Clever Geek Handbook

Keep getting the error "Arguments are not instantiated enough" can't figure out why

Keep getting error Arguments are not sufficiently instantiated

for the addition rule multiplication I wrote as shown below.

mult(_, 0, 0).                                   %base case for multiplying by 0
mult(X, 1, X).                                   % another base case
mult(X, Y, Z) :-
   Y > 1,
   Y1 is Y - 1,
   mult(X, Y1, Z1),
   Z is X + Z1.

I am new to Prolog and am really struggling with such simple problems.

Any recommendation for books or online tutorials would be great.

I am running it on SWI-Prolog on Ubuntu Linux.

+3


source to share


2 answers


I think you have removed the last two calls. You don't mean:

mult(X,Y,Z):- Y>1,Y1 is Y-1, Z1 is X+Z, mult(X,Y1,Z1).

Strike> Edit: It doesn't matter if you look at the code again, it doesn't make sense. I believe your original code is correct.

As for the reason behind this error, I need to know how you call the predicate. Can you provide an example of input?



The correct way to call the predicate is mult(+X, +Y, ?Z)

:

?- mult(5,0,X).
X = 0

?- mult(5,1,X).
X = 5

?- mult(5,5,X).
X = 25

?- mult(4,4,16).
yes

?- mult(3,3,10).
no

etc .. Calling this function with a free variable in the first two arguments will result in this error, because one of them will be used on the right side is

or on both sides of <

, and these predicates expect.

+2


source


The mult/3

first two arguments must be known in your definition . If one of them is still a variable, a creation error will occur. For example. mult(2, X, 6)

will result in a creation error, although this X = 3

is the correct answer; in fact, the only answer.

There are several options:

successor-arithmetics, constraints, or meta-word predicates.

Here's a starting point with successor arithmetic:



add(0,Y,Y).
add(s(X),Y,s(Z)) :- add(X,Y,Z).

Another approach would be to use integer constraints. YAP and SWI have library(clpfd)

which can be used in a very flexible way: both for regular integer computation and for more general constraints. Of course, the multiplication is already predefined:

? - A * B # = C.
A * B # = C.

? - A * B # = C, C = 6.
C = 6,
A in -6 .. -1 \ /1..6,
A * B # = 6,
B in -6 .. -1 \ /1..6.

? - A * B # = C, C = 6, A = 2.
A = 2,
B = 3,
C = 6.

Meta-logic predicates: I can not recommend this option, in which you would use var/1

, nonvar/1

, ground/1

, to distinguish between different cases and handle them differently. These are so prone to errors that I have rarely seen the correct program using them. In fact, even very famous textbooks contain serious errors!

+4


source






More articles:


All Articles