1 chapter 25: all-pairs shortest-paths. 2 some algorithms when no negative edges –using...
TRANSCRIPT
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Chapter 25: All-Pairs Shortest-Paths
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Some Algorithms• When no negative edges
– Using Dijkstra’s algorithm: O(V3)– Using Binary heap implementation: O(VE lg
V)– Using Fibonacci heap: O(VE + V2log V)
• When no negative cycles– Floyd-Warshall [1962]: O(V3) time
• When negative cycles– Using Bellman-Ford algorithm: O(V2 E) =
O(V4 )– Johnson [1977]: O(VE + V2log V) time based
on a clever combination of Bellman-Ford and Dijkstra
3
Notations
• G = (V, E) with w: E -> R• Input form: matrix W= (wij )
• wij =0 if i = j ,
• wij = the weight of the directed edge if i≠j and (i, j) є E,
• Otherwise wij = ∞
• Output: D = (dij),
• dij = δ(i,j) the shortest weight path from i to j
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A dynamic programming approach
1. characterize the structure of an optimal solution,2. recursively define the value of an optimal solution,3. compute the value of an optimal solution in a bottom-up fashion.
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The structure of an optimal solution
•Consider a shortest path p from vertex i to vertex j, and suppose that p contains at most m edges. Assume that there are no negative-weight cycles. Hence m ≤ n-1 is finite.
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The structure of an optimal solution
• If i = j, then p has weight 0 and no
edge.
• If i≠j, we decompose p into i ~ k -
> j where p’ contains at most
m-1 edges.
• Moreover, p’ is a shortest path from i to k and δ(i,j) = δ(i,k) + wkj
'p
7
Recursive solution to the all-pairs shortest-path problem
• Define: lij(m) = minimum weight of any path from i to j that contains at most m edges.
0 if i = j• lij(0) =
∞ if i ≠ j• Then lij(m) = min{ lij(m-1), min1≤k≤n
{lik(m-1) + wkj}} = min1≤k≤n {lik(m-1) + wkj} (why?)
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Recursive solution to the all-pairs shortest-path
problem• Since shortest path from i to j
contains at most n-1 edges, δ(i,j) = lij(n-1) = lij(n) = lij(n+1) = …• Computing the shortest-path weight
bottom up:– Compute L(1) ,L(2) ,…., L(n-1) where L(m)=(lij (m))– Note that L(1) = W.
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EXTENDED-SHORTEST-PATHS(L, W)
• Given matrices L(m-1) and W return L(m)
1 n <- L.row2 Let L’ = (l’ij) be a new n x n matrix
3 for i = 1 to n4 for j = 1 to n5 l’ij = ∞
6 for k = 1 to n7 l’ij = min(l’ij, lik + wkj)
8 return L’
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06
052
04
710
4830
)1(L
0618
20512
11504
71403
42830
)2(L
1
45
2
3
-4
17
2
6
-5
83
4
Example:
W =
11
06158
20512
115047
11403
42330
)3(L
06158
20512
35047
11403
42310
)4(L
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Matrix Multiplication
Let l(m-1) -> a w -> b l(m) -> c min -> + + -> ‧
time complexity : O(n3)
Cij = k=1 to n aik‧bkj
lij(m) = min1≤k≤n {lik(m-1) + wkj}
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SLOW-ALL-PAIRS-SHORTEST-PATHS(W)
L(1) = L(0) ‧ W = WL(2) = L(1) ‧ W = W2
L(3) = L(2) ‧ W = W3
‧ ‧ ‧
L(n-1) = L(n-2) ‧W = Wn-1
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SLOW-ALL-PAIRS-SHORTEST-PATHS(W)
1 n = W.rows2 L(1) = W3 for m = 2 to n-14 let L(m) be a new n x n matrix5 L(m) = EXTENDED-SHORTEST-
PATHS( L(m-1), W )6 return L(n-1)
Time complexity : O(n4)
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Improving the running time
L(1) = WL(2) = W2 =W．WL(4) =W4 = W2．W2
. . .
i.e., using repeating squaring!
