Question

我正在进行一项任务，我已经为前缀表示法算术语言创建了一个解析器。我需要编写一个谓词，它为任何给定的值V建立一个ast（即生成一个A，这样无论何时A被评估，它的值都是V）。我的想法很简单：

genAst(Val, Env, Ast) :-
   ev(Ast, Env, Val).

其中ev是evaluate-predicate。当我运行这个时，我得到关于ev-predicate的这一部分的标题错误：

ev(xer_(power(N)), Env, V) :-
   integer(N),
   V is Env^N. %THIS LINE

其中V和N都是未绑定的。我正在努力想到另一种优雅的方法来做到这一点，有谁知道如何让prolog为这两个变量生成整数？

我希望这是可以理解的：）

Answer 1

使用library(clpfd)。它包含这种功能。只要您不需要它们，就没有必要生成具体的值！

?- X #= Y^Z.
Y^Z#=X.

?- X #= Y^Z, [Y,Z]ins 1..3.
 Y^Z#=X,
Y in 1..3,
Z in 1..3.

?- X #= Y^Z, [Y,Z]ins 1..3, labeling([], [Y,Z]).
X = Y, Y = Z, Z = 1 ;
X = Y, Y = 1,
Z = 2 ;
X = Y, Y = 1,
Z = 3 ;
X = Y, Y = 2,
Z = 1 ;
X = 4,
Y = Z, Z = 2 ;
X = 8,
Y = 2,
Z = 3 ;
X = Y, Y = 3,
Z = 1 ;
X = 9,
Y = 3,
Z = 2 ;
X = 27,
Y = Z, Z = 3.

Answer 2

正如所提出的那样，你的问题似乎无法解决，那么我认为我会以不同的方式处理整个问题，生成所有 AST，最多可达一个令牌数。

genAst(Val, Env, Ast) :-
    length(Tokens, N),
    (N > 10, !, fail ; true),
    phrase(sum(Ast), Tokens),
    ev(Ast, Env, Val).

sum(sum(A,B)) --> [+], mul(A), sum(B).
sum(N) --> mul(N).

mul(mul(N,X)) --> [*], xer(X), num(N).
mul(N) --> xer(N).

xer(exp(x,N)) --> [^,x], num(N).
xer(var(x)) --> [x].
xer(N) --> num(N).

%num(num(X)) --> [X], {var(X) -> between(1,9,X) ; integer(X)}.
num(num(X)) --> [X], {X=2;X=3}.

产量

?- genAst(6,2,A).
A = sum(num(3), num(3)) ;
A = mul(num(3), var(x)) ;
A = mul(num(3), num(2)) ;
A = mul(num(2), num(3)) ;
A = sum(mul(num(2), var(x)), var(x)) ;
A = sum(mul(num(2), var(x)), num(2)) ;
A = sum(mul(num(2), num(2)), var(x)) ;
A = sum(mul(num(2), num(2)), num(2)) ;
A = sum(exp(x, num(2)), var(x)) ;
A = sum(exp(x, num(2)), num(2)) ;
A = sum(var(x), sum(var(x), var(x))) ;
A = sum(var(x), sum(var(x), num(2))) ;
A = sum(var(x), sum(num(2), var(x))) ;
A = sum(var(x), sum(num(2), num(2))) ;
A = sum(var(x), mul(num(2), var(x))) ;
A = sum(var(x), mul(num(2), num(2))) ;
A = sum(var(x), exp(x, num(2))) ;
A = sum(num(2), sum(var(x), var(x))) ;
A = sum(num(2), sum(var(x), num(2))) ;
A = sum(num(2), sum(num(2), var(x))) ;
A = sum(num(2), sum(num(2), num(2))) ;
A = sum(num(2), mul(num(2), var(x))) ;
A = sum(num(2), mul(num(2), num(2))) ;
A = sum(num(2), exp(x, num(2))) ;
false.

由于正确的递归非终端和// 1

，因此需要限制此DCG中输入的长度

Prolog是/ 2，参数没有充分实例化

2 个答案: