In this lab assignment, you will gain some experience using
Prolog. Specifically, you will use the SWI-Prolog interpreter, which
is available on the Mac OS X systems in the MCS lab by executing the
swipl command in a terminal window. You can also download
SWI-Prolog for your own computer.
You will be working with the same grammar as in Lab 3:
<blck> ::= begin <stmts> end <stmts> ::= <empty> | <stmt> <stmts> <stmt> ::= pass | declare <name> | use <name> | <blck>
One minor change is the specification of
<name>; now any Prolog atom is a
<name>. You can test whether
X is bound to an atom using
This lab is subdivided into four stages so that you will have less to work on at any one time and have a greater chance of success. When you have succeeded at a particular stage, be sure to save your program out under a different filename before starting work on the next stage. That way, if you aren't successful at the new stage, you will still have your working program from the previous stage. If you get the final stage to work, just turn in your code from that stage. If not, turn in the code from the most advanced stage that you got to work and, if you have an interesting but non-working attempt at the next stage, you can turn that in as well. Be sure to label what you turn in so that I know how to interpret it.
You will define a predicate for each of the three nonterminals (
<blck>) similar to what I will demonstrate in class with
In order that you can work on one predicate at a time and test each predicate before you move to the next one, we need to initially break the cycle of mutual recursion between them. Therefore, you will start with the following simplified grammar for
<stmt> ::= pass | declare <name> | use <name>
Define a two-argument predicate
stmt such that
stmt(List1, List2) is provable if List1 starts with atoms
that match this simplified version of the
<stmt> nonterminal and then continues with the atoms in
List2. For example, the following query should produce
In this stage, you are not to worry about issues such as undeclared names and illegal redeclarations. The string of atoms should just be checked for conformity with the grammar.
Once you are satsified with your test results on this predicate, use it to define a
stmts(List1, List2) is provable if List1 starts with atoms
that match the
<stmts> nonterminal and then continues with the atoms in
List2. For example, the following query should produce
After you've tested that sequences of
use statements work, define a
blck(List1, List2) is provable if List1 starts with atoms
that match the
<blck> nonterminal and then continues with the atoms in
List2. When testing this, remember that nested
blcks are still not possible because you are still using the
simplified version of
For example, the following query should produce
Finally, you can modify the
stmt predicate to take into account the third production,
that is, the possibility that a statement is a nested block. That should allow queries such as the following one to
blck predicate, it is easy to write a
predicate that can test whether a list of atoms is legal, that is,
whether it is in the language described by the grammar. Namely, a
legal list of atoms consists of a
<blck> followed by nothing else:
legal(L) :- blck(L,).
One of the goals of parsing a language is to tell legal input strings from illegal ones, as in stage 1. However, there is another goal as well, which is to uncover the hierarchical structure that is implicit in the input. For that reason, modify all your predicates to take one more argument, which is to be used as in the following examples:
?- stmt([declare,x,stuff],[stuff],P). P = declare(x) ?- stmts([pass,use,x,stuff],[stuff],P). P = [pass, use(x)] ?- blck([begin,declare,x,use,x,end,stuff],[stuff],P). P = [declare(x), use(x)] ?- legal([begin,declare,x,begin,use,x,declare,y,end,use,y,end],P). P = [declare(x), [use(x), declare(y)], use(y)]
Note that the string of atoms used in the final example is treated as
legal even though the use of the name
y appears outside the scope of that name's declaration;
just as in stage 1, we are ignoring this issue for the moment.
That is about to change in stage 3.
Your goal for this stage is to make the preceding example produce
false because the given list of atoms contains a use of an undeclared
name. For lists without any uses of undeclared names, you should
produce the same result as in the previous stage.
To do this, you should add two more arguments to each of the
predicates that correspond to a nonterminal
stmt). Each of these new arguments is a list of names.
The first one is a list of the names that can legally be used in this
portion of the input and the second one is a list of the names that
can legally be used after it. For example, consider the following
?- stmt([use,y,moreinput],Tail,P,[x,y,z],Vars). Tail = [moreinput], P = use(y), Vars = [x, y, z]
This query is parsing a statement, which from the
variable's binding we can see is a use of the name
y. This is legal
y is one of the three names that the query
specified could legally have used. (The other two were
z.) The fact that
Vars is bound to the
exact same list of three names reflects the fact that the same names
are legal for use after this statement as were legal for use before
it. Here are two further examples:
?- stmt([use,w,moreinput],Tail,P,[x,y,z],Vars). false. ?- stmt([declare,w,moreinput],Tail,P,[x,y,z],Vars). Tail = [moreinput], P = declare(w), Vars = [w, x, y, z]
The first of these two queries returns
false because the list of
atoms now does not start with a legal
<stmt>. Although the first two
atoms are legal so far as the grammar goes, the use of
is now recognized as illegal because this name is not in the list of
names that may legally be used. In the second example, you can see
how the list of legal names is built up; one more name is legal after
the declaration of
Note that I am suggesting that the the
blck predicates have the same two new arguments as I illustrated above with
stmt. In the case of
stmts, the first would be a list of the names already legally usable at the start of the sequence of statements and the second would be the list of names legally usable after the full sequence. In the case of
blck, the second new argument will always be the same as the first, because after the block has ended, no new names are usable, even if some were declared within the block. The only reason to add two arguments rather than one is for the sake of consistency.
After modifying the three predicates for the three nonterminals,
you will also have to make a minor modification to the
legal(L,P) :- blck(L,,P,,).
With this definition, you should now be able to execute queries such as these:
?- legal([begin,declare,x,begin,use,x,declare,y,end,use,y,end],P). false. ?- legal([begin,declare,x,begin,use,x,declare,y,end,use,x,end],P). P = [declare(x), [use(x), declare(y)], use(x)]
Finally, you should also reject any input that contains a redeclaration of a name that has already been declared at the same nesting level, while allowing redeclaration of names defined in outer levels.
You can do this by adding two more arguments to each of the nonterminal predicates. As in the previous stage, each is a list of names, one for the situation before the current portion of the input and one for the situation afterward. However, this time the lists should contain those names that would be illegal to declare at the current nesting level because they are already declared at this level.