Rule to calculate power of a number when the exponent is Negative in Prolog?

问题

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.

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

回答1:

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:

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).

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

I choose to look at it as being undefined.

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 )
  .

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 )
  .

回答2:

Don't forget that a^(2b) = (a^b)^2 and x^2 = x*x. It is ok to call a tail-recursive working predicate with accumulator, in a non-tail fashion, from a top-level "UI" predicate. That way you don't have to implement working predicate for negative powers but rather reuse the one for positive power, and alter its result in the top-level predicate (I see this has already been suggested):

pow(B, E, R):- E<0 -> ... ; E=:=0 -> ... ; E>0 -> ... .

回答3:

To start, your second clause is non tail recursive (you can read about the subject here). It means that eventually, you will run out of call stack memory when running it. A good thing would be to use an accumulator to make it tail recursive. You can achieve that as follows :

% we add an accumulator to poW/3, making it pow/4.
pow(B, E, Result) :- pow(B, E, 1, Result).

% when we hit 0, our accumulator holds B^E so we unify it with result.
pow(_, 0, Accu, Accu) :- !.

% at each step, we multiply our accumulator by B
pow(B, E, Accu, Result) :-
    NewE is E - 1,
    NewAccu is Accu * B,
    pow(B, NewE, NewAccu, Result).

Then, you can simply handle the negative case by adding this clause on top of the others (it simply tells prolog that a negative power is the inverse of the positive one) :

pow(B, E, Result) :-
    E < 0,
    PositiveE is - E,
    pow(B, PositiveE, 1, R),
    !,
    Result is 1 / R.

Note that you can do it directly with your code :

pow(B, E, Result) :-
    E < 0,
    PositiveE is - E,
    pow(B, PositiveE, R),
    !,
    Result is 1 / R.

Plus, we now introduced a very red cut (see here for the meaning of red cut if necessary). So it'd be better to turn into a green cut with this modification :

pow(B, E, Result) :-
    E < 0,
    PositiveE is - E,
    pow(B, PositiveE, 1, R),
    !,
    Result is 1 / R.

% we add an accumulator to poW/3, making it pow/4.
pow(B, E, Result) :-
    E >= 0, %************* HERE *****************
    pow(B, E, 1, Result).

% when we hit 0, our accumulator holds B^E so we unify it with result.
pow(_, 0, Accu, Accu) :- !.

% at each step, we multiply our accumulator by B
pow(B, E, Accu, Result) :-
    NewE is E - 1,
    NewAccu is Accu * B,
    pow(B, NewE, NewAccu, Result).

来源：https://stackoverflow.com/questions/8240952/rule-to-calculate-power-of-a-number-when-the-exponent-is-negative-in-prolog

标签

prolog

exponentiation