Fun Facts about Prolog
There is more to say about Prolog than can ever be said. At
the same time, you also need to collect your own experiences
with the language, and ponder certain aspects for yourself.
Here are a few true facts about Prolog to help you
begin your own observations and reflections.
Tail recursion often arises naturally
In Prolog, due to the power of logic variables, many
predicates can be naturally written in a
tail recursive way.
For example, consider
list_list_together(, Bs, Bs).
list_list_together([A|As], Bs, [A|Cs]) :-
list_list_together(As, Bs, Cs).
It is easy to see that list_list_together(As, Bs, Cs) is
a call in a tail position, because no further
Contrast this with a function like append in Lisp:
(defun append (x y)
(cons (car x) (append (cdr x) y))
This version of append is not tail recursive
in Lisp, because the recursive call is wrapped within a call
of cons and thus not the final
function call in this branch.
It is somewhat remarkable that such a basic function is not
naturally tail recursive in Lisp, but the corresponding
relation is tail recursive in Prolog!
In Prolog, many more predicates are naturally tail recursive.
Tail recursion is sometimes inefficient
In many cases, tail recursion is good for performance: When the
predicate is deterministic, a tail call means that the
Prolog system can automatically reuse the allocated space
of the environment on the local stack. In typical cases, this
measurably reduces memory consumption of your programs, from
O(N) in the number of recursive calls to O(1). Since
decreased memory consumption also reduces the stress on memory
allocation and garbage collection, writing tail recursive
predicates often improves both space and
time efficiency of your
However, take into account that many Prolog programs
are not deterministic. As long as choice points
remain, the current environment on the local stack cannot be
discarded, because it may still be needed on backtracking.
For example, let us define the
relation list_element_rest/3 between
a list, one of its elements, and the
list without that element:
list_element_rest([L|Ls], L, Ls).
list_element_rest([L|Ls0], E, [L|Ls]) :-
list_element_rest(Ls0, E, Ls).
- The relation quite obviously holds for the
list [L|Ls], its
first element L, and the remainder Ls.
- If the relation holds for the
list Ls0, one of its elements E,
and the remainder Ls, then the relation
also holds for [L|Ls0] and E, and the
remainder [L|Ls]. This rule is
clearly tail recursive, because the recursive
call is its only goal.
This predicate is quite versatile. Operationally, we can use it
to remove one element from a list, to add an element
to a list, and also in several other directions to either
generate, complete or test possible solutions. For example:
?- list_element_rest([a,b], E, Rest).
E = a,
Rest = [b] ;
E = b,
Rest = [a] ;
?- list_element_rest(Ls, c, [a,b]).
Ls = [c, a, b] ;
Ls = [a, c, b] ;
Ls = [a, b, c] ;
for Prolog draft and several Prolog systems, an almost
identical predicate is available under the
Using list_element_rest/3 as a building block, we now
define list_permutation1/2 to describe the relation
between a list and one of its permutations:
list_permutation1(Ls, [E|Ps]) :-
list_element_rest(Ls, E, Rs),
Note that that this predicate is also tail recursive.
Here is an example query:
?- list_permutation1([a,b,c], Ps).
Ps = [a, b, c] ;
Ps = [a, c, b] ;
Ps = [b, a, c] ;
Ps = [b, c, a] ;
Ps = [c, a, b] ;
Ps = [c, b, a] ;
Let us now run a few benchmarks. We generate all solutions
for lists of length 9, 10 and 11:
?- L in 9..11, indomain(L), portray_clause(L),
% 3,322,110 inferences, 0.440 CPU in 0.457 seconds (96% CPU, 7546288 Lips)
% 33,221,103 inferences, 4.447 CPU in 5.166 seconds (86% CPU, 7470928 Lips)
% 365,432,136 inferences, 49.162 CPU in 58.028 seconds (85% CPU, 7433257 Lips)
Now consider an alternative definition of this relation, which we
list_permutation2([L|Ls0], Ps) :-
list_element_rest(Ps, L, Ls).
This version is not tail recursive. Let us run the same
benchmarks with this new version:
?- L in 9..11, indomain(L), portray_clause(L),
% 772,005 inferences, 0.094 CPU in 0.099 seconds (94% CPU, 8252942 Lips)
% 7,666,725 inferences, 0.922 CPU in 1.044 seconds (88% CPU, 8313976 Lips)
% 83,871,526 inferences, 10.084 CPU in 13.271 seconds (76% CPU, 8317163 Lips)
Note that this version is several times faster in each of
the above cases! If you care a lot about performance, try to
understand why this is so.
