Related
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'm new to Prolog and I'm trying to write fully working magic square program, but to say the truth I don't really know how to do, I have started but I feel that I'm doing it wrong. I'm sharing my code and I hope someone will help me, now when numbers are good I get true, but when they are not I get like out of stack error... (here is only checking rows and columns I know about obliquely check)
thanks for your attention!
:- use_module(library(clpfd)).
:- use_module(library(lists)).
magicSq(List, N) :-
Number is N * N,
belongs(Number ,List), % check if numbers are correct.
all_different(List), % check if numbers not occur.
Suma is N*(N*N + 1)/2,
checkC(List,N,N,Suma), % check column
checkR(List,1,N,Suma). % check row
belongs(0, _).
belongs(N, List) :- member(N,List) , Index is N - 1 , belongs(Index, List).
consecutiveSum(_, 0 , _,0).
consecutiveSum(List, HowMuch , From,Sum):-
Index is HowMuch - 1,
From1 is From +1,
nth1(From, List,Element),
consecutiveSum(List,Index,From1,Z),
Sum is Z + Element,!.
sumObliCol(0,_, [], _,_). % sums by columns or obliquely
sumObliCol(X,Number, [H|T], Ind, Residue) :-
Index is Ind + 1,
Y is mod(Index,Number),
Y =:= Residue,
sumObliCol(Z,Number, T, Index,Residue),
X is Z + H, !.
sumObliCol(X,Number, [_|T], Ind,Residue) :-
Index is Ind + 1,
sumObliCol(X,Number, T, Index,Residue).
checkC(_,0,_,_). % check column
checkC(List,N, Number,Answ):-
N1 is N-1,
checkC(List,N1, Number,Answ),
sumObliCol(Ats,Number,List,0,N1),Ats is Answ,!.
checkR(_,N,Number,_):- N>(Number*Number). % check row
checkR(List,N,Number,Answ):-
consecutiveSum(List,Number,N,Sum), Sum is Answ,
N1 is N + Number,
checkR(List,N1, Number,Answ),!.
In programming one often assumes that
everything is deeply intertwingled ... since the cross-connections among the myriad topics of this world/program simply cannot be divided up neatly.1
But in Prolog, sometimes, we can divide things up much more neatly. In particular, if you concentrate on a single property like non-termination. So let's consider magic squares of size one — very magic indeed! Like so using a failure-slice:
?- magicSq(Xs,1), false.
magicSq(List, N) :-
Number is N * N,
belongs(Number ,List), false,
all_different(List),
Suma is N*(N*N + 1)/2,
checkC(List,N,N,Suma),
checkR(List,1,N,Suma).
belongs(0, _) :- false.
belongs(N1, List) :-
member(N1,List), false,
N2 is N1 - 1,
belongs(N2, List).
That's all you need to understand! Evidently, the List is unconstrained and thus the goal member(N1, List) cannot terminate. That's easy to fix, adding a goal length(List, Number). And still, the program does not terminate but in a different area:
?- magicSq(Xs,1), false.
magicSq(List, N) :-
Number is N * N,
length(List, Number),
belongs(Number ,List), false,
all_different(List),
Suma is N*(N*N + 1)/2,
checkC(List,N,N,Suma),
checkR(List,1,N,Suma).
belongs(0, _) :- false.
belongs(N1, List) :-
member(N1,List),
N2 is N1 - 1,
belongs(N2, List), false.
Now this does not terminate, for N1 may be negative, too. We need to improve that adding N1 > 0.
Now, considering the program with a false in front of all_different/1, I get:
?- time(magicSq(List, 3)).
% 8,571,007 inferences
That looks like an awful lot of inferences! In fact, what you are doing is to enumerate all possible configurations first. Thus, you do not use the powers of constraint programming. Please go through tutorials on this. Start here.
However, the problems do not stop here! There is much more to it, but the remaining program is very difficult to understand, for you are using the ! in completely unrelated places.
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?
Suppose I have such goals:
times(0,_,0). % zero times X is zero
times(X,Y,Z) :- times(Y,X,Z) ,!. % X * Y = Y * X
When I try to ask:
?- times(0,1,X).
I get the double answer :
X = 0 ;
X = 0.
Possibly because first answer is deduced from the fact and second - from the rule.
Question - how to make prolog to give only one answer instead of two ?
add a cut to 'confirm' the first choice:
times(0,_,0) :- !.
or ban the 0 from the second:
times(X,Y,Z) :- X \= 0, times(Y,X,Z).
I've deleted the cut, but leave it if there are more rules.
But I think the 'reflexivity' rule will put you in trouble, with undue recursion.