# Derandomizing Majority

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```					                    Derandomizing Majority
Simone Bova
Department of Mathematics and Computer Science
University of Siena, Italy
bova@unisi.it
November 30, 2006

Abstract
We present a research program aimed to derandomize Valiant’s elegant
probabilistic construction of O(n5.3 ) size monotone formulae computing
the majority function of n variables.

1    Introduction
The only deterministic construction of a monotone Boolean circuit comput-
ing the majority function is based on the celebrated sorting network of Ajtai,
o              e
Koml´s and Szemer´di [AKS83]. The size of the monotone Boolean formula,
which can be derived from such circuit, has been estimated [Pat90] to be roughly
O(n5000 ). But, Valiant’s probabilistic construction [Val84] guarantees that a
dramatically smaller formula, of size only O(n5.3 ), indeed exists. During the
last two decades, however, the problem to present explicit uniform monotone
formulae of small size for majority eluded every eﬀort, as emphasized in recent
work on majority circuits [HMP06].
We remark, as an interesting consequence in proof complexity, that a simple
(namely, not using expansion properties of graphs) deterministic polynomial size
construction of a formula computing majority, would improve the simulation of
[AGP01] from super-polynomial to polynomial.

2    Valiant’s Probabilistic Construction
Let x = (x1 , . . . , xn ) ∈ {0, 1}n, xi be the ith component of x and |x| = n xi .
i=1
Then, for all n, M AJn (x) : {0, 1}n → {0, 1} is deﬁned as M AJn (x) = 1 if and
only if |x| ≥ n/2. The monotone Boolean function M AJn is called the majority
function of n variables.
For d ≥ 2 even, let the Valiant formula V ALd be the monotone Boolean
formula of 2d inputs corresponding to the complete binary (rooted) tree of depth
d, with alternating layers of ∨ and ∧ gates from the root to the leaves. The

1
formula V ALd computes a monotone Boolean function of 2d variables in the
natural way.
The following statement [Val84] guarantees the existence of a O(n5.3 ) size
Boolean formula of n inputs, computing the majority function of n variables.
√
Theorem 1. Let x = (x1 , . . . , xn ) ∈ {0, 1}n and ψ = ( 5 − 1)/2. Assume,
without loss of generality, that d = 5.3 log n is even. For j = 1, . . . , 2d , label the
leaf lj of the Valiant formula V ALd with the jth independent execution of the
following statistical experiment:
(i) with probability 2ψ − 1, label lj with the constant 1;
(ii) with complementary probability 1−(2ψ −1), label lj with a variable chosen
uniformly at random from {x1 , . . . , xn }.
Then, the formula obtained computes M AJn with positive probability.
d
Rephrasing, at least one sample point x ∈ Ω = {1, x1 , . . . , xn }2 corresponds
to a sound labeling of the leaves of V ALd , that is, under such labeling V ALd
computes M AJn .
Corollary 2 (Valiant). There exist monotone Boolean formulae for M AJn of
size O(n5.3 ).
On the one hand, the original proof exploits the probabilistic method [AS92],
hence it does not provide any explicit construction of (a leaves labeling for)
a Valiant formula computing the majority function. On the other hand, an
exhaustive search of Ω is clearly unfeasible in the general case, since |Ω| =
d        5.3
(n + 1)2 ≤ 2O(n log n) .
In this context, the challenge of derandomizing Valiant’s construction arises
naturally. The problem is open since 1983.

3      Valiant’s Construction Derandomization
For x, y ∈ {0, 1}n, stipulate that x ≤ y if and only if xi ≤ yi for all i = 1, . . . , n
(notice that ≤ is a partial order). Say that x is maximal for d if V ALd (x) = 0
but, for all x ≤ y = x, V ALd (y) = 1, and x is minimal for d if V ALd (x) = 1
d
but, for all x = y ≤ x, V ALd (y) = 0. Let E1 , E0 ⊆ {0, 1}2 be deﬁned by
d  d

d                              d
E0 = {x : x maximal for d} and E1 = {x : x minimal for d}.
d       d
By the deﬁnition of V ALd , E0 and E1 can be deﬁned inductively on even d ≥ 2.
d  d
Below, |M | is the number of rows of a matrix M , and each x ∈ E0 , E1 corre-
d       d                    d    d                 d      d
sponds to a row (M0 )i , (M1 )i of the matrices M0 , M1 , for 1 ≤ i ≤ |M0 |, |M1 |.
d                    2          4×4
The inductive deﬁnition of E0 is the following. M0 ∈ {0, 1}       is
             
0 1 0 1
2
 0 1 1 0 
M0 =    1 0 0 1 


1 0 1 0

2
d+2                 d 2     d+2
and, for all even d ≥ 2, M0 ∈ {0, 1}4·|M0 | ×2               is
         d                                  d

(M0 )1         1                    M0         1
        .
.            .
.                    .
.         .
.