Together with the previous section, this example illustrates that
tail recursion does have its uses, yet you should not
overemphasize it. For beginners, it is more important to
and nontermination, and to focus
on clear declarative descriptions.
Most cuts are red
Almost every time you add a !/0 to your program, it will
be a so-called red cut. This means that it will make
your program wrong in that it incorrectly omits some
solutions only in some usage modes.
This is a tough truth to accept for most Prolog programmers. There
always seems hope that we can somehow outsmart the system and get
away with green cuts, which improve performance and
honor all usage patterns. This hope is quite unfounded, because
there typically are simply too many cases to consider, and you
will almost invariably forget at least one of them.
Especially if you are used to imperative programming, it is easy
to fall into the trap of ignoring general cases.
For example, many beginners can correctly write a Prolog predicate
that describes a list:
This is a very general relation that works in all directions. For example:
Ls =  ;
Ls = [_15346] ;
Ls = [_15346, _15352] .
After seeing they can actually affect the control flow of Prolog
with !/0, many beginners will incorrectly
apply this construct in cases where it cuts away solutions.
For example, from a quick first glance, it may appear that the two
clauses are mutually exclusive. After all, a list is
either empty or has at least one element, right? And
so the horror begins when
list() :- !. % incorrect!
A quick test case "confirms" that everything still works:
The flaw in this reasoning is that the clauses are in
fact not mutually exclusive. They are only exclusive if the
argument is sufficiently instantiated!
In Prolog, there are more general cases than you may be
used to from other programming languages. For example, with
the broken definition, we get:
Ls = . % incompleteness!
Thus, instead of an infinite set of solutions, the
predicate now only describes a single solution!
To truly benefit from declarative programming, stay in
the pure subset.
Use pattern matching or meta-predicates
like if_/3 to
efficiently express logical alternatives while retaining full
generality of your relations.
flatten/2 is no relation
In some introductory Prolog courses, you will be required to
implement a predicate called flatten/2.
In such cases, consider the following query:
?- flatten(Xs, Ys).
What is a valid answer in such situations? Suppose your predicate
answers as follows:
Ys = [Xs].
From a declarative point of view, this answer
is wrong! The reason is that Xs may as well
be a list, and in such cases, the result
is not flattened! Witness the evidence for yourself:
?- flatten(Xs, Ys), Xs = [a].
Xs = [a],
Ys = [[a]].
Thus, Y is not a flat list, but
a nested list! In contrast, exchanging the goals yields:
?- Xs = [a], flatten(Xs, Ys).
Xs = Ys, Ys = [a].
This means that exchanging the order of goals changes the
set of solutions.
Your instructor should be able to understand this fundamental
declarative shortcoming if you point it out. In practice,
use append/2 to remove precisely one level
of nesting in a sound way.
Iterative deepening is often a good strategy
Prolog's default search strategy is incomplete, and we can
often easily make it complete by turning it
into iterative deepening. This means that we search
for solutions of depth 0, then for those of depth 1, then
for those of depth 2, etc.
From a quick first glance, this may seem very inefficient to you,
because we are visiting the same solutions over and over, although
we need each of them only once.
Now consider a search tree of depth k, with a
branching factor of b. With iterative deepening
up to and including depth k, we visit the root
node k+1 times, since we start at depth 0, and
revisit the root in each iteration, up to k. Further,
we visit the b nodes at depth 1
exactly k times. In general, we
visit the bj nodes at
depth j (0≤j≤k)
exactly k+1−j times.
Let us now sum this up to calculate the total number of visits:
b0(k+1) + b1k +
b2(k−1) + …
Now the point: This sum is asymptotically dominated
(Exercise: Why?) by the number of visits at the final
depth k, where we visit
bk nodes, each
exactly once. This shows that iterative deepening is
in fact an asymptotically optimal search strategy
under very general assumptions, because every
uninformed search strategy must visit
these bk nodes at least once
It also shows that people usually underestimate what
exponential growth really means. If you find yourself in
a situation where you really need to traverse such a
search tree, iterative deepening is often a very good
strategy, combining the memory efficiency of
depth-first search with the completeness of
breadth-first search. In addition, iterative deepening is
easy to implement in Prolog via its built-in backtracking.
Note that iterative deepening requires monotonicity!
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