen's graph

101. 101 K 3,3 is hiding in Petersen's graph

102. 102 K 3,3 is hiding in Petersen's graph

103. 103 K 3,3 is hiding in Petersen's graph

104. 104 K 3,3 is hiding in Petersen's graph

105. 105 So? We have shown that K 3,3 is “contained” in Petersen's graph.

106. 106 So? We have shown that K 3,3 is “contained” in Petersen's graph. So what?

107. 107 So? We have shown that K 3,3 is “contained” in Petersen's graph. So what? So, that proves that Petersen's graph is nonplanar.

108. 108 How so? Say that Petersen's graph is planar.

109. 109 How so? Say that Petersen's graph is planar. Then we can draw it with no edge crossings.

110. 110 How so? Say that Petersen's graph is planar. Then we can draw it with no edge crossings. Now, remove the four vertices and six edges as we did above, and what have we got?

111. 111 How so? A planar drawing of K 3,3.

112. 112 How so? A planar drawing of K 3,3. BUT, K 3,3 is nonplanar, so a planar drawing does not exist, so we have a contra-diction.

113. 113 How so? A planar drawing of K 3,3. But, K 3,3 is nonplanar, so a planar drawing does not exist, so we have a contra- diction. Thus the original assumption that Petersen's graph is planar is false.

114. 114 A test for nonplanarity We can generalize the above proof that Petersen's graph is nonplanar:

115. 115 A test for nonplanarity We can generalize the above proof that Petersen's graph is nonplanar: If a graph G “contains” a nonplanar graph, then G is nonplanar also.

116. 116 A test for nonplanarity We can generalize the above proof that Petersen's graph is nonplanar: If a graph G “contains” a nonplanar graph, then G is nonplanar also. But, it gets even better...

117. 117 Kuratowski to the rescue In 1930 Kuratowski (1896-1980) proved: A graph is planar if and only if it does not “contain” K 3,3 or K 5.

118. 118 Kuratowski to the rescue So every nonplanar graph “contains” K 3,3 or K 5.

119. 119 Kuratowski to the rescue So every nonplanar graph “contains” K 3,3 or K 5. And no planar graph contains K 3,3 or K 5.

120. 120 K 3,3 and K 5

121. 121 Now prove these are nonplanar

122. 122 The Traveling Salesman Problem A salesman's territory covers the city he lives in and 4 other cities.

123. 123 The Traveling Salesman Problem A salesman's territory covers the city he lives in and 4 other cities. The salesman starts from his city and visits one of the other cities, and from that city he visits a third city, etc., until he arrives back home, having visited each of the 5 cities exacty once.

124. 124 The Traveling Salesman Problem A salesman's territory covers the city he lives in and 4 other cities. The salesman starts from his city and visits one of the other cities, and from that city he visits a third city, etc., until he arrives back home, having visited each of the 5 cities exacty once. How to pick a shortest tour?

125. 125 What is the shortest tour? 403 A B C D E 300 632 447 47 2 608 640 25 0 44 7

126. 126 Shortest tour = ABDCE (1855) 403 A B C D E 300 632 447 47 2 608 640 25 0 44 7

127. 127 How to find the shortest tour? Here's one way to do it: list all possible tours add up the edge distances for each tour pick the tour with smallest total distance

128. 128 Try ABCDE 403 A B C D E 300 632 447 47 2 608 640 25 0 44 7 300 472 250 447 1916

129. 129 Lengths of all tours ABCDE: 1916 ACBDE: 2378 ABCED: 2101 ACBED: 2595 ABDCE: 1855 ADBCE: 2410 ABDEC: 2009 ADBEC: 2534 ABECD: 2073 AEBCD: 2442 ABEDC: 2041 AEBDC: 2349

130. 130 Tours sorted by length ABDCE: 1855 AEBDC: 2349 ABCDE: 1916 ACBDE: 2378 ABDEC: 2009 ADBCE: 2410 ABEDC: 2041 AEBCD: 2442 ABECD: 2073 ADBEC: 2534 ABCED: 2101 ACBED: 2595