        .            .                    .         .         

 (M d ) d           1                    M0 d
1         
       0 |M0 |                                                
d                                            d
(M0 )1         1                     1        M0
                                                              
                                                              
        .
.            .
.                    .
.         .
.

.            .                    .         .
                                                              
                                                              
d)                                             d
                                                              
d+2
 (M0 |M d |         1                     1        M0         
M0 =                0
d                     d
.

        1       (M0 )1                   M0         1         

        .
.            .
.                    .
.         .
.


        .            .                    .         .         

d                        d

        1      (M0 )|M0 |
d                M0         1         

d                               d
1       (M0 )1                    1        M0
                                                              
                                                              
        .
.            .
.                    .
.         .
.

.            .                    .         .
                                                              
                                                              
d)
(M0 |M0 |                              d
1               d                 1        M0
2
The inductive deﬁnition of E1 is the following. M1 ∈ {0, 1}2×4 is
d

2         0 0 1 1
M1 =
1 1 0 0
d 2
d+2                        ×2d+2
and, for all even d ≥ 0, M1 ∈ {0, 1}2·|M1 |                  is
                                    d             d

(M1 )1         M1
.
.            .
.
0
                                                         

                                    .            .       

d                d
d+2
                                (M1 )|M1 |
d       M1       
M1     =                                                         .
                                                         
d          d
      (M1 )1      M1                                     
         .
.         .
.


         .         .                        0            

d             d
(M1 )|M1 |
d    M1

d        d                                                   2
Observe that |E0 | and |E1 | are described, respectively, by the recurrences |E0 | =
d+2         d 2        2
4 and, for all d ≥ 2 even, |E0 | = 4 · |E0 | and |E1 | = 2 and, for all d ≥ 2
d+2
even, |E1 | = 2 · |E1 |2 . If d = Ω(log n), then t The solutions of the recurrences
d
d/2                 d/2
are |E0 | = 22(2 −1) and |E1 | = 22 −1 , that is, if d = Ω(log n), |Eo |, |E1 | ≥
d                       d                                        d      d
Ω(n 1/2
)                       d      d                                    2d
2           . Observe also that E0 and E1 are antichains in the poset ({0, 1} , ≤).
d
Deﬁne (E0 )∗ , (E1 )∗ ⊆ {0, 1}2 as follows:
d      d

(E0 )∗ = {x : (∃y ∈ E0 ) x ≤ y ∧ 2d−1 ≤ |x| ≤ |y| },
d                  d

(E1 )∗ = {x : (∃y ∈ E1 ) y ≤ x ∧ |y| ≤ |x| < 2d−1 },
d                  d

and ﬁnally deﬁne E d = (E0 )∗ ∪ (E1 )∗ and S d = E d . Observe that, if d =
d         d
1/2
Ω(log n), then |E d | = |E0 | + |E1 | ≥ 2Ω(n ) .
d       d

3
d
Claim 3. For all x ∈ {0, 1}2 , x ∈ E d if and only if M AJ2d (x) = V ALd (x)
(equivalently, x ∈ S d if and only if M AJ2d (x) = V ALd (x)).
Proof. The claim follows immediately from the deﬁnition of E d , observing that
M AJ2d is monotone and that M AJ2d (x) = 1 if and only if |x| ≥ 2d−1 .
Say that E d excludes some string l = (l1 , . . . , lk ) ∈ {0, 1}k , for some even
k ≤ 2d , if there is a subset I = {i1 , . . . , ik } ⊆ [2d ] of coordinates such that, for
all x ∈ E d , xi = li for at least one i ∈ I (l ∈ E dI ). Say that l is balanced if
/
|l| = k/2.
Claim 4. If E d excludes some balanced string l of length k on coordinates I,
then V ALd , under the labeling corresponding to l on I and to (x1 , . . . , x2d −k )
on [2d ] \ I, computes M AJ2d −k .
d                       d
Proof. If E d excludes l on I, then all the 22 −k strings x ∈ {0, 1}2 corresponding
to l on I are in S, that is, M AJ2d (x) = V ALd (x) on all these x. But then, by
the balancing of l, the formula V ALd under the labeling corresponding to l on
I and to to (x1 , . . . , x2d −k ) on [2d ] \ I is a formula of 2d − k inputs computing
M AJ2d −k .
Experimental work enlightened interesting special cases.
Example 5 (d = 4). We found |E 4 | = 21, 124 and |S 4 | = 44, 412. Also, there
are 5, 759 combinations of balanced binary strings l of length 14 and subsets I ⊆
[16] of size 14 such that E 4 excludes l on I. By easy counting, the probability that
a uniform and independent random choice of l and I induces a sound labeling for
V AL4 is 5, 759/411, 840 ≈ 1.4%. Figures 1 and 2 sample the balanced strings
excluded on coordinates I by E 4 (each row corresponds to an excluded binary
string, black and white boxes correspond respectively to 1 and 0).

