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# Consistent Readers - PowerPoint PPT Presentation

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 :

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Read Consistently a value for arbitrary points

• We are going to use several consistency tests for Consistent Readers.

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 ).

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

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)

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?

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.

• 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/||

For an appropriate  (namely: cO(r/||)):

c + r/|| = O(c)

So, PPx-Test’s error-probability is

£ c’, for some 0<c’<1

Back to Plane-Vs.-Point:

• Let pplanes, x(points on p), such that:

• A(p) and A(x) are impermissible.

• Let llines 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 lx, p1 ( (A(p1))(x) = A(x) )

Prp, x ((A(p))(x) = A(x) )

= Prlx, P1 ( (A(p1))(x) = A(x) )

=* Elx ( Prp1 ( (A(p1))(x) = A(x)|xl ) )

=**Elx( (Prp1, p2 ( (A(p1))(x) = (A(p2))(x) = A(x) | xl ) )1/2 )

 ( Elx(Prp1, p2 ((A(p1))(x) = (A(p2))(x) = A(x) | xl ) )1/2

*( Prlx, 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) | xL ) ) = (p1,p2 are independent)

(Prp1 ( (A(p1))(x) = A(x) | xl ) )* (Prp1 ( (A(p2))(x) = A(x) | xl ) ) =

(Prp1 ( (A(p1))(x) = A(x) | xl ) )2

***PPx-Test

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.

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.

• 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.

• 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.

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.

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.

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.

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  ).

• 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.

• One local-reader for every cube C containingand 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.

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.

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.

• We saw some consistent readersand how “accurate” they are. They will be a useful tool in this proof.