13 ShortestPath II

8/2/2019 13 ShortestPath II

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Huo Hongwei 1

Design and Analysis of Algorithms

Shortest Paths II

Reference:CLRS Chapter 24

Topics:

Bellman-Ford algorithm

DAG shortest paths

Linear programming and difference constraints

VLSI layout compaction


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Huo Hongwei 2

Negative-weight cycles

Recall: If a graph G = (V,E) contains a negative-weightcycle, then some shortest paths may not exist.

Example:

u vxx

< 0 xx


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Huo Hongwei 3

Negative-weight cycles

Recall: If a graph G = (V,E) contains a negative-weightcycle, then some shortest paths may not exist.

Example:

Bellman-Ford algorithm: Finds all shortest-path lengths

from a source s Vto all v Vor determines thatnegative-weight cycle exists.

u vxx

< 0 xx


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Huo Hongwei 4

Bellman-Ford algorithm

Time = O(VE).

BELLMAN-FORD(G,w,s)

1 d[s] 02 for each v V {s}3 do d[v] 4 for i 1 to |V|-15 do for each edge (u,v) E6 do if d[v] > d[u] + w(u, v)7 then d[v] d[u] + w(u, v)8 for each edge (u,v) E9 do if d[v] > d[u] + w(u,v)

10 then report that a negative-weight cycle exists.

11 return d[v] // = (s, u), if no negative-weight cycles

BELLMAN-FORD's ALGORITHM

Relaxationstep

initialization


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Example of Bellman-Ford

A

C

B

E

D

1

-12

2

5

3

4 -3


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Initialization

A

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B

E

D

1

-12

2

5

3

4 -3

0


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Order of edge relaxation

A

C

B

E

D

1

-12

2

5

3

4 -3

0

4 71

52

6

8

3


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A

C

B

E

D

1

-12

2

5

3

4 -3

0

4 71

52

6

8

3


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A

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B

E

D

1

-12

2

5

3

4 -3

0

4 71

52

6

8

3


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A

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B

E

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1

-12

2

5

3

4 -3

0

4 71

52

6

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3


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A

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B

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D

1

-12

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4 -3

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-1

4 71

52

6

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3


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A

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B

E

D

1

-12

2

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3

4 -3

0

4

-1

4 71

52

6

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3


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A

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B

E

D

1

-12

2

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4 -3

0

4

-1

4 71

52

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3


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A

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B

E

D

1

-12

2

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4 -3

0

2

-1

4 71

52

6

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3


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A

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B

E

D

1

-12

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4 -3

0

2

-1

4 71

52

6

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3


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A

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B

E

D

1

-12

2

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3

4 -3

0

2

-1

4 71

52

6

8

3

End of pass 1


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A

C

B

E

D

1

-12

2

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4 -3

0

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-1

147

1

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3


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A

C

B

E

D

1

-12

2

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3

4 -3

0

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-1

147

1

52

6

8

3


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A

C

B

E

D

1

-12

2

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3

4 -3

0

2 1

-1

147

1

52

6

8

3


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A

C

B

E

D

1

-12

2

5

3

4 -3

0

2 1

-1

147

1

52

6

8

3


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A

C

B

E

D

1

-12

2

5

3

4 -3

0

2 1

-1

147

1

52

6

8

3


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A

C

B

E

D

1

-12

2

5

3

4 -3

0

2 1

-1

147

1

52

6

8

3


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A

C

B

E

D

1

-12

2

5

3

4 -3

0

2 1

-1

147

1

52

6

8

3


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A

C

B

E

D

1

-12

2

5

3

4 -3

0

2 -2

-1

147

1

52

6

8

3


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A

C

B

E

D

1

-12

2

5

3

4 -3

0

2 -2

-1

4 71

52

6

8

3

End of pass 2 (and 3 and 4)


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Correctness

Theorem. If G = (V,E) contains no negative-weightcycles, then after the Bellman-Ford algorithm executes,

d[v] = (s, v) for all v V.Proof. Let v Vbe any vertex, and consider a shortestpathp froms to v with the minimum number of edges.

Sincep is a shortest path, we have(s, vi) = (s, vi-1) + w(vi-1, vi)

v

v0v1

v2v3 vk

p:s


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Correctness (continued)

Initially,d[v0] = 0 = (s, v0) andd[v0] is unchanged by subsequentrelaxations (because of the the lemma from previous Lecture that

d[v] (s, v)).

