I am attempting a past paper question for a Prolog exam. I drew a 'tree' for how I believed Prolog ought to behave given the program and a certain goal. However, Prolog does not behave as I expected, and given a query for which I believed it would return 'true', it actually returned 'false'.
Here is my program:
sum(Term,N) :- Term = 0, N = 0.
sum(Term,N) :- Term = f(M,Subterm), number(M), sum(Subterm,N-M).
My query and search tree are as follows (goals are bracketed and in bold):
[ sum(f(1,0),1) ]
Using Rule 1, let Term = 0, N = 0, tries to unify [ 1 = 0, 1 = 0 ] fail.
Redo: using Rule 2, let Term = f(1,0), N=1 [ f(1,0) = f(M,Subterm), number(M), sum(Subterm,1-1) ]
Unifying, let M=1 and Subterm=0 [ number(1), sum(0,0) ]
Using Rule 1, this should succeed. However (SWI) Prolog says 'false'.
If someone can point out to me why my reasoning is flawed (and how I can learn from this in future), I would be very grateful.
Since your program is almost a pure1 one, you can locate the error in a systematic manner without using a debugger. The idea is to generalize your program by removing goals, one-by-one. I came up with the following pure generalization which I obtained by "commenting" out some goals like so:
:- op(950, fy, *).
*(_).
sum(Term,N) :-
Term = 0,
N = 0.
sum(Term,N) :-
* Term = f(M,Subterm),
* number(M),
sum(Subterm,N-M).
?- sum(Term, N).
Term = 0, N = 0
; false.
Also the query above is more general than yours. This is a very useful technique in Prolog: Instead of thinking about concrete solutions, we
first let Prolog do all the work for us.
The answer was quite clear: There is exactly one solution to this relation, even if the relation is now generalized.
So the problem must be somewhere in the remaining visible part. Actually, it's the -. Why not write instead:
:- use_module(library(clpfd)).
sum(0, 0).
sum(Term, N0) :-
Term = f(M, Subterm),
N0 #= M+N1,
sum(Subterm, N1).
I find that program much easier to understand. If I read a name sum, I immediately look for a corresponding +. Of course, if you insist, you could write N0-M #= N1 instead. It would be exactly the same, except that this requires a bit more thinking.
Fine print you don't need to read
1) Your original program used number/1 which is not pure. But since the problem persisted by removing it, it did not harm our reasoning.
To be more accurate, the first rule tries to unify f(1,0) = 0 and 1 = 0, which of course fails.
Analysis of rule 2 is also incorrect. Partly, it's because Prolog does not evaluate arithmetic expressions inline. The term N-M is just a term (short-hand for '-'(N, M). It does not result in M being subtracted from M unless the evaluation is done explicitly via is/2 or an arithmetic comparison (e.g., =:=/2, =</2, etc).
The analysis of rule 2 would go as follows. Step 5 is where your logic breaks down due to the above.
Call sum(f(1,0), 1) results in Term = f(1,0) and N = 1.
In rule 2, Term = f(M, Subterm) becomes f(1,0) = f(M, Subterm) which results in M = 1 and Subterm = 0.
number(N) becomes number(1) and succeeds (since 1 is a number)
The call sum(Subterm, N-M) becomes sum(0, 1-1).
Prolog matches sum(0, 1-1) with the head of rule 1 sum(Term, N) :- Term = 0, N = 0., but it fails because 1-1 = 0 (which is the same as '-'(1, 1) = 0 unification fails.
Prolog matches sum(0, 1-1) with the head of rule 2, and unifies Term = 0 and N = 1-1 (or N = '-'(1, 1)).
Term = f(M, Subterm) becomes 0 = f(M, Subterm) which fails because 0 cannot match the term f(M, Subterm).
No more rules to attempt, so the predicate call fails.
The easy fix here is a common, basic Prolog pattern to use a new variable to evaluate the expression explicitly:
sum(Term,N) :-
Term = f(M,Subterm),
number(M),
R is N - M,
sum(Subterm, R).
