As many things in the history of analysis of algorithms the all-pairs shortest path has a long history (From the point of view of Computer Science). We can see the initial results from the book “Studies in the Economics of Transportation” by Beckmann, McGuire, and Winsten (1956) where the notation that we use for the matrix multiplication alike was first used. In this slides, we go trough several all pairs Shortest path problem solutions from a slow version to the Johnson algorithm.

1. Analysis of Algorithms
All-Pairs Shortest Path
Andres Mendez-Vazquez
November 11, 2015
1 / 79

2. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
2 / 79

3. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
3 / 79

4. Problem
Definition
Given u and v, find the shortest path.
Now, what if you want ALL PAIRS!!!
Use as a source all the elements in V .
Clearly!!! you can fall back to the old algorithms!!!
4 / 79

5. Problem
Definition
Given u and v, find the shortest path.
Now, what if you want ALL PAIRS!!!
Use as a source all the elements in V .
Clearly!!! you can fall back to the old algorithms!!!
4 / 79

6. Problem
Definition
Given u and v, find the shortest path.
Now, what if you want ALL PAIRS!!!
Use as a source all the elements in V .
Clearly!!! you can fall back to the old algorithms!!!
4 / 79

7. Problem
Definition
Given u and v, find the shortest path.
Now, what if you want ALL PAIRS!!!
Use as a source all the elements in V .
Clearly!!! you can fall back to the old algorithms!!!
4 / 79

8. What can we use?
Use Dijkstra’s |V | times!!!
If all the weights are non-negative.
This has, using Fibonacci Heaps, O (V 2
log V + VE) complexity.
Which is equal O (V 3
) in the case of E = O(V 2
), but with a hidden large
constant c.
Use Bellman-Ford |V | times!!!
If negative weights are allowed.
Then, we have O (V 2
E).
Which is equal O (V 4
) in the case of E = O(V 2
).
5 / 79

9. What can we use?
Use Dijkstra’s |V | times!!!
If all the weights are non-negative.
This has, using Fibonacci Heaps, O (V 2
log V + VE) complexity.
Which is equal O (V 3
) in the case of E = O(V 2
), but with a hidden large
constant c.
Use Bellman-Ford |V | times!!!
If negative weights are allowed.
Then, we have O (V 2
E).
Which is equal O (V 4
) in the case of E = O(V 2
).
5 / 79

10. What can we use?
Use Dijkstra’s |V | times!!!
If all the weights are non-negative.
This has, using Fibonacci Heaps, O (V 2
log V + VE) complexity.
Which is equal O (V 3
) in the case of E = O(V 2
), but with a hidden large
constant c.
Use Bellman-Ford |V | times!!!
If negative weights are allowed.
Then, we have O (V 2
E).
Which is equal O (V 4
) in the case of E = O(V 2
).
5 / 79

11. What can we use?
Use Dijkstra’s |V | times!!!
If all the weights are non-negative.
This has, using Fibonacci Heaps, O (V 2
log V + VE) complexity.
Which is equal O (V 3
) in the case of E = O(V 2
), but with a hidden large
constant c.
Use Bellman-Ford |V | times!!!
If negative weights are allowed.
Then, we have O (V 2
E).
Which is equal O (V 4
) in the case of E = O(V 2
).
5 / 79

12. What can we use?
Use Dijkstra’s |V | times!!!
If all the weights are non-negative.
This has, using Fibonacci Heaps, O (V 2
log V + VE) complexity.
Which is equal O (V 3
) in the case of E = O(V 2
), but with a hidden large
constant c.
Use Bellman-Ford |V | times!!!
If negative weights are allowed.
Then, we have O (V 2
E).
Which is equal O (V 4
) in the case of E = O(V 2
).
5 / 79

13. What can we use?
Use Dijkstra’s |V | times!!!
If all the weights are non-negative.
This has, using Fibonacci Heaps, O (V 2
log V + VE) complexity.
Which is equal O (V 3
) in the case of E = O(V 2
), but with a hidden large
constant c.
Use Bellman-Ford |V | times!!!
If negative weights are allowed.
Then, we have O (V 2
E).
Which is equal O (V 4
) in the case of E = O(V 2
).
5 / 79

14. This is not Good For Large Problems
Problems
Computer Network Systems.
Aircraft Networks (e.g. flying time, fares).
Railroad network tables of distances between all pairs of cites for a
road atlas.
Etc.
6 / 79

15. This is not Good For Large Problems
Problems
Computer Network Systems.
Aircraft Networks (e.g. flying time, fares).
Railroad network tables of distances between all pairs of cites for a
road atlas.
Etc.
6 / 79

16. This is not Good For Large Problems
Problems
Computer Network Systems.
Aircraft Networks (e.g. flying time, fares).
Railroad network tables of distances between all pairs of cites for a
road atlas.
Etc.
6 / 79

17. This is not Good For Large Problems
Problems
Computer Network Systems.
Aircraft Networks (e.g. flying time, fares).
Railroad network tables of distances between all pairs of cites for a
road atlas.
Etc.
6 / 79

18. For more on this...
Something Notable
As many things in the history of analysis of algorithms the all-pairs
shortest path has a long history.
We have more from
“Studies in the Economics of Transportation” by Beckmann, McGuire, and
Winsten (1956) where the notation that we use for the matrix multiplication alike
was first used.
In addition
G. Tarry, Le probleme des labyrinthes, Nouvelles Annales de Mathématiques (3) 14 (1895)
187–190 [English translation in: N.L. Biggs, E.K. Lloyd, R.J. Wilson, Graph Theory
1736–1936, Clarendon Press, Oxford, 1976, pp. 18–20] (For the theory behind
depth-first search techniques).
Chr. Wiener, Ueber eine Aufgabe aus der Geometria situs, Mathematische Annalen 6
(1873) 29–30, 1873.
7 / 79

19. For more on this...
Something Notable
As many things in the history of analysis of algorithms the all-pairs
shortest path has a long history.
We have more from
“Studies in the Economics of Transportation” by Beckmann, McGuire, and
Winsten (1956) where the notation that we use for the matrix multiplication alike
was first used.
In addition
G. Tarry, Le probleme des labyrinthes, Nouvelles Annales de Mathématiques (3) 14 (1895)
187–190 [English translation in: N.L. Biggs, E.K. Lloyd, R.J. Wilson, Graph Theory
1736–1936, Clarendon Press, Oxford, 1976, pp. 18–20] (For the theory behind
depth-first search techniques).
Chr. Wiener, Ueber eine Aufgabe aus der Geometria situs, Mathematische Annalen 6
(1873) 29–30, 1873.
7 / 79

20. For more on this...
Something Notable
As many things in the history of analysis of algorithms the all-pairs
shortest path has a long history.
We have more from
“Studies in the Economics of Transportation” by Beckmann, McGuire, and
Winsten (1956) where the notation that we use for the matrix multiplication alike
was first used.
In addition
G. Tarry, Le probleme des labyrinthes, Nouvelles Annales de Mathématiques (3) 14 (1895)
187–190 [English translation in: N.L. Biggs, E.K. Lloyd, R.J. Wilson, Graph Theory
1736–1936, Clarendon Press, Oxford, 1976, pp. 18–20] (For the theory behind
depth-first search techniques).
Chr. Wiener, Ueber eine Aufgabe aus der Geometria situs, Mathematische Annalen 6
(1873) 29–30, 1873.
7 / 79

21. For more on this...
Something Notable
As many things in the history of analysis of algorithms the all-pairs
shortest path has a long history.
We have more from
“Studies in the Economics of Transportation” by Beckmann, McGuire, and
Winsten (1956) where the notation that we use for the matrix multiplication alike
was first used.
In addition
G. Tarry, Le probleme des labyrinthes, Nouvelles Annales de Mathématiques (3) 14 (1895)
187–190 [English translation in: N.L. Biggs, E.K. Lloyd, R.J. Wilson, Graph Theory
1736–1936, Clarendon Press, Oxford, 1976, pp. 18–20] (For the theory behind
depth-first search techniques).
Chr. Wiener, Ueber eine Aufgabe aus der Geometria situs, Mathematische Annalen 6
(1873) 29–30, 1873.
7 / 79

22. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
8 / 79

23. Assumptions Matrix Representation
Matrix Representation of a Graph
For this, we have that each weight in the matrix has the following values
wij =



