带有列表的Prolog程序

Question

>我必须解决 turbo prolog 中的以下问题作为家庭作业："确定列表中表示为数字的数字与给定数字的乘积。例如：[1 9 3 5 9 9] * 2 --> [3 8 7 1 9 8] "。

我

解决这个问题的思路是，我首先计算乘积，然后将其数字放入列表中。只是我真的无法弄清楚最后一部分。这是我到目前为止的源代码：

域

list=integer*

谓词

    length(list,integer)
powerten(integer,integer)
product(integer,list,integer) /* this predicate computes the product */
    /* the product,powerten and length are taken care of */

addDigit(integer,list) /* this predicate should decompose the number in its digits and put them in the list */

productList(integer,list,list)

第

     length([],0).
     length([_|T],L):-
            length(T,L1),
            L=L1+1.
powerten(0,1):-!.
powerten(L,N):-
    L1=L-1,
    powerten(L1,N1),
    N=N1*10.
product(_,[],0):-!.
product(NR,[H|T],RESULT ):-
    length([H|T],LEN),
    L2=LEN-1,
    powerten(L2,N),
    product(NR,T,R1),
    RESULT=R1+H*N*NR.


addDigit(0,[]):-!.

addDigit(NR,[NR|_]):-
    NR>0,
    DIGIT = NR MOD 10,
    NR1=NR DIV 10,
    addDigit(NR1,_).        

productList(NR,L1,L2):-
     /* this is the "main" predicate . Its arguments are NR - the first factor, L1- the 
         initial list, whose digits make the second factor, L2 - the result list which            
          contains  the digits of he result */
            product(NR,L1,RESULT),
            addDigit(RESULT,L2).

如您所见，在addDigit谓词之前一切都很好。我只是找不到将产品数字添加到最终列表中的方法。谁能帮我解决问题？谢谢。

Answer 1

这对我来说似乎很复杂。问题基本上是做长格式乘法，就像你在纸上做的那样。如果您首先反转列表（使用内置的reverse/2谓词），事情会变得简单得多：

%--------------------------------------------------------------%
% Mult/3: multiply a integer, represented as a list of digits, %
%         by an integer value N, producing an integer, also    %
%         represented as a lsit of digits.                     %
%--------------------------------------------------------------%
multiply( Ds , N , Result) :-
  reverse(Ds,Rs) ,
  multiply( Rs , N , 0 , T ) ,
  reverse( T , Result )
  .
%
% the worker predicate that does all the work
%
multiply( [] , _ , C , []      ) :-    % if the list is exhausted
  C =< 0                               %   and the carry is 0,
  .                                    %   we're done. C'est fini.
multiply( [] , _ , C , [C] ) :-        % if the list is exhausted
  C > 0 ,                              %   and the carry is 1-9, then
  C < 10                               %   The last digit in the result is the carry
  .                                    %   We're done. C'est fini.
multiply( [] , _ , C , [R|Rs] ) :-     % If the list is exhausted,
  C >= 10 ,                            %   and the carry is 10+,
  R is C rem 10 ,                      %   the next digit in the result is the carry modulo 10
  Q is C div 10 ,                      %   take the quotient
  multiply( [] , _ , Q , Rs )          %   and recurse down with the quotient as the carry
  .                                    %
multiply( [D|Ds] , N , C , [R|Rs] ) :- % if the list is NOT exhausted,
  T is D*N + C ,                       %   compute the product for this digit
  R is T rem 10 ,                      %   the next digit in the result is that product modulo 10
  Q is T div 10 ,                      %   the next carry is the quotient
  multiply( Ds , N , Q , Rs )          %   recurse down
  .                                    $ Easy!

Answer 2

addDigit 对我来说似乎是错误的：你只使用第一个，而没有指定尾巴！

尝试

addDigit(NR,[DIGIT|Ns]):-
    NR>0,
    DIGIT = NR MOD 10,
    NR1 = NR DIV 10,
    addDigit(NR1, Ns).