Abstract

Practice designing and working with NFAs.

1 Working with Automat

 

1. The NFA shown to the right is intended to accept the language $\{aa,aab\}^{*}\{b\}$.

   Although it has no $\epsilon$ transitions, it is, nonetheless, nondeterministic because state $q_0$ has two different transitions on the same input symbol, a.

   Try running it on each of the following inputs to see if it works as expected:

   • b
   • aa
   • aab
   • aabb
   • aaab
   • aaaab
   • aaaabb
   • aaaabbb
    
   

2. The NFA on the right uses an $\epsilon$ transition to help recognize the language $\{aab\}^*\{a,aba\}^*$. Look at the figure and compare to the structure of the set expression. Can you see how each element of the set expression is reflected in the arrangement of the states and transitions?

   Try running it on each of the following inputs to see if it works as expected:

   • a
   • aa
   • aab
   • aba
   • aaba
   • aabaab
   • aabaaba
   • aababa
   • abaaab

3. Revisit some of the DFA problems that you worked in the final step of the earlier DFA exercise. Try rewriting some of them as NFAs.

   • In many cases, an NFA will use fewer states and transitions than the corresponding DFA. Certainly, it should never need more states or transitions than a DFA (because every DFA is also an NFA that just happens to not use non-determinism).

   • You can think of this simplification as the pay-off for the slightly more complicated conceptual effort involved in understanding non-determinism.

4. Exercise 2.3.4 at the end of section 2.3 2 of your text suggests drawing a number of NFAs. Pick one or two of these and create them. Run a handful of inputs through each one to convince yourself that you have done so correctly.

1. the string 101

   Reveal

   

   (Editor)

   There’s not really anything that we can gain from the use of nondeterminism in this case. So the solution is the same as it was when we looked at DFAs for this language.

2. all strings beginning with 101

   Reveal

   

   (Editor)

   Again, non-determinism doesn’t help us here.

3. all strings ending with 101

   Reveal

   Here we can make good use of nondeterminism.

   

   (Editor)

   (video)

   Use the Editor link to try running this on input 101110101 and observe how state $q_0$ acts as a kind of “feeder” supplying every partially processed input to an “is followed by an ending 101” test.

    

   

   Compare that to our earlier DFA for this language. The non-deterministic version looks simpler. Once you start to wrap your mind around how the nondeterminism works, you’ll probably find the NFA version is easier to understand. It was certainly easier to write. But that’s after you get used to the idea of nondeterminism. Don’t be discouraged if that takes a little while.

4. all strings containing 101

   Reveal

   

   (Editor)

   (video)

   Use the Editor link to try running this on input 10011010101 and observe the effects of the non-determinism.

    

   

   Again, by comparison to our earlier DFA, the NFA version is much simpler.

1. all strings beginning with and ending with 101
   Reveal

   In essence, we need to merge these:

    

   

    

   

    

   and we get:

   

   (Editor)

   (video)

1. strings beginning with 101 or 110
   Reveal

   

   (Editor)

   (video)

2.2.3 Repetition

Design an FSA to recognize the language consisting of…

1. All strings containing one or more instances of 101

   Reveal

   Same as saying “all strings that contain 101”

    

   

2. All strings containing zero or more instances of 101

   Reveal

   Same as saying “all strings”

    

   

3. all strings composed of 1 or more (non-overlapping) occurrences of 101

   Reveal

   

   (Editor)

   (video)

4. all strings composed of zero or more occurrences of 101 only

   Reveal

   

   (Editor)

   (video)

5. all strings that do not contain 101

   Reveal

    

   

   This is the solution we obtained when doing a DFA for the language of strings that do not contain 101.

   In essence, the is the complement of the language of strings that do contain 101. Non-determinism doesn’t really help us with this problem. Intuitively, non-determinism helps us explore multiple possibilities at once. But taking the complement is a very restrictive operation. The ability to have a string be in multiple states, some of which might accept and some of which might reject, just isn’t what we need here.

   So we can’t do any better than to use our deterministic solution.

1. Convert the automaton below to a deterministic FA.
    
   
   Reveal

   

   (Editor)

   (video)

^

• Converting a Non-deterministic automaton to deterministic