I am re-writing the following function in Prolog:
V1:
f(X,Y):- X < 2, Y is X+1.
f(X,3):- 2 =< X, X < 5.
f(X,Y):- 5 =< X, Y is 8-X.
As V2:
f(X,Y) :-
X < 2,
Y is X + 1.
f(X,Y) :-
X >= 2,
X < 5,
Y is 3.
f(X,Y) :-
X >= 5,
Y is 8-X.
I then wanted to experiment with cuts. For green cuts (V3):
f(X,Y) :-
X < 2, !,
Y is X + 1.
f(X,Y) :-
X >= 2,
X < 5, !,
Y is 3.
f(X,Y) :-
X >= 5,
Y is 8-X.
For red cuts (V4):
f(X,Y) :-
X < 2, !,
Y is X + 1.
f(X,Y) :-
X < 5, !,
Y is 3.
f(X,Y) :-
Y is 8-X.
However, I don't understand their advantage, as deleting the cuts would allow the same behaviour of the code... Any help?
All your versions V1..V4 are observationally equivalent, so you got some reasoning right. Still, there are differences.
Avoiding superfluous choice points
In many implementations, V1 and V2 might be particularly less efficient, for, internally, they "leave open a choice point". This is so because such Prologs do not look any further to the other rules. So each goal f(1,X) consumes a bit of memory that can be freed only on backtracking (or using !). Here is a simple way to try this out yourself:
loop(Goal) :-
Goal,
loop(Goal).
Here is what I get in SWI:
?- time(loop(f1(1,2))).
% 5,991,554 inferences, 81.282 CPU in 81.443 seconds (100% CPU, 73713 Lips)
ERROR: Out of local stack
?- time(loop(f2(1,2))).
% 5,991,553 inferences, 85.032 CPU in 85.212 seconds (100% CPU, 70462 Lips)
ERROR: Out of local stack
Whereas V3 and V4 seem to run indefinitely - at least much longer than 85s. Experiments such as this one are funny for very tiny programs but are not very practical for bigger ones. Fortunately, there is a simple way to tell in many Prologs whether or not a query is executed determinately. To see if your system does this, enter:
?- X = 1.
X = 1.
For your variations:
?- f1(1,2).
true
; % <== Prolog asked for another answer
false. % <== only to conclude that there is none.
?- f2(1,2).
true
; false. % same again
?- f3(1,2).
true. % <== Prolog knows there will be no further answer
?- f4(1,2).
true.
Avoiding recalculations - making cuts red
While V3 avoids superfluous choice points, V4 now even avoids superfluous calculations. So it should be the most efficient. But it comes at the price of fixing the order of the clauses.
However, V3 was only possible, because two necessary conditions for green cuts coincided:
Non-overlapping conditions. That should be obvious to you.
Safe testing of instantiations. This is far from obvious. Please note that the goal X < 2 has an implicit test for a correct instantiation attached! It produces an instantiation error should X be an uninstantiated variable. It is because of this very test that the cut in V3 happens to be a green cut. Without that testing, it would be a red cut.
Note also that V1 and V2 would not be equivalent, if the second rule would be alone! For the goal f(X,5). would fail in V1 but it would produce an error in V2.
As you noted the first version shows green cuts and the second red cuts.
It is not necessary that you will feel the difference between these two versions.
a) one reason can be efficiency, but for toy codes with fast machines you hardly notice it.
b) shuffling the rules should not change code's behavior in case of green cuts, and that's true for the first code. But in the second code, if you put the second clause before the first one than the behavior changes: f(0,3) is true, but initially it was false. Therefore you would feel difference if you shuffle the rules.
Advantage of shuffling is that you don't care about order but content - that's one of the points declarative programing.
Related
A paper I'm reading says the following:
Plaisted [3] showed that it is possible to write formally correct
PROLOG programs using first-order predicate-calculus semantics and yet
derive nonsense results such as 3 < 2.
It is referring to the fact that Prologs didn't use the occurs check back then (the 1980s).
Unfortunately, the paper it cites is behind a paywall. I'd still like to see an example such as this. Intuitively, it feels like the omission of the occurs check just expands the universe of structures to include circular ones (but this intuition must be wrong, according to the author).
I hope this example isn't
smaller(3, 2) :- X = f(X).
That would be disappointing.
Here is the example from the paper in modern syntax:
three_less_than_two :-
less_than(s(X), X).
less_than(X, s(X)).
Indeed we get:
?- three_less_than_two.
true.
Because:
?- less_than(s(X), X).
