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Consistent Readers

Feb 02, 2016

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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 : - PowerPoint PPT Presentation
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Page 1: Consistent Readers

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IntroductionIntroduction

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

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Plane Vs. Point Test - Plane Vs. Point Test - RepresentationRepresentation

RepresentationRepresentation:One variable for each planeplane pp of

planes(), supposedly assigned the restriction of ƒƒ to p. (Values of the variables rang over all 2-dimensional, degree-r polynomials).

One variable for each pointpoint xx . (Values of the variables rang over the field ).

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Plane Vs. Point Test - TestPlane Vs. Point Test - Test

TestTest:

One local-test for every:

planeplane pp and a pointpoint xx on p.

AcceptAccept if – A’s value on x, and

– A’s value on p restricted to x are consistent.

Reminder:

AA: planes dimension-2 degree-r polynomial

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Plane Vs. Point Test: Error Plane Vs. Point Test: Error ProbabilityProbability

ClaimClaim: 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-permissible degree-r

polynomial (on the planes),

* fraction from the set of all combination of (point, plane)

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Plane Vs. Point Test: Error Plane Vs. Point Test: Error Probability - ProofProbability - Proof

ProofProof: By reduction to Plane-Vs.-Plane test: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-TestPPx-TestWhat is its error-probability?

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Plane Vs. Point Test: Error Plane Vs. Point Test: Error Probability - Proof Cont.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 & p2 agree on l, but– Have impermissible values (i.e. they do not

represent restrictions of 2 -permissible polynomials).

Second case:– p1 & p2 do not agree on l, but – Agree on the (randomly) chosen point x on l.

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Plane Vs. Point Test: Error Plane Vs. Point Test: Error Probability - Proof Cont.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)Pr(First-Case Error) cc

For the second case, recall that:– rr = #points, that two r-degree, 1-dimensional

polynomials can agree on.

– |||| = #points on the line l.

So Pr(Second-Case Error) Pr(Second-Case Error) r/|r/|||

PPx-Test’s error-probability c c + r/|+ r/|||

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Plane Vs. Point Test: Error Plane Vs. Point Test: Error Probability - Proof Cont.Probability - Proof Cont.

For an appropriate (namely: (namely: ccO(r/|O(r/|

|)|)))::

c c + r/|+ r/|| = O(| = O(cc))

So, PPx-Test’s error-probability is

c’c’, for some 0<c’<1

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Plane Vs. Point Test: Error Plane Vs. Point Test: Error Probability - Proof Cont.Probability - Proof Cont.Back to Plane-Vs.-PointBack to Plane-Vs.-Point:: Let ppplanesplanes, xx((pointspoints on on p p)), such that:

– A(p)A(p) and A(x)A(x) are impermissible. Let lllines lines such that x l Let p1, p2 be planes through l

Plane-Vs.-Point’s error probability is:

Pr Pr p, x p, x (( ((A(p)A(p)))(x) (x) = = A(x) A(x) ) =) =

= Pr Pr llx, p1 x, p1 ( (( (A(p1)A(p1)))(x) (x) = = A(x)A(x) ) )

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Plane Vs. Point Test: Error Plane Vs. Point Test: Error Probability - Proof Cont.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’c’)1/21/2

** 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 ) )22

****** PPx-Test

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Plane Vs. Point Test: Error Plane Vs. Point Test: Error Probability - Proof Cont.Probability - Proof Cont.ConclusionConclusion::We’ve established that:Plane-Vs.-Point error probability, i.e.,The probability that p (which is random) is

– Assigned an impermissible value, and– This value agrees with the value assigned to x

(which is also random),

is < < c’/2c’/2.

Note: This proof is only valid as long as the point x whose value we would like to read is randomrandom.

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Reading an Arbitrary PointReading an Arbitrary Point

Can we have similar procedure that

would work for any arbitraryarbitrary 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-readerconsistent-reader..

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Consistent Reader for Consistent Reader for Arbitrary Arbitrary PointPoint

Representation: As in Plane-Vs-Point test.local-readerslocal-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].

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33-Planes Consistent Reader -Planes Consistent Reader for a Point for a Point xx

Representation: One variable for each plane.

Consistent-Reader:

For a point x, [x] has one local-reader [p2, p3]

for every pair of planes p2 & p3 that intersect by a

line l.

Let p1 be the plane spanned by x and l, [p2, p3]

– rejects, unless A’s values on p1, p2 & p3 agree on l,

– otherwise: returns A’s value on p1 restricted to x.

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Consistency ClaimConsistency 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…”.

Note that randomly selecting X and using it with l to span p1 is

equal to randomly selecting l in p1.

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Consistency ProofConsistency Proof

Proof: The value A assigns l, according to p2 &

p3’s values, is permissible w.h.p. (1-c’).

On the other hand, l is a random line in

p1 and if p1 is assigned an impermissible

value (by A), then that value restricted to most l’s would be impermissible.

with high probability

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Consistent-Reader for Arbitrary Consistent-Reader for Arbitrary kk pointspoints

How can we read consistently How can we read consistently more more than one value 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 = {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 to x1, .., xk.

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Hyper-Cube-Vs.-Point Hyper-Cube-Vs.-Point Consistent-Reader For Consistent-Reader For kk Points 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 point x .

(Values of the variables rang over ).

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Hyper-Cube-Vs.-Point Hyper-Cube-Vs.-Point Consistent-Reader For Consistent-Reader For kk Points Points

Show that the following distribution:– Choose a random cube C of dimension

k+2 containing – Choose a random plane p in C– Return p

Produces a distribution very close to uniform over planes p

Also, p w.h.p. does not contain a point of .

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Consistent Reader For Consistent Reader For kk Values Values - - Cont.Cont.

Consistent-Reader:

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,

– otherwise: returns A’s values on C

restricted to x1, .., xk.

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Proof of ConsistencyProof of Consistency

Error Probability: c’/2

Suppose, We have, in addition, a variable for each

plane, The test compares A’s value on the cube C

– against A’s value on a plane p, and then

– against a point x on that plane.

The error probability doesn’t increase.

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Proof of Consistency - Cont.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.

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SummarySummary

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