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soln6

# soln6 - CS 151 Complexity Theory Spring 2011 Solution Set 6...

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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 VC-DIMENSION 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 VC-DIMENSION 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 = { 0 , 1 } n × { 1 , 2 , 3 , . . . , n } . We identify n -bit 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 VC-DIMENSION 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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