The Complexity of the Network Design Problem

Networks, 1978

Classic Paper Reading 99.12

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Outline

• Introduction• NDP is NP-complete• SNDP is NP-complete• Conclusion

Introduction

B96902094 傅莉雯

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• Combinatorial optimization is a topic in

-theoretical computer science -applied mathematics

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• Combinatorial optimization finding the least-cost solution to a mathematical problem in which each solution is associated with a numerical cost.Ex.-finding shortest path-the traveling salesman problem -the minimum spanning tree problem-…

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P

• The class P consists of all the problems that can be solved in polynomial time.– Sorting– Exact string matching– Primes

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P v.s. NP

• P consists of the problems that can be solved in (deterministically) polynomial time.

• NP consists of the problems that can be solved in non-deterministically polynomial time.

Deterministic Non-deterministic

Accept/Reject RejectAccept

Rejectf (n) = O ( n^k )

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P v.s. NP

• P NP⊆• Any problem in NP can be solved in

(deterministically) exponential time. • Could it be the case that any problem in NP

can also be solved in (deterministically) polynomial time?P = NP?

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NP-hard

• A problem is NP-hard if it is at least as hard as all the problems in NP.

• a problem X is NP-hard if the following condition holds:If X can be solved in (deterministic) polynomial time, then all the problems in NP can be solved in (deterministic) polynomial time.

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NP-complete

• A problem is NP-complete if the following holds.– it is NP-hard – it is in NP

• An NP-complete problem is a “hardest” problem in NP.

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NP-complete

• If one proves that a NP-complete broblem can be solved by a polynomial-time algorithm, then NP = P.

• If somebody proves that a NP-complete problem cannot be solved by any polynomial-time algorithm, then NP ≠ P.

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Reduction

• Problem A can be reduced (in polynomial time) to Problem B if the following condition holds:

-we can map input to A in polynomial time to an input to B, and get A’s answer when we get B’s.– if problem B has a polynomial-time algorithm, then so does problem A.

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Network Design Problem

Input: a weighted undirected graph

Output:a subgraph which connects all the

original vertices and minimizes the sum of the shortest path weights between all vertex pairs, subject to a budget constraint on the sum of its edge weights

NDP is NP-complete

R99943153 張允耀 R99943143 楊凱文

NDPNETWORK DESIGN PROBLEM (NDP)：Given an undirected graph EVG , , a weight function EL : , a budget B and a criterion threshold C CB, , does there

exist a subgraph ',' EVG of G with weight

',

,Eji

BjiL and criterion value

CGF ' , where 'GF denotes the sum of the weight of the shortest paths in 'G

between all vertex pairs?

Optimal v.s. Decision

Calculate↓

Optimal value

Calculate↓

Illegal or not?

Optimal Decision

96 100X

90O

95O

97X

96

Knapsack ProblemGiven positive integer baaat t ,,,,, 21 , does there exist a subset tTS ,,1

such that baSi

i

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2KG

NDP is NP-complete

knapsack NDPPolynomial-bounded

knapsack NP-complete NDP

reducible

NDP Knapsack

Given any instance of knapsack, we write

Ti

iaA and define an instance of NDP as follows：

btAC

bAB

TiaiiLiLiL

TiiiiiE

TiiiV

i

4

2

',',0,0

:',,',0,,0

:',0

Figure 1 illustrate this reduction. We claim that knapsack has a solution if and only if EVG ,

contains a subgraph with weight at most B and criterion value at most C. 7,64,53,32,21,4 t:KNAPSACK baaaa

example

1 1’

3’ 3

2

2’

4

4’

0

2

2 2

5

5 5

66

6 3

33

TiiiV :',0

TiiiiiE :',,',0,,0

iaiiLiLiL ',',0,0

7,64,53,32,21,4 t:KNAPSACK baaaa 166532 A3971622 bAB

249716444 btAC

KNAPSACK problem:

t = 4, a1 = 2, a2 = 3, a3 = 5, a4 = 6, b =7. t = 4, a1 = 2, a2 = 3, a3 = 5, a4 = 6, b =7.

Solution:

a1 + a3 = b

KNAPSACK has solutions <=> NDP has solutions??

