Derandomized DP

1. 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

2. 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

3. 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)

4. 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

5. 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

6. 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

7. Section 4 The proof of Theorem 1.2

8. 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)

9. 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, D1and D2,and(2)the subspace E is linearly independent from D1and D2and (3) E is a d0-dimensional subspace (as A0)

10. 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 

11. 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

12. 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) 

13. 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:

14. 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

15. Excellence implies Local Agreement Subsection 4.2

16. 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:

17. 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

18. 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

19. Sampling Procedure • For every d0-dimensional subspace A0, thus A0 U. The set of all the vectors orthogonal to A0is 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:

20. 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:

21. 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

22. 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

23. Sampling Procedure • Claim 4.5 - |Cons|/|B| = |Cons|/|B| • Proof:

24. 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

25. 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)

26. Sampling Procedure • Claim 4.7 - |ConsE|/|BE| = |(ConsE)|/|(BE)| • Proof: Pretty much the same as the others… so I’ll skip it 

27. Sampling Procedure • Say a pair (A0,B0) is good for a subspace B0linearly 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

28. 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

29. 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 and since also good (previews slide) then by assuming we will get a contradiction

30. 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 Bxand toConsxas (Consx)and proving it to at most fractions

31. 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:

32. 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 xA0+E such as that Consxis large.

33. 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…

34. 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 xA0+E, we pick E – a random d0-dimensinal linear subspace orthogonal to A0

35. Proof of lemma 4.4 • Proof:

36. Local Agreement implies Global Agreement Subsection 4.3 – The proof of theorem 1.2

37. 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: