Créer une présentation
Télécharger la présentation

Télécharger la présentation
## Consistent Readers

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Consistent Readers**Read Consistently a value for arbitrary points**Introduction**• We are going to use several consistency tests for Consistent Readers.**Plane Vs. Point Test - Representation**Representation: • One variable for each planepof planes(), supposedly assigned the restriction of ƒ to p.(Values of the variables rang over all 2-dimensional, degree-r polynomials). • One variable for eachpointx .(Values of the variables rangover the field ).**Plane Vs. Point Test - Test**Test: • One local-test for every: planepand a pointx on p. • Accept if • A’s value on x,and • A’s value on p restricted to x are consistent. Reminder:A: planes dimension-2 degree-r polynomial**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 <x, p> for a point x and plane p whose: • A’s value are consistent, and yet • Do not agree with any -permissible degree-r polynomial (on the planes), * fraction from the set of all combination of (point, plane)**Plane Vs. Point Test: Error Probability - Proof**Proof: By reduction to Plane-Vs.-Planetest: replace every • Local-test for p1 & p2 that intersect by a line l, by a • Set of local-tests, one for each point x on l, that compares p1’s & p2’s values on x. Let’s denote this test by PPx-Test What is its error-probability?**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: • p1& p2agree on l, but • Have impermissible values (i.e. they do not represent restrictions of 2 -permissiblepolynomials). • Second case: • p1& p2do not agreeon l, but • Agree on the (randomly) chosen point x on l.**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<1Pr(First-Case Error)£ c • For the second case, recall that: • r= #points, that two r-degree, 1-dimensional polynomials can agree on. • || = #points on the line l. So Pr(Second-Case Error) £ r/|| PPx-Test’s error-probability £ c + r/||**Plane Vs. Point Test: Error Probability - Proof Cont.**For an appropriate (namely: cO(r/||)): c + r/|| = O(c) So, PPx-Test’s error-probability is £ c’, for some 0<c’<1**Plane Vs. Point Test: Error Probability - Proof Cont.**Back to Plane-Vs.-Point: • Let pplanes, x(points on p), such that: • A(p) and A(x) are impermissible. • Let llines such that x l • Let p1, p2 be planes through l Plane-Vs.-Point’s error probability is: Pr p, x ((A(p))(x) = A(x) ) = = Pr lx, p1 ( (A(p1))(x) = A(x) )**Plane Vs. Point Test: Error Probability - Proof Cont.**Prp, x ((A(p))(x) = A(x) ) = Prlx, P1 ( (A(p1))(x) = A(x) ) =* Elx ( Prp1 ( (A(p1))(x) = A(x)|xl ) ) =**Elx( (Prp1, p2 ( (A(p1))(x) = (A(p2))(x) = A(x) | xl ) )1/2 ) ( Elx(Prp1, p2 ((A(p1))(x) = (A(p2))(x) = A(x) | xl ) )1/2 *( Prlx, p1, p2 ((A(p1))(x) = (A(p2))(x) = A(x))1/2 *** (c’)1/2 *event A, and random variable Y, Pr(A) = EY( Pr(A|Y) ) ** Prp1, p2 ( (A(p1))(x) = (A(p2))(x) = A(x) | xL ) ) = (p1,p2 are independent) (Prp1 ( (A(p1))(x) = A(x) | xl ) )* (Prp1 ( (A(p2))(x) = A(x) | xl ) ) = (Prp1 ( (A(p1))(x) = A(x) | xl ) )2 ***PPx-Test**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 israndom)is • Assigned an impermissible value, and • This value agrees with the value assigned to x(which is alsorandom), is < c’/2. Note: This proof is only valid as long as the point x whose value we would like to read is random.**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.**Consistent Reader for Arbitrary Point**• Representation: As in Plane-Vs-Point test. • local-readers: Insteadoflocal-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].**3-Planes Consistent Reader for a Point x**Representation: One variable for each plane. Consistent-Reader: • For a point x, [x]hasone local-reader[p2, p3] for every pair of planes p2 & p3 that intersect by a line l. • Let p1 be the plane spanned byxand l, [p2, p3] • rejects, unless A’s values on p1, p2 & p3agree on l, • otherwise: returnsA’s value on p1 restricted to x.**Consistency Claim**Claim:Withhighprobability ( 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 Ris used for “randomly select from…”. • Note that randomly selecting X and using it with l to spanp1is equal to randomly selecting l in p1.**with high probability**Consistency Proof Proof: • The value A assigns l, according to p2 & p3’s values, is permissible w.h.p. (1-c’). • On the other hand, lis a random line in p1 and if p1 is assigned an impermissible value (by A), then that value restricted to mostl’s would be impermissible.**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 differentpermissible polynomials. • Let = {x1, .., xk} be tuple of k point of the domain , • [ ] = { 1, .., m } is now set of functions, which can either reject or evaluate an assignment tox1, .., xk.**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 variablesrang over all degree-r, dimension k+2 polynomials ) • one variable for every point x .(Values of the variablesrang over ).**Hyper-Cube-Vs.-Point Consistent-Reader For k Points**• Show that the following distribution: • Choose a random cube C of dimension k+2containing • Choose a random plane p in C • Return p Produces a distribution very close to uniform over planes pAlso, p w.h.p. does not contain a point of .**Consistent Reader For k Values - Cont.**Consistent-Reader: • One local-reader for every cube C containingand a pointy C, which • rejectsifA’svalue for C restricted to ydisagrees with A’s value on y, • otherwise: returnsA’s values on C restricted to x1, .., xk.**Proof of Consistency**Error Probability: c’/2 Suppose, • We have, in addition, a variable for each plane, • The test compares A’s value on the cubeC • against A’s value on a planep, and then • against a pointx on that plane. The error probability doesn’t increase.**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.**Summary**• We saw some consistent readersand how “accurate” they are. They will be a useful tool in this proof.