Define a dcg for palindromes over the alphabet {a,b} of 2 symbols a and b such that the number of a's is one less than the number of b's

Question

I am trying to define a palindrome where the number of a's is one less than the number of b's. I cant seem to figure out how to write it properly

   please-->palindromes.
   palindromes-->[].
   palindromes-->[a].
   palindromes-->[b].
   palindromes--> [b],palindromes,[b].

your second and third base cases don't follow your rules. They are not palindromes where the number of `a`'s is one less than the number of `b`'s — joneshf, May 11 '13 at 22:48

score 0 · Answer 1 · answered May 09 '13 at 06:19

0

Think about this: where the surplus 'b' could stay ? In a palindrome, there is only one such place. Then change the symmetric definition, that in BNF (you already know as translate to DCG) would read

S :: P
P :: a P a | b P b | {epsilon}

answered May 09 '13 at 06:19

CapelliC

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The surplus b could stay in the middle but a b a b is a valid palindrome – user1850254 May 09 '13 at 06:29
Yes, in the middle. But abab is not a palindrome, should be abba. – CapelliC May 09 '13 at 06:30
of course, to get equal count of a and b, these must be in pairs, `P :: ab P ba | ...` – CapelliC May 09 '13 at 07:02
The number of b's has to be 1 greater than the number of a's. – user1850254 May 09 '13 at 07:11
"there is only one such place" - false. You seem to be implying the center letter must be `b`, but `bab` also satisfies the condition. – aschepler Jun 21 '14 at 17:31

score 0 · Answer 2 · answered Jun 21 '14 at 23:21

You're on the right track, you just need a way to deal with the difference in counts. You can do this by adding a numeric argument to your palindromes grammar term.

First I'll define an ordinary Prolog rule implementing "B is two more than A":

plus2(A,B) :- number(A), !, B is A+2.
plus2(A,B) :- number(B), !, A is B-2.
plus2(A,B) :- var(A), var(B), throw(error(instantiation_error,plus2/2)).

Then we'll say palindromes(Diff) means any palindrome on the given alphabet where the number of b letters minus the number of a letters is Diff. For the base cases, you know Diff exactly:

palindromes(0) --> [].
palindromes(-1) --> [a].
palindromes(1) --> [b].

For the recursive grammar rules, we can use a code block in {braces} to check the plus2 predicate:

palindromes(DiffOuter) --> [b], palindromes(DiffInner), [b],
    { plus2(DiffInner, DiffOuter) }.
palindromes(DiffOuter) --> [a], palindromes(DiffInner), [a],
    { plus2(DiffOuter, DiffInner) }.

To finish off, the top-level grammar rule is simply

please --> palindromes(1).

Define a dcg for palindromes over the alphabet {a,b} of 2 symbols a and b such that the number of a's is one less than the number of b's

2 Answers2