1. Honors Track: Competitive Programming & Problem Solving Avoiding negative edges
Steven Ge

2. Content Negative Cycles Dijkstra’s Algorithm with negative edges
Adding constant
Bellman-Ford Algorithm
Single source shortest path
Johnson’s Algorithm
All pair shortest path
Suurballe’s Aglorithm
Cheapest multiple, non-overlapping paths

3. Negative Cycles
Graph 1 1 2 S D
-10
Shortest Path?

4. Negative Cycles
Graph 1 1 2 S D
-10
Shortest Path?
No Shortest Path!

5. Dijkstra’s with negative edges
Graph A 10
-100 1 S B -2 D

6. Dijkstra’s with negative edges
Shortest Path A 10
-100 1 S B 1 D
Total length: -89

7. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: D B A A 10
-100 1 ∞ S B 1 D ∞

8. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: B A A 10
-100 1 ∞ S B 1 D ∞

9. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: B A A 10
-100 1 ∞ S B 1 D

10. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: A A 10
-100 1 ∞ S B 1 D

11. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: A A 10
-100 1 1 S B 1 D

12. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: D A A 10
-100 1 1 S B 1 D

13. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: A A 10
-100 1 1 S B 1 D

14. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: A A 10
-100 1 1 S B 1 D
No expansion on D

15. Dijkstra’s with negative edges
Dijkstra’s with closed set ∞
PQ: A 10
-100 1 1 S B 1 D

16. Dijkstra’s with negative edges
Dijkstra’s with closed set 10
PQ: A 10
-100 1 1 S B 1 D

17. Dijkstra’s with negative edges
Dijkstra’s with closed set 10
PQ: B A 10
-100 1 1 S B 1 D

18. Dijkstra’s with negative edges
Dijkstra’s with closed set 10
PQ: A 10
-100 1 1 S B 1 D

19. Dijkstra’s with negative edges
Dijkstra’s with closed set 10
PQ: A 10
-100 1 1 S B 1 D
B is in closed set, so no more expansions
0 > -89
Incorrect output, so don’t bother

20. Dijkstra’s with negative edges
Dijkstra’s without closed set A 10
-100 1 S B 1
Gigantic graph with
Node D T

21. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: T B A A 10
-100 1 ∞ S B 1
Gigantic graph with
Node D T ∞

22. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: B A A 10
-100 1 ∞ S B 1
Gigantic graph with
Node D T ∞

23. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ:
GgwND B A A 10
-100 1 ∞ S B 1
Gigantic graph with
Node D T

24. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: B A A 10
-100 1 ∞ S B 1
Gigantic graph with
Node D T

25. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: B A A 10
-100 1 ∞ S B 1
Gigantic graph with
Node D T

26. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: A A 10
-100 1 ∞ S B 1
Gigantic graph with
Node D T

27. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: A A 10
-100 1 1 S B 1
Gigantic graph with
Node D T

28. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: T A A 10
-100 1 1 S B 1
Gigantic graph with
Node D T

29. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: A A 10
-100 1 1 S B 1
Gigantic graph with
Node D T

30. Dijkstra’s with negative edges
Dijkstra’s without closed set ∞
PQ: A 10
-100 1 1 S B 1
Gigantic graph with
Node D T

31. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ: A 10
-100 1 1 S B 1
Gigantic graph with
Node D T

32. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ: B A 10
-100 1 1 S B 1
Gigantic graph with
Node D T

33. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ: A 10
-100 1 1 S B 1
Gigantic graph with
Node D T

34. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ: A 10
-100 1
-90 S B 1
Gigantic graph with
Node D T

35. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ: T A 10
-100 1
-90 S B 1
Gigantic graph with
Node D T

36. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ: A 10
-100 1
-90 S B 1
Gigantic graph with
Node D T

37. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ: A 10
-100 1
-90 S B 1
Gigantic graph with
Node D T
-89

38. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ:
GgwND A 10
-100 1
-90 S B 1
Gigantic graph with
Node D T
-89

39. Dijkstra’s with negative edges
Dijkstra’s without closed set 10
PQ: A 10
-100 1
-90 S B 1
Gigantic graph with
Node D T
-89
Correct output
Takes very long

