Prolog lists and recursion?

Use the accumulators to build up the lists in reverse order. Don't use add_number or you'll get a quadratic time algorithm, while you can solve this problem in linear time.

Use the accumulators to build up the lists in reverse order. Don't use add_number or you'll get a quadratic time algorithm, while you can solve this problem in linear time. Foo(, NumsR, Nums, Letters, Letters) :- reverse(NumsR, Nums).

Foo(X|Xs, NumsR, Nums, LettersR, Letters) :- % the following is the Prolog syntax for if-then-else; % you could also do this with two recursive clauses, % but this option is faster because of first-argument indexing (number(X) -> X1 is X+1, foo(Xs, X1|NumsR, Nums, LettersR, Letters) ; foo(Xs, NumsR, Nums, X|LettersR, Letters) ).

Foo(, Nums, Nums, Letters, Letters). Foo(X|Xs, Nums_1, Nums, Letters_1, Letters) :- number(X), X1 is X+1, add_number(Nums_1, X1, Nums_2), foo(Xs, Nums_2, Nums,Letters_1, Letters). Foo(X|Xs, Nums_1, Nums, Letters_1, Letters) :- letter(X), add_letter(Letters_1, X, Letters_2), foo(Xs, Nums_1, Nums, Letters_2, Letters).

Add_number(Nums_1,X,Nums_2) :- append(Numbs_1,X,nums_2). Add_letter(Letters_1,X,Letters_2) :- append(Letters_1,X,Letters_2).

I'd do it something like this: foo( List , Numbers , Letters ) :- worker( List , , Numbers , , Letters ). Worker( , Numbers , Numbers , Letters , Letters ). Worker( X|Xs , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :- digit(X), X1 is X+1 , append( NumberAccumulator , X1 , NumberAccumulator1 ) , worker( Xs , NumberAccumulator1 , Numbers , LetterAccumulator , Letters ).

Worker( X|Xs , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :- letter(X) , worker( Xs , NumberAccumulator , Numbers , X|LetterAccumulator , Letters ). Worker( X|Xs , NumberAccumulator , Numbers , LetterAccumulator , Letters ) :- not letter(X) , not digit(X) , worker( Xs , NumberAccumulator , Numbers , LetterAccumulator , Letters ). Letter( a ).

Letter( be ). Letter( c ). ... letter( z ).

Letter('A'). Letter('B'). Letter('C').... letter('Z').

Digit('0'). Digit('1'). Digit('2').... digit('9').

Since this is a learning exercise, I'd not defer the reversal of the list: I'd do the obvious and build the list in reverse sequence, despite the performance hit. I believe the point of the exercise is that you need to learn to build lists both ways.

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