3 Composition Theorem

10:8 A Composition Theorem for Randomized Query Complexity

≥ (1/2−δ/2)·P

y∈Cµ(y)

1/2 +δ/2 (From Equations (1) and (2)

= 1/2−δ/2 1/2 +δ/2 ·Pr

µ[C]

Thus,

Prµ[C]≤ 1/2 +δ/2 1/2−δ/2 ·Pr

µ_b[C]≤(1 + 4δ)·Pr

µ_b[C]. (since δ≤ ¹₂) J

A. Anshu et al. 10:9

Algorithm 1:Randomized query algorithmA⁰ forf Input: z∈ {0,1}ⁿ

1 Initializev ←root of the decision treeB,Q← ∅

2 while v is not a leaf do

3 Let a bit inx⁽ⁱ⁾be queried at v

4 if i6∈Qthen /* codim(C⁽ⁱ⁾v )<D^µ(g) if this is satisfied */

5 Setv←v_bwith probability Prµ[Cv⁽ⁱ⁾_b | C⁽ⁱ⁾v ]

6 if codim(C⁽ⁱ⁾v ) =D^µ(g)−1then

7 Queryzi

8 SetQ←Q∪ {i}

9 else

10 if Prµ_zi[Cv⁽ⁱ⁾] = 0 then

11 Output 0

12 else

13 Setv←vbwith probability Prµ_zi[Cv⁽ⁱ⁾b | Cv⁽ⁱ⁾]

14 Output label of v

From the definition of bias one can verify that the event in step 5 in Algorithm 1 that is being conditioned on, has non-zero probabilities under the respective distribution; hence, the probabilistic process is well-defined.

From the description ofA⁰it is immediate thatzi is queried only if the underlying simulation ofBqueries at leastR(g) locations inx⁽ⁱ⁾. Thus the worst-case query complexity ofA⁰ is at mostR_1/3(f◦gⁿ)/R(g).

We are left with the task of bounding the error of A⁰. Let L be the set of leaves of the decision treeB. Each leaf`∈ Lis labelled withb<sub>`</sub>∈ R; whenever the computation reaches`, b` is output.

For a vertexv, let the corresponding subcubeCvbeCv⁽¹⁾×. . .× C⁽ⁿ⁾v , whereCv⁽ⁱ⁾ is a subcube of the domain of thei-th copy ofg (corresponding to the inputx⁽ⁱ⁾). Recall from Section 2 that forb∈ {0,1},v_b denotes theb-th child ofv.

For each leaf`∈ Landi= 1, . . . , n, definesnip<sup>(i)</sup>(`) to be 1 if there is a nodetin the unique path from the root of B to ` such that codim(C<sub>t</sub><sup>(i)</sup>) <D<sup>µ</sup>(g) and bias<sup>µ</sup>(C<sub>t</sub><sup>(i)</sup>) ≥ <sub>n</sub><sup>2</sup>2, and 0 otherwise. Definesnip(`) :=∨ⁿ_i=1snip⁽ⁱ⁾(`).

For each ` ∈ L, define p<sup>z</sup><sub>`</sub> to be the probability that for an input drawn from γ^z, the computation ofB terminates at leaf`. We have,

x∼γPr^z[(x,B(x))∈f ◦gⁿ] = Pr

x∼γ^z[(z,B(x))∈f] = X

`∈L:(z,b<sub>`</sub>)∈f

p^z_`. (3)

From our assumption aboutBwe also have that,

x∼γPr[(x,B(x))∈f◦gⁿ] = E

z∼λ Pr

x∼γ^z[(x,B(x))∈f◦gⁿ]≥ 2

3. (4)

Now, consider a run ofA⁰ onz. For each `∈ L ofB, defineq<sup>z</sup><sub>`</sub> to be the probability that the computation ofA⁰ onzterminates at leaf`ofB. Note that the probability is over the internal randomness ofA⁰.

F S T T C S 2 0 1 7

10:10 A Composition Theorem for Randomized Query Complexity

To finish the proof, we need the following two claims. The first one states that the leaves

`∈ Lare sampled with similar probabilities byBandA⁰.

IClaim 10. For each`∈ Lsuch thatsnip(`) = 0, and for eachz∈ {0,1}ⁿ, ⁸₉·p^z_{`</sub> ≤q<sup>z</sup><sub>`} ≤ ¹⁰₉·p^z<sub>`</sub>. The next Claim states that for each z, the probability according toγ<sup>z</sup> of the leaves` for whichsnip(`) = 1 is small.

