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2 Introduction We are going to use several consistency tests for Consistent Readers. We are going to use several consistency tests for Consistent Readers.Consistent ReadersConsistent Readers

3 Plane Vs. Point Test - Representation Representation: One variable for each plane p of planes(  ), supposedly assigned the restriction of ƒ to p. (Values of the variables rang over all 2-dimensional, degree-r polynomials). One variable for each plane p of planes(  ), supposedly assigned the restriction of ƒ to p. (Values of the variables rang over all 2-dimensional, degree-r polynomials). One variable for each point x  . (Values of the variables rang over the field  ). One variable for each point x  . (Values of the variables rang over the field  ).

4 Plane Vs. Point Test - Test Test : One local-test for every: One local-test for every: plane p and a point x on p. Accept if Accept if A’s value on x, and A’s value on x, and A’s value on p restricted to x are consistent. A’s value on p restricted to x are consistent. Reminder: A : planes  dimension-2 degree-r polynomial

5 Plane Vs. Point Test: Error Probability Claim: The error probability of this test is very small, i.e. <  c’/2, for some known 0<c’<1. The error probability is the fraction * of pairs for a point x and plane p whose: A’s value are consistent, and yet A’s value are consistent, and yet Do not agree with any  -permissible degree-r polynomial (on the planes), Do not agree with any  -permissible degree-r polynomial (on the planes), * fraction from the set of all combination of (point, plane)

6 Plane Vs. Point Test: Error Probability - Proof Proof: By reduction to Plane-Vs.-Plane test : replace every Local-test for p 1 & p 2 that intersect by a line l, Local-test for p 1 & p 2 that intersect by a line l, by a Set of local-tests, one for each point x on l, that compares p 1 ’s & p 2 ’s values on x. Set of local-tests, one for each point x on l, that compares p 1 ’s & p 2 ’s values on x. Let’s denote this test by PPx-Test What is its error-probability?

7 Plane Vs. Point Test: Error Probability - Proof Cont. Proposition: The error-probability of PPx-Test is “almost the same“ as Plane-Vs.-Plane’s. Proof: The test errs in one of two cases: First case: First case: p1 & p2 agree on l, but p1 & p2 agree on l, but Have impermissible values (i.e. they do not represent restrictions of 2  -permissible polynomials). Have impermissible values (i.e. they do not represent restrictions of 2  -permissible polynomials). Second case : Second case : p1 & p2 do not agree on l, but p1 & p2 do not agree on l, but Agree on the (randomly) chosen point x on l. Agree on the (randomly) chosen point x on l.

8 Plane Vs. Point Test: Error Probability - Proof Cont. In the first case Plane-Vs.-Plane also errs, so according to [RaSa], for some constant 0<c<1 Pr(First-Case Error)  c In the first case Plane-Vs.-Plane also errs, so according to [RaSa], for some constant 0<c<1 Pr(First-Case Error)  c For the second case, recall that: For the second case, recall that: r = #points, that two r-degree, 1-dimensional polynomials can agree on. r = #points, that two r-degree, 1-dimensional polynomials can agree on. |  | = #points on the line l. |  | = #points on the line l. So Pr(Second-Case Error)  r/|  |  PPx-Test’s error-probability   c + r/|  |

9 Plane Vs. Point Test: Error Probability - Proof Cont. For an appropriate  (namely:  c  O(r/|  |)) :  c + r/|  | = O(  c ) So, PPx-Test’s error-probability is   c’, for some 0<c’<1

10 Plane Vs. Point Test: Error Probability - Proof Cont. Back to Plane-Vs.-Point: Let p  planes, x  (points on p), such that: Let p  planes, x  (points on p), such that: A(p) and A(x) are impermissible. A(p) and A(x) are impermissible. Let l  lines such that x  l Let l  lines such that x  l Let p1, p2 be planes through l Let p1, p2 be planes through l Plane-Vs.-Point’s error probability is: Pr p, x ( (A(p))(x) = A(x) ) = = Pr l  x, p1 ( (A(p1))(x) = A(x) )

11 Plane Vs. Point Test: Error Probability - Proof Cont. Pr p, x ( (A(p))(x) = A(x) ) = Pr l  x, P1 ( (A(p1))(x) = A(x) ) = * E l  x ( Pr p1 ( (A(p1))(x) = A(x) | x  l ) ) = ** E l  x ( (Pr p1, p2 ( (A(p1))(x) = (A(p2))(x) = A(x) | x  l ) ) 1/2 )  ( E l  x (Pr p1, p2 ( (A(p1))(x) = (A(p2))(x) = A(x) | x  l ) ) 1/2  * ( Pr l  x, p1, p2 ( (A(p1))(x) = (A(p2))(x) = A(x) ) 1/2  *** (  c’ ) 1/2 *  event A, and random variable Y, Pr(A) = E Y ( Pr(A|Y) ) *  event A, and random variable Y, Pr(A) = E Y ( Pr(A|Y) ) ** Pr p1, p2 ( (A(p1))(x) = (A(p2))(x) = A(x) | x  L ) ) = (p1,p2 are independent) (Pr p1 ( (A(p1))(x) = A(x) | x  l ) )* (Pr p1 ( (A(p2))(x) = A(x) | x  l ) ) = (Pr p1 ( (A(p1))(x) = A(x) | x  l ) ) 2 *** PPx-Test

12 Plane Vs. Point Test: Error Probability - Proof Cont. Conclusion: We’ve established that: Plane-Vs.-Point error probability, i.e., The probability that p (which is random) is Assigned an impermissible value, and Assigned an impermissible value, and This value agrees with the value assigned to x (which is also random), This value agrees with the value assigned to x (which is also random), is <  c’/2. Note: This proof is only valid as long as the point x whose value we would like to read is random.

13 Reading an Arbitrary Point Can we have similar procedure that would work for any arbitrary point x? i.e., a set of evaluating functions, where the function returns an impermissible value with only a small (<  c’ ) probability. Such procedure is called: consistent-reader.

14 Consistent Reader for Arbitrary Point Representation: As in Plane-Vs-Point test. Representation: As in Plane-Vs-Point test. local-readers : Instead of local-tests, we have a set of (non Boolean) functions,  [x] = {  1,...,  m }, referred to as: local-readers. local-readers : Instead of local-tests, we have a set of (non Boolean) functions,  [x] = {  1,...,  m }, referred to as: local-readers. A local reader, can either reject or return a value from the field . [supposedly the value is ƒ(x), with ƒ a degree-r polynomial].

15 3-Planes Consistent Reader for a Point x Representation: One variable for each plane. Consistent-Reader: For a point x,  [x] has one local-reader  [p 2, p 3 ] for every pair of planes p 2 & p 3 that intersects by a line l. For a point x,  [x] has one local-reader  [p 2, p 3 ] for every pair of planes p 2 & p 3 that intersects by a line l. Let p 1 be the plane spanned by x and l,  [p 2, p 3 ] Let p 1 be the plane spanned by x and l,  [p 2, p 3 ] rejects, unless A’s values on p 1, p 2 & p 3 agree on l, rejects, unless A’s values on p 1, p 2 & p 3 agree on l, otherwise: returns A’s value on p 1 restricted to x. otherwise: returns A’s value on p 1 restricted to x.

16 Consistency Claim Claim: With high probability (  1-  c’ )   R  [x] either rejects or returns a permissible value for x. [i.e., consistent with one of the permissible polynomials]. Remarks : The sign  R is used for “randomly select from…”. The sign  R is used for “randomly select from…”. Note that randomly selecting X and using it with l to span P 1 is equal to randomly selecting l in P 1. Note that randomly selecting X and using it with l to span P 1 is equal to randomly selecting l in P 1.

17 Consistency Proof Proof: The value A assigns l, according to p 2 & p 3 ’s values, is permissible w.h.p. (1-  c’ ). The value A assigns l, according to p 2 & p 3 ’s values, is permissible w.h.p. (1-  c’ ). On the other hand, l is a random line in p 1 and if p 1 is assigned an impermissible value (by A), then that value restricted to most l’s would be impermissible. On the other hand, l is a random line in p 1 and if p 1 is assigned an impermissible value (by A), then that value restricted to most l’s would be impermissible. with high probability

18 Consistent-Reader for Arbitrary k points How can we read consistently more than one value ? Note: Using the point-consistent-reader, we need to invoke the reader several times, and the received values may correspond to different permissible polynomials. Let  = {x 1,.., x k } be tuple of k point of the domain , Let  = {x 1,.., x k } be tuple of k point of the domain ,  [  ] = {  1,..,  m } is now set of functions, which can either reject or evaluate an assignment to x 1,.., x k.  [  ] = {  1,..,  m } is now set of functions, which can either reject or evaluate an assignment to x 1,.., x k.

19 Hyper-Cube-Vs.-Point Consistent- Reader For k Points Representation: One variable for every cube (affine subspace) of dimension k+2, containing . ( Values of the variables rang over all degree-r, dimension k+2 polynomials ) One variable for every cube (affine subspace) of dimension k+2, containing . ( Values of the variables rang over all degree-r, dimension k+2 polynomials ) one variable for every point x  . (Values of the variables rang over  ). one variable for every point x  . (Values of the variables rang over  ).

20 Hyper-Cube-Vs.-Point Consistent- Reader For k Points Show that the following distribution: Show that the following distribution: Choose a random cube C of dimension k+2 containing  Choose a random cube C of dimension k+2 containing  Choose a random plane p in C Choose a random plane p in C Return p Return p Produces a distribution very close to uniform over planes p Also, p w.h.p. does not contain a point of .

21 Consistent Reader For k Values - Cont. Consistent-Reader: One local-reader for every cube C containing  and a point y  C, which One local-reader for every cube C containing  and a point y  C, which rejects if A’s value for C restricted to y disagrees with A’s value on y, rejects if A’s value for C restricted to y disagrees with A’s value on y, otherwise: returns A’s values on C restricted to x 1,.., x k. otherwise: returns A’s values on C restricted to x 1,.., x k.

22 Proof of Consistency Error Probability:  c’/2 Suppose, We have, in addition, a variable for each plane, We have, in addition, a variable for each plane, The test compares A’s value on the cube C The test compares A’s value on the cube C against A’s value on a plane p, and then against A’s value on a plane p, and then against a point x on that plane. against a point x on that plane. The error probability doesn’t increase.

23 Proof of Consistency - Cont. Proposition: This test induces a distribution over the planes p which is almost uniform. Lemma: Plane-Vs.-Point test works the same if instead of assigning a single value, one assigns each plane with a distribution over values.

24 Summary We saw some consistent readers and how “accurate” they are. They will be a useful tool in this proof. We saw some consistent readers and how “accurate” they are. They will be a useful tool in this proof.