X = s(s(X)).
Specifically, this explains the choice of 3 and 2 in the query: Given X = s(s(X)) the value of s(X) is "three-ish" (it contains three occurrences of s if you don't unfold the inner X), while X itself is "two-ish".
Enabling the occurs check gets us back to logical behavior:
?- set_prolog_flag(occurs_check, true).
true.
?- three_less_than_two.
false.
?- less_than(s(X), X).
false.
So this is indeed along the lines of
arbitrary_statement :-
arbitrary_unification_without_occurs_check.
I believe this is the relevant part of the paper you can't see for yourself (no paywall restricted me from viewing it when using Google Scholar, you should try accessing this that way):
Ok, how does the given example work?
If I write it down:
sm(s(s(s(z))),s(s(z))) :- sm(s(X),X). % 3 < 2 :- s(X) < X
sm(X,s(X)). % forall X: X < s(X)
Query:
?- sm(s(s(s(z))),s(s(z)))
That's an infinite loop!
Turn it around
sm(X,s(X)). % forall X: X < s(X)
sm(s(s(s(z))),s(s(z))) :- sm(s(X),X). % 3 < 2 :- s(X) < X
?- sm(s(s(s(z))),s(s(z))).
true ;
true ;
true ;
true ;
true ;
true
The deep problem is that X should be Peano number. Once it's cyclic, one is no longer in Peano arithmetic. One has to add some \+cyclic_term(X) term in there. (maybe later, my mind is full now)
I'm currently learning SWI-Prolog. I want to implement a function factorable(X) which is true if X can be written as X = n*b.
This is what I've gotten so far:
isTeiler(X,Y) :- Y mod X =:= 0.
hatTeiler(X,X) :- fail,!.
hatTeiler(X,Y) :- isTeiler(Y,X), !; Z is Y+1, hatTeiler(X,Z),!.
factorable(X) :- hatTeiler(X,2).
My problem is now that I don't understand how to end the recursion with a fail without backtracking. I thought the cut would do the job but after hatTeilerfails when both arguments are equal it jumps right to isTeiler which is of course true if both arguments are equal. I also tried using \+ but without success.
It looks like you add cuts to end a recursion but this is usually done by making rule heads more specific or adding guards to a clause.
E.g. a rule:
x_y_sum(X,succ(Y,1),succ(Z,1)) :-
x_y_sum(X,Y,Z).
will never be matched by x_y_sum(X,0,Y). A recursion just ends in this case.
Alternatively, a guard will prevent the application of a rule for invalid cases.
hatTeiler(X,X) :- fail,!.
I assume this rule should prevent matching of the rule below with equal arguments. It is much easier just to add the inequality of X and Y as a conditon:
hatTeiler(X,Y) :-
Y>X,
isTeiler(Y,X),
!;
Z is Y+1,
hatTeiler(X,Z),
!.
Then hatTeiler(5,5) fails automatically. (*)
You also have a disjunction operator ; that is much better written as two clauses (i drop the cuts or not all possibilities will be explored):
hatTeiler(X,Y) :- % (1)
Y > X,
isTeiler(Y,X).
hatTeiler(X,Y) :- % (2)
Y > X,
Z is Y+1,
hatTeiler(X,Z).
Now we can read the rules declaratively:
(1) if Y is larger than X and X divides Y without remainder, hatTeiler(X,Y) is true.
(2) if Y is larger than X and (roughly speaking) hatTeiler(X,Y+1) is true, then hatTeiler(X, Y) is also true.
Rule (1) sounds good, but (2) sounds fishy: for specific X and Y we get e.g.: hatTeiler(4,15) is true when hatTeiler(4,16) is true. If I understand correctly, this problem is about divisors so I would not expect this property to hold. Moreover, the backwards reasoning of prolog will then try to deduce hatTeiler(4,17), hatTeiler(4,18), etc. which leads to non-termination. I guess you want the cut to stop the recursion but it looks like you need a different property.
Coming from the original property, you want to check if X = N * B for some N and B. We know that 2 <= N <= X and X mod N = 0. For the first one there is even a built-in called between/2 that makes the whole thing a two-liner:
hT(X,B) :-
between(2, X, B),
0 is (X mod B).
?- hT(12,X).
X = 2 ;
X = 3 ;
X = 4 ;
X = 6 ;
X = 12.
Now you only need to write your own between and you're done - all without cuts.
(*) The more general hasTeiler(X,X) fails because is (and <) only works when the right hand side (both sides) is variable-free and contains only arithmetic terms (i.e. numbers, +, -, etc).
