Derandomized DP Thus far, the DP-test was over sets of size k

For instance, the Z-Test required three random sets: a set of size k, a set of size k-k’ and a set of size k’

When the input size is exponential in d

dk q

Derandomized DP It is possible to have input size of d and still

get all the results (or almost all) as presented in section 3 (with input size of k)

The domain is The function stays the same: For example R could be for a Boolean-

function

mqU F

:f U R 0,1R

Derandomized DP The k-wise DP of a function f is:

Where k is all the points in a subspace A A is a d-dimensional linear subspace of U

Thus, we have points in A with input size of d Since now, we only need to know the d-vectors that

define the subspace A (and not k items of a set A)

kf A

dk q

Derandomized Z-Test The operator “+” is: A and B are subspace of U, a and b are vectors and

a+b is the component-wise addition of those vectors

The test requires four random subspaces;

d0-dimensional subspaces and (d-d0)-dimensional subspaces

Where

A B a b a A b B

025 and and 25ddd d k q

Derandomized Z-Test The test is (as presented in the paper)

Pick a random d0-dimensional subspace A0 and a random (d-d0)-dimensional subspace B0 of U that is linearly independent from A0

Pick a random (d-d0)-dimensional subspace B1 of U that is linearly independent from A0. Reject if . , else continue

Pick a random d0-dimensional subspace A1 of U that is linearly independent from B1. Reject if . , else accept

0 00 0 0 1A AC A B C A B

1 10 1 1 1B BC A B C A B

Derandomized Z-Test Theorem 1.2: (equivalent to Theorem 1.1 for the

non-derandomized Z-Test)

Suppose the derandomized Z-Test accepts with

probability , for

Then there is a function such that, for each of at least fraction of d-dimensional subspaces S from U, the oracle value C(S) agrees with the direct product for all but at most fraction of elements in S

11

k

:g U R4

kg S 1k

Section 4

The proof of Theorem 1.2

Derandomized Z-Test The proof of this Theorem is on section 4

B consistent with a pair (A0,B0)

Where the subspace A0 is linearly independent from B and B0

The set (defined as before) – its members are

all the B’s that are consistent with a pair (A0,B0)

0 00 0 0, ,A AC A B C A B

0 0,A BCons

Good/Excellent pair Pair (A0,B0) is good The size of the set ,

is at least ;i.e. has a measure at least Pair (A0,B0) is (,γ)-excellent The pair (A0,B0) is

good and

Where (1) the subspace A0 is linearly independent from E, D1 and D2, and (2) the subspace E is linearly independent from D1 and D2 and (3) E is a d0-dimensional subspace (as A0)

0 0,A BCons0

02

m d

d d

2

0 0

1 2

0 0

,

, ,

0 1 0 2

, 1, 2

Pri A B

E D D

A E A E

E D Cons i

C A E D C A E D

Excellence - Lemma 4.1 Lemma 4.1 (analogues of Lemma 3.2) –

Assume that

Then a random pair (A0,B0) is good with probability at least

Proof: Nothing really changed from the proof of Lemma 3.2

0 0 1 0 0, , 0 0 0 1PrA B B A AC A B C A B

2

Excellence - Lemma 4.2 Lemma 4.2 (analogues of Lemma 3.3) –

where

Proof: (1) Let the event be the event good

but not excellent; i.e. the pair (A0,B0) is good and

0 0, 0 0Pr , is good but not excellent 'A B A B

2' 'O q a k

1 0 0,A B

0 0

1 2

0 0

,

', ,

0 1 0 2

, 1, 2

Pri A B

E D D

A E A E

E D Cons i

C A E D C A E D

Excellence - Lemma 4.2 Proof – continue:

(2) Let the event (A0,B0) is good and and

(3) By using Lemma 2.2 we get

(4) According to (2) + (1)

2 0 0 1 2, , , ,A B E D D

0 0, , 1, 2i A BE D Cons i

0 0

'

0 1 0 2A E A EC A E D C A E D

2Pr '

2 1Pr

Excellence - Lemma 4.2 Proof – continue:

(5) According to (3) - then it is clear that

(6) Since , then according to (4) + (5) we’ll get:

2 1 2 1 1Pr Pr Pr

2Pr ' 2 1 2Pr Pr '

2 1 2 1 1 1

1

1

Pr Pr Pr ' Pr

' Pr

Pr '

Excellence - Corollary 4.3 Corollary 4.3 (analogues of Corollary 3.4) –

Assume that

Then we have

where and γ are such that Proof: By using Lemma 4.2 and where

0 0 1 0 0, , 0 0 0 1PrA B B A AC A B C A B

0 0

2, 0 0 0 0Pr , is , , is good 1A B A B excellent A B

1 3'k poly

201

2c

q

Excellence implies Local AgreementSubsection 4.2

Subsection 4.2 – Local Agreement In this section we set: (for some fixed )

Cons to be A0 and E are d0-dimensional subspaces

D1 and D2 are (d-2d0)-dimensional subspaces

The function g is now:

0 0,A BCons

0

0 0 0

0

:0

0

x

B Cons x A B

C A B x A

x U Ag x Plurality

B Cons x A B

else

0 0, ,excellent A B

Local Agreement - Lemma 4.4 Lemma 4.4 (analogues of Lemma 3.5) –

There are fewer than fraction of . such as that for more than fraction of

Proof by contradiction: By assuming that

2v O

B Cons 0 0 xC A B g x 40 0x A B

0 0PrB Cons C A B g A B v

Local Agreement - Lemma 4.4 Proof – continue: Since is excellent and therefore is good

then we can assume now that also

From now on, the sampling procedures are changing, thus instead of taking one subspace to be linearly independent from the other one we will take the subspace to be orthogonal to the other one

0 0,A B

0 0Pr ' 2B B cons C A B g A B v v

Sampling Procedure For every d0-dimensional subspace A0, thus A0 U.

