如何在 Prolog 中解决这个算术表达式难题？

我有一个编程问题 （https://blog.svpino.com/2015/05/08/solution-to-problem-5-and-some-other-thoughts-about-this-type-of-questions https://blog.svpino.com/2015/05/08/solution-to-problem-5-and-some-other-thoughts-about-this-type-of-questions):

编写一个程序，输出所有在 + 或 - 之间放置或不放置任何内容的可能性 数字 1、2、...、9（按此顺序），结果为 始终为 100。例如：1 + 2 + 34 – 5 + 67 – 8 + 9 = 100。

我用Python解决了这个问题11 个答案:

import itertools   
for operator in [p for p in itertools.product(['+','-',''], repeat=8)]:
    values = zip([str(x) for x in range(1, length+1)], operator) + ['9']
    code = ''.join(itertools.chain(*values))
    if 100 == eval(code):
        print "%s = %d" % (code, eval(code))

这是我的第二个更长的 Python 代码（https://gist.github.com/prosseek/41201d6508f01cf1643e https://gist.github.com/prosseek/41201d6508f01cf1643e):

[1, 2, 34, -5, 67, -8, 9]
[1, 23, -4, 56, 7, 8, 9]
[12, 3, -4, 5, 67, 8, 9]
[123, -4, -5, -6, -7, 8, -9]
[1, 23, -4, 5, 6, 78, -9]
[12, 3, 4, 5, -6, -7, 89]
[12, -3, -4, 5, -6, 7, 89]
[123, -45, -67, 89]
[123, 45, -67, 8, -9]
[1, 2, 3, -4, 5, 6, 78, 9]
[123, 4, -5, 67, -89]

我还在Prolog中找到了一个建议的解决方案 （http://www.reddit.com/r/programming/comments/358tnp/ Five_programming_problems_every_software_engineer/cr2dvsz http://www.reddit.com/r/programming/comments/358tnp/five_programming_problems_every_software_engineer/cr2dvsz):

sum([Head|Tail],Signs,Result) :-
   sum(Head,Tail,Signs,Result).
sum(X,[],[],X).

sum(First,[Second|Tail],['+'|Signs],Result) :-
   Head is First + Second,
   sum(Head,Tail,Signs,Result).
sum(First,[Second|Tail],['-'|Signs],Result) :-
   Head is First - Second,
   sum(Head,Tail,Signs,Result).
sum(First,[Second|[Third|Tail]],['+'|[''|Signs]],Result) :- 
   C    is Second*10+Third, 
   Head is First + C, 
   sum(Head,Tail,Signs,Result).
sum(First,[Second|[Third|Tail]],['-'|[''|Signs]],Result) :-
   C    is Second*10+Third,
   Head is First - C,
   sum(Head,Tail,Signs,Result).

然而，这给出了只有4个解决方案（不是预期的 11 个）：

?- sum([1,2,3,4,5,6,7,8,9],X,100).
X = [+, +, -,+, +, +,'',+] ;
X = [+, +,'',-, + '', -,+] ;
X = [+,'', -,+, +, +,'',-] ;
X = [+,'', -,+ '', +, +,+] ;
false.

这是因为''不显示为第一个列表项。所以解决方案[12,...] and [123,...]被跳过。

我尝试添加sum(First,[Second|Tail],[''|Signs],Result) :- Head is First*10 + Second, sum(Head,Tail,Signs,Result)., 但这样做会返回15 种解决方案，不是 11。

解释说，错误的解释1+23 to ((1)+2)*10+3.

?- sum([1,2,3], [+,''], Result). 
Result = 33.

那么，在Prolog中如何解决这个问题呢？如何教授 Prolog1 + 23 is 24在这个例子中？

Python eval 的对应部分可以通过以下方式实现read_term/3 and is/2, or

give_100(A) :-
    generate(1, S),
    atomic_list_concat(S, A),
    read_term_from_atom(A, T, []),
    T =:= 100.

generate(9, [9]).
generate(N, [N|Ns]) :-
    N < 9, sep(N, Ns).

sep(N, L) :-
    ( L = [+|Ns] ; L = [-|Ns] ; L = Ns ),
    M is N+1,
    generate(M, Ns).

示例查询：

?- give_100(X).
X = '1+2+3-4+5+6+78+9' ;
X = '1+2+34-5+67-8+9' ;
X = '1+23-4+5+6+78-9' ;
X = '1+23-4+56+7+8+9' ;
X = '12+3+4+5-6-7+89' ;
X = '12+3-4+5+67+8+9' ;
X = '12-3-4+5-6+7+89' ;
X = '123+4-5+67-89' ;
X = '123+45-67+8-9' ;
X = '123-4-5-6-7+8-9' ;
X = '123-45-67+89' ;
false.