floyd-warshall algorithm
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Department of Computer Science
and Information Engineering
1
Floyd-Warshall Algorithm
Instructor: Yao-Ting Huang
Bioinformatics Laboratory,
Department of Computer Science & Information Engineering,
National Chung Cheng University.
Dept. of Computer Science
& Information Engineering
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SLOW-ALL-PAIRS-SHORTEST-PATHS(W)
1 n row W [ ]
2 WL )1(
3 for m2 to n-1
4 do )(mL EXTENDED-SHORTEST-PATHS( ),)1( WL m
5 return )1( nL
Slow All pair shortest Path Algorithm
Dept. of Computer Science
& Information Engineering
3
FASTER-ALL-PAIRS-SHORTEST-PATHS(W)
1 n row W [ ]
2 WL )1(
3 m1
4 while n m 1
5 do )2( mL EXTENDED-SHORTEST-PATHS( ), )()( mm LL
6 m m2
7 return )(mL
Fast All pair shortest Path Algorithm
Dept. of Computer Science
& Information Engineering
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06
052
04
710
4830
)1(L
0618
20512
11504
71403
42830
)2(L
All pair shortest Path Algorithm
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1
5 4
3
3
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-4
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2
8
1
4
-5
Can we do better?
Dept. of Computer Science
& Information Engineering
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Robert Floyd
Robert W Floyd (1936 –2001) finished school at age
14.
At the University of Chicago, he received a Bachelor's
degree in liberal arts when still only 17 and a second
Bachelor's degree in physics in 1958.
Becoming a computer operator in the early 1960s,
he began publishing many noteworthy papers.
He was appointed an associate professor at Carnegie
Mellon University by the time he was 27.
He became a full professor at Stanford University six
years later without a Ph.D. degree.
Dept. of Computer Science
& Information Engineering
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Review of Bellman-Ford Algorithm
j
i
The shortest paths are computed on the basis of
number of path edges.
DP over the number of path edges.
e.g.,
i to a: one edge
i to b: two edges
i to c: three edges
…
a
b
c
Dept. of Computer Science
& Information Engineering
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Floyd-Warshall Algorithm
j
i
Idea: the shortest path must go through some vertices,
which are also optimal subpaths.
e.g., a->b->c must be the shortest path between a and c.
Both short paths of i to j and a to c must visit through b.
a
b
c
Dept. of Computer Science
& Information Engineering
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Floyd-Warshall Algorithm
j
i
Remember that not only paths i to j may go through b.
Why not computing the shortest paths through b first
(i.e., subproblem of b)?
a
b
c
Dept. of Computer Science
& Information Engineering
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Floyd-Warshall Algorithm
DP on the subset of intermediate vertices.
All vertices are numbered from 1 to n.
Let Vk={1, 2, …, k}.
Sk(i, j) = length of a shortest path between i and j
whose interior vertices are all inVk .
Sk(i, j) = ??
j
ia
b
c
Dept. of Computer Science
& Information Engineering
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Key Observation
i
k
j
P1 P2
P1:All intermediate vertices in {1,2,..,k-1} P2:All intermediate vertices in {1,2,..,k-1}
P: All intermediate vertices in {1,2,..,k}
Sk-1(i, k) Sk-1(k, j)
Sk (i, j)
Dept. of Computer Science
& Information Engineering
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Floyd-Warshall DP Algorithm
Let Vk={1, 2, …, k}.
Sk(i, j) = length of a shortest path between i and j
whose interior vertices are all inVk .
Sk(i, j) = ??