40. Dijkstra’s with negative cycles
Graph
PQ: A 1 1 2 ∞ S A B D ∞ ∞
-10

41. Dijkstra’s with negative cycles
Graph
PQ: 1 1 2 ∞ S A B D ∞ ∞
-10

42. Dijkstra’s with negative cycles
Graph
PQ: 1 1 2 ∞ S A B D ∞ 1
-10

43. Dijkstra’s with negative cycles
Graph
PQ: B 1 1 2 ∞ S A B D ∞ 1
-10

44. Dijkstra’s with negative cycles
Graph
PQ: 1 1 2 ∞ S A B D ∞ 1
-10

45. Dijkstra’s with negative cycles
Graph
PQ: 1 1 2 ∞ S A B D 1 2
-10

46. Dijkstra’s with negative cycles
Graph
PQ: A D 1 1 2 ∞ S A B D 1 2
-10

47. Dijkstra’s with negative cycles
Graph
PQ: D 1 1 2 ∞ S A B D 1 2
-10

48. Dijkstra’s with negative cycles
Graph
PQ: D 1 1 2 ∞ S A B D -8 2
-10

49. Dijkstra’s with negative cycles
Graph
PQ: B D 1 1 2 ∞ S A B D -8 2
-10

50. Dijkstra’s with negative cycles
Graph
PQ: D 1 1 2 ∞ S A B D -8 2
-10

51. Dijkstra’s with negative cycles
Graph
PQ: D 1 1 2 ∞ S A B D -8 -7
-10

52. Dijkstra’s with negative cycles
Graph
PQ: A D 1 1 2 ∞ S A B D -8 -7
-10
Etc.

53. Adding Constant
Graph 1 1 1
-100 10 10

54. Adding Constant Shortest Path Add constant 100 and run Dijkstra’s? 1 1
-100 10 10
Add constant 100 and run Dijkstra’s?

55. Adding Constant Shortest Path Add constant 100 and run Dijkstra’s?
101
101
101
110
110
Add constant 100 and run Dijkstra’s?
Difference in amount of edges

56. Bellman-Ford Similar to Dijkstra’s No priority queue
Slower than Dijkstra’s
O(V*E)
More versatile than Dijkstra’s (can detect negative cycles)
Used in Johnson’s Algorithm

57. Bellman-Ford Steps: 1) Initialize graph
2) Relax all edges v-1 times (v = amount vertices)
In other words: For each vertex, check if the distance to current vertex + edge distance < distance to child vertex
3) Check for negative cycles
For each edge, if the distance to current vertex + edge distance < distance to child vertex, return error

58. Bellman-Ford
Graph A
100 50
-100 10 20 S B D 20 1 20 C

59. Bellman-Ford ∞ ∞ ∞ ∞ Step 1 ∞ S A B C D A 100 50 -100 10 20 S B D 20 1∞ S A B C D ∞ A
100 50
-100 ∞ 10 20 ∞ S B D 20 1 20 C ∞

60. Bellman-Ford ∞ ∞ ∞ ∞ Step 2 ∞ S A B C D A 100 50 -100 10 20 S B D 20 1∞ S A B C D ∞ A
100 50
-100 ∞ 10 20 ∞ S B D 20 1 20 C ∞

61. Bellman-Ford ∞ ∞ ∞ Step 2 100 ∞ S A B C D 100 A 100 50 -100 10 20 S B
100 ∞ S A B C D
100 A
100 50
-100 ∞ 10 20 ∞ S B D 20 1 20 C ∞

62. Bellman-Ford ∞ ∞ ∞ Step 2 100 ∞ S A B C D 100 A 100 50 -100 10 20 S B
100 ∞ S A B C D
100 A
100 50
-100 ∞ 10 20 ∞ S B D 20 1 20 C ∞

63. Bellman-Ford ∞ ∞ Step 2 100 10 ∞ S A B C D 100 A 100 50 -100 10 10 20
100 10 ∞ S A B C D
100 A
100 50
-100 10 10 20 ∞ S B D 20 1 20 C ∞

64. Bellman-Ford ∞ ∞ Step 2 100 10 ∞ S A B C D 100 A 100 50 -100 10 10 20
100 10 ∞ S A B C D
100 A
100 50
-100 10 10 20 ∞ S B D 20 1 20 C ∞

65. Bellman-Ford ∞ Step 2 100 10 1 ∞ S A B C D 100 A 100 50 -100 10 10 20
100 10 1 ∞ S A B C D
100 A
100 50
-100 10 10 20 ∞ S B D 20 1 20 C 1

