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Plane-vs.-Plane Test. proof for d=3. Introduction. In last chapter we saw a few consistency tests. In this chapter we are going to prove the properties of Plane-vs.-Plane test:
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Plane-vs.-Plane Test proof for d=3
Introduction In last chapter we saw a few consistency tests. In this chapter we are going to prove the properties of Plane-vs.-Plane test: Thm[RaSa]:as long as ³||-c for some constant 1 > c > 0test errs (w.r.t. ) with very small probability, namely £c’for some constant c’ > 0. i.e., the plane-vs.-plane test, (with probability ³ 1 - c’), either accepts values of a -permissible polynomial or rejects.
Introduction Cont. • We will prove the theorem only for d = 3[i.e for a polynomials with 3 variables]. • We are going to construct a consistency-graphto represent an assignment, and show that it is almost a transitive-graph and the theorem will follow.
Consistency Graph - Definition • Let A be an assignment (namely assigning to each plane a degree-r 2-dimension polynomial). • Def: The consistency-graph of A, G[A], is an undirected graph that: • Has one vertex for each plane. • Two planes p1&p2 are connected if A’s value for p1is consistent with A’s value for p2. Note: • if p1 & p2 do not intersect by a line, their vertexes are connected. • Every vertex will have a self (cyclic) edge.
A Little Finite-Field Geometry • In case d=3 • Two planes cannot intersect by only a point;They are either parallel or intersect by a line. • For any given line l, only fraction ||-1 of planes do not intersect l.
Success Probability for A Notations: • (A)= the fraction of local-tests satisfied by A. • (G[A])= the fraction of connected ordered pairs of vertices. (i.e. planes assigned consistent values by A. ) Hence,(G[A]) = (A) + ||-2
Non-Transitive Triangles Def: A non-transitive triangle of an undirected graph G is a threesome (v1, v2, v3 )where • v1is connected tov2, • v2 is connected tov3, • however,v1 & v3 are not connected. v2 v1 v3 Def: An undirected graph with no non- transitive-triangle is a transitive graph.
If consistency-graph were transitive • Note that a transitive graph is a disjoint union of cliques. • For a clique C, we define: fc=#(vertices in C)/#(vertices in the graph) • In a transitive graph, (G) = cliques C in graph(fc)2 i.e (G) is the sum of squares of clique’sfractions
Large Clique Lemma Lemma(large-clique):In a transitive graph Gthere must be at least one clique of ³(G)•|G| vertices. Hint: Notice the square in the value of (G)on a transitive graph.
Large Clique Consistency On the other hand, Lemma(Large Clique’s Consistency): a large(> ||-½)clique agrees almost everywhere with a single degree-r polynomial Proof: Let us now proceed to show that G[A] is almost transitive.
The Parameter Def: For a non-edge (v1, v3)of G we let(v1, v3) be the fraction of vertices that form a non-transitive triangle with it, i.e.Prv2[ (v1, v2, v3) non-transitive ]. Def: We then let (G)be the maximum, over all non-edges(v1, v3)of G, of (v1, v3).
Clique’ing a Graph Lemma(-transitivity): From any graph G, a transitive graph can be obtain, by removing at most 3 • (G) • ½|G|2 edges. Proof(-transitivity): repeat following steps • if the degree of a vertex v < (G)•|G|,remove all v’s edges. • Otherwise, pick any vertex uand break G into neighbors of u and non neighbors of u;then remove all edges between these two sets. Once none of these step is applicable, G is transitive.
Clique’ing a Graph Show: • The above process removes at most 3 • (G) • ½|G|2of G‘s edges. • The graph thus obtained is indeed transitive.
Back to: Proof of Consistency We will next show that, for any assignment A to the planes, (G[A])is small. Hence, by the -transitivity lemma, one can obtain a sub-graph of G[A] of almost (the same) •|G|2 number of edges, which isneverthelesstransitive.
Back to: Proof of Consistency That is, by disregarding only a small fraction of the success probability of the test, the consistency graph becomes a disjoint union of cliques,hence has a large clique, (see exercise)which must be globally consistent(by the lemma of Large Clique’s Consistency) and the theorem ([RaSa]) follows.
(G) is Small Lemma: let A be an assignment to the planes; then (G[A])is small, namely (r+1) • ||-1. Proof: Consider a non-edge of G[A],that is, a pair ofplanes (p1, p3)that intersect by a line l,whoseAvalues on ldisagree.
Proof - Cont. First: Since d = 3, all but ||-1 fraction of planes p2 meet l [ See finite-field geometry above ] In that case,A’s value for p2 can agree with its values for bothp1&p3 only if p2intersectsl on a point at which the values for p1&p3agree,which is at most a small fraction, r • ||-1, of points on l.
Proof - Cont. Summing up the two fractions||-1 + r||-1 = (r+1)||-1 It now only remains to show that a non-negligible clique of planes agree with a single degree-r polynomial (lemma above) .
Home Assignment Prove the theorem for general d. By simple induction one can show for ³ d•||c (which would do). Bonus: try proving for³ ||c .
Summary • We have proven the Plane-vs.-Plane consistency test (by showing that a consistency-graph is almost a transitive-graph). • Next chapter we will use this test to construct better tests that will allow us consistent reading.