Sk(i, j) = min{ Sk-1(i, j), Sk-1(i, k)+Sk-1(k, j) }.
j
ki
Dept. of Computer Science
& Information Engineering
12
Illustration
The shortest path P consists of shortest
subpaths: Vk={v1}
The shortest path between v5 and v4 (through v1) is
found.
s v1 vv2 v3v4v5
Dept. of Computer Science
& Information Engineering
13
Illustration
The shortest path P consists of shortest
subpaths. Vk={v1, v2}
The shortest path between s and v5 (through v2) is found.
s v1 vv2 v3v4v5
Dept. of Computer Science
& Information Engineering
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Illustration
The shortest path P consists of shortest
subpaths. Vk={v1, v2, v3}
The shortest path between v4 and v (through v3) is found.
s v1 vv2 v3v4v5
Dept. of Computer Science
& Information Engineering
15
Illustration
The shortest path P consists of shortest
subpaths: Vk={v1, v2, v3, v4}
The shortest path between v5 and v (through v4) is found.
s v1 vv2 v3v4v5
Dept. of Computer Science
& Information Engineering
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Illustration
The shortest path P consists of shortest
subpaths: Vk={v1, v2, v3, v4, v5}
The shortest path between s and v (through v5) is found.
s v1 vv2 v3v4v5
Dept. of Computer Science
& Information Engineering
Higher-order Illustration
Vk={v1}
The shortest paths of all pairs that visit through v1
will be found.
17
v1 v4v5
v7
v6
v2
v3
Dept. of Computer Science
& Information Engineering
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Bottom-Up Dynamic Programming
Floyd(W[1:n,1:n],S[1:n,1:n])
For i =1 to n
For j =1 to n
if <i, j> in E then S(i, j) ← W(i, j) else S(i, j) ← ∞
For k = 1 to n
For i = 1 to n
For j = 1 to n
if S[i, j] > S[i, k] + S[k, j] then
S[i, j] ← S[i, k] + S[k, j]
Dept. of Computer Science
& Information Engineering
19
Comparison with DP by Edges
At most m-1
edges
0. since },{min
}}{min ,min{
)1(
1
)1(
1
)1()(
jjkj
m
iknk
kj
m
iknk
m
ij
m
ij
wwl
wlll
wkj
likm-1
i j
Dept. of Computer Science
& Information Engineering
20
Comparison with DP by Edges
06
052
04
710
4830
)1(L
EXTENDED-SHORTEST-PATHS(L, w)
1 ][Lrown
2 Let )( ijlL be an n n matrix
3 for 1i to n
4 do for 1j to n
5 do ijl
6 for 1k to n
7 do ),min(
kjikijijwlll
8 return L
SLOW-ALL-PAIRS-SHORTEST-PATHS(W)
1 n row W [ ]
2 WL )1(
3 for 2m to n-1
4 do )(mL EXTENDED-SHORTEST-PATHS( ),)1( WL m
5 return )1( nL
Dept. of Computer Science
& Information Engineering
21
Dynamic Programming
j
ki
Sk(i, j) = length of the shortest path between i
and j whose interior vertices are all inVk.
Sk(i, j) = min of
• Sk(i, j) = Sk-1(i, k) + Sk-1(k ,j).
• If k is on the shortest path from i to j
• Sk(i, j) = Sk-1(i, j).
• If k is not on the shortest path from i to j
How many subproblems?