If you put cut before the fail, it will be freeze the backtracking.
The cut operation freeze the backtracking , if prolog cross it.
Actually when prolog have failed, it backtracks to last cut.
for example :
a:- b,
c,!,
d,
e,!,
f.
Here, if b or c have failed, backtrack do not freeze.
if d or f have failed, backtrack Immediately freeze, because before it is a cut
if e have failed , it can backtrack just on d
I hope it be useful
I have understood the theory part of Recursion. I have seen exercises but I get confused. I've tried to solve some, some I understand and some I don't. This exercise is confusing me. I can't understand why, so I use comments to show you my weak points. I should have power (X,N,P) so P=X^N.
Some examples:
?- power(3,5,X).
X = 243
?- power(4,3,X).
X = 64
?- power(2,4,X).
X = 16
The solution of this exercise is: (See comments too)
power(X,0,1). % I know how works recursion,but those numbers 0 or 1 why?
power(X,1,X). % X,1,X i can't get it.
power(X,N,P) :- % X,N,P if only
N1 is N-1, % N1=N-1 ..ok i understand
power(X,N1,P1), % P1 is used to reach the the P
P is P1*X. % P = P1*X
What I know recursion, I use a different my example
related(X, Y) :-
parent(X, Z),
related(Z, Y).
Compare my example with the exercise. I could say that my first line, what I think. Please help me out with it is a lot of confusing.
related(X, Y) :- is similar to power(X,N,P) :- . Second sentence of my example parent(X, Z), is similar to N1 is N-1, and the third sentence is related(Z, Y). similar to power(X,N1,P1), and P is P1*X..
Let's go over the definition of the predicate step by step. First you have the fact...
power(X,0,1).
... that states: The 0th power of any X is 1. Then there is the fact...
power(X,1,X).
... that states: The 1st power of any X is X itself. Finally, you have a recursive rule that reads:
power(X,N,P) :- % P is the Nth power of X if
N1 is N-1, % N1 = N-1 and
power(X,N1,P1), % P1 is the N1th power of X and
P is P1*X. % P = P1*X
Possibly your confusion is due to the two base cases that are expressed by the two facts (one of those is actually superfluous). Let's consider the following queries:
?- power(5,0,X).
X = 1 ;
ERROR: Out of local stack
The answer 1 is certainly what we expect, but then the predicate loops until it runs out of stack. That's certainly not desirable. And this query...
?- power(5,1,X).
X = 5 ;
X = 5 ;
ERROR: Out of local stack
... yields the correct answer twice before running out of stack. The reason for the redundant answer is that the recursive rule can reduce any given N to zero and to one thus yielding the same answer twice. If you look at the structure of your recursive rule, it is obvious that the first base case is sufficient, so let's remove the second. The reason for looping out of stack is that, after N becomes zero, the recursive rule will search for other solutions (for N=-1, N=-2, N=-3,...) that do not exist. To avoid that, you can add a goal that prevents the recursive rule from further search, if N is equal to or smaller than zero. That leaves you with following definition:
power(X,0,1). % the 0th power of any X is 1
power(X,N,P) :- % P is the Nth power of X if
N > 0, % N > 0 and
N1 is N-1, % N1 = N-1 and
power(X,N1,P1), % P1 is the N1th power of X and
P is P1*X. % P = P1*X
Now the predicate works as expected:
?- power(5,0,X).
X = 1 ;
false.
?- power(5,1,X).
X = 5 ;
false.
?- power(5,3,X).
X = 125 ;
false.
I hope this alleviates some of your confusions.
I have an AI project in which I should make a sudoku solver in Prolog but without using the clpfd package. How should I write the code and is there any way to print whenever a variable gets a value?
Without clpfd, you have to write a classic generate-and-test search. The nice thing about clpfd is that the constraints limit the search space. You can emulate this without using clpfd though. Let's simplify the task a bit so I can illustrate. How can you find X and Y values that solve both of these equations: X + Y = 10, 2*X + Y - 1 = 15?
First, encode your two equations:
eq1(X,Y) :- 10 is X + Y.
eq2(X,Y) :- 15 is 2*X + Y - 1.
Now create your search space in a solver predicate:
solution(X, Y) :-
between(1, 10, X), between(1, 10, Y),
eq1(X,Y),
eq2(X,Y).
Now you can run it and see:
?- solution(X,Y).
X = 6,
Y = 4 ;
false.
This is less efficient than it would be with clpfd because clpfd would notice things about the equations that would help it constrain the search space:
?- [library(clpfd)].
true.