0 if i = j
w(i, j) if i = j and (i, j) ∈ E
∞ if i = j and (i, j) /∈ E
Then, we have W =







w11 w22 ... w1k−1 w1n
. . .
. . .
. . .
wn1 wn2 ... wnn−1 wnn







Important
There are not negative weight cycles.
9 / 79

24. Assumptions Matrix Representation
Matrix Representation of a Graph
For this, we have that each weight in the matrix has the following values
wij =



0 if i = j
w(i, j) if i = j and (i, j) ∈ E
∞ if i = j and (i, j) /∈ E
Then, we have W =







w11 w22 ... w1k−1 w1n
. . .
. . .
. . .
wn1 wn2 ... wnn−1 wnn







Important
There are not negative weight cycles.
9 / 79

25. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
10 / 79

26. Observations
Ah!!!
The next algorithm is a dynamic programming algorithm for
The all-pairs shortest paths problem on a directed graph G = (V , E).
At the end of the algorithm will generate the following matrix
D =







d11 d22 ... d1k−1 d1n
. . .
. . .
. . .
dn1 dn2 ... dnn−1 dnn







Each entry dij = δ (i, j).
11 / 79

27. Observations
Ah!!!
The next algorithm is a dynamic programming algorithm for
The all-pairs shortest paths problem on a directed graph G = (V , E).
At the end of the algorithm will generate the following matrix
D =







d11 d22 ... d1k−1 d1n
. . .
. . .
. . .
dn1 dn2 ... dnn−1 dnn







Each entry dij = δ (i, j).
11 / 79

28. Observations
Ah!!!
The next algorithm is a dynamic programming algorithm for
The all-pairs shortest paths problem on a directed graph G = (V , E).
At the end of the algorithm will generate the following matrix
D =







d11 d22 ... d1k−1 d1n
. . .
. . .
. . .
dn1 dn2 ... dnn−1 dnn







Each entry dij = δ (i, j).
11 / 79

29. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
12 / 79

30. Structure of a Shortest Path
Consider Lemma 24.1
Given a weighted, directed graph G = (V , E) with p =< v1, v2, ..., vk >
be a SP from v1 to vk. Then,
pij =< vi, vi+1, ..., vj > is a Shortest Path (SP) from vi to vj, where
1 ≤ i ≤ j ≤ k.
We can do the following
Consider the shortest path p from vertex i and j, p contains at most
m edges.
Then, we can use the Corollary to make a decomposition
i
p
k → j =⇒ δ(i, j) = δ(i, k) + wkj
13 / 79

31. Structure of a Shortest Path
Consider Lemma 24.1
Given a weighted, directed graph G = (V , E) with p =< v1, v2, ..., vk >
be a SP from v1 to vk. Then,
pij =< vi, vi+1, ..., vj > is a Shortest Path (SP) from vi to vj, where
1 ≤ i ≤ j ≤ k.
We can do the following
Consider the shortest path p from vertex i and j, p contains at most
m edges.
Then, we can use the Corollary to make a decomposition
i
p
k → j =⇒ δ(i, j) = δ(i, k) + wkj
13 / 79

32. Structure of a Shortest Path
Idea of Using Matrix Multiplication
We define the following concept based in the decomposition
Corollary!!!
l
(m)
ij =minimum weight of any path from i to j, it contains at most m
edges i.e.
l
(m)
ij could be min
k
l
(m−1)
ik + wkj
14 / 79

33. Structure of a Shortest Path
Idea of Using Matrix Multiplication
We define the following concept based in the decomposition
Corollary!!!
l
(m)
ij =minimum weight of any path from i to j, it contains at most m
edges i.e.
l
(m)
ij could be min
k
l
(m−1)
ik + wkj
14 / 79

34. Graphical Interpretation
Looking for the Shortest Path
15 / 79

35. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
16 / 79

36. Recursive Solution
Thus, we have that for paths with ZERO edges
l
(0)
ij =
0 if i = j
∞ if i = j
Recursion Our Great Friend
Consider the previous definition and decomposition. Thus
l
(m)
ij = min l
(m−1)
ij , min
1≤k≤n
l
(m−1)
ik + wkj
= min
1≤k≤n
l
(m−1)
ik + wkj
17 / 79

37. Recursive Solution
Thus, we have that for paths with ZERO edges
l
(0)
ij =
0 if i = j
∞ if i = j
Recursion Our Great Friend
Consider the previous definition and decomposition. Thus
l
(m)
ij = min l
(m−1)
ij , min
1≤k≤n
l
(m−1)
ik + wkj
= min
1≤k≤n
l
(m−1)
ik + wkj
17 / 79

38. Recursive Solution
Thus, we have that for paths with ZERO edges
l
(0)
ij =
0 if i = j
∞ if i = j
Recursion Our Great Friend
Consider the previous definition and decomposition. Thus
l
(m)
ij = min l
(m−1)
ij , min
1≤k≤n
l
(m−1)
ik + wkj
= min
1≤k≤n
l
(m−1)
ik + wkj
17 / 79

39. Recursive Solution
Thus, we have that for paths with ZERO edges
l
(0)
ij =
0 if i = j
∞ if i = j
Recursion Our Great Friend
Consider the previous definition and decomposition. Thus
l
(m)
ij = min l
(m−1)
ij , min
1≤k≤n
l
(m−1)
ik + wkj
= min
1≤k≤n
l
(m−1)
ik + wkj
17 / 79

40. Recursive Solution
Why? A simple notation problem
l
(m)
ij = l
(m−1)
ij + 0 = l
(m−1)
ij + wjj
18 / 79

41. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
19 / 79

42. Transforming it to a iterative one
What is δ (i, j)?
If you do not have negative-weight cycles, and δ (i, j) < ∞.
Then, the shortest path from vertex i to j has at most n − 1 edges
δ (i, j) = l
(n−1)
ij = l
(n)
ij = l
(n+1)
ij = l
(n+2)
ij = ...
Back to Matrix Multiplication
We have the matrix L(m) = l
(m)
ij .
Then, we can compute first L(1) then compute L(2) all the way to
L(n−1) which contains the actual shortest paths.
What is L(1)
?
First, we have that L(1) = W , since l
(1)
ij = wij.
20 / 79

43. Transforming it to a iterative one
What is δ (i, j)?
If you do not have negative-weight cycles, and δ (i, j) < ∞.
Then, the shortest path from vertex i to j has at most n − 1 edges
δ (i, j) = l
(n−1)
ij = l
(n)
ij = l
(n+1)
ij = l
(n+2)
ij = ...
Back to Matrix Multiplication
We have the matrix L(m) = l
(m)
ij .
Then, we can compute first L(1) then compute L(2) all the way to
L(n−1) which contains the actual shortest paths.
What is L(1)
?
First, we have that L(1) = W , since l
(1)
ij = wij.
20 / 79

44. Transforming it to a iterative one
What is δ (i, j)?
If you do not have negative-weight cycles, and δ (i, j) < ∞.
Then, the shortest path from vertex i to j has at most n − 1 edges
δ (i, j) = l
(n−1)
ij = l
(n)
ij = l
(n+1)
ij = l
(n+2)
ij = ...
Back to Matrix Multiplication
We have the matrix L(m) = l
(m)
ij .
Then, we can compute first L(1) then compute L(2) all the way to
L(n−1) which contains the actual shortest paths.
What is L(1)
?
First, we have that L(1) = W , since l
(1)
ij = wij.
20 / 79

45. Transforming it to a iterative one
What is δ (i, j)?
If you do not have negative-weight cycles, and δ (i, j) < ∞.
Then, the shortest path from vertex i to j has at most n − 1 edges
δ (i, j) = l
(n−1)
ij = l
(n)
ij = l
(n+1)
ij = l
(n+2)
ij = ...
Back to Matrix Multiplication
We have the matrix L(m) = l
(m)
ij .
Then, we can compute first L(1) then compute L(2) all the way to
L(n−1) which contains the actual shortest paths.
What is L(1)
?
First, we have that L(1) = W , since l
(1)
ij = wij.
20 / 79