Sum of weights = 2A = 32

Sum of the shortest path weights between all vertex pairs = 4tA = 256

2 x 2t ( a1 + a2 + a3 + a4 ) = 4tA

B = 2A+b = 39C = 4tA-b = 249

B = 2A+b = 39C = 4tA-b = 249

KNAPSACK problem:

t = 4, a1 = 2, a2 = 3, a3 = 5, a4 = 6, b =4.

We can’t find any solution for this KNAPSACK problem!

Another example:

B = 2A+b = 36C = 4tA-b = 252

This is not a solution.

There are not solutions.

B = 2A+b = 36C = 4tA-b = 252

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SNDP is NP-complete

D99922018 陳琨D98922013 吳彥緯

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Problem Formulation of SNDP

• unit edge weight• spanning tree• SNDPNP

• NDP with L({i, j}) = 1 for all {i, j}E and B = |V| 1

Simple Network Design Problem (SNDP):

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A Known NPC problem: Exact 3-cover

• Given a collection S = {1, 2, …, s} of 3-element subsets of a set T= {1, 2, …, 3t}, does there exist a subcollection S’ S of pairwise

disjoint sets such that• Example: – T = {1,2,3,4,5,6}, S={{1,2,3}, {1,2,4}, {1,2,5}, {1,2,6}}

– T = {1,2,3,4,5,6}, S={{1,2,3}, {3,4,5}, {1,2,5}, {1,2,6}}

S’={{3,4,5},{1,2,6}}

No!

Yes!

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Reduction• Given any instance of EXACT 3-COVER, we

define an instance of SNDP as follows:

= r

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s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s2 s3 s4

t2 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1

t3

s2 s3 s4

t2 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1

t3

s2 s3 s4

t2 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1

t3

s2 s3 s4

t2 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1

t3

Illustration of reduction

S

T

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s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

Illustration of reductionR

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Illustration of reduction

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

S

T

R

polynomial-time reduction

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Claim

has a feasible solution if and only if

G has a feasible solution

EXACT 3-COVER

SNDP

G

Sol. Sol. G

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Sol. GBeginning of the proof

• Let G’ = (V,E’) be a spanning tree of G.• Let FPQ(G’) denote the sum of the weights of

all shortest paths in G’ between vertex sets P and Q (P, Q V).

• Example:

G G’

P

Q

RP

Q

FPQ(G’)=1+1+2+2=6

SNDPG

Sol. G

SNDPG

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Sol. G

Observations• Under the mentioned reduction, any feasible G’

has 3 properties.① G’ contains all edges on 3rd floor② G’ contains all edges on 2nd floor③ G’ contains some edges on 1st floor in a specific way

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0RST

SNDPG

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

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s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

?

2nd Property• If {0, }E’ for some S, then

F(G’) = FRR(G’) + FRS(G’) + FRT(G’)+FSS(G’)+FST(G’)+FTT(G’)

> FRR(G’) + FRS(G’) + FRT(G’)

CRR + CRS + 2(r+1) + CRT

> CRR + CRS + r + CRT

= CRR + CRS + CRT + r

= CRR + CRS + CRT + CSS + CST + CTT

= C

• {0, }E’ for all S– Implication: In G’, each vertex in T is adjacent to exactly one vertex

in S– FRR(G’)=CRR , FRS(G’)=CRS , FSS(G’)=CSS ,

FRT(G’)=CRT , FST(G’)=CST F(G’)<=C iff FTT(G’)=CTT

CRR = r2

CRS = 2rs + sCRT = 9rt + 6tCSS = s2 – sCST = 9st – 6tCTT = 18t2 – 12t

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

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3rd Property• Denote the number of vertices in S being

adjacent in G’ to exactly h vertices in T by Sh (h=0,1,2,3)Example: S0=2, S1=1, S2=1, S3=0

Clearly, FTT(G’) CTT=18t2-12t iff S3=t, S0=s-t, S1=S2=0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r0

t s-t

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Proved

has a feasible solution if and only if

G has a feasible solution

EXACT 3-COVER SNDP

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

s1 s2 s3 s4

t2 t3 t4 t5t1 t6

r2r1 r119 r120‧ ‧ ‧

r0

Conclusion

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It is generally believed that PNP

• This justifies the development of

– Enumerative optimization methods with heuristics

– Approximation algorithms

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Applications• Common Issues of network construction– communication convenience– construction cost

Fire fighting network

MRT network

Sensor network