complexity 1 hardness of approximation. complexity 2 introduction objectives: –to show several...
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![Page 1: Complexity 1 Hardness of Approximation. Complexity 2 Introduction Objectives: –To show several approximation problems are NP-hard Overview: –Reminder:](https://reader030.vdocuments.us/reader030/viewer/2022032522/56649d6b5503460f94a4a1e8/html5/thumbnails/1.jpg)
Complexity1
Hardness of Approximation
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Complexity2
Introduction
• Objectives:– To show several approximation
problems are NP-hard• Overview:
– Reminder: How to show inapproximability?
– Probabilistic Checkable Proofs– Hardness of approximation for clique
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Complexity3
Optimization Problems
Consider an optimization problem P:
instances: x1,x2,x3,…
optimization measure
feasible solutions
all graphs
Example:
all cliques in that graph
the clique’s size (max)
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Complexity4
Each Instance Has an Optimal Solution
OPTx1 x2 x3x4
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Complexity5
Approximation (Max Version)
OPTxi
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Complexity6
How To Show Hardness of Approximation?
Hardness of distinguishing far off instances Hardness of approximation
OPT
A B
gap
xi
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Complexity7
Gap Problems (Max Version)
• Instance: …
• Problem: to distinguish between the following two cases:
The maximal solution B
The maximal solution ≤ A
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Complexity8
Formally:
Claim: If the [A,B]-gap version of a problem
is NP-hard, then that problem is NP-hard to
approximate within factor B/A.
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Complexity9
Formally:
Proof: Suppose there is an approximation algorithm that outputs C so that C/C*≤B/A
A proper distinguisher:* If CB, return ‘YES’* Otherwise return ‘NO’
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Complexity10
Proof
Since C*≥AC/B, (1) If C>B (we answer ‘YES’), then
necessarily C*>A (the correct answer cannot be ‘NO’).
(2) If C*≤A (the correct answer is ‘NO’), then necessarily C≤B (we answer ‘NO’)
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Complexity11
Idea
• We’ve shown “standard” problems are NP-hard by reductions from 3SAT.
• We want to prove gap-problems are NP-hard,
• Why won’t we prove some canonical gap-problem is NP-hard and reduce from it?
• If a reduction reduces one gap-problem to another we refer to it as approximation-preserving
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Complexity12
Gap-3SAT[]
Instance: a set of clauses {c1,…,cm} over variables v1,…,vn.
Problem: to distinguish between the following two cases:
There exists an assignment which
satisfies all clauses.No assignment can satisfy more than 7/8+ of the clauses.
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Complexity13
Gap-3SAT: Example
( x1 x2 x3 )
( x1 x2 x2 )
( x1 x2 x3 )
( x1 x2 x2 )
(x1 x2 x3 )
( x3 x3 x3 )
= { x1 F ; x2 T ; x3 F }satisfies 5/6 of the clauses
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Complexity14
Why 7/8?
Claim: For any set of clauses with exactly three independent literals,
there always exists an assignment which satisfies at least 7/8.
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Complexity15
The Probabilistic Method
Proof: Consider a random assignment.
x1 x2 x3 xn
. . .
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Complexity16
1. Find the Expectation
Let Yi be the random variable indicating the outcome of the i-th clause.
For any 1im, E[Yi]=0·1/8+1·7/8=7/8
E[ Yi] = E[Yi] = 7/8m
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Complexity17
2. Conclude Existence
Expectedly, the number of clauses satisfied is 7/8m.
Thus, there exists an assignment which satisfies at least that many.
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Complexity18
PCP (Without Proof)
Theorem (PCP): For any >0,
Gap-3SAT[] is NP-hard.
This is tight! Gap-3SAT[0] is polynomial time
decidable
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Complexity19
Approximation Preservation
A B
•YES
•don’t care
•NO
• YES
• don’t care
• NO
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Complexity20
Hardness of Approximation
• Do the reductions we’ve seen also work for the gap versions?
• We’ll revisit the CLIQUE example.
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Complexity21
CLIQUE Construction
.
.
.
a part for each
clause
a vertex for each literal
edge indicates
consistency
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Complexity22
Approximation Preservation
• If there is an assignment which satisfies all clauses, there is a clique of size m.
• If there is a clique of size (7/8+)m, there is an assignment which satisfies more than 7/8+ of the clauses.
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Complexity23
Gap-CLIQUE (Ver1)
The following problem is NP-hard for any >0:
Instance: a graph G=(V,E) composed of m independent sets of size 3.