You can also tidy up the code quite a bit by unifying in the heads of the clauses. So the clauses could be rewritten:
sum(0, 0).
sum(f(M, Subterm), N) :-
number(N),
R is N - M,
sum(Subterm, R).
EDIT: My answer is intended to guide you through a walk through of your existing logic. Other than correcting the misunderstanding regarding expression evaluation, I did not analyze your solution for overall correctness.
Related
So I've got a predicate that works for uninstantiated variables nat_cset(N,Cs) that relates a natural number with the counting set Sn = {1,2,...,N}.
nat_cset_(0,Acc,Acc).
nat_cset_(N,Acc,Cs) :- N #> 0, N_1 #= N - 1, nat_cset_(N_1, [N|Acc], Cs).
nat_cset(N,Cs) :- nat_cset_(N,[],Cs).
One thing I've noticed is that is the following query nat_cset(N,Cs), N = 6. enters an infinite loop while N = 6, nat_cset(N,Cs). does not:
?- nat_cset(N,Cs), N = 6, false.
...
?- N = 6, nat_cset(N,Cs), false.
false
This happens because in the second query, prolog it should only look for solutions for N = 6 while in the first query, prolog will search for infinite solutions such that N = 6.
My question is, is this 'adequate' behaviour for nat_cset/2 or is this a unwanted behaviour? I've read somewhere that the order of predicates changing the solutions makes it not monotonic and removes its purity, but at the same time I can't think of a way to make nat_cset/2 any different.
Sorry for the noob question, a lot of the concepts are new to me and I'm still trying to process it all.
The query
?- nat_cset(N,Cs), N = 6, false.
has exactly the same termination property as
?- nat_cset(N,Cs), false, N = 6.
So the goal N = 6 has in this case no influence on termination whatsoever,1. Only if you add it in front can it influence termination.
As already said, rather start with programs using successor-arithmetics and not clpfd/clpz. They are much simpler and still expose the relevant properties needed to understand failure slicing.
1 This assumes that N = 6 is always terminating, as it does with clpz in Scryer and clpfd in SWI. With general coroutining this is not necessarily the case. Think of freeze(N, inf), N = 6.
I've written some predicates which take the length of a list and attaches some constraints to it (is this the right vocabulary to be using?):
clp_length([], 0).
clp_length([_Head|Rest], Length) :-
Length #>= 0, Length #= Length1 + 1,
clp_length(Rest, Length1).
clp_length2([], 0).
clp_length2([_Head|Rest], Length) :-
Length #= Length1 + 1,
clp_length2(Rest, Length1).
The first terminates on this simple query, but the second doesn't:
?- Small in 1..2, clp_length(Little, Small).
Small = 1,
Little = [_1348] ;
Small = 2,
Little = [_1348, _2174] ;
false.
?- Small in 1..2, clp_length2(Little, Small).
Small = 1,
Little = [_1346] ;
Small = 2,
Little = [_1346, _2046] ;
% OOPS %
This is strange to me, because Length is pretty clearly greater than 0. To figure that out you could either search, find the zero, and deduce that adding from zero can only increase the number, or you could propagate the in 1..2 constraint down. It feels like the extra clause is redundant! That it isn't means my mental model of clpfd is pretty wrong.
So I think I have two questions (would appreciate answers to the second as comments)
Specifically, why does this additional constraint cause the query to work correctly?
Generally, is there a resource I can use to learn about how clpfd is implemented, instead of just seeing some examples of how it can be used? I'd prefer not to have to read Markus Triska's thesis but that's the only source I can find. Is that my only option if I want to be able to answer questions like this one?
1mo, there is the issue with naming. Please refer to previous answers by
mat
and me recommending relational names. You won't go far using inappropriate names. So list_length/2 or list_fdlength/2 would be an appropriate name. Thus we have list_fdlength/2 and list_fdlength2/2.
