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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: Thm[RaSa]: as long as | | -c for some constant 1 > c > 0 test errs (w.r.t. ) with very small probability, namely c’ for some constant c’ > 0. i.e., the plane-vs.-plane test, (with probability 1 - c’ ), either accepts values of a -permissible polynomial or rejects. i.e., the plane-vs.-plane test, (with probability 1 - c’ ), either accepts values of a -permissible polynomial or rejects.

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3 Introduction Cont. We will prove the theorem only for d = 3 [i.e for a polynomials with 3 variables]. 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. 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.

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4 Consistency Graph - Definition Let A be an assignment (namely assigning to each plane a degree-r 2-dimension polynomial). 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: Def: The consistency-graph of A, G[A], is an undirected graph that: Has one vertex for each plane. Has one vertex for each plane. Two planes p 1 & p 2 are connected if A’s value for p 1 is consistent with A’s value for p 2. Two planes p 1 & p 2 are connected if A’s value for p 1 is consistent with A’s value for p 2. Note: Note: if p 1 & p 2 do not intersect by a line, their vertexes are connected. if p 1 & p 2 do not intersect by a line, their vertexes are connected. Every vertex will have a self (cyclic) edge. Every vertex will have a self (cyclic) edge.

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5 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.

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6 Success Probability for A Notations: (A) = the fraction of local-tests satisfied by A. (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. ) (G[A]) = the fraction of connected ordered pairs of vertices. (i.e. planes assigned consistent values by A. ) Hence, (G[A]) = ( A) + | | -2

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7 Non-Transitive Triangles Def: A non-transitive triangle of an undirected graph G is a threesome (v 1, v 2, v 3 ) where v 1 is connected to v 2, v 1 is connected to v 2, v 2 is connected to v 3, v 2 is connected to v 3, however, v 1 & v 3 are not connected. however, v 1 & v 3 are not connected. v1v1 v3v3 Def: An undirected graph with no non- transitive graph transitive-triangle is a transitive graph. v2v2

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8 If consistency-graph were transitive Note that a transitive graph is a disjoint union of cliques. Note that a transitive graph is a disjoint union of cliques. For a clique C, we define: For a clique C, we define: f c = #(vertices in C) / #(vertices in the graph) In a transitive graph, In a transitive graph, (G) = cliques C in graph (f c ) 2 i.e (G) is the sum of squares of clique’s fractions

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9 Large Clique Lemma Lemma (large-clique): In a transitive graph G there must be at least one clique of (G) |G| vertices. Hint: Notice the square in the value of (G) on a transitive graph.

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10 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.

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11 The Parameter Def: For a non-edge (v 1, v 3 ) of G we let (v 1, v 3 ) be the fraction of vertices that form a non-transitive triangle with it, i.e. Pr v 2 [ (v 1, v 2, v 3 ) non-transitive ]. Def: We then let (G) be the maximum, over all non-edges (v 1, v 3 ) of G, of (v 1, v 3 ).

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12 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 u and 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.

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13 Clique’ing a Graph Show: The above process removes at most 3 (G) ½|G| 2 of G‘s edges. The above process removes at most 3 (G) ½|G| 2 of G‘s edges. The graph thus obtained is indeed transitive. The graph thus obtained is indeed transitive.

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14 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 is nevertheless transitive.

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15 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.

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16 (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 of planes (p 1, p 3 ) that intersect by a line l, whose A values on l disagree.

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17 Proof - Cont. First: Since d = 3, all but | | -1 fraction of planes p 2 meet l [ See finite-field geometry above ] See finite-field geometry above See finite-field geometry above In that case, A’s value for p 2 can agree with its values for both p 1 & p 3 only if p 2 intersects l on a point at which the values for p 1 & p 3 agree, which is at most a small fraction, r | | -1, of points on l.

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18 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).

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19 Home Assignment Prove the theorem for general d. By simple induction one can show for d| | c (which would do). Bonus: try proving for | | c.

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20 Summary We have proven the Plane-vs.-Plane consistency test ( by showing that a consistency-graph is almost a transitive-graph). We have proven the Plane-vs.-Plane consistency test ( by showing that a consistency-graph is almost a transitive-graph). consistency-graph transitive-graph consistency-graph transitive-graph Next chapter we will use this test to construct better tests that will allow us consistent reading. Next chapter we will use this test to construct better tests that will allow us consistent reading.consistent readingconsistent reading

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