Prolog中指数为负时计算数字幂的规则? [英] Rule to calculate power of a number when the exponent is Negative in Prolog?

问题描述

我有一个幂函数pow，试图将B的值计算为E的幂.到目前为止，我已经处理了案件-
1.指数为0
2.指数不为零

I have a power function pow that attempts to calculate the value of B to the power of E. So far I handle the cases-
1. exponent is 0
2. exponent is non-zero

pow(B,0,1).
pow(B,E,Result):-   E2 is E - 1,
                    pow(B,E2,Result2),
                    Result is B*Result2.

如何添加幂函数可以处理负指数的另一种情况?

How can I add another case where the power function can handle negative exponents?

推荐答案

首先，应该考虑如何定义0 ⁰ .从形式上来说，这是不确定.它可以是零，也可以是1.正如Wolfram的Mathworld在其有关权力的文章中以及其文章为零:

First, one should consider how to define 0⁰. Formally speaking it is indeterminate. It could be zero or it could be 1. As Wolfram's Mathworld says in its article on powers and in its article on zero:

0 ⁰ (从零到零的幂)本身是未定义的.缺乏相互矛盾的事实是因为 a ⁰ 始终为1，所以0 ⁰ 应该等于1，但0 ^a 始终为0(对于 a > 0)，因此0 ^a 应该等于0.尽管定义0 ⁰ = 1允许简单地表示一些公式，但通常定义为0 ⁰ 的定义选择是不确定的. 1992年Knuth； 1997年Knuth，第57页).

0⁰ (zero to the zeroth power) itself is undefined. The lack of a well-defined meaning for this quantity follows from the mutually contradictory facts that a⁰ is always 1, so 0⁰ should equal 1, but 0^a is always 0 (for a > 0), so 0^a should equal 0. The choice of definition for 0⁰ is usually defined to be indeterminate, although defining 0⁰ = 1 allows some formulas to be expressed simply (Knuth 1992; Knuth 1997, p. 57).

因此，您首先应该选择如何定义0 ⁰ 的特例:是0吗?是1吗?是未定义的吗?

So you should first choose how to define the special case of 0⁰: Is it 0? Is it 1? Is it undefined?

我选择将其视为未定义.

I choose to look at it as being undefined.

话虽如此，您可以看到正指数，如所示的重复相乘(例如10 ³ 为10 * 10 * 10或1,000)，可以看到负指数，如所示重复除法(例如10 ^-3 为(((1/10)/10)/10)或0.001).我的倾向，部分是因为我喜欢这种方法的对称性，部分是为了避免割伤(因为割伤通常是您未正确定义解决方案的信号)，所以会像这样:

That being said, you can look at a positive exponent as indicated repeated multiplication (e.g. 10³ is 10*10*10, or 1,000), and you can look at a negative exponent as indicating repeated division (e.g, 10^-3 is (((1/10)/10)/10), or 0.001). My inclination, partly because I like the symmetry of this approach and partly to avoid the cuts (since a cut is often a signal that you've not defined the solution properly), would be something like this:

% -----------------------------
% The external/public predicate
% -----------------------------
pow( 0 , 0 , _ ) :- ! , fail .
pow( X , N , R ) :-
  pow( X , N , 1 , R )
  .

% -----------------------------------
% the tail-recursive worker predicate
% -----------------------------------
pow( _ , 0 , R , R  ).
pow( X , N , T , R  ) :-
  N > 0 ,
  T1 is T * X ,
  N1 is N-1   ,
  pow( X , N1 , T1 , R )
  .
pow( _ , 0 , R , R  ) :-
  N < 0 ,
  T1 is T / X ,
  N1 is N+1   ,
  pow( X , N1 , T1 , R )
  .

另一种方法，正如其他人指出的那样，是将正指数定义为表示重复乘法，而将负指数定义为表示正指数的倒数，因此10 ³ 为10 * 10 * 10或1,000，而10 ^-3 是1/(10 ³ )或1/1,000或0.001.要使用此定义，我将再次避免削减并执行以下操作:

The other approach, as others have noted, is to define a positive exponent as indicating repeated multiplication, and a negative exponent as indicating the reciprocal of the positive exponent, so 10³ is 10*10*10 or 1,000, and 10^-3 is 1/(10³), or 1/1,000 or 0.001. To use this definition, I'd again avoid the cuts and do something like this:

% -----------------------------
% the external/public predicate
% -----------------------------
pow( 0 , 0 , _ ) :-  % 0^0 is indeterminate. Is it 1? Is it 0? Could be either.
  ! ,
  fail
  .
pow( X , N , R ) :-
  N > 0 ,
  pow( X , N , 1 , R )
  .
pow( X , N , R ) :-
  N < 0 ,
  N1 = - N ,
  pow( X , N1 , 1 , R1 ) ,
  R is 1 / R1
  .

% -----------------------------------
% The tail-recursive worker predicate
% -----------------------------------
pow( _ , 0 , R , R  ).
pow( X , N , T , R  ) :-
  N > 0 ,
  T1 is T * X ,
  N1 is N-1   ,
  pow( X , N1 , T1 , R )
  .

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