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Unformatted text preview: CS 151 Complexity Theory Spring 2011 Solution Set 6 Posted: May 23 Chris Umans 1. (a) We observe that the largest possible set shattered by a collection of 2 m subsets is m , since a set of size m + 1 has more than 2 m distinct subsets. The VC dimension of a collection of subsets succinctly encoded by a circuit C can therefore be at most  C  , since C can encode at most 2  C  subsets. Thus we can express VCDIMENSION as follows: { ( C,k ) : ∃ X ∀ X ′ ⊆ X ∃ i [  X  ≥ k and ∀ y ∈ X C ( i,y ) = 1 ⇔ y ∈ X ′ ] } Notice that  X  ,  X ′  , and  i  are all bounded by  C  (using the observation above), and that the expression in the square brackets is computable in poly(  C  ) time. Thus VCDIMENSION is in Σ p 3 . (b) Let ϕ ( a,b,c ) be an instance of QSAT 3 (so we are interested in whether ∃ a ∀ b ∃ c ϕ ( a,b,c )). We may assume by adding dummy variables if necessary that  a  =  b  =  c  = n . As suggested our universe is U = { , 1 } n × { 1 , 2 , 3 ,...,n } . We identify nbit strings with subsets of { 1 , 2 , 3 ,...,n } , and define our collection S of sets to be the sets S a,b,c = { { a } × b if ϕ ( a,b,c ) = 1 ∅ otherwise for all a,b,c . There is a small circuit C that succinctly encodes this collection of sets – given an element x = ( a ′ ,k ) ∈ U and a set name ( a,b,c ), determining whether x ∈ S a,b,c requires only that we check if ϕ ( a,b,c ) = 1 (if it is not, then the set is the empty set and clearly x ̸∈ S a,b,c ) and then check if x ∈ { a } × b (i.e., check whether a ′ = a and b k = 1). Our instance of VCDIMENSION is ( C,n ). If ϕ is a positive instance, i.e., ∃ a ∀ b ∃ c ϕ ( a,b,c ) = 1, then the set U a = { a } × { 1 , 2 , 3 ,...,n } of size n is shattered, because S contains sets of the form { a } × b for all b . Thus the VC dimension of S is at least n . Conversely, if the VC dimension of S is at least n , then there is a set X of size n that is shattered by S . We observe that X cannot contain elements of two different subsets U a and U a ′ because then the set consisting of these two elements cannot be expressed as the intersection of X with some set in S (all of our sets are subsets of some U a ). We conclude that X ⊆ U a for some a , and the fact that it is shattered implies that sets of the form { a } × b for all b must be present in S . This implies that ∀ b ∃ c ϕ ( a,b,c ), so we have a positive instance....
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 Fall '09

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