Time complexity: O(n3lg n)
12122)2( )1log()1log()1log()1log(
nnnn
WWWL
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FASTER-ALL-PAIRS-SHORTEST-PATHS(W)1. n =W.row2. L(1) =W3. m =14. while m < n-1 5. let L(2m) be a new n x n matrix
6. L(2m) = Extend-Shorest-Path(L(m), L(m))7. m = 2m8. return L(m)
FASTER-ALL-PAIRS-SHORTEST-PATHS
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The Floyd-Warshall algorithm
• A different dynamic programming formulation
‧The structure of a shortest path: Let V(G)={1,2,…,n}. For any pair of
vertices i, j єV, consider all paths from i to j whose intermediate vertices are drawn from {1, 2,…,k}, and let p be a minimum weight path among them.
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The structure of a shortest path
• If k is not in p, then all intermediate vertices are in {1, 2,…,k-1}.
• If k is an intermediate vertex of p,
then p can be decomposed into i ~ k
~ j where p1 is a shortest path from i
to k with all the intermediate vertices in {1,2,…,k-1} and p2 is a shortest path from k to j with all the intermediate vertices in {1,2,…,k-1}.
1p1 p 2 p
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A recursive solution to the all-pairs shortest path
• Let dijk =the weight of a shortest path
from vertex i to vertex j with all intermediate vertices in the set {1,2,…,k}.
dij(k) = wij if k = 0
= min(dij(k-1), dik
(k-1) + dkj(k-1)) if k ≥ 1
D(n) = (dij(n)) if the final solution!
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FLOYD-WARSHALL(W)
1. n = W.rows 2. D(0)= W 3. for k = 1 to n 4. Let D(k) = (dij
(k)) be a new n x n matrix 5. for i = 1 to n 6. for j = 1 to n 7. dij
(k) = min (dij(k-1), dik
(k-1) + dkj(k-
1)) 8. return D(n)
Complexity: O(n3)
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Constructing a shortest path
• π(0), π(1),…, π(n)
• πij(k) : is the predecessor of the vertex j
on a shortest path from vertex i with all intermediate vertices in the set {1,2,…,k}.
πij(0) = NIL if i=j or wij = ∞
= i if i ≠j and wij < ∞
πij(k) = πij
(k-1) if dij(k-1) ≤ dik
(k-1) + dkj(k-1)
= πkj(k-1) if dij
(k-1) > dik(k-1) + dkj
(k-1)
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Example
06
052
04
710
4830
)0(D
06
20552
04
710
4830
(1)D
NNNN
NNN
NNNN
NNN
NN
5
44
3
22
111
(0)
NNNN
N
NNNN
NNN
NN
5
1414
3
22
111
(1)
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06
20552
11504
710
44830
(2)D
NNNN
N
NN
NNN
N
5
1414
223
22
1211
(2)
06
20512
11504
710
44830
)3(D
NNNN
N
NN
NNN
N
5
1434
223
22
1211
(3)
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06158
20512
35047
11403
44130
(4)D
06158
20512
35047
11403
42310
(5)D
N
N
N
N
N
5434
1434
1234
1244
1241
(4)
N
N
N
N
N
5434
1434
1234
1244
1541
(5)
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Transitive closure of a directed graph
• Given a directed graph G = (V, E) with V = {1,2,…, n}
• The transitive closure of G is G*= (V, E*) where E*={(i, j)| there is a path from i to j in G}.
Modify FLOYD-WARSHALL algorithm:tij
(0) = 0 if i≠j and (i,j) ∉ E
1 if i=j or (i,j) є Efor k ≥ 1 tij
(k) = tij(k-1) (tik
(k-1) tkj(k-1))
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TRANSITIVE-CLOSURE(G)1 n = |G.V|2 Let T(0) = (tij
(0)) be a new n x n matrix
3 for i = 1 to n 4 for j =1 to n5 if i == j or (i, j) ∈ G.E6 tij
(0) = 1
7 else tij(0) = 0
8 for k =1 to n9 Let T(k) = (tij
(k)) be a new n x n matrix
10 for i = 1 to n11 for j =1 to n12 tij
(k) = tij(k-1) (tik
(k-1) tkj(k-1))
123 return T(n)
Time complexity: O(n3)
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Example
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
① ②
③④
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Some Algorithms• When no negative edges
– Using Dijkstra’s algorithm: O(V3)– Using Binary heap implementation: O(VE lg
V)– Using Fibonacci heap: O(VE + V2log V)
• When no negative cycles– Floyd-Warshall [1962]: O(V3) time
• When negative cycles– Using Bellman-Ford algorithm: O(V2 E) =
O(V4 )– Johnson [1977]: O(VE + V2log V) time based
on a clever combination of Bellman-Ford and Dijkstra
29
Johnson’s algorithm for sparse graphs
• If all edge weights in G are nonnegative,
we can find all shortest paths in O(V2lg V
+ VE) by using Dijkstra’s algorithm with
Fibonacci heap
• Bellman-Ford algorithm takes O(VE)
• Using reweighting technique
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Reweighting technique
• If G has negative-weighted edge, we compute a new set of nonnegative weight that allows us to use the same method. The new set of edge weight ŵ satisfies:
• 1. For all pairs of vertices u, v єV, a shortest path from u to v using weight function w is also a shortest path from u to v using the weight function ŵ
• 2. ∀(u,v) є E(G), ŵ(u,v) is nonnegative
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Lemma: (Reweighting doesn’t change shortest paths)
• Given a weighted directed graph G = (V, E) with weight function w:E→R , let h:V →R be any function mapping vertices to real numbers. For each edge (u,v) є E, ŵ(u,v) = w(u,v) + h(u) – h(v)
• Let P=<v0,v1,…,vk> be a path from vertex v0 to vk Then w(p) = δ(v0,vk) if and only if ŵ(p) = (v0,vk) Also, G has a negative-weight cycle using weight function w iff G has a negative weight cycle using weight function ŵ.
