Prolog in Python. Version 2

This second version of our Prolog interpretor implements only a single new feature. But that feature will enable the use of trees and lists and is necessary for more interesting programs.

The feature is simply this. In addition to constants and variables, arguments in a term may be other terms. This complicates parsing and unification, because both will now need to be recursive, but leaves other parts of the program pretty much intact.

In this chapter, we'll look at compound terms (terms that contain other terms) only in regard to building Prolog lists.

Building lists with compound terms

Prolog does not support lists directly but builds them as a set of nested terms The predicate "." is used to bind the elements of the list together. The term ".(a,b)" may be represented as a mini tree

.__ b

and .(a, .(b, .(c, .()))) becomes

|  |  |
a  b  c

The first tree may also be represented in Prolog with the syntax "[a|b]" and the second, which is a proper list may be represented by the familiar "[a,b,c]". The term ".()" is the empty list represented by "[]". With the square bracket syntax, Prolog lists look just like Python lists. Prolog lists may contain constants, variables, and of course other lists.

Compound terms, and therefore lists, unify in the same way that we are already familiar. That is, the predicates must match, they must have the same number of arguments, and each argument must unify with the corresponding one in the other term. But since arguments themselves may be terms the process must be done recursively. Here are some examples. assume all variables in the second term are unbound.

Unification ...
[a,b,c]  with X        binds X to the [a,b,c]
[a,b,c]  with [X,b,c]  binds X to a
[a,b,c]  with [X|Y]    binds X to a, Y to [b,c]

It is important to remember that internally in the program lists are nested terms with the "." predicate, but that they are translated to and from the square bracket notation for input and display. The pipe character "|" effectively splits a list into its first element and the remaining elements.

A better parser for terms

Click here to access the Python code for You may find it convenient to save it into a file and access it with a text editor.

In the first version of our Prolog interpreter we used the split function in the string module to break a term like "mother(mary,bill)" first into a predicate "mother" and arguments "mary,bill" (by splitting on the "(" character), and later splitting the arguments apart (on the commas). Terms in a rule are also separated by commas but we got around that by using re.sub to turn just those commas into semicolons, and then splitting the terms apart.

Alas, this approach just won't work anymore. If we have nested list like "[a,[b,c,d],e]", we want to extract three elements, "a", "[b,c,d]", and "e". Not "a","[b","c","d]","e".

To accommodate this contains its own split function that splits only when a separator is at a zero nesting level. As a string is scanned, the nesting level is increased by one whenever a "(", or "[" is encountered, and decreased by one whenever a ")" or "]" is encountered. In addition an optional "All" parameter, when 0, lets us split off just the first predicate in the term.

As in terms are objects and compiled from text. There is also a mode where a predicate and argument list may be compiled into a term. (line 35). This is used by the eval function. With text however, either a list (line 39) or a standard term may be compiled. But now that arguments are also terms, Term initialization is recursive. The Python map function builds all the argument terms with a single call. (lines 42 and 52). Finally a simple constant or variable is a term with no arguments, like it was in (line 55).

The Term __repr__ method will determine how terms are displayed when printed or passed to the builtin str function. Since lists are really nested terms using the "." predicate, they are now translated back to the square bracket notation. (line 59). Other terms are output the same way they were in

Let's play with the Term class a bit

>>> from prolog2 import *
>>> a = Term("bill")
>>> print a
>>> a.pred
>>> a.args

Here we have compiled the simplest possible term, a constant. It prints its name and in order to see the insides (pred and args) we need to look at the attributes directly.

>>> b = Term("tasty(food(meat))")
>>> b
>>> b.args

This is a compound term showing the argument of "tasty" as a nested term.

Next we'll construct a list with the "." operator directly.

>>> c = Term(".(a,.(b,.()))")
>>> c

But of course, it is more natural to create lists from the list syntax itself.

>>> d = Term("[x,y,[a,b,c],z]")
>>> d

We can also use the "|" operator to prepend 'x' onto the list "[y,z]".

>>> e = Term("[x|[y,z]]")
>>> e

An improved unify function.