?- X + Y #= 10, 2*X + Y - 1 #= 15.
2*X+Y#=16,
X+Y#=10.
?- X + Y #= 10, 2*X + Y - 1 #= 15, X in 1..10.
X = 6,
Y = 4.
See, it has already figured out that there is only one solution and I didn't have to constrain Y at all. That's very impressive! But you can still use Prolog without clpfd, it's just worse. :) In this example, we probably tried all 10x10=100 possible combinations to find this one solution. Less efficient. But not impossible.
So, what are the constraints you have to figure out for sudoku? Probably something like this:
unique(Row) :- sort(Row, Sorted), length(Row, Length), length(Sorted, Length).
all_digits(Row) :- forall(between(1,9,X), memberchk(X, Row)).
And so forth.
Edit: Combinatorial complexity estimation
Suppose we do a totally unprincipled search: that is, we even try obviously wrong cases like a grid of all 9s. We have 9x9=81 cells and 9 possible values (1-9). This yields 9^81 = a very large number, unlikely to be checked in your lifetime. And most of those grids are going to be fruitless permutations.
Suppose you constrain your search so that each row is a permutation of 1-9. There are 9! permutations of 1-9; with nine of those, you can multiply that outcome by nine, so there should be 9!^9. This is still awful! clpfd is probably able to further winnow this down by combining the 3x3 grid constraints; I'm not sure how I would go about doing that manually, other than in a semi-procedural way, choosing a 3x3 grid permutation and then passing each 3x1 row of that on to the next 3x3 grid selection.
It's worth noting that most of the optimization in classic Prolog programs comes down to either making the generate step generate better-qualified candidates or making the test step less expensive. A more obvious implementation can still be useful for checking a more complex implementation.
Thanks to #mat for checking my math and recommending this excellent article on the combinatorial problem of Sudoku.
To grok green cuts in Prolog I am trying to add them to the standard definition of sum in successor arithmetics (see predicate plus in What's the SLD tree for this query?). The idea is to "clean up" the output as much as possible by eliminating all useless backtracks (i.e., no ... ; false) while keeping identical behavior under all possible combinations of argument instantiations - all instantiated, one/two/three completely uninstantiated, and all variations including partially instantiated args.
This is what I was able to do while trying to come as close as possible to this ideal (I acknowledge false's answer to how to insert green cuts into append/3 as a source):
natural_number(0).
natural_number(s(X)) :- natural_number(X).
plus(X, Y, X) :- (Y == 0 -> ! ; Y = 0), (X == 0 -> ! ; true), natural_number(X).
plus(X, s(Y), s(Z)) :- plus(X, Y, Z).
Under SWI this seems to work fine for all queries but those with shape ?- plus(+X, -Y, +Z)., as for SWI's notation of predicate description. For instance, ?- plus(s(s(0)), Y, s(s(s(0)))). yields Y = s(0) ; false.. My questions are:
How do we prove that the above cuts are (or are not) green?
Can we do better than the above program and eliminate also the last backtrack by adding some other green cuts?
If yes, how?
First a minor issue: the common definition of plus/3 has the first and second argument exchanged which allows to exploit first-argument indexing. See Program 3.3 of the Art of Prolog. That should also be changed in your previous post. I will call your exchanged definition plusp/3 and your optimized definition pluspo/3. Thus, given
plusp(X, 0, X) :- natural_number(X).
plusp(X, s(Y), s(Z)) :- plusp(X, Y, Z).
Detecting red cuts (question one)
How to prove or disprove red/green cuts? First of all, watch for "write"-unifications in the guard. That is, for any such unifications prior to the cut. In your optimized program:
pluspo(X, Y, X) :- (Y == 0 -> ! ; Y = 0), (X == 0 -> ! ; true), ...
I spot the following:
pluspo(X, Y, X) :- (...... -> ! ; ... ), ...
So, let us construct a counterexample: To make this cut cut in a red manner, the "write unification" must make its actual guard Y == 0 true. Which means that the goal to construct must contain the constant 0 somehow. There are only two possibilities to consider. The first or third argument. A zero in the last argument means that we have at most one solution, thus no possibility to cut away further solutions. So, the 0 has to be in the first argument! (The second argument must not be 0 right from the beginning, or the "write unification would not have a detrimental effect.). Here is one such counterexample:
?- pluspo(0, Y, Y).
which gives one correct solution Y = 0, but hides all the other ones! So here we have such an evil red cut!