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Proof• ŵ(p) = w(p) + h(v0) – h(vk)
ŵ(p) = ŵ(vi-1 ,vi)
= (w(vi-1 ,vi) + h(vi-1)-h(vi))
= w(vi-1 ,vi) + h(v0)-h(vk)
=w(p) + h(v0) – h(vk)
k
i 1
k
i 1
k
i 1
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Proof
• Because h(v0) and h(vk) do not depend on the path, if one path from v0 to vk is shorter than another using weight function w, then it is also shorter using ŵ. Thus,
w(p) = δ(v0,vk) if and only if ŵ(p) = (v0,vk)
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Proof
• G has a negative-weight cycle using w iff G has a negative-weight cycle using ŵ.
• Consider any cycle C=<v0,v1,…,vk> with v0=vk . Then ŵ(C) = w(C) + h(v0) – h(vk) = w(C) .
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Producing nonnegative weight by reweighting
• Given a weighted directed graph G = (V, E)
• We make a new graph G’= (V’,E’), V’ = V ∪ {s}, E’ = E ∪{(s,v): v є V} and w(s,v) = 0, for all v in V
• Let h(v) = δ(s, v) for all v V’• We have h(v) ≤ h(u) + w(u, v) (why?)• ŵ(u, v) = w(u, v) + h(u) – h(v) ≥ 0
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-41
7
2
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-5
8
3 42
3
45
1
Example:
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-4
1
7
2
6
-5
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3 4-1
-5
-4 0
0
2
3
45
1
0S
0
0
0
0
0
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0
0
10
2
2
0
13
4 0-1
-5
-4 0
0
2
3
45
1
0S
4
0
0
1
5
h(v) = δ(s, v)ŵ(u, v) = w(u, v) + h(u) – h(v)
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JOHNSON algorithm
1 Computing G’, where G’.V = G.V ∪ {s}and G’.E= G.E ∪{(s, v): v є G.V} and w(s, v) = 0.
2 if BELLMAN-FORD(G’, w, s)= FALSE3 print “the input graph contains negative
weight cycle”4 else for each vertex v є G’.V5 set h(v) to be the value of δ(s, v) computed
by the BF algorithm6 for each edge (u, v) є G’.E, ŵ(u, v) = w(u, v)
+ h(u) – h(v)
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JOHNSON algorithm
7 Let D = (duv) be a new n x n matrix
8 for each vertex u є G.V run DIJKSTRA (G, ŵ, u) to compute (u, v)
for all v є V[G] .10 for each vertex v є G.V 11 duv = (u, v) + h(v) – h(u)
12 return DComplexity: O(V2lgV + VE)
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00
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2/1
2/-3
2/20/-4
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1 0/0
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2/7
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0 010
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12/5
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0/0
2/1
2/6
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Homework
• Practice at home: Exercises: 25.1-5, 25.1-6, 25.1-7
• Practice at home : 25.2-4, 25.2-5• Exercises: 25.2-8 (Due: Jan. 4)• Practice at home : 25.3-3• Exercises: 25.3-5 (Due: Jan. 4)
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Quiz 5• Please show the ordering of the
following vertices produced by Topological Sort Algorithm
under-shorts
pants
belt
shirt
tie
socks
shoes
jacket