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l24 - 6.896 Sublinear Time Algorithms May 3 2007 Lecture 24...

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6.896 Sublinear Time Algorithms May 3, 2007 Lecture 24 Lecturer: Ronitt Rubinfeld Scribe: Ivaylo Riskov 1 Recap f : { 0 , 1 } n → { 0 , 1 } x < y if i, x i y i f is monotone if x y, f ( x ) f ( y ) We define a violating edge ( x, y ) such that y = x e i , x < y, f ( x ) > f ( y ). We can use the following tester: Pick O ( n ) edges and reject if any violate. Finally, we started the proof of the following theorem: Theorem 1 The above tester passes monotone functions and fails -far from monotone functions (must chnage f on at least fraction of the inputs) with probability at least 3 / 4 . 2 Proof of Theorem Claim 2 If f is -far from monotone it has at least 2 n violating edges. Proof of claim: By contrapositive: assume f has less than 2 n violating edges. Let f ( i ) be f after the i -th phase. That is, phase i fixes edges in i -th direction without ruining lower direction edges. Inductive assumption: f ( i 1) has no violating edges in directions 1 . . . i 1. Definition 3 Consider the following definitions: H 0 nodes x such that x i = 0 H 1 nodes x such that x i = 1 M i violating edges in i -th direction.
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