Adventures in Prolog - Computations with lists VII

Again, as I am quite busy these days, I will present some not recent predicates, just to keep the flow going. Quite recently, I have seen the world of depth, breadth and intelligent search in Prolog. For this, I am working on an implementation of the 8-queens problem. Unfortunately, this is nothing to show off at the moment. But probably in 2 weeks or so, I will be able to show present it here and demonstrate how elegantly one may solve problems using Prolog.

Today's post is about the detection of a palindrome using Prolog. A palindrome is defined as some word that can be read from start to end or end to start and it will not change. For example, the word ANNA is a palindrome.

Now, using the idea that a palindromic word is a word that can be read back and forth and does not change, we come with the following ...

Let X be some word. If X is the empty word then it is a palindrome. If X is not the empty word then let Y be the reverse of X. If Y equals X, X is a palindromic word. Instead of focussing on words, we focus on lists.

%palin/1 X is a palindrome
palin([]).
palin(X) :- revList(X, Y), X = Y.

A more succinct version of the above would avoid the additional test of equality, i.e. lead to the following predicate:

palin(X) :- revList(X, X).

Which clearly is the the same, i.e. if reversing X leads to X, then clearly, X has to be palindromic. revList is a binary predicate that takes a list as input and outputs the reverse list.

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