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Large Clique Consistency
Lemma (large clique consistency): A large (> ||-½)
clique agrees almost everywhere with a single
degree r polynomial.
Def: Let a+=a+| |-1 denote the fraction of points
that two distinct encodings of an a-code cannot
agree on.
In out case, the set of all d-variables polynomials of
total degree r is an a-code where:
And hence:
r +1
a =
| F|
+
r
a=
| F|
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Definitions
Def: Let n=a| |+1
Claim (interpolation):
For any (i+1)-dimensional sub-space of | |d and
exactly n parallel i-dimensional sub-spaces of it:
S1 ,...,Sn  S | F |d
i  j .Si  S j = 
| S |=| F |i +1 ,| S j |=| F |i
For every set of i-dimensional degree-r
polynomials over the small subspaces, there
exists exactly one (i+1)-dimensional degree-r
polynomial that agrees with the small
polynomials over the subspaces.
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Definitions
Proof (interpolation): The proof of this claim is left
as homework. (Recall previous exercises from the
“encoding” section...)
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Proof
Now we return to the lemma:
Proof (large clique consistency):
Consider a large clique in the graph defined
previously:
T  V(G) = {S | F |d ;| S |=| F |d -1}
| T | a +  | V(G) |
That is, T, out clique, is a set of vertices in the
graph (recall each vertex was an assignment to
some affine subspace of dimension d-1,
supposedly a restriction of some polynomial).
The fractional size of T is large (a+ ).
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Proof
The main point of the proof:
Since the fractional size of T is larger then a+ ,
then there exists a set of n parallel (d-1)dimensional sub spaces:
S1 ,...,Sn | F |d
i  j .Si  S j = 
| S j |=| F |d -1
In addition, there exists another set of n parallel
(d-1)-dimensional sub spaces that are not
parallel to the previous set:
S1' ,...,Sn' | F |d
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Proof
Now consider the first set of parallel subspaces.
According to the interpolation claim discussed
previously, it follows that exactly one ddimensional degree-r polynomial, call it H, agrees
with the values of the small polynomials on all
subspaces.
In addition, for every d-1 dimensional subspace
which is not parallel to the set:
d
S1 ,...,Sn | F |
Must intersect each of these subspaces in some d2 dimensional subspace.
Since all these subspaces (of the intersection) are
parallel, it follows, again from the interpolation
property, that the polynomial on this subspace
must be H.
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Proof
In particular, the polynomial on the subspaces:
S1' ,...,Sn' | F |d
Is also H !
Since every d-1 dimensional subspace must
intersect either all of the first set or all of the
second set, we have proven consistency for all d1 dimensional subspaces, and thus for all the
space.
Q.E.D
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