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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 k q the input size is exponential in d d 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 U Fqm The function stays the same: f : U R For example R could be R 0,1 for a Booleanfunction Derandomized DP The k-wise DP of a function f is: f Where k is all the points in a subspace A A is a d-dimensional linear subspace of U k A Thus, we have k q d 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) Derandomized Z-Test The operator “+” is: A B a b a A b B 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 d 25 and d0 d and k q d 25 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 C . A0 B0 A C A0 B1 A , else continue 0 0 Pick a random d0-dimensional subspace A1 of U that is linearly independent from B1. Reject if . A0 B1 B1 C A1 B1 B1 , else accept C 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 1 1 k Then there is a function g : U R such that, for each of at least 4 fraction of d-dimensional subspaces S from U, the oracle value C(S) agrees with the direct product g k S for all but at most k 1 fraction of elements in S 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) C A0 , B0 A0 C A0 , B A0 Where the subspace A0 is linearly independent from B and B0 The set Cons A0 , B0 (defined as before) – its members are all the B’s that are consistent with a pair (A0,B0) Good/Excellent pair Pair (A0,B0) is good The size of the set Cons A0 , B0 , m d0 2 d d 0 is at least ;i.e. has a measure at least 2 Pair (A0,B0) is (,γ)-excellent The pair (A0,B0) is good and E D Cons i PrE , D1 , D2 C A0 E D1 A0 , B0 , i 1, 2 A0 E C A0 E D2 A0 E 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) Excellence - Lemma 4.1 Lemma 4.1 (analogues of Lemma 3.2) – Assume that PrA , B , B C A0 B0 A C A0 B1 A Then a random pair (A0,B0) is good with probability at least 2 0 0 1 0 0 Proof: Nothing really changed from the proof of Lemma 3.2 Excellence - Lemma 4.2 Lemma 4.2 (analogues of Lemma 3.3) – PrA0 , B0 A0 , B0 is good but not excellent ' 2 ' O q a k ' where Proof: (1) Let the event 1 A0 , B0 be the event good but not excellent; i.e. the pair (A0,B0) is good and E Di Cons A , B , i 1, 2 PrE , D1 , D2 C A0 E D1 0 0 ' A0 E C A0 E D2 A0 E Excellence - Lemma 4.2 Proof – continue: (2) Let the event 2 A0 , B0 , E, D1, D2 (A0,B0) is good and E Di ConsA , B , i 1, 2 and C A0 E D1 ' A0 E 0 0 C A0 E D2 A0 E (3) By using Lemma 2.2 we get Pr 2 ' (4) According to (2) + (1) Pr 2 1 Excellence - Lemma 4.2 Proof – continue: (5) According to (3) - Pr 2 ' then it is clear that Pr 2 1 Pr 2 ' (6) Since Pr 2 1 Pr 2 1 Pr 1 , then according to (4) + (5) we’ll get: Pr 2 1 Pr 2 1 Pr 1 ' Pr 1 ' Pr 1 Pr 1 ' Excellence - Corollary 4.3 Corollary 4.3 (analogues of Corollary 3.4) – Assume that PrA , B , B C A0 B0 A C A0 B1 A Then we have PrA , B A0 , B0 is , excellent A0 , B0 is good 1 2 where and γ are such that k ' poly 1 3 0 0 0 1 0 0 0 Proof: By using Lemma 4.2 and where c0 1 2q 2 Excellence implies Local Agreement Subsection 4.2 Subsection 4.2 – Local Agreement In this section we set: (for some fixed , excellent A , B ) 0 0 Cons to be Cons A , B 0 0 A0 and E are d0-dimensional subspaces D1 and D2 are (d-2d0)-dimensional subspaces The function g is now: C A0 B0 x g x PluralityBCons:x A0 B 0 x A0 x U A0 B Cons x A0 B else Local Agreement - Lemma 4.4 Lemma 4.4 (analogues of Lemma 3.5) – 2 v O fraction of There are fewer than .B Cons such as that C A0 B0 x g x for more than 40 fraction of x A0 B Proof by contradiction: By assuming that PrBCons C A0 B g A0 B v Local Agreement - Lemma 4.4 Proof – continue: Since A0 , B0 is excellent and therefore is good then we can assume now that also PrB B cons C A0 B g A0 B v ' v 2 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 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 A0 Denote B A0 a subspace orthogonal to A0 Every subspace B has an orthogonal basis B’ where .span B ' span B and thus we may define Si as the equivalence class: Si B B A0 span Bi' 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 span B span B ' span B Therefore – A0 B A0 B For those subspaces B, we may define Ti as the equivalence class: Ti B span B span B B Si 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 x A0 B Defining Consx Cons Bx Claim 4.6 - |Consx|/|Bx| = |(Consx)|/|(Bx)| Proof: Proved for every x U \ A0 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 .A0 B0 A0 B , 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 Sampling Procedure For an , excellent A0 , B pair, the measure of Cons stays the same, and by using claim 4.7 the probability in the equation E Di Cons A , B , i 1, 2 PrE , D1 , D2 C A0 E D1 0 0 ' A0 E C A0 E D2 A0 E The definition of function g(x) stays the same by using claim 4.6 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, assuming PrBCons C A0 B g A0 B v and since also good (previews slide) then by assuming PrB B cons C A0 B g A0 B v ' v 2 we will get a contradiction 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 1 k 1 5 fraction of the input x U \ A0, we have |Consx|/|Bx| 6 Proof: Referring to (Bx) as Bx and to Consx as (Consx) and proving it to at most 1 k 0.20 fractions 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 O 1 k 1 4 fraction of linear subspace E orthogonal to A0 as such that x A0 E, we get that PrBConsE C A0 B x g x 1 10 Proof: Proof of lemma 4.4 Claim 4.10 – For O 1 k 1 5 , PrE , x A0 E \ A0 PrBConsE C A0 B x g x 1 1 10 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. Proof of lemma 4.4 Claim 4.11 – For ' O 1 k 3 25 , x A0 E \ A0 1 ' PrE PrBConsE C A0 B x g x 110 where E is a random d0-dimensinal linear subspace orthogonal to A0 Proof: If we have enough time… Proof of lemma 4.4 Claim 4.12 – For all but at most 1 k 1 4 fraction of d0-dimensinal linear subspaces E, orthogonal to A0, we have |ConsE|/|BE| 6 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 Proof of lemma 4.4 Proof: Local Agreement implies Global Agreement Subsection 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 x S Proof: