calans - CSE 200 Spring 2010 Calibration Homework Solutions...

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CSE 200 Spring 2010 Calibration Homework Solutions Chris Calabro and Russell Impagliazzo April 11, 2010 1 Regular languages Prove that Reg k +1 is a proper superset of Reg k . It is immediate from the deFnition that Reg k +1 is a superset of Reg k . We need to show that the inclusion is proper. To do this, we need to exhibit a language L k +1 in Reg k +1 , i.e., that is accepted by a k + 1 state Fnite automaton, but not in Reg k , that is, cannot be accepted by any k state Fnite automaton. Let L k +1 = { x ∈ { 0 } * || x | = 0 mod k + 1 } of strings over a one-symbol alphabet whose length is evenly divisible by k +1. Let A be the automaton with states 0 , ..k , with 0 as the start state and only accepting state, and transitions δ ( q, 0) = q + 1 mod k + 1, i.e., upon reading the next symbol, we go to the next state, unless we are in the last, in which case we go to the initial state. We can see by induction that 0 l ends in state l mod k + 1. since this is true for l = 0, and if 0 l is in state l mod k +1, 0 l +1 ends in state ( l mod ( k +1))+1 mod ( k + 1) = ( l + 1) mod ( k + 1). Since the only accepting state is 0, then 0 l is accepted if and only if l mod ( k + 1) = 0, which is by deFnition if and only if 0 l L k +1 . Thus, automaton A is a k + 1 state machine that accepts L k +1 , so L k +1 Reg k +1 . Let
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calans - CSE 200 Spring 2010 Calibration Homework Solutions...

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