66. Bellman-Ford ∞ Step 2 100 10 1 ∞ S A B C D 100 A 100 50 -100 10 10 20
100 10 1 ∞ S A B C D
100 A
100 50
-100 10 10 20 ∞ S B D 20 1 20 C 1

67. Bellman-Ford Step 2 100 10 1 150 S A B C D 100 A 100 50 -100 10 10 20
100 10 1
150 S A B C D
100 A
100 50
-100 10 10 20 S B D
150 20 1 20 C 1

68. Bellman-Ford Step 2 100 10 1 150 S A B C D 100 A 100 50 -100 10 10 20
100 10 1
150 S A B C D
100 A
100 50
-100 10 10 20 S B D
150 20 1 20 C 1

69. Bellman-Ford Step 2 100 1 150 S A B C D 100 A 100 50 -100 10 20 S B D
100 1
150 S A B C D
100 A
100 50
-100 10 20 S B D
150 20 1 20 C 1

70. Bellman-Ford Step 2 100 1 150 S A B C D 100 A 100 50 -100 10 20 S B D
100 1
150 S A B C D
100 A
100 50
-100 10 20 S B D
150 20 1 20 C 1

71. Bellman-Ford Step 2 100 1 20 S A B C D 100 A 100 50 -100 10 20 S B D
100 1 20 S A B C D
100 A
100 50
-100 10 20 S B D 20 20 1 20 C 1

72. Bellman-Ford Step 2 100 1 20 S A B C D 100 A 100 50 -100 10 20 S B D
100 1 20 S A B C D
100 A
100 50
-100 10 20 S B D 20 20 1 20 C 1

73. Bellman-Ford Step 2 100 1 20 S A B C D 100 A 100 50 -100 10 20 S B D
100 1 20 S A B C D
100 A
100 50
-100 10 20 S B D 20 20 1 20 C 1

74. Bellman-Ford Step 2 100 1 20 S A B C D 100 A 100 50 -100 10 20 S B D
100 1 20 S A B C D
100 A
100 50
-100 10 20 S B D 20 20 1 20 C 1

75. Bellman-Ford
Step 2
100 1 20 S A B C D
100 A
100 50
-100 10 20 S B D 20 20 1 20 C 1
After the second iteration, the distances don’t change
This also means no negative cycles

76. Bellman-Ford Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2

77. Bellman-Ford ∞ ∞ ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 1 1 1 1 S D ∞ ∞ ∞

78. Bellman-Ford ∞ ∞ ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 1 1 1 1 S D ∞ ∞ ∞

79. Bellman-Ford ∞ ∞ ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 1 1 1 1 S D ∞ ∞ ∞

80. Bellman-Ford ∞ ∞ ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 1 1 1 1 S D ∞ ∞ ∞

81. Bellman-Ford ∞ ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 1 1 1 1 S D ∞ ∞ 1

82. Bellman-Ford ∞ ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 2 1 1 1 S D ∞ ∞ 1

83. Bellman-Ford ∞ ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 2 1 1 1 S D ∞ ∞ 1

84. Bellman-Ford ∞ ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 2 1 1 1 S D ∞ ∞ 1

85. Bellman-Ford ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 2 1 1 1 S D ∞ 1 2

86. Bellman-Ford ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 2 1 1 1 S D ∞ 1 2

87. Bellman-Ford ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 3 1 1 1 S D ∞ 1 2

88. Bellman-Ford ∞ Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 3 1 1 1 S D ∞ 1 2

89. Bellman-Ford Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 3 1 1 1 S D 1 2 3

90. Bellman-Ford Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 3 1 1 1 S D 1 2 3

91. Bellman-Ford Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 3 1 1 1 S D 1 2 3

92. Bellman-Ford Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 3 1 1 1 S D 1 2 3

93. Bellman-Ford Justification:
Step 2 needs to be done at most v-1 times, because the optimal path has at most v-1 edges, otherwise it contains a cycle
Step 3 follows from 2
Iteration: 3 1 1 1 S D 1 2 3
Contains cycle, because optimal path contains at most all the nodes
All pairs, O(V2 * E)

94. Johnson’s Algorithm Reweight all the edges effectively
Makes use of both Dijkstra’s and Bellman-Ford

95. Johnson’s Algorithm Steps: 1) Add new node q to graph
2) Connect node q to all other nodes in graph (directed towards other nodes)
3) Initialize graph with q as source
4) Run Bellman-Ford
5) Remove q and reweight the edges from u to v with the following formula
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
6) Run Dijkstra’s
7) Use the formula to get the original weight
Wold(u, v) = Wnew(u, v) - h(u) + h(v)

