1

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 matrix3 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’

10

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061820512

115047140342830

)2(L

1

45

2

3

-41

7

2

6

-5

83

4

Example:W =

11

0615820512

1150471140342330

)3(L

0615820512

350471140342310

)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

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24

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(4)D

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NN

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(4)

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NN

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54341434123412441541

(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) = 17 else tij

(0) = 08 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

1101011011100001

(0)T

1101011011100001

(1)T

1101111011100001

(2)T

1111111011100001

(3)T

1111111111110001

(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

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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

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-5

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3 42

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45

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Example:

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-4

17

2

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-5

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3 4-1

-5

-4 0

0

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45

1

0S

0

0

0

0

0

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0

010

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 matrix8 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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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