The Rule and Goal classes are the same as in but unification is more complex since looking up variables needs to be done recursively. The function unify returns one if the source term can be unified to the destination term. Variables in each are looked up in their own environments and destination variables may be bound to constants in the source.

The new function eval is used to "look up" variables. (line 206). eval is recursive so that if a term is evaluated, each argument is evaluated internally. eval then returns a term with all variable references converted to constants or the value None if this is impossible.

The function unify returns true (1) if the source (or parts of it) are still variables. Otherwise it returns true if destination can be matched to it piece by piece in basically the same manner as in

Notice that all returns from the function unify are channeled through the function sts. This is for tracing purposes and results in nested unifications being indented in a trace. Let's play with this.

First we'll set up an empty destination environment "e" and unify a constant list with a similar list with variables.

>>> import prolog2
>>> e = {}
>>> prolog2.unify(Term("[a,b,c]"),{},Term("[A,B,C]"),e)
>>> e
{'B': b, 'C': c, 'A': a}

Now we'll unify again (just matching values) since e is already set. But this time we'll turn on the trace in the module prolog2.

>>> prolog2.trace = 1
>>> prolog2.unify(Term("[a,b,c]"),{},Term("[A,B,C]"),e)
>>> prolog2.unify(Term("[a,b,c]"),{},Term("[A,B,C]"),e)
 Unify [a,b,c] {} to [A,B,C] {'A': a, 'C': c, 'B': b}
   Unify a {} to A {'A': a, 'C': c, 'B': b}
     Unify a {} to a {'A': a, 'C': c, 'B': b}
     Yes All args unify
   Yes Unify to Dest value
   Unify [b,c] {} to [B,C] {'A': a, 'C': c, 'B': b}
     Unify b {} to B {'A': a, 'C': c, 'B': b}
       Unify b {} to b {'A': a, 'C': c, 'B': b}
       Yes All args unify
     Yes Unify to Dest value
     Unify [c] {} to [C] {'A': a, 'C': c, 'B': b}
       Unify c {} to C {'A': a, 'C': c, 'B': b}
         Unify c {} to c {'A': a, 'C': c, 'B': b}
         Yes All args unify
       Yes Unify to Dest value
       Unify [] {} to [] {'A': a, 'C': c, 'B': b}
       Yes All args unify
     Yes All args unify
   Yes All args unify
 Yes All args unify

Two examples with lists

Finally we'll examine two classic operations with lists, element membership and the appending of lists.

? member(X,[X|T])
? member(X,[H|T]) :- member(X,T)

Two rules determine membership. If X is the head of a list or if X is a member of the tail of the list. In the first case it doesn't matter what the tail "T" of the list is. In the second case it doesn't matter what the head "H" of the list is.

With these two definitions we can test for membership

? member(a,[a,b,c])?
? member(b,[a,b,c])?

And finally Prolog can compute "backwards" so to speak to find all the members of the list. In this mode we have something similar to the Python "for" statement.

? member(X,[a,b,c])?
{'X': a}
{'X': b}
{'X': c}

The rules for append are a bit trickier. It's always wise to remember that the rules constrain the solution but don't specify the computation. Here are the two rules for append. The first two arguments are the input lists and the third argument is the result.

? append([],L,L)
? append([X|A],B,[X|C]) :- append(A,B,C)

The first rule says that any list "L" appended to the empty list results in the same list "L". The second rules says that if C is A appended to B, then it's also true if X is prepended to both A and C.

Let's run some tests computing in both directions.

? append([a,b],[c,d],X)?
{'X': [a,b,c,d]}

? append([a,b],Y,[a,b,c,d])?
{'Y': [c,d]}

? append(X,Y,[a,b,c])?
{'X': [], 'Y': [a,b,c]}
{'X': [a], 'Y': [b,c]}
{'X': [a,b], 'Y': [c]}
{'X': [a,b,c], 'Y': []}

In the last example, Prolog gives us all possible ways the list "[a,b,c]" can be constructed. I think this is kind of wild.

Copyright © 2009-2015 Chris Meyers