46. Transforming it to a iterative one
What is δ (i, j)?
If you do not have negative-weight cycles, and δ (i, j) < ∞.
Then, the shortest path from vertex i to j has at most n − 1 edges
δ (i, j) = l
(n−1)
ij = l
(n)
ij = l
(n+1)
ij = l
(n+2)
ij = ...
Back to Matrix Multiplication
We have the matrix L(m) = l
(m)
ij .
Then, we can compute first L(1) then compute L(2) all the way to
L(n−1) which contains the actual shortest paths.
What is L(1)
?
First, we have that L(1) = W , since l
(1)
ij = wij.
20 / 79

47. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
21 / 79

48. Algorithm
Code
Extended-Shortest-Path(L, W )
1 n = L.rows
2 let L = lij be a new n × n
3 for i = 1 to n
4 for j = 1 to n
5 lij = ∞
6 for k = 1 to n
7 lij = min lij, lik + wkj
8 return L
22 / 79

49. Algorithm
Code
Extended-Shortest-Path(L, W )
1 n = L.rows
2 let L = lij be a new n × n
3 for i = 1 to n
4 for j = 1 to n
5 lij = ∞
6 for k = 1 to n
7 lij = min lij, lik + wkj
8 return L
22 / 79

50. Algorithm
Code
Extended-Shortest-Path(L, W )
1 n = L.rows
2 let L = lij be a new n × n
3 for i = 1 to n
4 for j = 1 to n
5 lij = ∞
6 for k = 1 to n
7 lij = min lij, lik + wkj
8 return L
22 / 79

51. Algorithm
Code
Extended-Shortest-Path(L, W )
1 n = L.rows
2 let L = lij be a new n × n
3 for i = 1 to n
4 for j = 1 to n
5 lij = ∞
6 for k = 1 to n
7 lij = min lij, lik + wkj
8 return L
22 / 79

52. Algorithm
Complexity
If |V | == n we have that Θ V 3 .
23 / 79

53. Algorithm
Complexity
If |V | == n we have that Θ V 3 .
23 / 79

54. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
24 / 79

55. Look Alike Matrix Multiplication Operations
Mapping That Can be Thought
L =⇒ A
W =⇒ B
L =⇒ C
min =⇒ +
+ =⇒ ·
∞ =⇒ 0
25 / 79

56. Look Alike Matrix Multiplication Operations
Using the previous notation, we can rewrite our previous algorithm as
Square-Matrix-Multiply(A, B)
1 n = A.rows
2 let C be a new n × n matrix
3 for i = 1 to n
4 for j = 1 to n
5 cij = 0
6 for k = 1 to n
7 cij = cij + aik · bkj
8 return C
26 / 79

57. Complexity
Thus
The complexity of the Extended-Shortest-Path is equal to O n3
27 / 79

58. Using the Analogy
Returning to the all-pairs shortest-paths problem
It is possible to compute the shortest path by extending such a path edge
by edge.
Therefore
If we denote A · B as the “product” of the Extended-Shortest-Path
28 / 79

59. Using the Analogy
Returning to the all-pairs shortest-paths problem
It is possible to compute the shortest path by extending such a path edge
by edge.
Therefore
If we denote A · B as the “product” of the Extended-Shortest-Path
28 / 79

60. Using the Analogy
We have that
L(1)
= L(0)
· W = W
L(2)
= L(1)
· W = W 2
...
L(n−1)
= L(n−2)
· W = W n−1
29 / 79

61. Using the Analogy
We have that
L(1)
= L(0)
· W = W
L(2)
= L(1)
· W = W 2
...
L(n−1)
= L(n−2)
· W = W n−1
29 / 79

62. Using the Analogy
We have that
L(1)
= L(0)
· W = W
L(2)
= L(1)
· W = W 2
...
L(n−1)
= L(n−2)
· W = W n−1
29 / 79

63. The Final Algorithm
We have that
Slow-All-Pairs-Shortest-Paths(W )
1 n ← W .rows
2 L(1)
← W
3 for m = 2 to n − 1
4 L(m)
←EXTEND-SHORTEST-PATHS L(m−1)
, W
5 return L(n−1)
30 / 79

64. With Complexity
Complexity
O V 4
(1)
31 / 79

65. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
32 / 79

66. Example
We have the following
1
2
3
45
3
4
8
2
7
1
6
-4
-5
1 2 3 4 5
1 0 2 8 ∞ -4
2 ∞ 0 ∞ 1 7
3 ∞ 4 0 ∞ ∞
4 2 ∞ -5 0 ∞
5 ∞ ∞ ∞ 6 0
L(1) = L(0)W
33 / 79

67. Example
We have the following
1
2
3
45
3
4
8
2
7
1
6
-4
-5
1 2 3 4 5
1 0 2 8 2 -4
2 3 0 -4 1 7
3 ∞ 4 0 5 11
4 2 -1 -5 0 -2
5 8 ∞ 1 6 0
L(2) = L(1)W
34 / 79

68. Here, we use the analogy of matrix multiplication
D1
W
35 / 79

69. Thus, the update of an element lij
Example
l
(2)
14 = min



0 3 8 ∞ −4 +







∞
1
∞
0
6










= min ∞ 4 ∞ ∞ 2
= 2
36 / 79

70. Example
We have the following
1
2
3
45
3
4
8
2
7
1
6
-4
-5
1 2 3 4 5
1 0 3 -3 2 -4
2 3 0 -4 1 -1
3 7 4 0 5 11
4 2 -1 -5 0 -2
5 8 5 1 6 0
L(3) = L(2)W
37 / 79

71. Example
We have the following
1
2
3
45
3
4
8
2
7
1
6
-4
-5
1 2 3 4 5
1 0 1 -3 2 -4
2 3 0 -4 1 -1
3 7 4 0 5 3
4 2 -1 -5 0 -2
5 8 5 1 6 0
L(4) = L(3)W
38 / 79

72. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
39 / 79

73. Recall the following
We are interested only
In matrix L(n−1)
In addition
Remember, we do not have negative weight cycles!!!
Therefore, given the equation
δ (i, j) = l
(n−1)
ij = l
(n)
ij = l
(n)
ij = ... (2)
40 / 79

74. Recall the following
We are interested only
In matrix L(n−1)
In addition
Remember, we do not have negative weight cycles!!!
Therefore, given the equation
δ (i, j) = l
(n−1)
ij = l
(n)
ij = l
(n)
ij = ... (2)
40 / 79

75. Recall the following
We are interested only
In matrix L(n−1)
In addition
Remember, we do not have negative weight cycles!!!
Therefore, given the equation
δ (i, j) = l
(n−1)
ij = l
(n)
ij = l
(n)
ij = ... (2)
40 / 79

76. Thus
It implies
L(m)
= L(n−1)
(3)
For all
m ≥ n − 1 (4)
41 / 79

77. Thus
It implies
L(m)
= L(n−1)
(3)
For all
m ≥ n − 1 (4)
41 / 79

78. Something Faster
We want something faster!!! Observation!!!
L(1)
= W
L(2)
= W · W = W 2
L(4)
= W 2
· W 2
= W 4
L(8)
= W 4
· W 4
= W 8
...
L(2 log(n−1)
) = W 2 log(n−1) −1
· W 2 log(n−1) −1
= W 2 log(n−1)
Because
2 lg(n−1)
≥ n − 1 =⇒ L(2 lg(n−1)
) = L(n−1)
42 / 79

79. Something Faster
We want something faster!!! Observation!!!
L(1)
= W
L(2)
= W · W = W 2
L(4)
= W 2
· W 2
= W 4
L(8)
= W 4
· W 4
= W 8
...
L(2 log(n−1)
) = W 2 log(n−1) −1
· W 2 log(n−1) −1
= W 2 log(n−1)
Because
2 lg(n−1)
≥ n − 1 =⇒ L(2 lg(n−1)
) = L(n−1)
42 / 79

80. Something Faster
We want something faster!!! Observation!!!
L(1)
= W
L(2)
= W · W = W 2
L(4)
= W 2
· W 2
= W 4
L(8)
= W 4
· W 4
= W 8
...
L(2 log(n−1)
) = W 2 log(n−1) −1
· W 2 log(n−1) −1
= W 2 log(n−1)
Because
2 lg(n−1)
≥ n − 1 =⇒ L(2 lg(n−1)
) = L(n−1)
42 / 79

81. Something Faster
We want something faster!!! Observation!!!
L(1)
= W
L(2)
= W · W = W 2
L(4)
= W 2
· W 2
= W 4
L(8)
= W 4
· W 4
= W 8
...
L(2 log(n−1)
) = W 2 log(n−1) −1
· W 2 log(n−1) −1
= W 2 log(n−1)
Because
2 lg(n−1)
≥ n − 1 =⇒ L(2 lg(n−1)
) = L(n−1)
42 / 79

