prolog pascal triangle

hi is there anybody know how can i do the pascal nth row when i开发者_如何学运维 ask for :? pascal(2,Row). i get Row=[1,2,1] ??

please help me

Here is the code to compute the nth row.

The first part scans a row, to compute the next row. The first row must be prefixed with a 0, so that the first "1" in the next row is a sum, like the other elements. It recurses on the 2 lists:

pascal_next_row([X],[X]).
pascal_next_row([H,H2|T],[A|B]):-
    pascal_next_row([H2|T],B),
    A is H + H2.

The second part computes all the rows until the one which was asked. It recurses on N:

pascal(0, [1]) :- !.
pascal(N, R) :-
    N1 is N-1,
    pascal(N1, R1),
    pascal_next_row([0|R1], R).

If you need the full triangle, all you have to do is change the second parameter to handle a list of rows, and collect them:

pascal(0, [[1]]) :- !.
pascal(N, [R, R1 | RN]) :-
    N1 is N-1,
    pascal(N1, [R1 | RN]),
    pascal_next_row([0|R1], R).

This answer to a code golf has the implementation in prolog, just expand the names:

The Pascal Triangle is also known as the Tartaglia Triangle:

sumC([X,Y],[Z]) :- Z is X + Y.
sumC([X,Y|L], Z):- H is X + Y,
                    sumC([Y|L],L2),
                    Z = [H|L2].

tartaglia(1,[1]) :- ! .
tartaglia(2,[1,1]) :- !.
tartaglia(N, L) :- Ant is N - 1,
                   tartaglia(Ant,L2),
                   sumC(L2,R),
                   append([1|R],[1],L), !.

Using the helper predicate sumC, you can get it easily:

?- tartaglia(3,R).
R = [1, 2, 1].

?- tartaglia(2,R).
R = [1, 1].

?- tartaglia(1,R).
R = [1].

?- tartaglia(6,R).
R = [1, 5, 10, 10, 5, 1].

As said in my comment. You ask for the nth row. [1,2,1] from your example is the 3rd row.