2do, consider the rule of list_fdlength2/2. Nothing suggests that 0 is of relevance to you. So that rule will be exactly the same if you are using 0 or 1 or -1 or whatever as base case. So how should this poor rule ever realize that 0 is the end to you? Better, consider a generalization:
list_fdlength2(fake(N), N) :- % Extension to permit fake lists
N #< 0.
list_fdlength2([], 0).
list_fdlength2([_Head|Rest], Length) :-
Length #= Length1 + 1,
list_fdlength2(Rest, Length1).
This generalization shows all real answers plus fake answers. Note that I have not changed the rule, I added this alternative fact only. Thus the fake solutions are actually caused by the rule:
?- list_fdlength2(L, 1).
L = [_A]
; L = [_A, _B|fake(-1)]
; L = [_A, _B, _C|fake(-2)]
; ... .
?- list_fdlength2(L, 0).
L = []
; L = [_A|fake(-1)]
; L = [_A, _B|fake(-2)]
; ... .
Each clause tries to contribute to the solutions just in the scope of the clause. But there is no way to derive (by the built-in Prolog execution mechanism) that some rules are no longer of relevance. You have to state that explicitly with redundant constraints as you did.
Now, back to your original solution containing the redundant constraint Length #>= 0. There should not be any such fake solution at all.
list_fdlength(fake(N), N) :-
N #< 0.
list_fdlength([], 0).
list_fdlength([_Head|Rest], Length) :-
Length #>= 0,
Length #= Length1 + 1,
list_fdlength(Rest, Length1).
?- list_fdlength(L, 1).
L = [_A]
; L = [_A, _B|fake(-1)] % totally unexpected
; false.
?- list_fdlength(L, 0).
L = []
; L = [_A|fake(-1)] % eek
; false.
There are fake answers, too! How ugly! At least, they are finite in number. But, you could have done it better by using
Length #>= 1 in place of Length #>=0. With this little change, there are no longer any fake solutions when N is non-negative and thus also your original program will be better.
I'm playing around with recursion in Prolog, and I'm confused. I am trying to write rules that can determine if a number is even or odd. I know that there are other stackoverflow questions about this, but I don't care about having a working solution, I am more interested in knowing why mine doesn't work.
Here are my rules:
even(0).
even(N) :- N>0, N1 is N-1, odd(N1).
odd(N) :- N>0, N1 is N-1, even(N1).
When I query even(0), I get returned 2 results. The first result is true, the 2nd is false. This also happens with odd(1), even(2), odd(3), etc. Why am I getting 2 return results? Shouldn't I just get 1?
When you query even(0), it succeeds as you have seen. But you've also seen it prompts you for more results because it left a choicepoint, which is a place in the logic where Prolog decides it can come back and explore other alternatives for a potentially successful query. Upon going back to the choicepoint and attempting to find more solutions, it does not find more, so it comes back "false" since it found no more solutions. So it did just find one solution, but the choice point caused backtracking after which it found no additional solutions. This is the case with your other successful queries as well.
You'll note that if you make a more general query, it gives an error (example taken from GNU Prolog):
| ?- even(N).
N = 0 ? ;
uncaught exception: error(instantiation_error,(>)/2)
| ?-
This is because you are using specific arithmetic expression operators that require that the variables be instantiated. These are relational operators like (>)/2 and the is/2 operator. You can make the solution more relational by using the CLP(FD) operators which are designed for reasoning with integers:
even(0).
even(N) :-
N #> 0,
N1 #= N-1,
odd(N1).
odd(N) :-
N #> 0,
N1 #= N-1,
even(N1).
Then you get a more general solution, which is more complete and more useful:
| ?- even(N).
N = 0 ? ;
N = 2 ? ;
N = 4 ? ;
N = 6 ? ;
...
| ?- odd(N).
N = 1 ? ;
N = 3 ? ;
N = 5 ? ;
N = 7 ?
...
If you know there is at most one answer, or if you only care about the first possible answer, you can use once/1 (examples taken from SWI Prolog here):
2 ?- even(2).
true ;
false.
3 ?- once(even(2)).
true.
4 ?- even(N).
N = 0 ;
N = 2 ;
N = 4 ;
...
5 ?- once(even(N)).
N = 0.
6 ?-
As expected, once(even(N)) terminates after finding the first solution.
