soln3 - CS 151 Complexity Theory Spring 2011 Solution Set 3...

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Unformatted text preview: CS 151 Complexity Theory Spring 2011 Solution Set 3 Posted: April 25 Chris Umans 1. (a) Note: it is most convenient to think of as the permutation k 7 ( ( k )) rather than the more conventional k 7 ( ( k )) the two notions are equivalent by taking inverses; however the second is somewhat more cumbersome notationally for this problem. We start with m + 1 levels 1 , 2 ,..., m +1 of 5 nodes each. We describe the edges directed from level i to level i + 1 based on the i-th instruction ( i j , j , j ): connect the outgoing 0 edges from node k to ( k ) for k { 1 , 2 , 3 , 4 , 5 } , and the outgoing 1 edges from node k to ( k ) for k { 1 , 2 , 3 , 4 , 5 } . Suppose on input x { , 1 } n the instructions yield S 5 . Then the path in the branching program starting at node k in level 1 and dictated by x leads to node ( k ) in level m + 1. Since = e , we can find some k { 1 , 2 , 3 , 4 , 5 } for which ( k ) = k . We designate node k in the first level as the start node, node ( k ) in level m +1 as the accept node, and node k in level m +1 as the reject node, and we discard the other nodes in level m + 1. The result is a width 5 branching program with m levels. For every x A , the path dictated by x from the start node leads to the accept node (formerly node ( k ) in level m + 1), and for every x A , the path dictated by x from the start node leads to the reject node (formerly node e ( k ) = k in level m + 1), as required. (b) For every pair of 5-cycles and we can find an element S 5 for which 1 = . We replace each instruction ( i j , j , j ) with the instruction ( i j , j 1 , j 1 ). (c) We replace the last instruction ( i m , m , m ) with the instruction ( i m , m 1 , m 1 ). The resulting sequence of m instructions yields e on inputs x A and 1 on inputs x A . Thus the modified sequence of m instructions 1-accepts the complement of A . Since is a 5-cycle, 1 is a 5-cycle, and we can apply the previous part to obtain a sequence of m instructions that -accept the complement of A as required. (d) We concatenate the following 4 sequences: (1) a sequence of m instructions that - accepts A , obtained using part (b); (2) a sequence of m instructions that -accepts B , obtained using part (b); (3) a sequence of m instructions that 1-accepts A , obtained using part (b); (2) a sequence of m instructions that 1-accepts B , obtained using part (b). We claim that this sequence 1...
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soln3 - CS 151 Complexity Theory Spring 2011 Solution Set 3...

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