1. Hardness of Approximation
Complexity ©D.Moshkovits

2. 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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3. Promise Problems Sometimes you can promise something about the input
It doesn’t matter what you say for inputs that do not satisfy the promise
I know my graph has clique of size n/4! Does it have a clique of size n/2?
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4. Promise Problems & Approximation
We’ll see promise problems of a certain type, called gap problems, can be utilized to prove hardness of approximation.
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5. Optimization Problems
Consider an optimization problem P:
Example:
all graphs
instances: x1,x2,x3,…
all cliques in that graph
feasible solutions
the clique’s size (max)
optimization measure
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6. Each Instance Has an Optimal Solutionx4 x1 x2 x3
OPT
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7. Approximation (Max Version)
OPT
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8. How To Show Hardness of Approximation?
Hardness of distinguishing far off instances  Hardness of approximation A B xi
OPT
gap
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9. Gap Problems (Max Version)
Instance: …
Problem: to distinguish between the following two cases:
The maximal solution  B
The maximal solution ≤ A
YES NO
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10. Formally: Claim: If the [A,B]-gap version of a problem is NP-hard,
then that problem is NP-hard to approximate to within factor B/A.
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11. Formally:
Proof (for maximization): Suppose there is an approximation algorithm that outputs C≤C* so that C*/C ≤ B/A
A proper distinguisher:
* If CA, return ‘YES’
* Otherwise return ‘NO’
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12. Proof
Since C ≥ C*·A/B,
(1) If C* > B (the correct answer is ‘YES’), then necessarily C ≥ C*·A/B > B·A/B = A (we answer ‘YES’)
(2) If C*≤A (the correct answer is ‘NO’), then necessarily C≤C*≤A (we answer ‘NO’).
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13. Translating To Decision Problems To Prove Hardness
Optimization Problems
Approximation Problems
Threshold Problems
Gap Problems
Is the size of the max clique > ½n?
Is the size of the max clique > ¾n or < ¼n?
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14. 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 gap-preserving
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15. Gap-3SAT[]
Instance: a set of 3-clauses {c1,…,cm} over variables v1,…,vn.
Problem: to distinguish between the following two cases:
There exists an assignment that
satisfies all clauses.
No assignment can satisfy more
than 7/8+ of the clauses.
YES NO
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16. Gap-3SAT: Example ( x1  x2  x3 ) ( x1  x2  x2 )
 = { x1  F ;
x2 T ;
x3 F }
satisfies 5/6 of the clauses
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17. Why 7/8?
Claim: For any set of clauses with exactly three independent literals,
there always exists an assignment that satisfies at least 7/8 clauses.
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18. The Probabilistic Method
Proof: Consider a random assignment. x1 x2 x3 xn
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19. 1. Find the Expectation
Let Yi be the random variable indicating the outcome of the i-th clause.
For any 1im,   
 F T
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20. 1. Find the Expectation
The number of clauses satisfied is a random variable Y=Yi.
By the linearity of the expectation:
E[Y] = E[ Yi] =  E[Yi] = 7/8m
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21. 2. Conclude Existence
Thus, there exists an assignment which satisfies at least the expected number of clauses. 
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22. This is tight! Gap-3SAT[0] is polynomial time decidable
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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23. Why Is It Called PCP? (Probabilistically Checkable Proofs)
3SAT has a polynomial membership proof checkable in polynomial time.
Prove it!
My formula is satisfiable!
This assignment satisfies it! x1 x2 x3 x4 x5 x6 x7 x8
xn-3
xn-2
xn-1 xn
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24. Why Is It Called PCP? (Probabilistically Checkable Proofs)
…Now our verifier has to check the assignment satisfies all clauses…
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25. Why Is It Called PCP? (Probabilistically Checkable Proofs)
But gap-3SAT also has a polynomial membership proof checkable in polynomial time.
Prove it!
My formula is satisfiable!
This assignment satisfies it! x1 x2 x3 x4 x5 x6 x7 x8
xn-3
xn-2
xn-1 xn
Complexity ©D.Moshkovits

26. Why Is It Called PCP? (Probabilistically Checkable Proofs)
In a NO instance of gap-3SAT, 1/8 of the clauses are not satisfied!