96. Johnson’s Algorithm
Graph A
100 50
-100 10 20 S B D 20 1 20 C

97. Johnson’s Algorithm
Step 1 A
100 50
-100 10 20 S B D 20 1 20 C q

98. Johnson’s Algorithm
Step 2 A
100 50
-100 10 20 S B D 20 1 20 C q

99. Johnson’s Algorithm Step 3 + 4 A 100 50 -100 10 -100 20 -80 S B D 20 1A
100 50
-100 10
-100 20
-80 S B D 20 1 20
-80 C q

100. Johnson’s Algorithm Step 3 + 4 A 100 50 -100 10 -100 20 -80 S B D 20 1A
100 50
-100 10
-100 20
-80 S B D 20 1 20
-80 C q

101. Johnson’s Algorithm Step 5 A 100 50 -100 10 -100 20 -80 S B D 20 1 20A
100 50
-100 10
-100 20
-80 S B D 20 1 20
-80 C q

102. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v) A 100A
100 50
-100 10
-100 20
-80 S B D 20 1 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)

103. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v) A 100A
100 50
-100 10
-100 20
-80 S B D 20 1 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)

104. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
-100 10
-100 20
-80 S B D 20 1 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– 0 = 100

105. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
-100 10
-100 20
-80 S B D 20 1 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -100 = 110

106. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
-100
110
-100 20
-80 S B D 20 1 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -100 = 110

107. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
-100
110
-100 20
-80 S B D 20 1 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
1 + 0 – -80 = 81

108. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
-100
110
-100 20
-80 S B D 20 81 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
1 + 0 – -80 = 81

109. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
-100
110
-100 20
-80 S B D 20 81 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -100 = 0

110. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
110
-100 20
-80 S B D 20 81 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -100 = 0

111. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
110
-100 20
-80 S B D 20 81 20
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -80 = 0

112. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
110
-100 20
-80 S B D 20 81
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -80 = 0

113. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100 50
110
-100 20
-80 S B D 20 81
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -80 = 130

114. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100
130
110
-100 20
-80 S B D 20 81
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -80 = 130

115. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100
130
110
-100 20
-80 S B D 20 81
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -80 = 0

116. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100
130
110
-100
-80 S B D 20 81
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -80 = 0

117. Johnson’s Algorithm Step 5 Wnew(u, v) = Wold(u, v) + h(u) – h(v)A
100
130
110
-100
-80 S B D 20 81
-80 C
Wnew(u, v) = Wold(u, v) + h(u) – h(v)
– -80 = 20

118. Johnson’s Algorithm
Step 6 A
100
130
110
-100
-80 S B D 20 81
-80 C

119. Johnson’s Algorithm Step 6, shortest path to AA
100
130
110
-100
-80 S B D 20 81
-80 C
Wold(u, v) = Wnew(u, v) - h(u) + h(v)
100 – = 100

120. Johnson’s Algorithm Step 6, shortest path to A
100 A
100
130
110
-100
-80 S B D 20 81
-80 C
Wold(u, v) = Wnew(u, v) - h(u) + h(v)
100 – = 100

121. Johnson’s Algorithm Step 6, shortest path to B
100 A
100
130
110
-100
-80 S B D 20 81
-80 C
Wold(u, v) = Wnew(u, v) - h(u) + h(v)
( ) – = 0

122. Johnson’s Algorithm Step 6, shortest path to B
100 A
100
130
110
-80 S B D 20 81
-80 C
Wold(u, v) = Wnew(u, v) - h(u) + h(v)
( ) – = 0

123. Johnson’s Algorithm Step 6, shortest path to C
100 A
100
130
110
-80 S B D 20 81
-80 C
Wold(u, v) = Wnew(u, v) - h(u) + h(v)
81 – = 1

124. Johnson’s Algorithm Step 6, shortest path to C
100 A
100
130
110
-80 S B D 20 81 1 C
Wold(u, v) = Wnew(u, v) - h(u) + h(v)
81 – = 1

125. Johnson’s Algorithm Step 6, shortest path to D
100 A
100
130
110
-80 S B D 20 81 1 C
Wold(u, v) = Wnew(u, v) - h(u) + h(v)
( ) – = 20

126. Johnson’s Algorithm Step 6, shortest path to D
100 A
100
130
110 20 S B D 20 81 1 C
Wold(u, v) = Wnew(u, v) - h(u) + h(v)
( ) – = 20