The return values you have are correct. The point is how Prolog is evaluating predicates. When you query i.e.
even(2)
Prolog firstly evaluate that this predicate is Yes / true. When going through next possibility it return No / false, because it cannot find any more.
To check what exactly is performed under the hood go to:
https://swish.swi-prolog.org
on the left side type rules (i.e. odd/even) and on the query window type like 'odd(2)', but just before running click 'solutions'->'debug(trace)'. It will let you go step by step of what Prolog is doing.
Also please take a look at the successor example in tutorial below.
http://www.learnprolognow.org/lpnpage.php?pagetype=html&pageid=lpn-htmlse9
from a link above, try such code for a reversed example:
numeral(0).
numeral(succ(X)) :- numeral(X).
Now evaluating numeral(0) for the first time return succ(0), another time succ(succ(0)) etc.
Each time next evaluation brings another possible solution for a query.
What Prolog does is a "depth-first search", which means Prolog walks through a decision tree until it either finds a solution and succeeds OR it fails. In either case a process called "backtracking" kicks in. Along the way, going through the tree of choices, Prolog keeps track of where it has MULTIPLE possible routes that could potentially satisfy the goal. Such a point in the decision tree is called a "choice point".
This means Prolog will
search ->
succeed or fail ->
go back to the last choice point ->
repeat until all possible paths have been tried
Given your program:
even(0).
even(N) :- N>0, N1 is N-1, odd(N1).
odd(N) :- N>0, N1 is N-1, even(N1).
We can clearly see TWO ways to satisfy even(0).. The first is the fact even(0) and the second is the recursive rule even(N). Prolog reads top to bottom, left to right so the first encounter is even(0). which is true, and the second is even(N). which goes through N-1 making the result N1 = -1, then goes through odd(N) making the result N1 = -2, which in unequal to even(0). so it fails and then calls even(N) again. Your specific version of Prolog likely sees that it is an infinitely recursive predicate and doesn't even try to satisfy it even though it's a valid declarative path , but not a valid procedural path.
If you know that the mode is (+), you can place a cut,
to suppress the unnecessary choice point:
even(0) :- !.
even(N) :- N > 0, N1 is N-1, odd(N1).
odd(N) :- N > 0, N1 is N-1, even(N1).
The above is better than wrapping a query with
once/1 since it allows the Prolog interpreter to
use last call optimization. There is now no more
problem with an extra choice point:
?- even(3).
false.
?- even(4).
true.
But if the mode is not fixed, you have to be more careful
with cuts. Probably write a separate carefully crafted
predicate for each mode.
CLP(FD) itself seems not to help, it cannot avoid the need
to place cuts, but can sometimes avoid the need to code
different variants for different modes.
I have a List with Integers and anonymous variables and I try to find the index of a special values. Problem is as soon I'm using nth1/3 to find the indices Prolog assigns values to the anonymous variables and therefore I find way too indices.
Example:
List = [1,\_,1], where I want as result X = 1, X = 3 from nth1(X,List,1), but as stated before I get X = 1, X = 2, X = 3.
There is a somewhat problematic issue hidden in your requirements: They violate an important declarative property called monotonicity. By this we mean that adding constraints can at most make the solution more specific, never more general.
For example, with the solution you posted, we get:
?- list_el_index([_], 1, N).
false.
Now I add a constraint by imposing an additional requirement on the hitherto free anonymous variable:
?- Var = 1, list_el_index([Var], 1, N).
Var = 1,
N = 0 .
I mean: Come on! We have added a constraint, and as a result get more solutions than before? Such a result is unfortunate and prevents us from reasoning in a logical way about this program.
The program also fails us in other respects. For example, let us ask: Which solutions are there at all?
?- list_el_index(Ls, El, I).
nontermination
Ideally, we would like the program to generate solutions in such cases! This generality is one of the foremost attractions of logic programming, and distinguishes it from more low-level paradigms.
One way to solve such issues is to symbolically distinguish the different kinds of elements that appear in your list.
For example, let us use:
u for an unknown value.
i(I) for an integer I.
