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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
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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 ).
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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
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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)
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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?
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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.
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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/| |
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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
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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) )
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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
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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.
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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.
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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].
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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.
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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.
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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
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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.
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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 ).
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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 .
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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.
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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.
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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.
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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.
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