82. Something Faster
We want something faster!!! Observation!!!
L(1)
= W
L(2)
= W · W = W 2
L(4)
= W 2
· W 2
= W 4
L(8)
= W 4
· W 4
= W 8
...
L(2 log(n−1)
) = W 2 log(n−1) −1
· W 2 log(n−1) −1
= W 2 log(n−1)
Because
2 lg(n−1)
≥ n − 1 =⇒ L(2 lg(n−1)
) = L(n−1)
42 / 79

83. Something Faster
We want something faster!!! Observation!!!
L(1)
= W
L(2)
= W · W = W 2
L(4)
= W 2
· W 2
= W 4
L(8)
= W 4
· W 4
= W 8
...
L(2 log(n−1)
) = W 2 log(n−1) −1
· W 2 log(n−1) −1
= W 2 log(n−1)
Because
2 lg(n−1)
≥ n − 1 =⇒ L(2 lg(n−1)
) = L(n−1)
42 / 79

84. The Faster Algorithm
Complexity of the Previous Algorithm
Slow-All-Pairs-Shortest-Paths(W )
1 n ← W .rows
2 L(1)
← W
3 m ← 1
4 while m < n − 1
5 L(2m)
←EXTEND-SHORTEST-PATHS L(m)
, L(m)
6 m ← 2m
7 return L(m)
Complexity
If n = |V | we have that O V 3 lg V .
43 / 79

85. The Faster Algorithm
Complexity of the Previous Algorithm
Slow-All-Pairs-Shortest-Paths(W )
1 n ← W .rows
2 L(1)
← W
3 m ← 1
4 while m < n − 1
5 L(2m)
←EXTEND-SHORTEST-PATHS L(m)
, L(m)
6 m ← 2m
7 return L(m)
Complexity
If n = |V | we have that O V 3 lg V .
43 / 79

86. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
44 / 79

87. The Shortest Path Structure
Intermediate Vertex
For a path p = v1, v2, ..., vl , an intermediate vertex is any vertex of p
other than v1 or vl.
Define
d
(k)
ij =weight of a shortest path between i and j with all intermediate
vertices are in the set {1, 2, ..., k}.
45 / 79

88. The Shortest Path Structure
Intermediate Vertex
For a path p = v1, v2, ..., vl , an intermediate vertex is any vertex of p
other than v1 or vl.
Define
d
(k)
ij =weight of a shortest path between i and j with all intermediate
vertices are in the set {1, 2, ..., k}.
45 / 79

89. The Recursive Idea
Simply look at the following cases cases
Case I k is not an intermediate vertex, then a shortest path from i to
j with all intermediate vertices {1, ..., k − 1} is a shortest path from i
to j with intermediate vertices {1, ..., k}.
=⇒ d
(k)
ij = d
(k−1)
ij
Case II if k is an intermediate vertice. Then, i
p1
k
p2
j and we can
make the following statements using Lemma 24.1:
p1 is a shortest path from i to k with all intermediate vertices in the
set {1, ..., k − 1}.
p2 is a shortest path from k to j with all intermediate vertices in the
set {1, ..., k − 1}.
=⇒ d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
46 / 79

90. The Recursive Idea
Simply look at the following cases cases
Case I k is not an intermediate vertex, then a shortest path from i to
j with all intermediate vertices {1, ..., k − 1} is a shortest path from i
to j with intermediate vertices {1, ..., k}.
=⇒ d
(k)
ij = d
(k−1)
ij
Case II if k is an intermediate vertice. Then, i
p1
k
p2
j and we can
make the following statements using Lemma 24.1:
p1 is a shortest path from i to k with all intermediate vertices in the
set {1, ..., k − 1}.
p2 is a shortest path from k to j with all intermediate vertices in the
set {1, ..., k − 1}.
=⇒ d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
46 / 79

91. The Recursive Idea
Simply look at the following cases cases
Case I k is not an intermediate vertex, then a shortest path from i to
j with all intermediate vertices {1, ..., k − 1} is a shortest path from i
to j with intermediate vertices {1, ..., k}.
=⇒ d
(k)
ij = d
(k−1)
ij
Case II if k is an intermediate vertice. Then, i
p1
k
p2
j and we can
make the following statements using Lemma 24.1:
p1 is a shortest path from i to k with all intermediate vertices in the
set {1, ..., k − 1}.
p2 is a shortest path from k to j with all intermediate vertices in the
set {1, ..., k − 1}.
=⇒ d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
46 / 79

92. The Recursive Idea
Simply look at the following cases cases
Case I k is not an intermediate vertex, then a shortest path from i to
j with all intermediate vertices {1, ..., k − 1} is a shortest path from i
to j with intermediate vertices {1, ..., k}.
=⇒ d
(k)
ij = d
(k−1)
ij
Case II if k is an intermediate vertice. Then, i
p1
k
p2
j and we can
make the following statements using Lemma 24.1:
p1 is a shortest path from i to k with all intermediate vertices in the
set {1, ..., k − 1}.
p2 is a shortest path from k to j with all intermediate vertices in the
set {1, ..., k − 1}.
=⇒ d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
46 / 79

93. The Recursive Idea
Simply look at the following cases cases
Case I k is not an intermediate vertex, then a shortest path from i to
j with all intermediate vertices {1, ..., k − 1} is a shortest path from i
to j with intermediate vertices {1, ..., k}.
=⇒ d
(k)
ij = d
(k−1)
ij
Case II if k is an intermediate vertice. Then, i
p1
k
p2
j and we can
make the following statements using Lemma 24.1:
p1 is a shortest path from i to k with all intermediate vertices in the
set {1, ..., k − 1}.
p2 is a shortest path from k to j with all intermediate vertices in the
set {1, ..., k − 1}.
=⇒ d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
46 / 79

94. The Recursive Idea
Simply look at the following cases cases
Case I k is not an intermediate vertex, then a shortest path from i to
j with all intermediate vertices {1, ..., k − 1} is a shortest path from i
to j with intermediate vertices {1, ..., k}.
=⇒ d
(k)
ij = d
(k−1)
ij
Case II if k is an intermediate vertice. Then, i
p1
k
p2
j and we can
make the following statements using Lemma 24.1:
p1 is a shortest path from i to k with all intermediate vertices in the
set {1, ..., k − 1}.
p2 is a shortest path from k to j with all intermediate vertices in the
set {1, ..., k − 1}.
=⇒ d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
46 / 79

95. The Graphical Idea
Consider
All possible intermediate vertices in {1, 2, ..., k}
Figure: The Recursive Idea
47 / 79

96. The Recursive Solution
The Recursion
d
(k)
ij =



wij if k = 0
min d
(k−1)
ij , d
(k−1)
ik + d
(k−1)
kj if k ≥ 1
Final answer when k = n
We recursively calculate D(n) = d
(n)
ij or d
(n)
ij = δ (i, j) for all i, j ∈ V .
48 / 79

97. The Recursive Solution
The Recursion
d
(k)
ij =



wij if k = 0
min d
(k−1)
ij , d
(k−1)
ik + d
(k−1)
kj if k ≥ 1
Final answer when k = n
We recursively calculate D(n) = d
(n)
ij or d
(n)
ij = δ (i, j) for all i, j ∈ V .
48 / 79

98. Thus, we have the following
Recursive Version
Recursive-Floyd-Warshall(W )
1 D(n)
the n × n matrix
2 for i = 1 to n
3 for j = 1 to n
4 D(n)
[i, j] =Recursive-Part(i, j, n, W )
5 return D(n)
49 / 79

99. Thus, we have the following
Recursive Version
Recursive-Floyd-Warshall(W )
1 D(n)
the n × n matrix
2 for i = 1 to n
3 for j = 1 to n
4 D(n)
[i, j] =Recursive-Part(i, j, n, W )
5 return D(n)
49 / 79

100. Thus, we have the following
Recursive Version
Recursive-Floyd-Warshall(W )
1 D(n)
the n × n matrix
2 for i = 1 to n
3 for j = 1 to n
4 D(n)
[i, j] =Recursive-Part(i, j, n, W )
5 return D(n)
49 / 79

