1. 1 Building The Ultimate Consistent Reader

2. 2 Introduction We’ve already built a consistent reader (cube-Vs.-point)... Except it had variables ranging over a set of polynomials instead of over the field. In this lecture we’ll use that construction to build a perfected consistent reader.

3. 3 Starting Point: Cube-Vs.-Point cube var. point var. ? supposedly assigned the restriction of the polynomial to that cube supposedly assigned the value of the polynomial in that point Each cube is actually a new domain!

4. 4 General Framework: Sketch consistent reader “better” consistent reader extension composition repeat this process replacing the cube variables by variables with lower degree adding new readers

5. 5 Extensions We’ll introduce two extension procedures: –Power SubstitutionPower Substitution –Linearization ExtensionLinearization Extension Both embed the domains in new domains with higher dimension, but much lower degree.

6. 6 Extensions All our transformations, but the last one, will use the power substitution extension procedure. The final reader will be created using the linearization extension.

7. 7 Power Substitution - Example (b=3) x 31 + x 15 = 0 (31) 3 =1011 (15) 3 =120 i=0,1,... x i :=x 3 i x 0 x 1 x 3 + x 1 2 x 2 = 0 How many new variables do we need at most? t:=  log 3 (s+1)  Bound the degree: 2t Extend this idea to general b (base) and d (dimension) Now, say the total degree of the polynomial deg satisfies r  deg  s.  deg  r/3 t

8. 8 Embedding Extension dd  dt (x 1,...,x d ) (x 1 b 0,..,x 1 b t,....,x d b t ) Apply this transformation to every cube, where (the parameter s associated with the domain)

9. 9 Composition: How To Build A Consistent Reader For The New Domains? For each local reader (having variables from exactly one domain) generate a Cube-Vs.-Point reader. Replace each occurrence of these variables in the original reader with the proper evaluation. Put the new local test in conjunction with the existing local test. verify this procedure takes polynomial time Make sure you can prove correctness. We only need to read one point! Hence the dimension of the new domains is constant

10. 10 What Next? Repeat this process until When will this happen? Recall that b=(s+1) 1/log 1-  n (as long as this is not less than 2)  t=log 1-  n. s new =dt(b-1)=polylog(n)s 1/log 1-  n Thus in the i’th iteration, s=2 O(log  -i(1-  ) n) (when  -i(1-  )>0). (*)

11. 11 Degree Decreases Rapidly Let i 0 :=  /(1-  ) . 1-  -(i 0 -1)(1-  )>0 Hence in the (i 0 -1)’th iteration, s=2 O(log  -(i 0 -1)(1-  ) n). Then b=O(1). Consequently, t=O(log 1-  n). In the (i 0 +1)’th iteration (*) should hold, since the dimension is constant.

12. 12 Obtaining Linear Polynomials When ( s+ d d ) is small enough, we can apply another technique, called linearization extension, to obtain linear polynomials. Which means our consistent reader relies on constant number of representation variables.

13. 13 Linerization - Example x 2 yz + xy 2 + z = 0 u x 2 yz + u xy 2 + u z = 0 How many new variables do we need at most? u x 2 yz := x 2 yz u xy 2 := xy 2 u z := z Now, say the total degree of the polynomial deg satisfies r  deg  s. The dimension of the polynomial is d. Linear polynomial!

14. 14 Linearization Embedding dd MM (x 1,...,x d ) (m 1 (x 1,..,x d ),..,m M (x 1,..)) Apply this transformation to every cube (where m 1,...,m M are all the degree-s dimension-d monomials):

15. 15 Summary Using the Cube-Vs.-Point consistent reader as a black box, we’ve managed to build an adequate consistent reader: –Each local reader depends on constant number of variables. –All variables range over the field. –The error probability is small.

16. 16 Appendix

17. 17 Linearization: How Many Degree-s, Dimension-d Monomials? Or equivalently - how many partitions of at most s balls into d boxes? +1... s balls d mobile partitions d identical partitions, s identical balls s+d objects arranged in a row