IClaim 11.

∀z∈ {0,1}ⁿ, X

`∈L,snip(`)=1

p^z_` ≤ 4 n.

We first finish the proof of Theorem 1 assuming Claims 10 and 11, and then prove the claims.

For a fixed inputz∈ {0,1}ⁿ, the probability thatA⁰, when run onz, outputs anrsuch that (z, r)∈f, is at least

X

`∈L, (z,b`)∈f,snip(`)=0

q^z_` ≥ X

`∈L, (z,b`)∈f,snip(`)=0

8

9 ·p^z_` (By Claim 10)

= 8 9





 X

`∈L, (z,b`)∈f

p^z_`− X

`∈L, (z,b`)∈f,snip(`)=1

p^z_`







≥ 8 9





 X

`∈L, (z,b`)∈f

p^z_`− 4 n





. (By Claim 11) (5)

Thus, the success probability ofA⁰ is at least

z∼λE

X

`∈L, (z,b`)∈f,snip(`)=0

q_`^z≥8 9 ·





 E

z∼λ

X

`∈L, (z,b`)∈f

p^z_`− 4 n





 (By Equation (5))

≥8 9 ·

2 3 −4

n

(By Equations (3) and (4))

≥5

9. (For large enoughn) We now give the proofs of Claims 10 and 11.

Proof of Claim 10. We will prove the first inequality. The proof of the second inequality is similar¹.

Fix az∈ {0,1}ⁿ and a leaf`∈ L such thatsnip(`) = 0. For eachi= 1, . . . , n, assume that codim(C<sub>`</sub><sup>(i)</sup>) =d<sup>(i)</sup>, and in the path from the root ofBto`the variablesx⁽ⁱ⁾₁ , . . . , x⁽ⁱ⁾_d(i) are set to bitsb₁, . . . , b_d(i) in this order.

We first observe that ifv is a vertex ofBfor which the condition in step 10 ofA⁰ is satisfied, then co-dimension ofCv⁽ⁱ⁾ is at mostD^µ(g)−1, andbias(Cv⁽ⁱ⁾) = 1; hence each leaf`<sup>0</sup> of B reachable fromv hassnip(`⁰) = 1.

1 Note that only the first inequality is used in the proof of Theorem 1.

A. Anshu et al. 10:11

The computation ofA⁰ terminates at leaf`if the values of the different bitsx<sup>(i)</sup><sub>j</sub> sampled by A<sup>0</sup> agree with the leaf`. The probability of that happening is given by

q_`^z=

n

Y

i=1

PrA⁰[x⁽ⁱ⁾₁ =b1, . . . , x⁽ⁱ⁾_d(i) =b_d(i) |z] (6)

=

n

Y

i=1

x∼µPr[x⁽ⁱ⁾₁ =b1, . . . , x⁽ⁱ⁾_Dµ

(g)−1=b_D^µ_(g)−1]·

x∼µPr_zi[x⁽ⁱ⁾_Dµ

(g)=b_D^µ

(g), . . . , x⁽ⁱ⁾_d(i)=b_d(i) |x⁽ⁱ⁾₁ =b₁, . . . , x⁽ⁱ⁾_Dµ

(g)−1=b_D^µ

(g)−1]. (7) The second equality above follows from the observation that in Algorithm 1, the firstD^µ(g)−1 bits ofx⁽ⁱ⁾are sampled from their marginal distributions with respect toµ, and the subsequent bits are sampled from their marginal distributions with respect to µ_zi. In equation (7), the term Pr_x∼µzi[x⁽ⁱ⁾_Dµ

(g) = b_D^µ_(g), . . . , x⁽ⁱ⁾_d(i) = b_d(i) | x⁽ⁱ⁾₁ = b1, . . . , x⁽ⁱ⁾_Dµ

(g)−1 = b_D^µ_(g)−1] is interpreted as 1 ifd⁽ⁱ⁾<D^µ(g).