And contrast it to the unoptimized program which gave infinitely many solutions:
Y = 0
; Y = s(0)
; Y = s(s(0))
; Y = s(s(s(0)))
; ... .
So, your program is incomplete, and any questions about further optimizing it are thus not relevant. But we can do better, let me restate the actual definition we want to optimize:
plus(0, X, X) :- natural_number(X).
plus(s(X), Y, s(Z)) :- plus(X, Y, Z).
In practically all Prolog systems, there is first-argument indexing, which makes the following query determinate:
?- plus(s(0),0,X).
X = s(0).
But many systems do not support (full) third argument indexing. Thus we get in SWI, YAP, SICStus:
?- plus(X, Y, 0).
X = Y, Y = 0
; false.
What you probably wanted to write is:
pluso(X, Y, Z) :-
% Part one: green cuts
( X == 0 -> ! % first-argument indexing
; Z == 0 -> ! % 3rd-argument indexing, e.g. Jekejeke, ECLiPSe
; true
),
% Part two: the original unifications
X = 0,
Y = Z,
natural_number(Z).
pluso(s(X), Y, s(Z)) :- pluso(X, Y, Z).
Note the differences to pluspo/3: There are now only tests prior to the cut! All unifications are thereafter.
?- pluso(X, Y, 0).
X = Y, Y = 0.
The optimizations so far relied only on investigating the heads of the two clauses. They did not take into account the recursive rule. As such, they can be incorporated into a Prolog compiler without any global analysis. In O'Keefe's terminology, these green cuts might be considered blue cuts. To cite The Craft of Prolog, 3.12:
Blue cuts are there to alert the Prolog system to a determinacy it should have noticed but wouldn't. Blue cuts do not change the visible behavior of the program: all they do is make it feasible.
Green cuts are there to prune away attempted proofs that would succeed or be irrelevant, or would be bound to fail, but you would not expect the Prolog system to be able to tell that.
However, the very point is that these cuts do need some guards to work properly.
Now, you considered another query:
?- pluso(X, s(s(0)), s(s(s(0)))).
X = s(0)
; false.
or to put a simpler case:
?- pluso(X, s(0), s(0)).
X = 0
; false.
Here, both heads apply, thus the system is not able to determine determinism. However, we know that there is no solution to a goal plus(X, s^n, s^m) with n > m. So by considering the model of plus/3 we can further avoid choicepoints. I'll be right back after this break:
Better use call_semidet/1!
It gets more and more complex and chances are that optimizations might easily introduce new errors in a program. Also optimized programs are a nightmare to maintain. For practical programming purposes use rather call_semidet/1. It is safe, and will produce a clean error should your assumptions turn out to be false.
Back to business: Here is a further optimization. If Y and Z are identical, the second clause cannot apply:
pluso2(X, Y, Z) :-
% Part one: green cuts
( X == 0 -> ! % first-argument indexing
; Z == 0 -> ! % 3rd-argument indexing, e.g. Jekejeke, ECLiPSe
; Y == Z, ground(Z) -> !
; true
),
% Part two: the original unifications
X = 0,
Y = Z,
natural_number(Z).
pluso2(s(X), Y, s(Z)) :- pluso2(X, Y, Z).
I (currently) believe that pluso2/3 is the optimal usage of green/blue cuts w.r.t. leftover choicepoints. You asked for a proof. Well, I think that is well beyond an SO answer...
The goal ground(Z) is necessary to ensure the non-termination properties. The goal plus(s(_), Z, Z) does not terminate, that property is preserved by ground(Z). Maybe you think it is a good idea to remove infinite failure branches too? In my experience, this is rather problematic. In particular, if those branches are removed automatically. While at first sight it seems to be a good idea, it makes program development much more brittle: An otherwise benign program change might now disable the optimization and thus "cause" non-termination. But anyway, here we go:
Beyond simple green cuts
pluso3(X, Y, Z) :-
% Part one: green cuts
( X == 0 -> ! % first-argument indexing
; Z == 0 -> ! % 3rd-argument indexing, e.g. Jekejeke, ECLiPSe
; Y == Z -> !
; var(Z), nonvar(Y), \+ unify_with_occurs_check(Z, Y) -> !, fail
; var(Z), nonvar(X), \+ unify_with_occurs_check(Z, X) -> !, fail
; true
),
% Part two: the original unifications
X = 0,
Y = Z,
natural_number(Z).
pluso3(s(X), Y, s(Z)) :- pluso3(X, Y, Z).
Can you find a case where pluso3/3 does not terminate while there are finitely many answers?