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The Complexity of the Network Design Problem. Networks, 1978 Classic Paper Reading 99.12. Outline. Introduction NDP is NP-complete SNDP is NP-complete Conclusion. Introduction. B96902094 傅莉雯. - PowerPoint PPT Presentation
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The Complexity of the Network Design Problem
Networks, 1978
Classic Paper Reading 99.12
2
Outline
• Introduction• NDP is NP-complete• SNDP is NP-complete• Conclusion
Introduction
B96902094 傅莉雯
4
• Combinatorial optimization is a topic in
-theoretical computer science -applied mathematics
5
• 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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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
