4.2
Maximum Edge Biclique
In this section we present a reduction from the Quasi-random PCP construction given by Theorem 4.1.4 to the maximum edge biclique problem so that in the completeness case the graph has an edge biclique with “large” value, whereas in the soundness case all edge bicliques have “small” value (see Section 4.2.5 for details on the achieved gap). We first present the construction (Section 4.2.1) followed by an important property of the constructed graph (Section 4.2.2). We then present the completeness and soundness analyses (Section 4.2.3 and Section 4.2.4).
Since the reduction and analysis are relatively easy, this section serves as a good starting point before continuing to the more complex reductions (that follow the same general pattern) in Sections 4.3 and 4.4.
4.2.1
Construction
Let N be the proof size and M be the total number of tests of the PCP verifier in Theorem 4.1.4. Both N and M are bounded by 2O(nε), where n is the size of the original SAT formula. Let d be the integer as in Theorem 4.1.4. Select w to beβ−α12
·d
2
(very small), whereβ := (1 − O(1/d))2d1−1 andα := 1
2d +
1
220d are the
bounds given by the completeness and soundness of Theorem 4.1.4.
Construct a bipartite graph G(V, W, E) with |V | = |W| as follows (for an overview of the construction see Figure 4.3). The right-hand side (RHS) consists of N bit-vertices corresponding to the bits in the PCP proof and M test-vertices corresponding to the tests of the PCP verifier. The left-hand side (LHS) consists of N bit-vertices corresponding to the bits in the PCP proof and M slack-vertices to keep the bipartite graph balanced. (The slack-vertices are not adjacent to any vertices and are thus not included in any bipartite clique). Connect a LHS bit-vertex to all RHS bit-vertices except the one corresponding to the same bit of the proof. Furthermore, connect it to a RHS test-vertex if and only if the bit is
not queriedby the test. Finally, assume that wN2 = M. (This can be achieved by simply copying vertices: every bit-vertex is replaced by cN copies of itself, and every test-vertex is replaced by cM copies of it such that now wN/2 = M holds. Copies are connected if and only if the original vertices were. Any maximal biclique must take none or all the copies of a vertex on either side of G.)
The intuition behind the construction is the following. As there are many more bit-vertices than test-vertices, any maximum edge biclique must include approximately half of the bit-vertices of the LHS and the remaining bit-vertices of the RHS (see Section 4.2.2). We then use Theorem 4.1.4 together with the fact that bit-vertices are partitioned into two sets of approximately equal size to
69 4.2 Maximum Edge Biclique
LHS
A LHS bit-vertex is adjacent toRHS
all RHS bit-vertices except the one corresponding to the same bit
A LHS bit-vertex is adjacent to a RHS test-vertex if and only if the bit is not accessed by the test
N bit-vertices N bit-vertices
M slack-vertices M test-vertices
Figure 4.3: An example of the construction. Only the edges incident to the dark gray bit-vertex are depicted.
analyze the completeness and soundness (see Sections 4.2.3 and 4.2.4 respec- tively).
4.2.2
An Optimal Edge Biclique is Quasi-Balanced
Given a biclique let L and R denote respectively the number of bit-vertices of LHS and bit-vertices of RHS that are included in the biclique. Note that in any maximal edge biclique L+ R = N. We say that a biclique is quasi-balanced if |L − R| ≤ β−α6d N.
The following lemma follows in a straightforward manner from the fact that we have many more bit-vertices than test-vertices in our constructed biclique instance.
Lemma 4.2.1 Any optimal edge biclique is quasi-balanced.
Proof. Any balanced biclique of G, i.e., a biclique with L = R = N/2, has value
at least N22, which serves as a lower bound on the optimal solution. Now consider a biclique with L = 1+b2 N and R= 1−b2 N where |b| > β−α6d . Taking all
70 4.2 Maximum Edge Biclique
test-vertices in the biclique gives us the upper bound:
L(R + M) = 1+ b 2 N 1 − b 2 N+ M = 1+ b 2 N 1− b 2 N+ w N 2 = (1 − b2+ bw + w)1 4N 2.