101. Thus, we have the following
The Recursive-Part
Recursive-Part(i, j, k, W )
1 if k = 0
2 return W [i, j]
3 if k ≥ 1
4 t1 =Recursive-Part(i, j, k − 1, W )
5 t2 =Recursive-Part(i, k, k − 1, W )+...
6 Recursive-Part(k, j, k − 1, W )
7 if t1 ≤ t2
8 return t1
9 else
10 return t2
50 / 79

102. Thus, we have the following
The Recursive-Part
Recursive-Part(i, j, k, W )
1 if k = 0
2 return W [i, j]
3 if k ≥ 1
4 t1 =Recursive-Part(i, j, k − 1, W )
5 t2 =Recursive-Part(i, k, k − 1, W )+...
6 Recursive-Part(k, j, k − 1, W )
7 if t1 ≤ t2
8 return t1
9 else
10 return t2
50 / 79

103. Thus, we have the following
The Recursive-Part
Recursive-Part(i, j, k, W )
1 if k = 0
2 return W [i, j]
3 if k ≥ 1
4 t1 =Recursive-Part(i, j, k − 1, W )
5 t2 =Recursive-Part(i, k, k − 1, W )+...
6 Recursive-Part(k, j, k − 1, W )
7 if t1 ≤ t2
8 return t1
9 else
10 return t2
50 / 79

104. Thus, we have the following
The Recursive-Part
Recursive-Part(i, j, k, W )
1 if k = 0
2 return W [i, j]
3 if k ≥ 1
4 t1 =Recursive-Part(i, j, k − 1, W )
5 t2 =Recursive-Part(i, k, k − 1, W )+...
6 Recursive-Part(k, j, k − 1, W )
7 if t1 ≤ t2
8 return t1
9 else
10 return t2
50 / 79

105. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
51 / 79

106. Now
We want to use a storage to eliminate the recursion
For this, we are going to use two matrices
1 D(k−1) the previous matrix.
2 D(k) the new matrix based in the previous matrix
Something Notable
With D(0) = W or all weights in the edges that exist.
52 / 79

107. Now
We want to use a storage to eliminate the recursion
For this, we are going to use two matrices
1 D(k−1) the previous matrix.
2 D(k) the new matrix based in the previous matrix
Something Notable
With D(0) = W or all weights in the edges that exist.
52 / 79

108. Now
We want to use a storage to eliminate the recursion
For this, we are going to use two matrices
1 D(k−1) the previous matrix.
2 D(k) the new matrix based in the previous matrix
Something Notable
With D(0) = W or all weights in the edges that exist.
52 / 79

109. Now
We want to use a storage to eliminate the recursion
For this, we are going to use two matrices
1 D(k−1) the previous matrix.
2 D(k) the new matrix based in the previous matrix
Something Notable
With D(0) = W or all weights in the edges that exist.
52 / 79

110. In addition, we want to rebuild the answer
For this, we have the predecessor matrix Π
Actually, we want to compute a sequence of matrices
Π(0)
, Π(1)
, ..., Π(n)
Where
Π = Π(n)
53 / 79

111. In addition, we want to rebuild the answer
For this, we have the predecessor matrix Π
Actually, we want to compute a sequence of matrices
Π(0)
, Π(1)
, ..., Π(n)
Where
Π = Π(n)
53 / 79

112. What are the elements in Π(k)
Each element in the matrix is as follow
π
(k)
ij = the predecessor of vertex j on a shortest path from vertex i with all
intermediate vertices in the set {1, 2, ..., k}
Thus, we have that
π
(0)
ij =



NULL if i = j or wij = ∞
i if i = j and wij < ∞
54 / 79

113. What are the elements in Π(k)
Each element in the matrix is as follow
π
(k)
ij = the predecessor of vertex j on a shortest path from vertex i with all
intermediate vertices in the set {1, 2, ..., k}
Thus, we have that
π
(0)
ij =



NULL if i = j or wij = ∞
i if i = j and wij < ∞
54 / 79

114. Then
We have the following
For k ≥ 1, if we take the path i k j where k = j.
Then, if d
(k−1)
ij > d
(k−1)
ik + d
(k−1)
kj
For the predecessor of j, we chose k on a shortest path from k with all
intermediate vertices in the set {1, 2, ..., k − 1}
Otherwise, if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
We choose the same predecessor of j that we chose on a shortest path
from i with all all intermediate vertices in the set {1, 2, ..., k − 1}.
55 / 79

115. Then
We have the following
For k ≥ 1, if we take the path i k j where k = j.
Then, if d
(k−1)
ij > d
(k−1)
ik + d
(k−1)
kj
For the predecessor of j, we chose k on a shortest path from k with all
intermediate vertices in the set {1, 2, ..., k − 1}
Otherwise, if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
We choose the same predecessor of j that we chose on a shortest path
from i with all all intermediate vertices in the set {1, 2, ..., k − 1}.
55 / 79

116. Then
We have the following
For k ≥ 1, if we take the path i k j where k = j.
Then, if d
(k−1)
ij > d
(k−1)
ik + d
(k−1)
kj
For the predecessor of j, we chose k on a shortest path from k with all
intermediate vertices in the set {1, 2, ..., k − 1}
Otherwise, if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
We choose the same predecessor of j that we chose on a shortest path
from i with all all intermediate vertices in the set {1, 2, ..., k − 1}.
55 / 79

117. Formally
We have then
π
(k)
ij =



π
(k−1)
ij if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
π
(k−1)
kj if d
(k−1)
ij > d
(k−1)
ik + d
(k−1)
kj
56 / 79

118. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
57 / 79

119. Final Iterative Version of Floyd-Warshall (Correction by Diego - Class Tec 2015)
Floyd-Warshall(W)
1. n = W.rows
2. D(0) = W
3. for k = 1 to n − 1
4. let D(k) = d
(k)
ij
be a new
n × n matrix
5. let Π(k) be a new
predecessor
n × n matrix
Given each k, we update using D(k−1)
6. for i = 1 to n
7. for j = 1 to n
8. if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
9. d
(k)
ij = d
(k−1)
ij
10. π
(k)
ij = π
(k−1)
ij
11. else
12. d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
13. π
(k)
ij = π
(k−1)
kj
14. return D(n) and Π(n)
58 / 79

120. Final Iterative Version of Floyd-Warshall (Correction by Diego - Class Tec 2015)
Floyd-Warshall(W)
1. n = W.rows
2. D(0) = W
3. for k = 1 to n − 1
4. let D(k) = d
(k)
ij
be a new
n × n matrix
5. let Π(k) be a new
predecessor
n × n matrix
Given each k, we update using D(k−1)
6. for i = 1 to n
7. for j = 1 to n
8. if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
9. d
(k)
ij = d
(k−1)
ij
10. π
(k)
ij = π
(k−1)
ij
11. else
12. d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
13. π
(k)
ij = π
(k−1)
kj
14. return D(n) and Π(n)
58 / 79

121. Final Iterative Version of Floyd-Warshall (Correction by Diego - Class Tec 2015)
Floyd-Warshall(W)
1. n = W.rows
2. D(0) = W
3. for k = 1 to n − 1
4. let D(k) = d
(k)
ij
be a new
n × n matrix
5. let Π(k) be a new
predecessor
n × n matrix
Given each k, we update using D(k−1)
6. for i = 1 to n
7. for j = 1 to n
8. if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
9. d
(k)
ij = d
(k−1)
ij
10. π
(k)
ij = π
(k−1)
ij
11. else
12. d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
13. π
(k)
ij = π
(k−1)
kj
14. return D(n) and Π(n)
58 / 79

122. Final Iterative Version of Floyd-Warshall (Correction by Diego - Class Tec 2015)
Floyd-Warshall(W)
1. n = W.rows
2. D(0) = W
3. for k = 1 to n − 1
4. let D(k) = d
(k)
ij
be a new
n × n matrix
5. let Π(k) be a new
predecessor
n × n matrix
Given each k, we update using D(k−1)
6. for i = 1 to n
7. for j = 1 to n
8. if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
9. d
(k)
ij = d
(k−1)
ij
10. π
(k)
ij = π
(k−1)
ij
11. else
12. d
(k)
ij = d
(k−1)
ik + d
(k−1)
kj
13. π
(k)
ij = π
(k−1)
kj
14. return D(n) and Π(n)
58 / 79