After 1 pass throughE, we haved[v1] = (s, v1). After 2 pass throughE, we haved[v2] = (s, v2).

Afterk pass throughE, we haved[vk] = (s, vk).Since G contains no negative-weight cycles,p is simple.

Longest simple path has |V| -1 edges.

v

v0v1

v2v3 vk

p:

s


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Detection of negative-weight cycles

Corollary. If a valued[v] fails to converge after |V| -1passes, there exists a negative-weight cycle in G

reachable froms.


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Linear programming

LetA be anm n matrix,b be anm-vector, andc be ann-vector. Find ann-vectorx that maximizescTx subject to

Ax b, or determine that no such solution exists.n

m

A x b

.

cT x

.maximizing


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Linear programming algorithm

Algorithms for the general problem

Simplex methods practical, but worst-case

exponential time.

Interior-point methods polynomial time andcompetes with simplex.

Feasibility problem: No optimization criterion. Just findx such thatAx b. In general, just as hard as ordinary LP.


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Solving a system of difference constrains

Linear programming where each row ofA containsexactly one 1, one 1, and the rest 0s.

Example:

x1 x2 3x2 x3 -2x1 x3 2

xj xi wij


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Example:

x1 x2 3x2 x3 -2x1 x3 2

xj xi wijx1 = 3

x2 = 0

x3 = 2

Solution:


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Example:

Constraint graph:

x1 = 3

x2 = 0

x3 = 2

Solution:

xj xi wij vi wij vj(The A

matrix hasdimensions

|E| |V|.)

x1 x2 3x2 x3 -2x1 x3 2

xj xi wij


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Unsatisfiable constraints

Theorem. If the constraint graph contains a negative-weight cycle, then the system of differences is

unsatisfiable.Proof. Suppose that the negative-weight cycle is v1

v2 vkv1. Then, we havex2 x1 w12x3 x2 w23

xk xk-1 wk-1,kx1 xk wk1


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Unsatisfiable constraints

Theorem. If the constraint graph contains a negative-weight cycle, then the system of differences is

unsatisfiable.Proof. Suppose that the negative-weight cycle is v1

v2 vkv1. Then, we havex2 x1 w12x3 x2 w23

xk xk-1 wk-1,kx1 xk wk1

0 weight of cycle< 0

Therefore, novalues for thexican satisfy the

constraints.


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Satisfying the constraints

Theorem. Suppose no negative-weight cycle exists inthe constraint graph. Then, the constraint is satisfiable.

Proof. Add a new vertexs to Vwith a 0-weight edge toeach vertex vi V.

v1

v4

v9

v7

v3


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Huo Hongwei 37

Satisfying the constraints

Theorem. Suppose no negative-weight cycle exists inthe constraint graph. Then, the constraint is satisfiable.

Proof. Add a new vertexs to Vwith a 0-weight edge toeach vertex vi V.

s

v1

v4

v9

v7

v3

Note: no negative-weight introduced shortest paths exist.


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Proof (continued)

Claim: The assignmentxi = (s, vi) solves the constraint.Consider any constraintxj xi wij, and consider theshortest paths froms to vj and vi:

The triangle inequality gives us (s, vj) (s, vi) + wij.Sincexi = (s, vi) andxj = (s, vj), the constraintxj xi wijis satisfied.

s vi

vj

(s, vi)

wij(s, vj)

B ll F d d li i


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Huo Hongwei 39

Bellman-Ford and linear programming

Corollary. The Bellman-Ford algorithm can solve asystem ofm difference constraints onn variable in

O(mn) time. Single-source shortest paths is a simple LP problem.

In fact, Bellman-Ford maximizesx1 +x2 + + xn subject

to the constraints xj xi

wij andxi

0. Bellman-Ford also minimizes maxi{xi} mini{xi}.

A li i VLSI l i


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Application to VLSI layout compaction

Problem: Compact (in one dimension) the spacebetween the features of a VLSI layout without bringingany features too close together.

minimum separation

Integrated-circuit features

A li ti t VLSI l t ti


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Application to VLSI layout compaction

Constraint: x2 x1 d1 + Bellman-Ford minimizes maxi{xi} mini{xi}, which

compacts the layout in thex-dimension.

1

d1

x1 x2

2

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13 ShortestPath II