{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}