123. Explanation
Lines 1 and 2
Initialization of variables n and D(0)
Line 3
In the loop, we solve the smaller problems first with k = 1 to k = n − 1
Remember the largest number of edges in any shortest path
Line 4 and 5
Instantiation of the new matrices D(k) and (k)
to generate the shortest
pats with at least k edges.
59 / 79

124. Explanation
Lines 1 and 2
Initialization of variables n and D(0)
Line 3
In the loop, we solve the smaller problems first with k = 1 to k = n − 1
Remember the largest number of edges in any shortest path
Line 4 and 5
Instantiation of the new matrices D(k) and (k)
to generate the shortest
pats with at least k edges.
59 / 79

125. Explanation
Lines 1 and 2
Initialization of variables n and D(0)
Line 3
In the loop, we solve the smaller problems first with k = 1 to k = n − 1
Remember the largest number of edges in any shortest path
Line 4 and 5
Instantiation of the new matrices D(k) and (k)
to generate the shortest
pats with at least k edges.
59 / 79

126. Explanation
Lines 1 and 2
Initialization of variables n and D(0)
Line 3
In the loop, we solve the smaller problems first with k = 1 to k = n − 1
Remember the largest number of edges in any shortest path
Line 4 and 5
Instantiation of the new matrices D(k) and (k)
to generate the shortest
pats with at least k edges.
59 / 79

127. Explanation
Line 6 and 7
This is done to go through all the possible combinations of i’s and j’s
Line 8
Deciding if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
60 / 79

128. Explanation
Line 6 and 7
This is done to go through all the possible combinations of i’s and j’s
Line 8
Deciding if d
(k−1)
ij ≤ d
(k−1)
ik + d
(k−1)
kj
60 / 79

129. Example
Graph
1
2
3
45
3
4
8
2
7
1
6
-4
-5
61 / 79

130. Example
D(0)
and Π(0)
1
2
3
45
3
4
8
2
7
1
6
-4
-5
D(0)
=






0 3 8 ∞ −4
∞ 0 ∞ 1 7
∞ 4 0 ∞ ∞
2 ∞ −5 0 ∞
∞ ∞ ∞ 6 0






Π(0)
=






NIL 1 1 NIL 1
NIL NIL NIL 2 2
NIL 3 NIL NIL NIL
4 NIL 4 NIL NIL
NIL NIL NIL 5 NIL






62 / 79

131. Example
D(1)
and Π(1)
1
2
3
45
3
4
8
2
7
1
6
-4
-5
D(1)
=






0 3 8 ∞ −4
∞ 0 ∞ 1 7
∞ 4 0 ∞ ∞
2 5 −5 0 −2
∞ ∞ ∞ 6 0






Π(1)
=






NIL 1 1 NIL 1
NIL NIL NIL 2 2
NIL 3 NIL NIL NIL
4 1 4 NIL 1
NIL NIL NIL 5 NIL






63 / 79

132. Example
D(2)
and Π(2)
1
2
3
45
3
4
8
2
7
1
6
-4
-5
D(2)
=






0 3 8 4 −4
∞ 0 ∞ 1 7
∞ 4 0 5 11
2 5 −5 0 −2
∞ ∞ ∞ 6 0






Π(2)
=






NIL 1 1 2 1
NIL NIL NIL 2 2
NIL 3 NIL 2 2
4 1 4 NIL 1
NIL NIL NIL 5 NIL






64 / 79

133. Example
D(3)
and Π(3)
1
2
3
45
3
4
8
2
7
1
6
-4
-5
D(3)
=






0 3 8 4 −4
∞ 0 ∞ 1 7
∞ 4 0 5 11
2 −1 −5 0 −2
∞ ∞ ∞ 6 0






Π(3)
=






NIL 1 1 2 1
NIL NIL NIL 2 2
NIL 3 NIL 2 2
4 3 4 NIL 1
NIL NIL NIL 5 NIL






65 / 79

134. Example
D(4)
and Π(4)
1
2
3
45
3
4
8
2
7
1
6
-4
-5
D(4)
=






0 3 −1 4 −4
3 0 −4 1 −1
7 4 0 5 3
2 −1 −5 0 −2
8 5 1 6 0






Π(4)
=






NIL 1 4 2 1
4 NIL 4 2 1
4 3 NIL 2 1
4 3 4 NIL 1
4 3 4 5 NIL






66 / 79

135. Example
D(5)
and Π(5)
1
2
3
45
3
4
8
2
7
1
6
-4
-5
D(5)
=






0 1 −3 2 −4
3 0 −4 1 −1
7 4 0 5 3
2 −1 −5 0 −2
8 5 1 6 0






Π(5)
=






NIL 3 4 5 1
4 NIL 4 2 1
4 3 NIL 2 1
4 3 4 NIL 1
4 3 4 5 NIL






67 / 79

136. Remarks
Something Notable
Because the comparison in line 8 takes O (1)
Complexity of Floyd-Warshall is
Time Complexity Θ V 3
We do not have elaborate data structures as Binary Heap or
Fibonacci Heap!!!
The hidden constant time is quite small:
Making the Floyd-Warshall Algorithm practical even with
moderate-sized graphs!!!
68 / 79

137. Remarks
Something Notable
Because the comparison in line 8 takes O (1)
Complexity of Floyd-Warshall is
Time Complexity Θ V 3
We do not have elaborate data structures as Binary Heap or
Fibonacci Heap!!!
The hidden constant time is quite small:
Making the Floyd-Warshall Algorithm practical even with
moderate-sized graphs!!!
68 / 79

138. Remarks
Something Notable
Because the comparison in line 8 takes O (1)
Complexity of Floyd-Warshall is
Time Complexity Θ V 3
We do not have elaborate data structures as Binary Heap or
Fibonacci Heap!!!
The hidden constant time is quite small:
Making the Floyd-Warshall Algorithm practical even with
moderate-sized graphs!!!
68 / 79

139. Remarks
Something Notable
Because the comparison in line 8 takes O (1)
Complexity of Floyd-Warshall is
Time Complexity Θ V 3
We do not have elaborate data structures as Binary Heap or
Fibonacci Heap!!!
The hidden constant time is quite small:
Making the Floyd-Warshall Algorithm practical even with
moderate-sized graphs!!!
68 / 79

140. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
69 / 79

141. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

142. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

143. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

144. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

145. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

146. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

147. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

148. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

149. Johnson’s Algorithm
Observations
Used to find all pairs in a sparse graphs by using Dijkstra’s algorithm.
It uses a re-weighting function to obtain positive edges from negative
edges to deal with them.
It can deal with the negative weight cycles.
Therfore
It uses something to deal with the negative weight cycles.
Could be a Bellman-Ford detector as before?
Maybe, we need to transform the weights in order to use them.
What we require
A re-weighting function w (u, v)
A shortest path by w is a shortest path by w.
All edges are not negative using w.
70 / 79

150. Proving Properties of Re-Weigthing
Lemma 25.1
Given a weighted, directed graph G = (D, V ) with weight function
w : E → R, let h : V → R be any function mapping vertices to real
numbers. For each edge (u, v) ∈ E, define
w (u, v) = w (u, v) + h (u) − h (v)
Let p = v0, v1, ..., vk be any path from vertex 0 to vertex k. Then:
1 p is a shortest path from 0 to k with weight function w if and only if
it is a shortest path with weight function w. That is w(p) = δ (v0, vk)
if and only if w(p) = δ (v0, vk).
2 Furthermore, G has a negative-weight cycle using weight function w
if and only if G has a negative-weight cycle using weight function w.
71 / 79

151. Proving Properties of Re-Weigthing
Lemma 25.1
Given a weighted, directed graph G = (D, V ) with weight function
w : E → R, let h : V → R be any function mapping vertices to real
numbers. For each edge (u, v) ∈ E, define
w (u, v) = w (u, v) + h (u) − h (v)
Let p = v0, v1, ..., vk be any path from vertex 0 to vertex k. Then:
1 p is a shortest path from 0 to k with weight function w if and only if
it is a shortest path with weight function w. That is w(p) = δ (v0, vk)
if and only if w(p) = δ (v0, vk).
2 Furthermore, G has a negative-weight cycle using weight function w
if and only if G has a negative-weight cycle using weight function w.
71 / 79

