8
Proof of Thm 6.6 (cont.): (1) Construct dual solution: I Kruskal constructs MST T with edge set I E (T )= {f 1 , f 2 ,..., f n-1 }, and c (f 1 ) c (f 2 ) ... c (f n-1 ) I Every edge f k creates a new connected component X k V by joining two smaller connected components. Note that X n-1 = V . 174 - 2 ties : :D ;¥na Xyy , ire . fe E f ( Xu ) Define : px . = Ce Cfe ) - Ce Cfa ) 2 O Pv = - C ( fu . , ) Ps = O for all other SEV

Xyy - University of Liverpool

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Proof of Thm 6.6 (cont.):

(1) Construct dual solution:

I Kruskal constructs MST T with edge set

I E (T ) = {f1, f2, . . . , fn�1}, and c(f1) c(f2) . . . c(fn�1)

I Every edge fk creates a new connected component Xk ✓ V by joining two smaller

connected components. Note that Xn�1 = V .

174 - 2

ties::D;¥naXyy

,

ire. fe E f ( Xu )

Define : px .= Ce Cfe ) - Ce Cfa ) 2 O

Pv =- C ( fu . , )

Ps = O for all other SEV

Proof of Thm 6.6 (cont.):

(2) Show that p is feasible for the dual:

I Sign constraints fulfilled by construction. minX

;6=S✓V

(|S| � 1)pS

s.t.X

S :e2�(S)

pS � �c(e) 8e 2 E

pS � 0 8; 6= S ⇢ V

pV free

174 - 3

-

s:{ eyes,Ps = %ec.ge ,

Ph

'

=Pxi t Pee

.

+ Pxe,

t. . .

t PuE sets containing e growing w

. v. t E

let Xi be the smallest such sets.

= Eifel- dfi )) t I effed - c Cfe.

)) -1. . .

+ [ dfa.it]

-

= - c Kfi ) fi either e -- fi Cie

. e will being , ,Z - e Ce )

⇒ Cle ) -- c Cf : )

or et fi⇒ e Cee) 's c Cfi )⇒ dual constraints are fulfilled

=) p is dual feasible

Proof of Thm 6.6 (cont.):

(3) Show that x⇤

and p fulfill complementary slackness conditions:

174 - 4

• XE so Cee ECT ) )⇒ E ps = - de )

seeks ,

• Ps to ⇒ S= Xk for some k

DEF'

S = vertex set of tree edges that

form a subtree of the finalMST

⇒ x CHS ) ) = 1st - I

⇒ primal slack = O

=) p and KA are optimal

Shortest Path Problem

Given: digraph D = (V ,A), node r 2 V , arc costs ca, a 2 A.

Task: for each v 2 V , find dipath from r to v of least cost (if one exists)

r

d

b c

a e

f g

7

85

9�3

6

4

6 8�9

11

3

7

175

Shortest Path Problem

Given: digraph D = (V ,A), node r 2 V , arc costs ca, a 2 A.

Task: for each v 2 V , find dipath from r to v of least cost (if one exists)

r

d

b c

a e

f g

7

85

9�3

6

4

6 8�9

11

3

7

175

Shortest Path Problem

Given: digraph D = (V ,A), node r 2 V , arc costs ca, a 2 A.

Task: for each v 2 V , find dipath from r to v of least cost (if one exists)

Remarks:

I Existence of r -v -dipath can be checked, e. g., by breadth-first search.

I Ensure existence of r -v -dipaths: add arcs (r , v) of suffic. large cost.

Basic idea behind all algorithms for solving shortest path problem:

If yv , v 2 V , is the least cost of a dipath from r to v , then

yv + c(v ,w) � yw for all (v ,w) 2 A. (6.3)

Remarks:

I More generally, subpaths of shortest paths are shortest paths!

I If there is a shortest r -v -dipath for all v 2 V , then there is a shortest path tree,

i. e., a directed spanning tree T rooted at r such that the unique r -v -dipath in T is

a least-cost r -v -dipath in D.

176

Shortest Path Problem

Given: digraph D = (V ,A), node r 2 V , arc costs ca, a 2 A.

Task: for each v 2 V , find dipath from r to v of least cost (if one exists)

Remarks:

I Existence of r -v -dipath can be checked, e. g., by breadth-first search.

I Ensure existence of r -v -dipaths: add arcs (r , v) of suffic. large cost.

Basic idea behind all algorithms for solving shortest path problem:

If yv , v 2 V , is the least cost of a dipath from r to v , then

yv + c(v ,w) � yw for all (v ,w) 2 A. (6.3)

Remarks:

I More generally, subpaths of shortest paths are shortest paths!

I If there is a shortest r -v -dipath for all v 2 V , then there is a shortest path tree,

i. e., a directed spanning tree T rooted at r such that the unique r -v -dipath in T is

a least-cost r -v -dipath in D.

176

Shortest Path Problem

Given: digraph D = (V ,A), node r 2 V , arc costs ca, a 2 A.

Task: for each v 2 V , find dipath from r to v of least cost (if one exists)

Remarks:

I Existence of r -v -dipath can be checked, e. g., by breadth-first search.

I Ensure existence of r -v -dipaths: add arcs (r , v) of suffic. large cost.

Basic idea behind all algorithms for solving shortest path problem:

If yv , v 2 V , is the least cost of a dipath from r to v , then

yv + c(v ,w) � yw for all (v ,w) 2 A. (6.3)

Remarks:

I More generally, subpaths of shortest paths are shortest paths!

I If there is a shortest r -v -dipath for all v 2 V , then there is a shortest path tree,

i. e., a directed spanning tree T rooted at r such that the unique r -v -dipath in T is

a least-cost r -v -dipath in D.

176

M-

yaoi ↳ triangular inequalityros fccu.nl

¥70 w