131. 131 Longest tour = ACBED (2595) 403 A B C D E 300 632 447 47 2 608 640 25 0 44 7

132. 132 How many tours are there? n # edges = n(n-1)/2 # tours = (n-1)!/2 ------------------------------------------------------------ 5 10 12 6 15 60 7 21 360 8 28 2,520 9 36 20,160 10 45 181,440 15 105 43,589,145,600 20 190 60,822,550,204,416,000 25 300 310,224,200,866,619,719,680,000

133. 133 How long to check all the tours? Fastest computer today does 360*10 12 operations per second.

134. 134 How long to check all the tours? Fastest computer today does 360*10 12 operations per second. Round this up to 1000*10 12 = 10 15 operations per second.

135. 135 How long to check all the tours? Fastest computer today does 360*10 12 operations per second. Round this up to 1000*10 12 = 10 15 operations per second. Assume 10 15 tours can be checked in 1 second.

136. 136 How long to check all the tours? How long will it take to check 310,224,200,866,619,719,680,000 tours?

137. 137 How long to check all the tours? How long will it take to check 310,224,200,866,619,719,680,000 tours? 310,224,200,866,619,719,680,000 / 10 15

138. 138 How long to check all the tours? How long will it take to check 310,224,200,866,619,719,680,000 tours? 310,224,200,866,619,719,680,000 / 10 15 = 310,224,201 seconds

139. 139 How long to check all the tours? How long will it take to check 310,224,200,866,619,719,680,000 tours? 310,224,200,866,619,719,680,000 / 10 15 = 310,224,201 seconds = 9.837 years

140. 140 How long to check all the tours? How long will it take to check 310,224,200,866,619,719,680,000 tours? 310,224,200,866,619,719,680,000 / 10 15 = 310,224,201 seconds = 9.837 years = 9 years, 10 months

141. 141 Bad news Obviously the method we have used to solve the TSP is no good at all for rather small problems, e.g., 25 cities.

142. 142 Bad news Obviously the method we have used to solve the TSP is no good at all for rather small problems, e.g., 25 cities. We want an efficient procedure, which means a process that does not grow exponentially with the number of cities.

143. 143 Bad news Obviously the method we have used to solve the TSP is no good at all for rather small problems, e.g., 25 cities. We want an efficient procedure, which means a process that does not grow exponentially with the number of cities. Here's the bad news: No such procedure is known.

144. 144 Bad news Obviously the method we have used to solve the TSP is no good at all for rather small problems, e.g., 25 cities. We want an efficient procedure, which means a process that does not grow exponentially with the number of cities. Here's the bad news: No such procedure is known. Finding an efficient procedure for solving TSP in general is an unsolved problem

145. 145 Good news However, this does not mean that we can't handle 25 cities.

146. 146 Good news However, this does not mean that we can't handle 25 cities. There are clever algorithms which make it possible to handle very large cases.

147. 147 Good news However, this does not mean that we can't handle 25 cities. There are clever algorithms which make it possible to handle very large cases. For example, consider this figure from a book by David Applegate, Robert Bixby, Vasek Chvatal and William Cook:

148. 148 A large solved tour 13,509 towns in the US with pop. > 500

149. 149 Bad news, again Despite success on various practical cases, from a mathematical point of view, the problem of finding an efficient process for solving the Traveling Salesman Problem is unsolved.

150. 150 Good news, again According to David Hilbert (1862-1943): “As long as a branch of science offers an abundance of problems, so long is it alive; a lack of problems foreshadows extinction or the cessation of independent development.”

151. 151 Good news, again According to David Hilbert (1862-1943): “As long as a branch of science offers an abundance of problems, so long is it alive; a lack of problems foreshadows extinction or the cessation of independent development.” According to the Everly Brothers: “Problems, problems, problems all day long”