152. Proving Properties of Re-Weigthing
Lemma 25.1
Given a weighted, directed graph G = (D, V ) with weight function
w : E → R, let h : V → R be any function mapping vertices to real
numbers. For each edge (u, v) ∈ E, define
w (u, v) = w (u, v) + h (u) − h (v)
Let p = v0, v1, ..., vk be any path from vertex 0 to vertex k. Then:
1 p is a shortest path from 0 to k with weight function w if and only if
it is a shortest path with weight function w. That is w(p) = δ (v0, vk)
if and only if w(p) = δ (v0, vk).
2 Furthermore, G has a negative-weight cycle using weight function w
if and only if G has a negative-weight cycle using weight function w.
71 / 79

153. Proving Properties of Re-Weigthing
Lemma 25.1
Given a weighted, directed graph G = (D, V ) with weight function
w : E → R, let h : V → R be any function mapping vertices to real
numbers. For each edge (u, v) ∈ E, define
w (u, v) = w (u, v) + h (u) − h (v)
Let p = v0, v1, ..., vk be any path from vertex 0 to vertex k. Then:
1 p is a shortest path from 0 to k with weight function w if and only if
it is a shortest path with weight function w. That is w(p) = δ (v0, vk)
if and only if w(p) = δ (v0, vk).
2 Furthermore, G has a negative-weight cycle using weight function w
if and only if G has a negative-weight cycle using weight function w.
71 / 79

154. Proving Properties of Re-Weigthing
Lemma 25.1
Given a weighted, directed graph G = (D, V ) with weight function
w : E → R, let h : V → R be any function mapping vertices to real
numbers. For each edge (u, v) ∈ E, define
w (u, v) = w (u, v) + h (u) − h (v)
Let p = v0, v1, ..., vk be any path from vertex 0 to vertex k. Then:
1 p is a shortest path from 0 to k with weight function w if and only if
it is a shortest path with weight function w. That is w(p) = δ (v0, vk)
if and only if w(p) = δ (v0, vk).
2 Furthermore, G has a negative-weight cycle using weight function w
if and only if G has a negative-weight cycle using weight function w.
71 / 79

155. Using This Lemma
Select h
Such that w (u, v) + h (u) − h (v) ≥ 0.
Then, we build a new graph G
It has the following elements
V = V ∪ {s}, where s is a new vertex.
E = E ∪ {(s, v) |v ∈ V }.
w(s, v) = 0 for all v ∈ V , in addition to all the other weights.
Select h
Simply select h(v) = δ(s, v).
72 / 79

156. Using This Lemma
Select h
Such that w (u, v) + h (u) − h (v) ≥ 0.
Then, we build a new graph G
It has the following elements
V = V ∪ {s}, where s is a new vertex.
E = E ∪ {(s, v) |v ∈ V }.
w(s, v) = 0 for all v ∈ V , in addition to all the other weights.
Select h
Simply select h(v) = δ(s, v).
72 / 79

157. Using This Lemma
Select h
Such that w (u, v) + h (u) − h (v) ≥ 0.
Then, we build a new graph G
It has the following elements
V = V ∪ {s}, where s is a new vertex.
E = E ∪ {(s, v) |v ∈ V }.
w(s, v) = 0 for all v ∈ V , in addition to all the other weights.
Select h
Simply select h(v) = δ(s, v).
72 / 79

158. Using This Lemma
Select h
Such that w (u, v) + h (u) − h (v) ≥ 0.
Then, we build a new graph G
It has the following elements
V = V ∪ {s}, where s is a new vertex.
E = E ∪ {(s, v) |v ∈ V }.
w(s, v) = 0 for all v ∈ V , in addition to all the other weights.
Select h
Simply select h(v) = δ(s, v).
72 / 79

159. Using This Lemma
Select h
Such that w (u, v) + h (u) − h (v) ≥ 0.
Then, we build a new graph G
It has the following elements
V = V ∪ {s}, where s is a new vertex.
E = E ∪ {(s, v) |v ∈ V }.
w(s, v) = 0 for all v ∈ V , in addition to all the other weights.
Select h
Simply select h(v) = δ(s, v).
72 / 79

160. Using This Lemma
Select h
Such that w (u, v) + h (u) − h (v) ≥ 0.
Then, we build a new graph G
It has the following elements
V = V ∪ {s}, where s is a new vertex.
E = E ∪ {(s, v) |v ∈ V }.
w(s, v) = 0 for all v ∈ V , in addition to all the other weights.
Select h
Simply select h(v) = δ(s, v).
72 / 79

161. Example
Graph G with original weight function w with new source s and
h(v) = δ(s, v) at each vertex
0
0 0
0
0
0
0
-1
1
2
3
4
5
-4
7
3
4
8
2
1
6
s -5
-5
-4 0
73 / 79

162. Proof of Claim
Claim
w (u, v) + h (u) − h (v) ≥ 0 (5)
By Triangle Inequality
δ (s, v) ≤ δ(s, u) + w(u, v)
Then by the way we selected h, we have:
h(v) ≤ h(u) + w (u, v) (6)
Finally
w (u, v) + h (u) − h (v) ≥ 0 (7)
74 / 79

163. Proof of Claim
Claim
w (u, v) + h (u) − h (v) ≥ 0 (5)
By Triangle Inequality
δ (s, v) ≤ δ(s, u) + w(u, v)
Then by the way we selected h, we have:
h(v) ≤ h(u) + w (u, v) (6)
Finally
w (u, v) + h (u) − h (v) ≥ 0 (7)
74 / 79

164. Proof of Claim
Claim
w (u, v) + h (u) − h (v) ≥ 0 (5)
By Triangle Inequality
δ (s, v) ≤ δ(s, u) + w(u, v)
Then by the way we selected h, we have:
h(v) ≤ h(u) + w (u, v) (6)
Finally
w (u, v) + h (u) − h (v) ≥ 0 (7)
74 / 79

