LECTURE
IIT Kanpur
Computational Complexity Theory
16
Instructor: Raghunath Tewari
Scribe: M.Arunothia
Date: 9th March, 2016
Toda’s Theorem Proof
16.1
Results from Previous Lecture
1. ⊕ · ⊕ · ζ = ⊕ · ζ
2. BP · BP · ζ = BP · ζ
3. ⊕ · BP · ζ ⊆ BP · ⊕ · ζ
where ζ is any class that belongs to the polynomial hierarchy.
16.2
Introduction
In this lecture, we prove T oda0 s T heorem.
P H ⊆ P #SAT
That is, we intend to show that any problem in P H can be solved by a polynomial time
machine that has oracle access to #SAT . In fact, our proof will show that just one #SAT
query is sufficient for solving any problem in P H.
The proof is split into two parts.
1. P H ⊆ BP · ⊕ · ζ
2. BP · ⊕ · ζ ⊆ P #SAT
We achieve the first part by showing that both ∃ and ∀ operator is contained in BP · ⊕,
which would imply any alternation of theirs will also be contained and hence, the whole of
P H would be contained.
For the second part, we device a special function f , that will enhance any parity (mod
2) function to become mod 2p(·) . This enhancement, enables us to convert the problem
in BP · ⊕ · P to a problem of computing the number of accepting paths, which using
parsimonious reductions can then be mapped to a single #SAT query.
16.3
Proving Toda’s Theorem
Lemma 16.1.
BP · ⊕ · ζ = BP · ⊕ · BP · ⊕ · ζ
Proof.
BP · ⊕ · ζ ⊆ BP · ⊕ · BP · ⊕ · ζ
is trivial as (r.h.s) is clearly at least as powerful as the (l.h.s).
BP · ⊕ · BP · ⊕ · ζ ⊆ BP · BP · ⊕ · ⊕ · ζ
using (16.13) from last lecture.
BP · BP · ⊕ · ⊕ · ζ ⊆ BP · ⊕ · ζ
using (16.11) and (16.12) from last lecture.
Lemma 16.2.
∃ · ζ ⊆ BP · ⊕ · ζ
∀ · ζ ⊆ BP · ⊕ · ζ
Proof. It is sufficient to prove the first part as BP operator is closed under compliment,
implying the second part will follow.
Let L ∈ ∃ · ζ
This implies that ∃ a language L0 ∈ ζ such that
x ∈ L ≡ ∃y s.t (x, y) ∈ L0 , therefore by using Valiant - Vazirani’s Theorem we get,
x ∈ L implies
P rh [|{y|(x, y) ∈ L0 ∧ h(y) = 0̄| = 1] ≥
1
8n
x 6∈ L implies
P rh [|{y|(x, y) ∈ L0 ∧ h(y) = 0̄| = 1] = 0
which implies that ∃ L̃ ∈ ζ such that x ∈ L implies
P rh [|{y|(x, y, h) ∈ L̃| = 1] ≥
1
8n
This should work as h being a hash function should be polynomial in |y| which in turn
means it is polynomial in the input size, |x|.
x 6∈ L implies
P rh [|{y|(x, y, h) ∈ L̃| = 1] = 0
Notice that we are dealing with the parity operator and hence, the cardinality being 1 is
not as important as it being odd.
We perform repeated trials to boost this probability. (i.e)., consider 2m + 1 trials of this
16-2
experiment where m is polynomially bound. Now take union (∪) of all these trials. The
bound on the probability of getting the cardinality as odd will be arrived using the inclusionexclusion principle. Ti below denotes the cardinality of trial i
X
X
Ti = odd −
Ti ∩ Tj = odd + ..
i
≥
i,j
X
Ti = odd −
i
X
Ti ∩ Tj = odd
i,j
2m+1
2m + 1
2
−
≥
8n
64n2
which can be made 23 by choosing m appropriately.
Hence, this finishes our proof as
x ∈ ∃ · L0 , if and only if x ∈ BP · ⊕ · L̃. Hence,
we prove (16.5) and (16.6).
Now, as this holds for any alternations, we get the corollary
Corollary 16.3.
P H ⊆ BP · ⊕ · ζ
Lemma 16.4.
BP · ⊕ · ζ ⊆ P #SAT
For a NDTM M on an input x, let
χM (x) = # of accepting paths of M on x.
For NDTMs M1 and M2 for a language L, ∃ NDTMs MP and MQ such that
χMP = χM1 + χM2
χMQ = χM1 ∗ χM2
MP non-deterministically chooses from M1 , M2 and runs that branch (running M1 and M2
in parallel) while, MQ runs M2 whenever we reach the accept state of M1 (running M1 and
M2 in series.)
Claim 16.5. If L ∈ ⊕ · P , then ∀ polynomial q(·), ∃ a polytime NDTM M̃ such that ∀x,
x ∈ L implies χM̃ (x) ≡ −1 mod 2q(|x|)
x 6∈ L implies χM̃ (x) ≡ 0 mod 2q(|x|)
Since, L ∈ ⊕P , ∃ a NP machine M such that
x ∈ L implies χM (x) ≡ −1 mod 2
x 6∈ L implies χM (x) ≡ 0 mod 2
Now, suppose we find a suitable∗ function f such that
16-3
n ≡ 0 mod 2k implies f (n) ≡ 0 mod 22k
n ≡ −1 mod 2k implies f (n) ≡ −1 mod 22k
By suitable∗ , we mean that the function should be
• A polynomial function in n.
• The resultant NDTM should be easily constructable using the concept of MP and
MQ discussed earlier.
Hence, starting with our NDTM M , in < log(q(n)) constructions of NDTMs in series using
the function f , we can arrive at the desired result.
Consider the function f as follows
f (n) = 3n4 + 4n3
• n ≡ 0 mod 2k implies
f (n) = n ∗ (3n2 + 4n) and n2 ≡ 0 mod 22k and hence,
f (n) ≡ 0 mod 22k .
2
• n ≡ −1 mod 2k implies
2
n + 1 ≡ 0 mod 2k and hence (n + 1) ≡ 0 mod 22k
2
2
and f (n) + 1 = (n + 1) ∗ (3n − 2n + 1) and hence,
f (n) ≡ −1 mod 22k .
Let L ∈ BP · ⊕ · P , then ∃L1 ∈ ⊕P such that
x ∈ L implies P ry [(x, y) ∈ L1 ] ≥
x 6∈ L implies P ry [(x, y) ∈ L1 ] <
2
3
1
3
where |y| = p(|x|) Let,
g(x) = |{y|(x, y) ∈ L1 , |y| = p(|x|)}|
Set q(n) = 3p(n), then from Claim-16.5, we know that ∃ a non-deterministic machine N
such that ∀(x, y),
(x, y) ∈ L1 implies χN (x, y) ≡ −1 mod 2q(n)
(x, y) 6∈ L1 implies χN (x, y) ≡ 0 mod 2q(n)
Consider,
h(x) =
X
χN (x, y)
y:|y|=p(n)
X
=
χN (x, y) +
X
χN (x, y)
y:(x,y)6∈L1
y:(x,y)∈L1
Hence,
h(x) ≡
P
P
(−1) + y:(x,y)6∈L1 (0) mod 2q(n)
h(x) ≡ −g(x) mod 2q(n)
y:(x,y)∈L1
But, by bounding h(x) based on its definition we have
16-4
h(x) ≤ 2p(n) ∗ 22p(n) ≤ 23p(n) ≤ 2q(n)
and hence we get, h(x) = 2q(n) − g(x)
Now, design an NP machine Ñ as follows Input - x
• Guess a string y of length p(|x|).
• Simulate N on (x, y).
Clearly, χÑ (x) = h(x).
We can use the standard parsimonious reduction [1] (that uses the map between number
of accepting paths in an NDTM to the number of satisfying assignments of the constructed
boolean formula) to reduce χÑ (x) to a #SAT instance. Hence, with one single query to
#SAT , we can estimate the value of h(x) and hence that of g(x) = 2q(n) − h(x).
Finally, from definition of BP operator and g(x), we have x ∈ L happening if and only
p(|x|)
if g(x) ≥ 2 2 . This check can be performed in P #SAT with just one oracle query as
discussed above. Hence, we have
BP · ⊕ · ζ ⊆ P #SAT
This, along with corollary-16.3 proves Toda’s Theorem, (i.e),
P H ⊆ P #SAT
References
[1] S. Arora and B. Barak. Computational Complexity – A Modern Approach. Cambridge
University Press, 2009.
16-5