1.7 Practical Session

Make sure that you understand how the Prolog programs that we saw in this section work.

We will start out with a couple of simple keyboard exercises so make sure that you understand how the Prolog programs that we saw in this section work.

Note: The file recognize1.pl contains a version of our recognizer/generator that that only works for FSAs without jump arcs. It defines the same predicates as recognize.pl , but all predicate names end in a 1.

1. Start Prolog and consult recognize.pl, recognize1.pl, and haha.pl. Using both test1 (which does not deal with jump arcs) and test (which does) see whether various strings are accepted e.g

   test1(a1,[h,a,!]).

   test(a1,[h,a,!]).

   test1(a1,[h,a,h,a,h,a,!]).

   Make sure you understand why these strings are accepted. That is, use trace to step through the execution so that you understand exactly what is going on.

2. Now let's have a look at the automaton a3. The difference between a1 and a3 is that haha5 is not deterministic. Try giving the following queries:

   test(a3,[h,a,!]).

   test1(a3,[h,a,!]).

   test(a3,[h,a,h,a,h,a,!]).

   Make sure you understand why these strings are accepted. That is, use trace to step through the execution so that you understand exactly what is going on.

3. We now turn to a2. The big difference between a1 and a2 is that a2 contains a jump arc #. Now, recognize1 does not handle jump arcs so it should not be able to handle a2 correctly.

   Try out the following experiments, and make sure you understand why recognize1 gives the responses it gives:

   test1(a2,[h,a,!]).

   test1(a2,[h,a,h,a,!]).

   test1(a2,[h,a,h,a,h,a,!]).

   Now try these examples with test instead of test1. Again, carry out traces.

4. Finally, make sure that you understand how the generate predicates are defined. Try using generate1 to generate using a2. Now try using generate with a2. Why does generate1 give a wrong response?