With this new representation, your solution becomes:
list_el_index([i(I)|_], I, 0).
list_el_index([_|Tail], Element, Index) :-
list_el_index(Tail, Element, Index0),
Index #= Index0+1.
I have also taken the liberty to replace (is)/2 by (#=)/2, to advertise and stick to more general integer arithmetic that lets us more freely reorder the goals, if necessary. Depending on your Prolog implementation, you may have to import a library to benefit from (#=)/2.
With this representation, your initial case becomes:
?- list_el_index([i(1),u,i(1)], 1, Index).
Index = 0 ;
Index = 2 ;
false.
This works as desired!
Importantly, we can use the predicate also more generally, namely to generate possible answers:
?- list_el_index(Ls, El, I).
Ls = [i(El)|_2994],
I = 0 ;
Ls = [_2992, i(El)|_3000],
I = 1 ;
Ls = [_2992, _2998, i(El)|_3006],
I = 2 ;
Ls = [_2992, _2998, _3004, i(El)|_3012],
I = 3 .
Due to the program's monotonicity, we can fairly enumerate solutions by iterative deepening:
?- length(Ls, _), list_el_index(Ls, El, I).
Ls = [i(El)],
I = 0 ;
Ls = [i(El), _4812],
I = 0 ;
Ls = [_4806, i(El)],
I = 1 ;
Ls = [i(El), _4812, _4818],
I = 0 ;
etc.
This has become possible by using a representation that lets us distinguish the cases by pattern matching. Consider using this approach to make your programs usable in all directions, and to make logical reasoning applicable. It is quite easy to apply by using the appropriate wrapper or constant, and greatly increases the generality of your programs.
This works :
- L = [1,_,1], nth1(X, L, Y), ground(Y), Y= 1.
L = [1,_310914,1],
X = Y, Y = 1 ;
L = [1,_310914,1],
X = 3,
Y = 1.
Thanks to lurkers hint, I came up with this solution.
list_el_index([El1|_], El2, 0) :-
El1 == El2.
list_el_index([_|Tail], Element, Index) :-
list_el_index(Tail, Element, Index1),
Index is Index1+1.
I'm new to Prolog and I'm trying to write a piece of code that calculates factorial of a number.
This code works fine:
fact(0,1).
fact(N, R) :- N > 0, N1 is N - 1, fact(N1, R1), R is R1 * N.
But this one doesn't:
fact(0, 1).
fact(N, R) :- N > 0, fact(N - 1, R1), R is R1 * N.
Can someone please explain?
The issue is that prolog primarily uses unification to do computation. To get it to do arithmetic operations you need to tell it to do so explicitly using the is operator.
So, in your first program you explicitly tell it to perform subtraction with the clause N1 is N - 1, so that works as expected.
But in your second program you are not asking for arithmetic computation, but unification, when you wrote fact(N - 1, R1).
If I had the fact fact(5 - 1, foo). defined, then I could query for ?- fact(N - 1, Y), write([N, Y]). and prolog would happily unify N with 5 and Y with foo. This query would output [5, foo].
So, to go one step further, if I had the fact fact(foo - bar). then the query ?- fact(X - Y), write([X, Y]). would happily unify and return [foo, bar]. The - doesn't denote subtraction - it's part of the structure of the fact being represented.
When passing around arithmetic expressions (instead of numbers), you need to evaluate expressions at certain times.
Arithmetic operators like (>)/2 automatically do that, so the goal 1 > (0+0) succeeds, just like 1 > 0 does.
Implicit unification (in clause heads) and explicit unification with (=)/2 goals expresses equality of arbitrary Prolog terms, not just arithmetic expressions. So the goal 0 = 0 succeeds, but 0 = (1-1) fails.
With arithmetic equality (=:=)/2, both 0 =:= 0 and 0 =:= (1-1) succeed.
In your second definition of fact/2, you could make the first clause more general by writing fact(N,1) :- N =:= 0. instead of fact(0,1).. As an added bonus, you could then run queries like ?- fact(5+5,F). :)