The statement follows by recalling w=β−α12
·d
2
and observing that
1. maximum of f(x) = −x2+ xw + w is achieved when x = w2 <β−α6d ; and 2. f(b) = −b2+ bw + w ≤ −(β−α6d )2+β−α6d (β−α12d)2+ (β−α12d)2< 0.
We have thus that the value of f(b) is always less than 0 when |b| > β−α6d .
4.2.3
Completeness
We will see that there is an edge biclique of size at least (1 + βw)
N 2
2
. (4.1)
This will be achieved by constructing a “balanced” solution, that is a biclique where the bit-vertices are partitioned into two equal sized sets. By Theorem 4.1.4, half the bits in the proof, namely the 1-bits in a correct proof, are such that a fractionβ of tests do not query any of them. Let Γ denote the set of all such tests with|Γ| = β M = β wN2. Now consider the biclique (see also Figure 4.4), where the LHS consists of the bit-vertices corresponding to the 1-bits in the proof and the RHS consists of the remaining bit-vertices (corresponding to the 0-bits in the proof) and the test-vertices corresponding to the tests in Γ. This gives an edge biclique of size N2 ·N2 + β M = N2 ·N2 + βwN2 = (1 + βw) N22.
4.2.4
Soundness
We will see that there is no edge biclique of size 1+α + β 2 w N 2 2 . (4.2)
By Lemma 4.2.1, it is enough to bound the value of quasi-balanced edge bicliques. Consider such a quasi-balanced biclique and let L, R, and T denote
71 4.2 Maximum Edge Biclique LHS RHS N/2 bit-vertices corresponding to 1-bits βM test-vertices corresponding to tests that only query 0-bits
N/2 bit-vertices corresponding to 0-bits
Figure 4.4: The edge biclique in the completeness case. The vertices included in the edge biclique are depicted in dark gray and only the edges incident to those vertices are depicted.
respectively the number of bit-vertices of LHS, bit-vertices of RHS, and test- vertices of RHS that are included in the biclique.
Note that a test-vertex can be included in a biclique only if it is adjacent to all bit-vertices in the LHS of the biclique. In other words, a test-vertex can be included in a biclique only if the corresponding test only queries bits that correspond to bit-vertices included in the RHS of the biclique. The soundness of Theorem 4.1.4 says that, for any given set of a fraction 1/2 + p of the bits, at most a fraction α + p · d of the tests only query those bits. Hence, any edge biclique with L=1−b2 N and R=1+b2 N has T ≤α +|b|2 dM ≤ (α + |b|d)wN2.
Assuming|b| ≤ β−α6d (Lemma 4.2.1), we have the following (rough) bound on the value of any edge biclique of G:
L(R + T) ≤ 1− b 2 N 1+ b 2 N+ (α + |b|d)w N 2 ≤ (1 + (1 + |b|)(α + |b|d)w) N 2 2 ≤ (1 + (α + |b|(2d + α)) w) N 2 2 < 1+α + β 2 w N 2 2 .
72 4.2 Maximum Edge Biclique
The last inequality holds because
α + |b|(2d + α) < α + β
2 ⇔
2|b|(2d + α) < β − α,
which is easily seen to be true by recalling that|b| ≤β−α6d andα < d.
4.2.5
Inapproximability Gap
By using Theorem 4.1.4, we have provided a probabilistic reductionΓ from SAT to maximum edge biclique. For any fixed ε > 0, given an instance φ of SAT of size n,Γ produces an edge biclique instance G in time 2O(nε)satisfying with high probability:
• (Completeness) Ifφ is satisfiable then G has an edge biclique of value (1 + βw)
N 2
2 .
• (Soundness) Ifφ is not satisfiable then all edge bicliques of G have value at most 1+α + β 2 w N 2 2 .
The claimed hardness of approximation result now follows by recalling that (i)
α, β, and w are all functions of parameter d of Theorem 4.1.4, which in turn is
73 4.3 Sparsest Cut