Dept. of Computer Science
& Information Engineering
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Bottom-Up Dynamic Programming
j
j
i
i
k
k
Sk (k=1,2,3,4,5)
Dept. of Computer Science
& Information Engineering
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Bottom-Up Dynamic Programming
5
2
4
3
1
1
43
3
6
4
1
2
2
S0=W=
i=1
j=1 2
5
2
5
Dept. of Computer Science
& Information Engineering
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Bottom-Up Dynamic Programming
5
2
4
3
1
1
43
3
6
4
1
2
2S0=W=
i=1
j=1 2
5
2
5
0
0
0
0
0
4 3
6 2
1
4 2 3
1∞∞
∞
∞∞∞
∞∞
∞∞
∞
For k = 1 to n
For i = 1 to n
For j = 1 to n
if S[i, j] > S[i, k] + S[k, j] then
S[i, j] ← S[i, k] + S[k, j]
Dept. of Computer Science
& Information Engineering
25
Bottom-Up Dynamic Programming
5
2
4
3
1
1
43
3
6
4
1
2
2
i=1
5
2
0
0
0
0
0
4 3
6 2
1
4 2 3
1∞∞
∞
∞∞
∞∞
∞
S1=
For k = 1 to n
For i = 1 to n
For j = 1 to n
if S[i, j] > S[i, k] + S[k, j] then
S[i, j] ← S[i, k] + S[k, j]
S0=W=
i=1
j=1 2
5
2
5
0
0
0
0
0
4 3
6 2
1
4 2 3
1∞∞
∞
∞∞∞
∞∞
∞∞
∞
Dept. of Computer Science
& Information Engineering
26
Bottom-Up Dynamic Programming
5
2
4
3
1
1
43
3
6
4
1
2
2
i=1
5
2
0
0
0
0
0
4 3
6 2
1
4 2 3
1∞∞
8
∞45
∞∞
∞∞
∞
S1=
For k = 1 to n
For i = 1 to n
For j = 1 to n
if S[i, j] > S[i, k] + S[k, j] then
S[i, j] ← S[i, k] + S[k, j]
Dept. of Computer Science
& Information Engineering
27
5
2
4
3
1
1
43
3
6
4
1
2
2
i=1
j=1 2
5
2
5
0
0
0
0
0
4 3
6 2
1
4 2 3
1∞∞
∞
∞∞∞
∞∞
∞∞
∞
i=1
5
2
0
0
0
0
0
4 3
3 2
1
3 2 3
162
7
745
74
65
5
S5=
S0=W=
Dept. of Computer Science
& Information Engineering
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Analysis
Running time is clearly O(|V|3).
Faster than all previous algorithms.
O(|V|4) and O(|V|3lg|V|).
Dept. of Computer Science
& Information Engineering
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Remarks
Is negative edges allowed?
Is it possible to detect negative cycle
using Floyd-Warshall’s algorithm?
Dept. of Computer Science
& Information Engineering
Remarks
Is it possible to detect negative cycle
using Floyd-Warshall’s algorithm?
30
s
u v
-1 -1
-1
Dept. of Computer Science
& Information Engineering
31
Extracting the Shortest Paths
How can we retrieve the shortest paths?
Maintain a predecessor p[i, j].
If the shortest path between i and j does not pass
through any intermediate vertex, then the
predecessor p[i, j] = 0.
0
0
0
0
0
4 3
3 2
1
3 2 3
162
7
745
74
65
5
Dept. of Computer Science
& Information Engineering
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How do we backtrack the actual path?
Maintain a predecessor p[i, j].
If the shortest path between i and j does not pass
through any intermediate vertex, then the predecessor
p[i, j] = 0.
Define Pk[i, j] as follows:
P0[i, j] = 0 for all i, j in V
Pk[i, j] = Pk-1[i, j], if Sk[i, j] = Sk-1[i, j]
Pk[i, j] = k, if Sk[i, j] ≠ Sk-1[i, j]
Dept. of Computer Science
& Information Engineering
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How do we retrieve the actual path?
We maintain Pk[i,j] as follows:
P0[i, j] = 0 for all i,j in V
Pk[i, j] = Pk-1[i,j], if Sk[i,j] = Sk-1[i,j]
Pk[i, j] = k, if Sk[i,j] ≠ Sk-1[i,j]
To reconstruct the shortest path in Pn[i, j]
If Pn[i, j]=0, edge <i, j> is the shortest path.
If Pn[i, j]=k, then k is an interior vertex on the path, other interior vertices can be obtained by checking Pn[i, k] and Pn[k, j].
