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Unformatted text preview: 6.896 Sublinear Time Algorithms May 3, 2007 Lecture 24 Lecturer: Ronitt Rubinfeld Scribe: Ivaylo Riskov 1 R e c a p f : { , 1 } n { , 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 failsfar 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 isfar 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 ith phase. That is, phase i fixes edges in ith 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 nodes x such that x i = 0 H 1 nodes x such that x i = 1 M i violating edges in ith direction....
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This note was uploaded on 04/02/2010 for the course CS 6.896 taught by Professor Ronittrubinfeld during the Fall '04 term at MIT.
 Fall '04
 RonittRubinfeld
 Algorithms

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