Problem: to distinguish between:
There’s a clique of size m
Every clique is of size at most (7/8+)m
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Complexity24
Corollary
Theorem: for any >0,CLIQUE is hard to approximate
within a factor of 1/(7/8+)
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Complexity25
Amplification
• The bigger the gap is, the better the hardness result.
• We’ll see how a gap can be amplified.
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Complexity26
.
.
.
...
...
Amplification
A part for every k vertices
vertex for each Boolean
assignment
edge indicates
consistency
Given an instance of the Gap-CLIQUE problem and a constant k:
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Complexity27
Boolean assignments
• A Boolean assignment over k vertices {v1,…,vk} is a function A:{v1,…,vk}{0,1}.
• Think about it as if it indicates whether each vertex belongs to the clique.
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Complexity28
Good Assignments
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Complexity29
Consistency
• Two assignments are inconsistent, when they give the same vertex different truth-values.
. . .
n
. . .. . .
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Complexity30
Consistency
• They are also inconsistent, if they both assign 1 to two vertices not connected by an edge.
non-edge
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Complexity31
Correctness
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Complexity32
Chromatic Number
• Instance: a graph G=(V,E).• Problem: To minimize k, so that
there exists a function f:V{1,…,k}, for which
(u,v)E f(u)f(v)
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Complexity33
Chromatic Number
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Complexity34
Chromatic NumberObservation: Each color group is an
independent set
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Complexity35
Clique Cover Number (CCN)
• Instance: a graph G=(V,E).• Problem: To minimize k, so that
there exists a function f:V{1,…,k}, for which
(u,v)E f(u)=f(v)
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Complexity36
Clique Cover Number (CCN)
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Complexity37
Reduction Idea
.
.
.
CLIQUE CCN
.
.
.
q
• cyclic shift-morphic
• clique preserving
m.
.
.
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Complexity38
Correctness
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Complexity39
Transformation
T:V[q]
for any v1,v2,v3,v4,v5,v6,
T(v1)+T(v2)+T(v3) T(v4)+T(v5)+T(v6) (mod q)
{v1,v2,v3}={v4,v5,v6}T is unique for triplets
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Complexity40
Observations
• Such T is unique for pairs and for single vertices as well:
• If T(x)+T(u)=T(v)+T(w), then {x,u}={v,w}
• If T(x)=T(y) (mod q), then x=y
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Complexity41
feasible values
Greedy Constructionv6
v6
v2
v2
v1
v1
v5
v5v
3
v3
v4
v4
vertices we determined
forbidden values
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Complexity42
Greedy Construction - Analysis
At most values are ruled out totally, so for q=n5 the greedy construction works.
Corollary: There exists a polynomial time algorithm which constructs a triplet unique transformation with q=n5
5
n
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Complexity43
Using the Transformation
0 1 2 3 4 … (q-1)
vi
vj
T(vi)=1
T(vj)=4
CLIQUE
CCN
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Complexity44
Completing the CCN Graph Construction
T(s)
T(t)
(s,t)ECLIQUE
(T(s),T(t))ECCN
![Page 45: Complexity 1 Hardness of Approximation. Complexity 2 Introduction Objectives: –To show several approximation problems are NP-hard Overview: –Reminder:](https://reader030.vdocuments.us/reader030/viewer/2022032522/56649d6b5503460f94a4a1e8/html5/thumbnails/45.jpg)
Complexity45
Completing the CCN Graph Construction
T(s)
T(t)
Close the set of edges under shift:
For every (x,y)E,
if x’-y’=x-y (mod q), then (x’,y’)E
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Complexity46
Max Clique of G-clique and G-ccn
• Lemma:Max-Clique(G-clique) = Max-Clique(G-CCN)
• Corollary: – MAX-clique(G-clique) = m CCN(G-ccn)=q– MAX-clqiue(G-clique) < m CCN(G-ccn)>
q
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Complexity47
Edge Origin Unique
T(s)
T(t)
First Observation: This edge comes
only from (s,t)
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Complexity48
Triangle Consistency
Second Observation: A
triangle only come from a triangle
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Complexity49
Clique Preservation
Corollary: {c1,…,ck} is a clique in the CCN graph
iff {T(c1),…,T(ck)} is a clique in the CLIQUE graph.
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Complexity50
Summary
• We’ve seen how to show hardness of approximation results in general,
• and even proven several such using the PCP theorem:– CLIQUE– CHROMATIC NUMBER