127. Bellman-Ford Step 2 100 1 20 S A B C D 100 A 100 50 -100 10 20 S B D
100 1 20 S A B C D
100 A
100 50
-100 10 20 S B D 20 20 1 20 C 1

128. Johnson’s Algorithm
Using node q guaranties that all edges and nodes are checked
Node q does not affect any properties of the graph
Single source:
Dijkstra: O((V + E) * log V)
Bellman-Ford: O(V * E)
Johnson: O(V * E)
All pairs:
Dijkstra: O((V2 + E) log V)
Bellman-Ford: O(V2 * E)
Johnson: O((V2 + E) log V)

129. Johnson’s Algorithm Correctness reweighting path:
Claim: Every path with the same source and destination will be shifted equally
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

130. Johnson’s Algorithm Correctness reweighting:
Claim: Every path with the same source and destination will be shifted equally
Let P be a path from S to D
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

131. Johnson’s Algorithm Correctness reweighting:
Claim: Every path with the same source and destination will be shifted equally
Let P be a path from S to D
(Wold (S, V1) + h(S) – h(V1)) + (Wold (V1, V2) + h(V1) – h(V2)) + … (Wold (Vn, D) + h(Vn) – h(D)) = Wnew(S, D)
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

132. Johnson’s Algorithm Correctness reweighting:
Claim: Every path with the same source and destination will be shifted equally
Let P be a path from S to D
(Wold (S, V1) + h(S) – h(V1)) + (Wold (V1, V2) + h(V1) – h(V2)) + … (Wold (Vn, D) + h(Vn) – h(D)) = Wnew(S, D)
+ h(Vi) and – h(Vi) cancel each other out.
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

133. Johnson’s Algorithm Correctness reweighting:
Claim: Every path with the same source and destination will be shifted equally
Let P be a path from S to D
(Wold (S, V1) + h(S) – h(V1)) + (Wold (V1, V2) + h(V1) – h(V2)) + … (Wold (Vn, D) + h(Vn) – h(D)) = Wnew(S, D)
+ h(Vi) and – h(Vi) cancel each other out.
(Wold (S, V1) + Wold (V1, V2) + … + Wold (Vn, D) + h(S) – h(D)) = Wnew(S, D)
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

134. Johnson’s Algorithm Correctness reweighting:
Claim: Every path with the same source and destination will be shifted equally
Let P be a path from S to D
(Wold (S, V1) + h(S) – h(V1)) + (Wold (V1, V2) + h(V1) – h(V2)) + … (Wold (Vn, D) + h(Vn) – h(D)) = Wnew(S, D)
+ h(Vi) and – h(Vi) cancel each other out.
(Wold (S, V1) + Wold (V1, V2) + … + Wold (Vn, D) + h(S) – h(D)) = Wnew(S, D)
Each path from S to T gets shifted a fixed amount, unaffected by the amount of edges
Each path from S to T gets shifted the same amount. (Only with h(S) – h(D))
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

135. Johnson’s Algorithm Correctness reweighting edge:
Claim: Wnew(u, v) = Wold(u, v) + h(u) – h(v) always creates non-negative edges
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

136. Johnson’s Algorithm Correctness reweighting:
Claim: Wnew(u, v) = Wold(u, v) + h(u) – h(v) always creates non-negative edges
Pick a random edge (u, v) from the original graph
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

137. Johnson’s Algorithm Correctness reweighting:
Claim: Wnew(u, v) = Wold(u, v) + h(u) – h(v) always creates non-negative edges
Pick a random edge (u, v) from the original graph
Use the q node from step 2
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

138. Johnson’s Algorithm Correctness reweighting:
Claim: Wnew(u, v) = Wold(u, v) + h(u) – h(v) always creates non-negative edges
Pick a random edge (u, v) from the original graph
Use the q node from step 2
h(u) = length of a shortest path from q to u
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

139. Johnson’s Algorithm Correctness reweighting:
Claim: Wnew(u, v) = Wold(u, v) + h(u) – h(v) always creates non-negative edges
Pick a random edge (u, v) from the original graph
Use the q node from step 2
h(u) = length of a shortest path from q to u
h(v) = length of a shortest path from q to v
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

140. Johnson’s Algorithm Correctness reweighting:
Claim: Wnew(u, v) = Wold(u, v) + h(u) – h(v) always creates non-negative edges
Pick a random edge (u, v) from the original graph
Use the q node from step 2
h(u) = length of a shortest path from q to u
h(v) = length of a shortest path from q to v
Length of a path from q to u to v is:
Wold(u, v) + h(u)
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