We invoke Claim 9(b) with C set to the subcube{x∈ {0,1}^m: x⁽ⁱ⁾₁ =b1, . . . , x⁽ⁱ⁾_Dµ (g)−1= b_D^µ_(g)−1} and δ set to _n²2. To see that the claim is applicable here, note that from the assumptionsnip(`) = 0 we have thatbias(C)< δ= _n²₂ < ¹₂, where the last inequality holds for large enoughn. Also, sinceD^µ(g)>0, by Proposition 6 the bias of{0,1}^m is at most

2

n⁴ < _n²₂ =δ. Continuing from Equation (7), by invoking Claim 9(b) we have, q_`^z≥

n

Y

i=1

(1−8/n²) Pr

x∼µ_zi[x⁽ⁱ⁾₁ =b1, . . . , x⁽ⁱ⁾_Dµ

(g)−1=b_D^µ_(g)−1]·

x∼µPr_zi[x⁽ⁱ⁾_Dµ

(g)=b_D^µ

(g), . . . , x⁽ⁱ⁾_d(i)=b_d(i)|x⁽ⁱ⁾₁ =b₁, . . . , x⁽ⁱ⁾_Dµ

(g)−1=b_D^µ

(g)−1]

= (1−8/n²)ⁿ

n

Y

i=1

x∼µPr_zi[x⁽ⁱ⁾₁ =b₁, . . . , x⁽ⁱ⁾_d(i) =b_d(i)]

≥ 8

9 ·p^z_`. (For large enoughn) J

Proof of Claim 11. Fix az∈ {0,1}ⁿ. We shall prove that for eachi, P

`∈L,snip<sup>(i)</sup>(`)=1p^z_` ≤

4

n². That will prove the claim, sinceP

`∈L,snip(`)=1p^z_` ≤Pn i=1

P

`∈L,snip<sup>(i)</sup>(`)=1p^z_`.

To this end, fix ani∈ {1, . . . , n}. For a randomxdrawn fromγ^z, letpbe the probability that in strictly less than D^µ(g) queries the computation of B reaches a node t such that bias(C_t⁽ⁱ⁾) is at least _n²₂. Note that this probability is over the choice of the differentx^(j)’s.

We shall show thatp≤ _n⁴2. This is equivalent to showing thatP

`∈L,snip<sup>(i)</sup>(`)=1p^z_` ≤ _n⁴2. Note that eachx^(j)is independently distributed according toµ_zj. By averaging, there exists a choice ofx^(j) for eachj6=isuch that for a randomx⁽ⁱ⁾chosen according toµz_i, a nodet as above is reached within at mostD^µ(g)−1 steps with probability at leastp. Fix such a setting for eachx^(j), j 6=i. Claim 11 follows from Claim 8 (note that = ¹₂−_n¹4 ≥ ¹₄ for

large enoughn). J

This completes the proof of Theorem 1. J

3.1 Hardness Amplification Using XOR Lemma

In this section we prove Theorem 2.

F S T T C S 2 0 1 7

10:12 A Composition Theorem for Randomized Query Complexity

Theorem 1 is useful only when the functiong is hard against randomized query algorithms even for error 1/2−1/n⁴. In this section we use an XOR lemma to show a procedure that, given anyg that is hard against randomized query algorithms with error 1/3, obtains another function on a slightly larger domain that is hard against randomized query algorithms with error 1/2−1/n⁴. This yields the proof of Theorem 2.

Letg :{0,1}^m→ {0,1} be a function. Letg_t^⊕ : ({0,1}^m)^t → {0,1} be defined as follows.

Forx= (x⁽¹⁾, . . . , x^(t))∈({0,1}^m)^t, g_t^⊕(x) =⊕^t_i=1g(x⁽ⁱ⁾).

The following theorem is obtained by specializing Theorem 3 of Andrew Drucker’s paper [5]

to this setting.

ITheorem 12(Drucker 2011 [5] Theorem3).

R_1/2−2−Ω(t)(g_t^⊕) = Ω(t·R_1/3(g)).

Theorem 2 (restated below) follows by settingt= Θ(logn) and combining Theorem 12 with Theorem 1.

ITheorem 2. Letf ⊆ {0,1}ⁿ× Rbe any relation. Let g:{0,1}^m→ {0,1} be a function.

Let g^⊕_t : ({0,1}^m)^t → {0,1} be defined as follows: for x = (x⁽¹⁾, . . . , x^(t)) ∈ ({0,1}^m)^t, g^⊕_t(x) =⊕^t_i=1g(x⁽ⁱ⁾). Then,

R_1/3 f◦

g^⊕_O(logn)n

= Ω(logn·R_4/9(f)·R_1/3(g)).

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Verification of Asynchronous Programs with

No documento 37th IARCS Annual Conference on Foundations of Software (páginas 109-115)