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6SpencerNotesOn6inP

6SpencerNotesOn6inP - Rough Notes on Bansal’s Algorithm...

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Unformatted text preview: Rough Notes on Bansal’s Algorithm Joel Spencer A quarter century ago I proved that given any S 1 ,... ,S n ⊆ { 1 ,... ,n } there was a coloring χ : { 1 ,... ,n } → {− 1 , +1 } so that disc ( S j ) ≤ 6 √ n for all 1 ≤ j ≤ n where we define χ ( S ) = summationdisplay i ∈ S χ ( i ) (1) and disc ( S ) = | χ ( 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 χ 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, Bansal’s algorithm uses Semidefinite Programming . Here is the context: Let S 1 ,... ,S m ⊆ { 1 ,... ,n } . Assume that there exists a map χ : { 1 ,... ,n } → {− 1 , , +1 } (such χ are called partial colorings and i with χ ( i ) = 0 are called uncolored) such that 1. At least nα of the i have χ ( i ) negationslash = 0. 2. For 1 ≤ j ≤ m disc ( S j ) ≤ β 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 ≥ nα 1 3. For 1 ≤ j ≤ m [ summationdisplay i ∈ S j vector v i ] · [ summationdisplay i ∈ S j vector v i ] ≤ [ β j √ n ] 2 The Semidefinite Program has a solution in R 1 by setting vector v i = χ ( 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 χ . (Well, I’m lying a bit here. You need an ǫ of space – that the conditions hold with nα and β j √ n replaced by nα (1 + ǫ ) and β j √ n (1 − ǫ ) respectively for some fixed ǫ . 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 β > 0 define COST( β ) = + ∞ summationdisplay i = −∞ − p i log 2 p i (3) with p i = Pr[ i − 1 2 ≤ N β ≤ i + 1 2 ], N standard normal That is, COST( β ) is the entropy of the roundoff of N to the nearest multiple of β . Theorem 2.1 Let S 1 ,... ,S m ⊆ { 1 ,... ,n } . Let β 1 ,... ,β m be such that m summationdisplay j =1 COST ( β j ) ≤ cn (4) where c < 1 . Write c = 1 − H ( α ) with α ∈ (0 , 1 2 ) . Then there is a χ : { 1 ,... ,n } → {− 1 , , 1 } with 1. At least 2 αn of the i have χ ( i ) negationslash = 0 ....
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6SpencerNotesOn6inP - Rough Notes on Bansal’s Algorithm...

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