Tips on Writing Recursive Procedures in Prolog

The Basics - Base Case and Recursive Case(s)

When writing a recursive procedure in Prolog, there will always be at least two rules to code: at least one rule for the base case, or non-recursive case, and at least one rule for the recursive case.

The Base Case

Typically the base-case rule(s) will deal with the smallest possible example(s) of the problem that you are trying to solve - a list with no members, or just one member, for example. If you are working with a tree structure, the base case might deal with an empty tree, or a tree with just one node in it (like tree(empty, a, empty)).

The Recursive Case

To write the recursive case rule(s), you need to think of how the current case of the problem could be solved assuming you had already solved it for all smaller cases. For example, if you were adding a list of 10 numbers, and you had a way of summing the last 9 numbers in the list, then you would do so, then add on the first number on the list. (It might seem more natural to add up the first 9 numbers and then add the last number to the subtotal, but in Prolog it is easy to access the first item in the list, but not the last.)

Example 1: adding up a list of numbers

Here's an example of how to apply this to adding up a list of numbers. The comments beginning %% would not normally appear - they are there this time to help you match the scheme described in the previous paragraphs to the code.

% addup(List, Sum)
% Binds Sum to the sum of the numbers in the list.
% Assumes that each member of the list really is a number.
% The List must be instantiated at the time addup is called.
% Example of use:
% ?- addup([1,2,3,4], X).
% X = 10

%% base case
addup([], 0). % sum of the empty list of numbers is zero

%% recursive case: if the base-case rule does not match, this one must:
addup([FirstNumber | RestOfList], Total) :-
	addup(RestOfList, TotalOfRest),   % add up the numbers in RestOfList
	Total is FirstNumber + TotalOfRest.
So the recursive call addup(RestOfList, TotalOfRest), two lines above, adds up all the items in the list except the first (FirstNumber), and binds the sum to TotalOfRest, and then the line after the recursive call, Total is FirstNumber + TotalOfRest, binds Total to the sum of the FirstNumber and the number TotalOfRest returned by the recursive call.

In one sense, that's all you need to know about the operation of addup. However, you may wonder how it all works when the first recursive call results in a second recursive call, and so on, and how the Prolog interpreter keeps track of all the different TotalOfRest instances, with recursive calls nested inside one another. You can get an idea of this by using Prolog's trace facility to add up the numbers in the list [3, 5, 7]. The material in bold is what the user types:

?- trace.
[trace]  ?- addup([3, 5, 7], Total).
   Call: (7) addup([3, 5, 7], _G322)
   Call: (8) addup([5, 7], _L177)
   Call: (9) addup([7], _L197)
   Call: (10) addup([], _L217)
   Exit: (10) addup([], 0)
^  Call: (10) _L197 is 7+0
^  Exit: (10) 7 is 7+0
   Exit: (9) addup([7], 7)
^  Call: (9) _L177 is 5+7
^  Exit: (9) 12 is 5+7
   Exit: (8) addup([5, 7], 12)
^  Call: (8) _G322 is 3+12
^  Exit: (8) 15 is 3+12
   Exit: (7) addup([3, 5, 7], 15)
Total = 15.

[debug]  ?- notrace.

Note that Prolog actually stops at the end of each line, prints a ? prompt, and you press Return to get the next line of output. This has been edited out of the transcript above to reduce the clutter.

Note that Prolog replaces Total with _G322 in the first line of tracing, labelled Call: (7).

In the next line, labelled Call: (8), it is executing the recursive rule for addup.

For each recursive call to addup, Prolog replaces the name TotalOfRest with a new variable name, guaranteed to be different from any other variable name currently in use.

So in the first recursive call: addup([3, 5, 7], _L177), Prolog replaces TotalOfRest with _L177: that is, Prolog uses the name _L177 for the instance of TotalOfRest in the first recursive call to addup. It's as though the recursive rule being executed has been replaced, just for this call, by:

addup([FirstNumber | RestOfList], Total) :-
	addup(RestOfList, _L177),   % add up the numbers in RestOfList
	Total is FirstNumber + _L177.

Similarly, in the second recursive call, addup([5, 7], _L197), Prolog is using the name _L197 for the instance of TotalOfRest in this nested recursive call to addup, and in the third recursive call, addup([7], _L217), Prolog uses the name _L217 for the instance of TotalOfRest in this doubly-nested recursive call to addup.

There is no fourth recursive call to addup, as the next goal, addup([], TotalOfRest), triggers the base case rule.

Further down the trace, as the recursive calls complete and the recursion "unwinds", you can see Prolog implicitly binding _L217 to 0 (addup([], 0) - using the base case rule) and explicitly binding _L197 to the value of 7+0, and binding _L177 to the value of 5+7 = 12, and finally binding _G322 to the value of 3+12.

