6SpencerNotesOn6inP - Rough Notes on Bansals Algorithm Joel...

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Unformatted text preview: Rough Notes on Bansals Algorithm Joel Spencer A quarter century ago I proved that given any S 1 ,... ,S n { 1 ,... ,n } there was a coloring : { 1 ,... ,n } { 1 , +1 } so that disc ( S j ) 6 n for all 1 j n where we define ( S ) = summationdisplay i S ( i ) (1) and disc ( S ) = | ( S ) | . Here 6 is an absolute constant whose precise value will not here be relevant. The question remained: Is there a polynomial time algorithm to find a coloring with this property. I frequently conjectured that there was no such algorithm. But now Nikhil Bansal has found one! His result (which has much more) can be found on ArXiv at http://arxiv.org/abs/1002.2259 Here I give a rough description, but emphasizing that the result is entirely his. 1 Semidefinite Programming At its core, Bansals algorithm uses Semidefinite Programming . Here is the context: Let S 1 ,... ,S m { 1 ,... ,n } . Assume that there exists a map : { 1 ,... ,n } { 1 , , +1 } (such are called partial colorings and i with ( i ) = 0 are called uncolored) such that 1. At least n of the i have ( i ) negationslash = 0. 2. For 1 j m disc ( S j ) j n Consider the Semidefinite Program to find vector v 1 ,... , vector v n R n satisfying (so this is a feasibility program): 1. vector v i vector v i 1 for 1 i n . 2. n i =1 vector v i vector v i n 1 3. For 1 j m [ summationdisplay i S j vector v i ] [ summationdisplay i S j vector v i ] [ j n ] 2 The Semidefinite Program has a solution in R 1 by setting vector v i = ( i ). There- fore, Semidefinite Programming will find a solution (though probably not that one!) in time polynomial in n,m . Critically, we do not need to know the original . (Well, Im lying a bit here. You need an of space that the conditions hold with n and j n replaced by n (1 + ) and j n (1 ) respectively for some fixed . This will always clearly hold.) 2 My Result I need to state my result in more general form. For completeness I give the argument, far better than my original proof, due to Ravi Boppana. Let H ( x ) = x log 2 x (1 x )log 2 (1 x ) (2) be the usual entropy function. For > 0 define COST( ) = + summationdisplay i = p i log 2 p i (3) with p i = Pr[ i 1 2 N i + 1 2 ], N standard normal That is, COST( ) is the entropy of the roundoff of N to the nearest multiple of . Theorem 2.1 Let S 1 ,... ,S m { 1 ,... ,n } . Let 1 ,... , m be such that m summationdisplay j =1 COST ( j ) cn (4) where c < 1 . Write c = 1 H ( ) with (0 , 1 2 ) . Then there is a : { 1 ,... ,n } { 1 , , 1 } with 1. At least 2 n of the i have ( i ) negationslash = 0 ....
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This note was uploaded on 10/23/2011 for the course CS 7520 taught by Professor Staff during the Spring '08 term at Georgia Institute of Technology.

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6SpencerNotesOn6inP - Rough Notes on Bansals Algorithm Joel...

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