The set of all the vectors orthogonal to A0 is a subspace of U, we denote this set as

Denote a subspace orthogonal to A0

Every subspace B has an orthogonal basis B’ where . and thus we may define Si as the equivalence class:

0B A

0A

'span B span B

'0i iS B B A span B span B

Sampling Procedure For every subspace B linearly independent from A0,

let B denote the dual of A0 inside A0+B Since B is also a subspace it has as well an

orthogonal basis B’, thus Therefore –

For those subspaces B, we may define Ti as the equivalence class:

0 0A B A B 'span B span B span B

i iT B span B span B B S

Sampling Procedure The claims 4.5-4.7 refer to any random event E(B)

which depends on the subspace A0+B rather then B itself

Thus the probability of the this event is the same when- We uniformly choose a subspace B linearly

independent from A0

Or when we uniformly choose a subspace B orthogonal to A0

Sampling Procedure Set B: all the (d-d0)-dimensional subspaces

linearly independent from A0 thus Cons B

Set B: all the (d-d0)-dimensional subspaces orthogonal to A0

Defining Cons= Cons B

Sampling Procedure Claim 4.5 - |Cons|/|B| = |Cons|/|B|

Proof:

Sampling Procedure Set Bx: all the (d-d0)-dimensional subspaces

linearly independent from A0 such as that

Defining Consx Cons Bx

Claim 4.6 - |Consx|/|Bx| = |(Consx)|/|(Bx)|

Proof: Proved for every

0x A B

0\x U A

Sampling Procedure Set BE: all the (d-d0)-dimensional subspaces linearly

independent from A0 and contain the subspace E

Set (BE): all the (d-d0)-dimensional subspaces orthogonal to A0 that contain E

Defining ConsE Cons BE

Defining (ConsE) Cons (BE)

Sampling Procedure Claim 4.7 - |ConsE|/|BE| = |(ConsE)|/|(BE)|

Proof: Pretty much the same as the others… so I’ll skip it

Sampling Procedure Say a pair (A0,B0) is good for a subspace B0 linearly

independent from A0 then Cons has a measure of μ

By using claim 4.5, since there is B such as that . , we will get that Cons has also a measure of μ

Thus, if the pair (A0,B0) is good then also the pair (A0,B) is good

0 0 0A B A B

Sampling Procedure For an pair, the measure of

Cons stays the same, and by using claim 4.7 the probability in the equation

The definition of function g(x) stays the same by using claim 4.6

0, ,excellent A B

0 0

1 2

0 0

,

', ,

0 1 0 2

, 1, 2

Pri A B

E D D

A E A E

E D Cons i

C A E D C A E D

Sampling Procedure Since good, excellent and g(x) remains the same we

may use this method of sampling in order to prove Lemma 4.4

Thus, assumingand since also good (previews slide) then by assuming

we will get a contradiction

0 0PrB Cons C A B g A B v

0 0Pr ' 2B B cons C A B g A B v v

Proof of lemma 4.4 Claims 4.8 – 4.12 are needed for this proof and they

are analogue of Claims 3.8-3.12 respectively

Claim 4.8 – For all but at most fraction of the input , we have |Consx|/|Bx|

Proof: Referring to (Bx) as Bx and to Consx as (Consx) and proving it to at most fractions

1 51 k

0\x U A 6

0.201 k

Proof of lemma 4.4 Claim 4.9 – Let x be any input such as that

Consx has measure at least /6 in Bx. Then for all but at most fraction of linear subspace E orthogonal to A0 as such that

, we get that

Proof:

1 41O k

0x A E

01Pr 10EB Cons xC A B g x

Proof of lemma 4.4 Claim 4.10 – For ,

where E is a random d0-dimensinal linear subspace orthogonal to A0

Proof: By using claim 4.8 and 4.9, for x where xA0+E such as that Consx is large.

1 51O k

0 0 0, \

1Pr Pr 110EB Cons xE x A E A C A B g x

Proof of lemma 4.4 Claim 4.11 – For ,

where E is a random d0-dimensinal linear subspace orthogonal to A0

Proof: If we have enough time…

3 25' 1O k

0 0

0

\Pr 1 '

1Pr 10E

E

B Cons x

x A E A

C A B g x

Proof of lemma 4.4 Claim 4.12 – For all but at most

fraction of d0-dimensinal linear subspaces E, orthogonal to A0, we have |ConsE|/|BE|

Proof: Analogue of Claim 4.8, almost the same proof, but except of choosing x such as xA0+E, we pick E – a random d0-dimensinal linear subspace orthogonal to A0

1 41 k

6

Proof of lemma 4.4 Proof:

Local Agreement implies Global AgreementSubsection 4.3 – The proof of theorem 1.2

Global Agreement - Lemma 4.13 Lemma 4.13 (analogues of Lemma 3.13) – If

the derandomized Z-test accepts with probability at least , then there is a function g:U→R such as that for at least ’= /4 fraction of all subspaces S, the oracle C(S) agrees with g(S) in all but at most ’=81 fraction of points

Proof:

x S