我在这个语法中有一个左递归的小问题.我正在尝试在Prolog中编写它,但我不知道如何删除左递归.
<expression> -> <simple_expression>
<simple_expression> -> <simple_expression> <binary_operator> <simple_expression>
<simple_expression> -> <function>
<function> -> <function> <atom>
<function> -> <atom>
<atom> -> <number> | <variable>
<binary_operator> -> + | - | * | /
expression(Expr) --> simple_expression(SExpr), { Expr = SExpr }.
simple_expression(SExpr) --> simple_expression(SExpr1), binary_operator(Op), simple_expression(SExpr2), { SExpr =.. [Op, SExpr1, SExpr2] }.
simple_expression(SExpr) --> function(Func), { SExpr = Func }.
function(Func) --> function(Func2), atom(At), { Func = [Func2, atom(At)] }.
function(Func) --> atom(At), { Func = At }.
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我写过类似的东西,但它根本不起作用.如何更改它以使该程序正常工作?
I am curious about how tabling works to improve efficiency of Prolog programs. How is it implemented? Both explanation and references are welcome.
我用b-prolog版本8.1 的表格功能做了一些实验, 并且对我观察到的性能感到非常惊讶.
这是我使用的代码.它计算将一些正整数减少到以下所需的Collatz步N
数:I
1
%:- table posInt_CollatzSteps/2. % remove comment to enable tabling
posInt_CollatzSteps(I,N) :-
( I == 1
-> N = 0 % base case
; 1 is I /\ 1
-> I0 is I*3+1, posInt_CollatzSteps(I0,N0), N is N0+1 % odd
; I0 is I>>1, posInt_CollatzSteps(I0,N0), N is N0+1 % even
).
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要确定从所有整数需要减少最大步数I0
来I
:
i0_i_maxSteps0_maxSteps(I0,I,M0,M) :-
( I0 > I
-> M0 = M
; posInt_CollatzSteps(I0,N0),
I1 is …
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