And for gap-3SAT the verifier would be right with high probability, even if he
picks at random a constant number of clauses and
checks only them
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27. Why Is It Called PCP? (Probabilistically Checkable Proofs)
Since gap-3SAT is NP-hard, All NP problems have probabilistically checkable proofs.
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28. Gap Preserving Reductions
YES
YES
don’t care
don’t care NO NO
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29. Hardness of Approximation
Do the reductions we’ve seen also work for the gap versions (i.e approximation preserving)?
We’ll revisit the CLIQUE example.
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30. CLIQUE Construction a vertex for each literal a part for each clause
edge indicates consistency: one is not the negation of the other .
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31. Cliques & Truth-Assignments.
A Clique CV corresponds to the assignment A:V{T,F} s.t C  A()=T.
An edge between two vertices implies the corresponding literals can be both assigned T.
Thus each clique corresponds to a satisfying truth-assignment.
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32. Gap Preservation
If there is an assignment that satisfies all clauses, there is a clique of size m.
If there is a clique of size m (for some 0<<1) there is an assignment that satisfies at least  of the clauses.
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33. Gap-CLIQUE (Ver1) YES NO
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=|V|/3
Every clique is of size at most (7/8+)m
YES NO
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34. Corollary Theorem: for any >0,
CLIQUE is hard to approximate within a factor of 1/(7/8+)
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35. Can We Do Better?
The bigger the gap is, the better the hardness result.
We’ll see an improved result for CLIQUE.
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36. Amplification
Given an instance of the Gap-3SAT problem and a constant k (to be determined later):
A part for every k clauses .
...
edge indicates consistency
vertex for each satisfying assignment to the k clauses
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37. Boolean Assignments Each clause has at most 7 satisfying assignments.
Thus k clauses have at most 7k satisfying assignments.   
 F T
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38. Consistency
Two assignments are inconsistent, if they give the same variable different truth-values.
x y z w
x  y  z F F T
x  w  y F T T
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39. The Graph G=<V, E>
Given  = {C1, …, Cm} over variables y1, …, yn denote Y(C1, …, Ck) the set of variables which appear in C1, …, Ck
Vertices
Edges between every two consistent assignments
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40. Cliques & Assignments
Observation: A clique on  of the parts corresponds to an assignment which satisfies all relevant clauses. .
... .
...
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41. Correctness (1)
If there is a satisfying assignment, then picking the corresponding assignment in each of the parts yields a clique of size
read: “m choose k”
i.e. m!/k!(m-k)!
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42. Observation
Fix an assignment. If 1/8 of the clauses are false, then only (7/8)k of the sets of k clauses are satisfiable.
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43. Correctness (2) For any 0<<1, set k so (7/8+)k < 
If there is a clique with representatives in ≥ of the parts
There is an assignment satisfying ≥ fraction of the k-tuples of clauses
Ruling out the NO case, in which no assignment satisfies more than 1/8- of the clauses. 
Complexity ©D.Moshkovits

44. Gap-CLIQUE (Ver2) YES NO
The following problem is NP-hard for any 0<<1:
Instance: a graph G=(V,E) composed of m independent sets of size r.
Problem: to distinguish between:
There’s a clique of size m = |V|/r
Every clique is of size at most m
YES NO
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45. Corollary
Theorem: MAX-CLIQUE is NP-hard to approximate to within any constant factor.
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46. Chromatic Number Instance: a graph G=(V,E).
Problem: To minimize , so that there exists a function f:V{1,…, }, for which
(u,v)E  f(u)f(v)
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47. Chromatic Number
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48. Observation: Each color class is an independent set
Chromatic Number
Observation: Each color class is an independent set
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49. Clique Cover Number (CCN)
Instance: a graph G=(V,E).
Problem: To minimize , so that there exists a function f:V{1,…, }, for which
(u,v)E  f(u)=f(v)
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50. Clique Cover Number (CCN)
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51. Observation
Claim: The CCN problem on graph G is the CHROMATIC-NUMBER problem of the complement graph Gc.
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52. Reduction Idea . . CLIQUE CCN G G’ m same under cyclic shift
clique preserving q
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53. Correctness Given such transformation: MAX-CLIQUE(G) = m  CCN(G’) = q
to be determined later
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54. T is unique for triplets
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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55. Observations
Such T is unique for pairs and for single vertices as well:
If T(x)+T(u)=T(v)+T(w) (mod q), then {x,u}={v,w}
If T(x)=T(y) (mod q), then x=y
T is unique for:
triplets, pairs and single vertex
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56. Vertices of CCN
Each vertex v Gclique in layer i is mapped to vertex T(v)GCCN in layer i. q
... . . G G’ m
CLIQUE
CCN
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57. Using the Transformation
CLIQUE
GCCN has m layers each of 0….q-1 vertices. vi vj
T(vj)=4
T(vi)=1
CCN
… (q-1)
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58. Edges of CCN (1) (s,t)ECLIQUE  (T(s),T(t))ECCN T(s) T(t)
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59. Edges of CCN (2) Closing edges under cyclic shift
For every (x,y)E, s.t. xlayer_i and ylayer_j
if x’-y’=x-y (mod q), and x’layer_i and y’layer_j
Then (x’,y’)E
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60. Completing the CCN Graph Construction
x=T(s)=3
x’+y’=x+y (mod q) x’
...
...
...
...
Y=T(t)=4 Y’ . . G G’
CLIQUE
CCN
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61. Edge Origin Unique First Observation: This edge comes only from (s,t)
T(s)
First Observation: This edge comes only from (s,t)
T(t)
(x,y)E’ iff  (s,t)E r{0..q-1} s.t x=T(s)+r y=T(b)+r1
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62. Triangle Consistency
Second Observation: A triangle only comes from a triangle.
T(a)=x a x
T(b)=y b y
T(c)=z c z
CLIQUE
CCN
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63. Proof - Triangle Consistencyx z y a c b
claim: <x,y,z> is a triangle in Gclique iff <a,b,c> is a triangle in GCCN.
Proof  ): (trivial- by def. of T)
(x,y)E’ iff  (a1,b1)E r1 [q] s.t x=T(a1)+r y=T(b1)+r1
(y,z)E’ iff  (b2,c2)E r2 [q] s.t y=T(b2)+r z=T(c2)+r2
(z,x)E’ iff  (c3,a3)E r3 [q] s.t z=T(c3)+r x=T(a3)+r3
x+y+z T(a1)+ T(b2)+T(c3)+r1 +r2 +r3 T(a3)+ T(b1)+T(c2)+r3 +r1 +r2
{a1, a2, a3} = {b1, b2, b3} and because {ai} , {bi} , {ci} are from 3 different layers then by the definition of T:
a1= b3 , a2 =b1 , a3 =b2 and r1 = r2 =r3 
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64. Clique Preservation Corollary:
{c1,…,ck} is a clique in the CLIQUE graph
iff
{T(c1)+r,…,T(ck) +r} r{0…q-1} are q different cliques in the CCN graph.
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65. What did we get by now? MAX-CLIQUE(G) = m
By the last corollary
MAX-CLIQUE(G) = m
 CCN(G’) has q distinct cliques of size m  CCN(G’) = q
MAX-CLIQUE(G) < m
 CCN(G’) > q/
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66. What Remains?
It remains to show how to construct the transformation T in polynomial time.
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67. vertices we determined
Greedy Construction
feasible values v6
vertices we determined
forbidden values v2 v1 v5 v3 v4 v1 v2 v6 v1 v6 v4 v6 v5 v6 v2 v6 v3 v6
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68. Greedy Construction - Analysis
At most values are ruled out totally (choose a triple and a couple), so for q=n5 the greedy construction works.
Corollary: There exists a polynomial time algorithm which constructs a triplet unique transformation with q=n5
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69. Corollaries
Theorem: CCN is NP-hard to approximate within any constant factor.
Theorem: CHROMATIC-NUMBER is NP-hard to approximate within any constant factor.
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70. 
Summary
We saw how to show hardness of approximation and explained the concept of gap problems.
We presented the PCP theorem, stating that 3SAT is hard to approximate within some constant factor.
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71. 
Summary
We saw that some of the reductions we know were approximation preserving.
That was the case for the 3SATpCLIQUE reduction.
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72.  Summary However, that reduction gave us a weak result for CLIQUE,
So we showed how to amplify it.
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73.  Summary Then we introduced a new problem, called CHROMATIC-NUMBER.
We reduced gap-CLIQUE to its gap version, showing it was in fact NP-hard to approximate.
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