141. Johnson’s Algorithm Correctness reweighting:
Claim: Wnew(u, v) = Wold(u, v) + h(u) – h(v) always creates non-negative edges
Pick a random edge (u, v) from the original graph
Use the q node from step 2
h(u) = length of a shortest path from q to u
h(v) = length of a shortest path from q to v
Length of a path from q to u to v is:
Wold(u, v) + h(u)
Length of a path ≥ length of a shortest path
Wold(u, v) + h(u) ≥ h(v)
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

142. Johnson’s Algorithm Correctness reweighting:
Claim: Wnew(u, v) = Wold(u, v) + h(u) – h(v) always creates non-negative edges
Pick a random edge (u, v) from the original graph
Use the q node from step 2
h(u) = length of a shortest path from q to u
h(v) = length of a shortest path from q to v
Length of a path from q to u to v is:
Wold(u, v) + h(u)
Length of a path ≥ length of a shortest path
Wold(u, v) + h(u) ≥ h(v)
Rewrite
Wold(u, v) + h(u) – h(v) ≥ 0
Wnew ≥ 0
Each path from S to T gets shifted the same amount.
So the shortest path of the set of paths from S to T will be
the same in the new graph as in the old graph.

143. Suurballe’s Algorithm
Finding disjoint paths in a network
The paths have the shortest length compared to other set of paths with the same amount.

144. Suurballe’s Algorithm
Steps:
1) Find the shortest path with Dijksta’s
2) Create a shortest path tree T from source S with Dijkstra’s
3) Use: Wnew(u, v) = Wold(u, v) + h(u) – h(v) to reweight the edges and reverse the edges of the shortest path
4) Rerun Dijkstra’s to find the second path
5) Remove the overlapping edges from the first and second path
6) connect the ends of the graphs together

145. Suurballe’s Algorithm
Graph: 99 A E 73 8 17 23 5 33 S B D 20 12 3 21 C

146. Suurballe’s Algorithm
Step 1: 99 A E 73 8 17 23 5 33 S B D 20 12 3 21 C

147. Suurballe’s Algorithm
Step 2: 23 20 A E 8 23 5 15 S B D 28 12 3 C 12

148. Suurballe’s Algorithm
Step 3: 99 23 20 A E 73 8 17 23 5 33 15 S B D 20 28 12 3 21 C 12

149. Suurballe’s Algorithm
Step 3: 79 23 20 A E 68 25 25 15 S B D 33 28 5 C 12

150. Suurballe’s Algorithm
Step 2 for comparison:
Notice that all the edges in this graph becomes 0
Because of the formula 23 20 A E 8 23 5 15 S B D 28 12 3 C 12

151. Suurballe’s Algorithm
Step 3: 79 A E 68 25 25 S B D 33 5 C

152. Suurballe’s Algorithm
Step 4: 79 A E 68 25 25 S B D 33 5 C

153. Suurballe’s Algorithm
Step 5: 99 A E 73 8 17 23 5 33 S B D 20 12 3 21 C

154. Suurballe’s Algorithm
Step 5: 99 A E 73 8 17 23 5 33 S B D 20 12 3 21 C

155. Suurballe’s Algorithm
Step 6: 99 A E 73 8 17 23 5 33 S B D 20 12 3 21 C

156. Suurballe’s Algorithm
Extra constraint:
Paths cannot go through the same node

157. Suurballe’s Algorithm
Extra constraint:
Paths cannot go through the same node

158. Suurballe’s Algorithm
Extra constraint:
Paths cannot go through the same node
Split the node in
out

159. Suurballe’s Algorithm
Justification:
Reweighting technique proven at Johnson’s
Reversed edges of the original graph used as backward edges
Similar to finding a minimum flow from S to D

160. Conclusion Single source:
Graph without negative edges: Dijkstra’s algorithm
Graph with negative edges: Bellman-Ford
All pairs:
Graph with negative edges: Johnson’s algorithm
Finding multiple, non-overlapping shortest paths in a graph
Suurballe’s algorithm

161. References Bellman-Ford: https://www.youtube.com/watch?v=W2fKGISUAtM
Johnson’s:
Suurballe’s:
Incorrect naming, but maybe nice to look at:
NWERC 2012 A - Admiral