Figure 1: I = [16] \ {1, 12}.                  Figure 2: I = [16] \ {5, 9}.

Notice that 4 < 5.3 = 5.3 log 2 (the upper bound of Theorem 1).

4
In the light of Claim 4 and Example 5, we list a series of open problems
equivalent (or related) to the main problem of Valiant’s construction deran-
domization.
A ﬁrst interesting issue is to ﬁnd a combinatorial proof of Valiant’s existential
statement (a promising notion is that of density [Alo83]), instead of the original
analytic argument based on the ampliﬁcation method.
Problem 6 (Combinatorial Proof ). Without loss of generality, let n = 2d
for some d. Show that, for some constant c and some k such that nc −k ≥ Ω(n),
there exist a balanced binary string l of length k and I ⊆ [nc ] of size k such that
l and I induce a sound labeling on the Valiant formula Vc·d .
It would be interesting to tighten the union bound argument applied by
Valiant. For x ∈ Ω chosen uniformly at random, by Theorem 1 we have only that
5.3
Pr[x sound for Vd ] > 0. As a worst case, Pr[x sound for Vd ] = 1/2O(n log n) ,
5.3
and hence a super-exponential number of 2O(n log n) independent random choices
of x ∈ Ω would be necessary to pick a sound point with probability ∼ 1 − e−1
(and actually ≥ 1 − e−1 ). In order to describe a randomized algorithm, we
should improve the bound to Pr[x sound for Vd ] ≥ 1/poly(n). This will allow
the application of derandomization strategies, based on pseudorandom explicit
combinatorial constructions such as expander graphs [Din05, Rei05].
Problem 7 (Randomized Algorithm). Without loss of generality, let n = 2 d
for some d. Show that, for some constant c and some k such that nc −k ≥ Ω(n),
the lower bound on the probability that E c·d excludes l on I is Ω(poly(n)−1 ), for
uniform and independent random choices of balanced binary string l of length k
and I ⊆ [nc ] of size k.
It would be interesting to explore an approximated scenario, where the ac-
curacy (with respect to majority) of the function computed by the underlying
Valiant formula increases asymptotically with the depth and decreases with the
number of inputs. Say that the formula V ALd of n inputs is an -approximator
of M AJn if, for random x ∈ {0, 1}n, Pr[V ALd (x) = M AJn (x)] ≤ .
Problem 8 (Approximator Formulae). Without loss of generality, let n =
2d for some d. Show that, for some constant c and some k such that nc − k ≥
Ω(n), the upper bound on the probability that V ALc·d (x) = M AJnc −k (x) is
= o(c, k), for uniform and independent random choices of l, I (as above) and
c
x ∈ {0, 1}n −k .
Finally, the following problem is equivalent to a uniform explicit construction
of a polynomial size monotone Boolean formula computing the majority function
of n bits.
Problem 9 (Majority Derandomization). Without loss of generality, let
n = 2d for some d. Describe a poly(n) time algorithm, that receives in input n
and returns in output a constant c, a binary string l of length k and I ⊆ [nc ] of
size k, where nc − k ≥ Ω(n), such that l on I induces a sound labeling on the
Valiant formula Vc·d .

5
References
a
[AGP01] Albert Atserias, Nicola Galesi, and Pavel Pudl´k. Monotone simula-
tions of nonmonotone proofs. In IEEE Conference on Computational
Complexity, pages 36–41, 2001.
o           a        o                     e
[AKS83] Mikl´s Ajtai, J´nos Koml´s, and Endre Szemer´di. Sorting in c log n
parallel sets. Combinatorica, 3(1):1–19, 1983.
[Alo83]   Noga Alon. On the density of sets of vectors. Discrete Math., 46:199–
202, 1983.

[AS92]    Noga Alon and Joel H. Spencer. The probabilistic method. Wiley, New
York, 1992.
[Din05]   Irit Dinur. The PCP theorem by gap ampliﬁcation. Electronic Col-
loquium on Computational Complexity (ECCC), (046), 2005.
[HMP06] Shlomo Hoory, Avner Magen, and Toniann Pitassi. Monotone circuits
for the majority function. In APPROX-RANDOM, 2006.

[Pat90]   Mike Paterson. Improved sorting networks with o(log n) depth. Al-
gorithmica, 5(1):65–92, 1990.
[Rei05]   Omer Reingold.   Undirected ST-connectivity in log-space. In
Harold N. Gabow and Ronald Fagin, editors, STOC, pages 376–385.
ACM, 2005.
[Val84]   Leslie G. Valiant. Short monotone formulae for the majority function.
J. Algorithms, 5(3):363–366, 1984.

6

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