## Homework 3 Solutions

#### Problem #1

We need to look at which assignment for the transitions from p and q on inputs 0 and 1 don't allow us to distinguish p from q. Consider the 9 possible assignments of transitions from p and q on input 0 (3 possible states from p x 3 from q = 9). If both states go to r on input 0, we cannot distinguish them. If they both go to p and / or q, we cannot distinguish them. Thus, 5 of the 9 assignments do not allow us to distinguish p and q on input 0. Similar analysis gives the same result for input 1.

So any combination of the 5 of 9 assignments on inputs 0 and 1 do not allow us to distinguish them, which is 25 out of the 81 possible assignments. We can combine these with any of the 9 assignments of transitions from r on input 0 or 1, so there are 225 assignments in which p and q are equivalent.

#### Problem #2

First we may test to see if the language of the DFA is infinite. If it is, it accepts at least 100 strings.

If it is not infinite, this means the DFA contains no loops on any path from the start state to a final state. So by the pigeonhole principle, any string accepted by the DFA must have length less than n, where n is the number of states in the DFA, since it can never go through the same state twice and still reach the final state.

We can thus systematically check every string of length 0, then those of length 1, then those of length 2, up to those of length n, and count how many strings have been accepted. If at any point the count of accepted strings reaches 100, we stop. If we test every string whose length is less than n, and find we have not accepted 100 strings, we stop and indicate that the DFA does not accept at least 100 different strings.

#### Problem #3

S -> aSa | bSb | aRb | bRa

R -> aRa | bRb | aRb | bRa | c

For w and x not to the be reverse of each other, at at least one position equidistant from the start of one string and the end of the other, the characters must be different.

S appears in strings for which the above property does not hold (at every position equidistant from the start of w and the end of x, the characters are the same). R appears in strings where the property does hold. S generates strings containing S in the middle which do not satisfy the above property, and strings containing R in the middle which do satisfy the above property. R generates strings of the form ycz, where y and z are strings of the same length containing a's and b's.