Plane-vs.-Plane Test

1
Plane-vs.-Plane Test

proof for d3
2
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
• ThmRaSaas long as ? ³?-c for some constant
  1 gt c gt 0test errs (w.r.t. ?) with very small
  probability, namely ?c for some constant c gt
  0.
• i.e., the plane-vs.-plane test, (with
  probability ³ 1 - ?c), either accepts values of
  a ?-permissible polynomial or rejects.

3
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-graph to
  represent an assignment, and show that it is
  almost a transitive-graph and the theorem will
  follow.

4
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, GA, is an
  undirected graph that
• Has one vertex for each plane.
• Two planes p1 p2 are connected if As value for
  p1 is consistent with As 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.

5
A Little Finite-Field Geometry

• In case d3
• Two planes cannot intersect by only a pointThey
  are either parallel or intersect by a line.
• For any given line l, only fraction ?-1 of
  planes do not intersect l.

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Success Probability for A

• Notations
• ?(A) the fraction of local-tests satisfied by
  A.
• ?(GA) the fraction of connected ordered pairs
  of vertices. (i.e. planes assigned consistent
  values by A. )
• Hence, ?(GA) ?(A) ?-2

7
Non-Transitive Triangles

• Def A non-transitive triangle of an undirected
  graph G is a threesome (v1, v2, v3 ) where
• v1 is connected to v2,
• v2 is connected to v3,
• however, v1 v3 are not connected.

v2
v1
v3

• Def An undirected graph with no non-
• transitive-triangle is a transitive graph.

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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 cliques
  fractions

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Large Clique Lemma

• Lemma (large-clique)In a transitive graph G
  there must be at least one clique of ³ ?(G) G
  vertices.
• Hint Notice the square in the value of ?(G) on a
  transitive graph.

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Large Clique Consistency

• On the other hand,
• Lemma (Large Cliques Consistency)
• a large (gt ?-½) clique agrees almost everywhere
  with a single degree-r polynomial
• Proof
• Let us now proceed to show that GA is almost
  transitive.

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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).

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Cliqueing a Graph

• Lemma (?-transitivity)
• From any graph G, a transitive graph can be
  obtain, by removing at most 3 ??(G) ½G2
  edges.
• Proof (?-transitivity) repeat following steps
• if the degree of a vertex v lt ??(G)G, remove
  all vs edges.
• Otherwise, pick any vertex u and break G into
  neighbors of u and non neighbors of uthen
  remove all edges between these two sets.
• Once none of these step is applicable, G is
  transitive.

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Cliqueing a Graph

• Show
• The above process removes at most 3 ??(G)
  ½G2of Gs edges.
• The graph thus obtained is indeed transitive.

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Back to Proof of Consistency

• We will next show that, for any assignment A to
  the planes, ?(GA) is small. Hence, by the
  ?-transitivity lemma, one can obtain a sub-graph
  of GA of almost (the same) ?G2 number of
  edges, which is nevertheless transitive.

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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 Cliques Consistency)
• and the theorem (RaSa) follows.

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?(G) is Small

• Lemma let A be an assignment to the planes then
  ?(GA) is small, namely (r1) ?-1 .
• Proof Consider a non-edge of GA,that is, a
  pair of planes (p1, p3)that intersect by a line
  l,whose A values on l disagree.

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Proof - Cont.

• First Since d 3, all but ?-1 fraction of
  planes p2 meet l See finite-field geometry
  above
• In that case,As value for p2 can agree with its
  values for both p1 p3 only if p2 intersects l
  on a point at which the values for p1 p3
  agree,which is at most a small fraction, r
  ?-1, of points on l .

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Proof - Cont.

• Summing up the two fractions ?-1 r?-1
  (r1) ?-1
• It now only remains to show that a non-negligible
  clique of planes agree with a single degree-r
  polynomial (lemma above) .

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Home Assignment

• Prove the theorem for general d.
• By simple induction one can show for ? ³ d?c
  (which would do).
• Bonus try proving for ? ³ ?c .

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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.