To find out more about the trace facility, and/or to see another example of tracing recursive calls, check out the Prolog Dictionary entry on tracing.

Example 2: finding the last item of a list

This example shows how to find the last item in a list:

% lastitem(List, Last)
% Binds Last to the item at the end of List.
% List must be instantiated at the time of call, and must not be empty.
% Example of use:
% ?- lastitem([a,b,c,d], X).
% X = d

%% base case: list with just one item.
lastitem([OnlyOne], OnlyOne).

%% recursive case: ignore first item, seek last item of rest of list
lastitem([First | Rest], Last) :-
	lastitem(Rest, Last).

Example 3: Squaring each item in a list of numbers

This example shows how to build a result that is a list, i.e. we are transforming a list to produce a new list.

This time we'll start with a version with bugs (= errors) in it and also a stylistic error, and then we'll correct the errors. With each version of the code, you should look carefully at it before reading on, and try to spot the bug (or stylistic error). Spotting the bugs and stylistic errors is something you will have to do for yourself when you write your own programs, so start on it now!

Every programmer makes errors like these at some point. My aim in working through several different versions of this procedure is to give you an idea of how Prolog will respond to programmer errors. Prolog's messages are part of the information that the programmer can use to figure out what is wrong with their program.

NB: square_1, square_2, square_3, and square_4 are all wrong. Only square_5 is correct.

For the sake of brevity in this exposition, I've left out the comments in all but the final version. In practice you would develop the comments as you wrote the code. However, remember that just because you say something is true about the code, in your comments, doesn't mean it is. And don't forget to review and if necessary correct your comments when you find and fix a bug.

Different dialects of Prolog may produce different error messages: the messages below come from SWI-Prolog.

square_1([First | Rest], [First * First | SquareRest]) :-
	square_1(Rest, SquareRest).
Try it out:
?- square_1([1, 3, 5], Squares).

Oops - we left out the base case. If you turn on tracing before you execute the query above, you will be able to see 1 * 1, 3 * 3, and 5 * 5 being produced, but because there is no base case, Prolog cannot complete the query.

square_2([], []).
square_2([First | Rest], [First * First | SquareRest]) :-
        square_2(Rest, SquareRest).
Try it out again:
?- square_2([1, 3, 5], Squares).
Squares = [1*1, 3*3, 5*5].

That's better, but why didn't it work out that 1 * 1 = 1, 3 * 3 = 9, etc.?
Answer: because we didn't ask it to. In Prolog, in most cases, evaluation of an arithmetic expression must be explicitly called for using the built-in predicate is.

square_3([], []).  
square_3([First | Rest], [Square | SquareRest]) :-
        FirstSquared is First * First,
        square_3(Rest, SquareRest),
        Square = FirstSquared.
Try it:
?- square_3([1, 3, 5], Squares).
Squares = [1, 9, 25].

It works. But it contains a stylistic problem. The final goal Square = FirstSquared is redundant - it can be done better by simply writing FirstSquared instead of Square in the head of the rule:

square_4([], []).  
square_4([First | Rest], [Firstsquared | SquareRest]) :-
        FirstSquared is First * First,
        square_4(Rest, SquareRest).
Always try it out after any change:
?- square_4([1, 3, 5], Squares)
Squares = [_G263, _G269, _G275].

Why doesn't it work any more?
Answer: We mistyped FirstSquared as Firstsquared in the head of the rule. Prolog cares desperately about upper and lower case letters, and the "s" of Squared is lower case in the head of the rule, but upper case in the first goal of the body. This can lead to a range of mystifying error messages - the type above, with one or more unresolved _G… variables, is one possibility.

When you re-loaded your code into Prolog, you would have received a warning message too (sometimes the warning is hard to notice because of Prolog's welcome message). The warning message is about "singleton variables" - that is, one or more variables that are only used once. In this case, the singleton variables were Firstsquared and FirstSquared. This could help you notice the problem with the capitalisation of the two instances of this variable.

%% base case
% nothing to square; result is empty list.
square_5([], []).  
%% recursive case
% square the First item, recursively square the Rest. 
square_5([First | Rest], [FirstSquared | SquareRest]) :-
        FirstSquared is First * First,
        square_5(Rest, SquareRest).
Always try it out - even if you just added comments, in case you forgot to put a % sign at the front of a comment line.
?- sqare_5([1, 3, 5], Squares).
Correct to: "square_5([1, 3, 5], Squares)"? yes
Squares = [1, 9, 25].
We mis-typed square_5 as sqare_5 in the query. Fortunately, SWI-Prolog has a feature called DWIM (Do What I Mean) which notices that there is no procedure called sqare_5 and attempts spelling correction - in this case successfully. It suggests square_5, we agree by typing "y", and then Prolog prints the answer out.
© Bill Wilson, 2004-2005. Modified 2009 for consistency with SWI-Prolog behaviour.
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