165. Proof of Claim
Claim
w (u, v) + h (u) − h (v) ≥ 0 (5)
By Triangle Inequality
δ (s, v) ≤ δ(s, u) + w(u, v)
Then by the way we selected h, we have:
h(v) ≤ h(u) + w (u, v) (6)
Finally
w (u, v) + h (u) − h (v) ≥ 0 (7)
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166. Example
The new Graph G after re-weighting G
0
5 1
0
4
0
0
-1
1
2
3
4
5
0
10
4
0
13
2
0
2
s 0
-5
-4 0
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167. Final Algorithm
Pseudo-Code
1. Compute G , where: G .V = G.E ∪ {(s, v) |v ∈ G.V } andw (s, v) = 0 for all
v ∈ G.V
2. If Bellman-Ford(G , w, s) == FALSE
3. print “Graphs contains a Neg-Weight Cycle”
4. else for each vertex v ∈ G .V
5. set h (v) = v.d computed by Bellman-Ford
6. for each edge (u, v) ∈ G .E
7. w (u, v) = w (u, v) + h (u) − h (v)
8. Let D = (duv) be a new n × n matrix
9. for each vertex u ∈ G.V
10. run Dijkstra(G, w, u) to compute δ (u, v) for all v ∈ G.V
11. for each vertex v ∈ G.V
12. duv = δ (u, v) + h (v) − h (u)
13. return D
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168. Final Algorithm
Pseudo-Code
1. Compute G , where: G .V = G.E ∪ {(s, v) |v ∈ G.V } andw (s, v) = 0 for all
v ∈ G.V
2. If Bellman-Ford(G , w, s) == FALSE
3. print “Graphs contains a Neg-Weight Cycle”
4. else for each vertex v ∈ G .V
5. set h (v) = v.d computed by Bellman-Ford
6. for each edge (u, v) ∈ G .E
7. w (u, v) = w (u, v) + h (u) − h (v)
8. Let D = (duv) be a new n × n matrix
9. for each vertex u ∈ G.V
10. run Dijkstra(G, w, u) to compute δ (u, v) for all v ∈ G.V
11. for each vertex v ∈ G.V
12. duv = δ (u, v) + h (v) − h (u)
13. return D
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169. Final Algorithm
Pseudo-Code
1. Compute G , where: G .V = G.E ∪ {(s, v) |v ∈ G.V } andw (s, v) = 0 for all
v ∈ G.V
2. If Bellman-Ford(G , w, s) == FALSE
3. print “Graphs contains a Neg-Weight Cycle”
4. else for each vertex v ∈ G .V
5. set h (v) = v.d computed by Bellman-Ford
6. for each edge (u, v) ∈ G .E
7. w (u, v) = w (u, v) + h (u) − h (v)
8. Let D = (duv) be a new n × n matrix
9. for each vertex u ∈ G.V
10. run Dijkstra(G, w, u) to compute δ (u, v) for all v ∈ G.V
11. for each vertex v ∈ G.V
12. duv = δ (u, v) + h (v) − h (u)
13. return D
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170. Final Algorithm
Pseudo-Code
1. Compute G , where: G .V = G.E ∪ {(s, v) |v ∈ G.V } andw (s, v) = 0 for all
v ∈ G.V
2. If Bellman-Ford(G , w, s) == FALSE
3. print “Graphs contains a Neg-Weight Cycle”
4. else for each vertex v ∈ G .V
5. set h (v) = v.d computed by Bellman-Ford
6. for each edge (u, v) ∈ G .E
7. w (u, v) = w (u, v) + h (u) − h (v)
8. Let D = (duv) be a new n × n matrix
9. for each vertex u ∈ G.V
10. run Dijkstra(G, w, u) to compute δ (u, v) for all v ∈ G.V
11. for each vertex v ∈ G.V
12. duv = δ (u, v) + h (v) − h (u)
13. return D
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171. Final Algorithm
Pseudo-Code
1. Compute G , where: G .V = G.E ∪ {(s, v) |v ∈ G.V } andw (s, v) = 0 for all
v ∈ G.V
2. If Bellman-Ford(G , w, s) == FALSE
3. print “Graphs contains a Neg-Weight Cycle”
4. else for each vertex v ∈ G .V
5. set h (v) = v.d computed by Bellman-Ford
6. for each edge (u, v) ∈ G .E
7. w (u, v) = w (u, v) + h (u) − h (v)
8. Let D = (duv) be a new n × n matrix
9. for each vertex u ∈ G.V
10. run Dijkstra(G, w, u) to compute δ (u, v) for all v ∈ G.V
11. for each vertex v ∈ G.V
12. duv = δ (u, v) + h (v) − h (u)
13. return D
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172. Final Algorithm
Pseudo-Code
1. Compute G , where: G .V = G.E ∪ {(s, v) |v ∈ G.V } andw (s, v) = 0 for all
v ∈ G.V
2. If Bellman-Ford(G , w, s) == FALSE
3. print “Graphs contains a Neg-Weight Cycle”
4. else for each vertex v ∈ G .V
5. set h (v) = v.d computed by Bellman-Ford
6. for each edge (u, v) ∈ G .E
7. w (u, v) = w (u, v) + h (u) − h (v)
8. Let D = (duv) be a new n × n matrix
9. for each vertex u ∈ G.V
10. run Dijkstra(G, w, u) to compute δ (u, v) for all v ∈ G.V
11. for each vertex v ∈ G.V
12. duv = δ (u, v) + h (v) − h (u)
13. return D
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173. Final Algorithm
Pseudo-Code
1. Compute G , where: G .V = G.E ∪ {(s, v) |v ∈ G.V } andw (s, v) = 0 for all
v ∈ G.V
2. If Bellman-Ford(G , w, s) == FALSE
3. print “Graphs contains a Neg-Weight Cycle”
4. else for each vertex v ∈ G .V
5. set h (v) = v.d computed by Bellman-Ford
6. for each edge (u, v) ∈ G .E
7. w (u, v) = w (u, v) + h (u) − h (v)
8. Let D = (duv) be a new n × n matrix
9. for each vertex u ∈ G.V
10. run Dijkstra(G, w, u) to compute δ (u, v) for all v ∈ G.V
11. for each vertex v ∈ G.V
12. duv = δ (u, v) + h (v) − h (u)
13. return D
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174. Final Algorithm
Pseudo-Code
1. Compute G , where: G .V = G.E ∪ {(s, v) |v ∈ G.V } andw (s, v) = 0 for all
v ∈ G.V
2. If Bellman-Ford(G , w, s) == FALSE
3. print “Graphs contains a Neg-Weight Cycle”
4. else for each vertex v ∈ G .V
5. set h (v) = v.d computed by Bellman-Ford
6. for each edge (u, v) ∈ G .E
7. w (u, v) = w (u, v) + h (u) − h (v)
8. Let D = (duv) be a new n × n matrix
9. for each vertex u ∈ G.V
10. run Dijkstra(G, w, u) to compute δ (u, v) for all v ∈ G.V
11. for each vertex v ∈ G.V
12. duv = δ (u, v) + h (v) − h (u)
13. return D
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175. Complexity
The Final Complexity
Times:
Θ(V + E) to compute G
O(VE) to run Bellman-Ford
Θ(E) to compute w
O (V 2
lg V + VE) to run Dijkstra’s algorithm |V | time using
Fibonacci Heaps
O (V 2
) to compute D matrix
Total : O(V 2 lg V + VE)
If E = O V 2 =⇒ O V 3
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176. Complexity
The Final Complexity
Times:
Θ(V + E) to compute G
O(VE) to run Bellman-Ford
Θ(E) to compute w
O (V 2
lg V + VE) to run Dijkstra’s algorithm |V | time using
Fibonacci Heaps
O (V 2
) to compute D matrix
Total : O(V 2 lg V + VE)
If E = O V 2 =⇒ O V 3
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177. Complexity
The Final Complexity
Times:
Θ(V + E) to compute G
O(VE) to run Bellman-Ford
Θ(E) to compute w
O (V 2
lg V + VE) to run Dijkstra’s algorithm |V | time using
Fibonacci Heaps
O (V 2
) to compute D matrix
Total : O(V 2 lg V + VE)
If E = O V 2 =⇒ O V 3
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178. Complexity
The Final Complexity
Times:
Θ(V + E) to compute G
O(VE) to run Bellman-Ford
Θ(E) to compute w
O (V 2
lg V + VE) to run Dijkstra’s algorithm |V | time using
Fibonacci Heaps
O (V 2
) to compute D matrix
Total : O(V 2 lg V + VE)
If E = O V 2 =⇒ O V 3
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179. Complexity
The Final Complexity
Times:
Θ(V + E) to compute G
O(VE) to run Bellman-Ford
Θ(E) to compute w
O (V 2
lg V + VE) to run Dijkstra’s algorithm |V | time using
Fibonacci Heaps
O (V 2
) to compute D matrix
Total : O(V 2 lg V + VE)
If E = O V 2 =⇒ O V 3
77 / 79

180. Complexity
The Final Complexity
Times:
Θ(V + E) to compute G
O(VE) to run Bellman-Ford
Θ(E) to compute w
O (V 2
lg V + VE) to run Dijkstra’s algorithm |V | time using
Fibonacci Heaps
O (V 2
) to compute D matrix
Total : O(V 2 lg V + VE)
If E = O V 2 =⇒ O V 3
77 / 79

181. Complexity
The Final Complexity
Times:
Θ(V + E) to compute G
O(VE) to run Bellman-Ford
Θ(E) to compute w
O (V 2
lg V + VE) to run Dijkstra’s algorithm |V | time using
Fibonacci Heaps
O (V 2
) to compute D matrix
Total : O(V 2 lg V + VE)
If E = O V 2 =⇒ O V 3
77 / 79

182. Complexity
The Final Complexity
Times:
Θ(V + E) to compute G
O(VE) to run Bellman-Ford
Θ(E) to compute w
O (V 2
lg V + VE) to run Dijkstra’s algorithm |V | time using
Fibonacci Heaps
O (V 2
) to compute D matrix
Total : O(V 2 lg V + VE)
If E = O V 2 =⇒ O V 3
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183. Outline
1 Introduction
Definition of the Problem
Assumptions
Observations
2 Structure of a Shortest Path
Introduction
3 The Solution
The Recursive Solution
The Iterative Version
Extended-Shoertest-Paths
Looking at the Algorithm as Matrix Multiplication
Example
We want something faster
4 A different dynamic-programming algorithm
The Shortest Path Structure
The Bottom-Up Solution
Floyd-Warshall Algorithm
Example
5 Other Solutions
The Johnson’s Algorithm
6 Exercises
You can try them
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184. Excercises
25.1-4
25.1-8
25.1-9
25.2-4
25.2-6
25.2-9
25.3-3
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