Dept. of Computer Science
& Information Engineering
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Floyd’s Algorithm (with matrix P)
Floyd(W[1:n,1:n],S[1:n,1:n],P[1:n,1:n])
For i =1 to n
For j =1 to n
P[i,j] ← 0
if <i,j> in E then S[i,j] ← W[i,j] else S[i,j] ← ∞
For k = 1 to n
For i = 1 to n
For j = 1 to nif S[i,j] > S[i,k] + S[k,j] then
S[i,j] ← S[i,k] + S[k,j]
P[i,j] ← k
Dept. of Computer Science
& Information Engineering
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A detailed example
5
2
4
3
1
1
43
3
6
4
1
2
2
i=1
j=1 2
5
2
5
0
0
0
0
0
4 0
5 0
0
3 0 0
033
3
211
55
20
3
i=1
5
2
0
0
0
0
0
4 3
3 2
1
3 2 3
162
7
745
74
65
5
S5=
P5=
How do we retrieve the
shortest path from 3 to 5?
Dept. of Computer Science
& Information Engineering
36
A detailed example
5
2
4
3
1
1
43
3
6
4
1
2
2
i=1
j=1 2
5
2
5
0
0
0
0
0
4 0
5 0
0
3 0 0
033
3
211
55
20
3
i=1
5
2
0
0
0
0
0
4 3
3 2
1
3 2 3
162
7
745
74
65
5
S5=
P5=
How do we retrieve the
shortest path from 3 to 5?
P[3,5]=2, so 3…2…5
Dept. of Computer Science
& Information Engineering
37
A detailed example
5
2
4
3
1
1
43
3
6
4
1
2
2
i=1
j=1 2
5
2
5
0
0
0
0
0
4 0
5 0
0
3 0 0
033
3
211
55
20
3
i=1
5
2
0
0
0
0
0
4 3
3 2
1
3 2 3
162
7
745
74
65
5
S5=
P5=
How do we retrieve the
shortest path from 3 to 5?
P[3,5]=2, so 3…2…5
P[3,2]=1,so 3…1…2…5
P[3,1]=P[2,5]=P[1,2]=0
So: 3→1→2→5.
Dept. of Computer Science
& Information Engineering
38
Application: Transitive Closure
Given directed graph G = (V, E)
Compute transitive closure G* = (V, E* )
E* = {(i, j) : there is path from i to j in G}
How to apply the previous algorithm to solve this
problem?
3
2
1
3
2
1
Dept. of Computer Science
& Information Engineering
39
Application: Transitive Closure
Given directed graph G = (V, E)
Compute transitive closure G* = (V, E* )
E* = {(i, j) : there is path from i to j in G}
How to apply the previous algorithm to solve this
problem?
1
4 3
2
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& Information Engineering
40
Transitive Closure – Solution1
Using Floyd-Warhshall Algorithm
Assign weight of 1 to each edge, then run FLOYD-
WARSHALL with this weight.
If dij(n) < n, there is a transitive path from i to j.
Dept. of Computer Science
& Information Engineering
41
Transitive Closure – Solution2
Instead of D(k), we have T(k) = (tij(k))
tij(0) = 0 (if i != j and (i, j) not in E)
1 (if i = j or (i, j) in E)
tij(k) = 1 (if there is a path from i to j with all
intermediate vertices in {1, 2,…, k})
(tij(k-1) is 1) or (tik
(k-1) is 1 and tkj(k-1) is 1)
0 (otherwise)
Dept. of Computer Science
& Information Engineering
42
Transitive Closure – Solution2
TRANSITIVE-CLOSURE(E, n)
for i = 1 to n
do for j = 1 to n
do if i=j or (i, j) in E
then tij(0) = 1
else tij(0) = 0
for k = 1 to n
do for i = 1 to n
do for j = 1 to n
do tij(k) = tij
(k-1) || ( tik(k-1) && tkj
(k-1) )
return T(n)
Dept. of Computer Science
& Information Engineering
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1
4 3
2
1101
0110
1110
0001
)0(T
1101
0110
1110
0001
)1(T
1101
1110
1110
0001
)2(T
1111
1110
1110
0001
)3(T
1111
1111
1111
0001
)4(T
Dept. of Computer Science
& Information Engineering
44
Conclusion
Floyd-Warshall runs in O(V3) time, which is faster than previous algorithms and easy to implement.
The running time is independent of the number of edges.
The dynamic programming focuses on the vertices instead of edges.
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