Related
What I have to do is, write a predicate Multiplication/3, whose first argument is an integer, second argument is a list, and the third argument is the result of multiplying the integer with the list, for example:
?-Multiplication(3,[2,7,4],Result).
should return
Result = [6,21,12].
Here's my code:
Multiplication(X,[],Result).
Multiplication(X,[Head|Tail],Result) :-
Y is X*Head,
append([Result], [Y], L),
append([],L,Result), // HERE
Multiplication(X,Tail,Result).
And I get the following error:
Domain error: 'acyclic_term ' expected, found '#(lists:append([],S_1,S_1),[S_1=[S_1,1]])'
on the second append call.
If anyone knows why I receive the error, how to fix it or another way to solve this, I'm open to ideas.
Your two goals append([Result], [Y], L), append([],L,Result) are exactly the same as:
L = [Result,Y], L = Result.
or even simpler:
L = [L,Y]
which would result either in silent failure or an infinite term. Instead, your Prolog produces an error, so that you can correct your program.
In your original code:
Multiplication(X,[Head|Tail],Result) :-
Y is X*Head,
append([Result], [Y], L),
append([],L,Result), // HERE
Multiplication(X,Tail,Result).
You're getting a "cycle" because you're appending Result to something to get L, then appending something to L to get Result. That's not good. You also have a capitalized predicate name, which is a syntax error. (I assume that, since you ran your code, it wasn't capitalized in the original version.)
You're new proposed solution is overly complicated. Why do you need the 4th argument? Also, your base case for return (which is return(X, [], Result) doesn't make sense, as it has to singleton variables. The use of append/3 is overkill since recursion handles the iteration through the list elements for you.
Starting from the top, you have a common pattern in Prolog where you want to run a query on corresponding elements of two or more lists. A simple recursive solution would look something like this:
multiplication(_, [], []). % Multiplying anything by the empty list is the empty list
multiplication(M, [X|Xs], [XX|XXs]) :-
XX is M * X,
multiplication(M, Xs, XXs).
Another way to implement this kind of pattern in Prolog is with maplist/3. You can first define the query on corresponding elements:
multiply(X, Y, Product) :- Product is X * Y.
Then use maplist/3:
multiplication(M, List, Result) :-
maplist(multiply(M), List, Result).
Maplist will do a call(multiply(M), ...) on each corresponding pair of elements of List and Result.
I edited the code and came up with this:
multiplication(X,[],Result,Result).
multiplication(X,[Head|Tail],List,Result) :-
Y is X*Head,
append(List, [Y], L),
multiplication(X,Tail,L,Result).
return(X,[],Result).
return(X,L,Result) :-
multiplication(X,L,_,Result).
and the query:
return(2,[1,2],Result).
After the first run, it seems to return Result as it should be, but it runs forever.
This is a question provoked by an already deleted answer to this question. The issue could be summarized as follows:
Is it possible to fold over a list, with the tail of the list generated while folding?
Here is what I mean. Say I want to calculate the factorial (this is a silly example but it is just for demonstration), and decide to do it like this:
fac_a(N, F) :-
must_be(nonneg, N),
( N =< 1
-> F = 1
; numlist(2, N, [H|T]),
foldl(multiplication, T, H, F)
).
multiplication(X, Y, Z) :-
Z is Y * X.
Here, I need to generate the list that I give to foldl. However, I could do the same in constant memory (without generating the list and without using foldl):
fac_b(N, F) :-
must_be(nonneg, N),
( N =< 1
-> F = 1
; fac_b_1(2, N, 2, F)
).
fac_b_1(X, N, Acc, F) :-
( X < N
-> succ(X, X1),
Acc1 is X1 * Acc,
fac_b_1(X1, N, Acc1, F)
; Acc = F
).
The point here is that unlike the solution that uses foldl, this uses constant memory: no need for generating a list with all values!
Calculating a factorial is not the best example, but it is easier to follow for the stupidity that comes next.
Let's say that I am really afraid of loops (and recursion), and insist on calculating the factorial using a fold. I still would need a list, though. So here is what I might try:
fac_c(N, F) :-
must_be(nonneg, N),
( N =< 1
-> F = 1
; foldl(fac_foldl(N), [2|Back], 2-Back, F-[])
).
fac_foldl(N, X, Acc-Back, F-Rest) :-
( X < N
-> succ(X, X1),
F is Acc * X1,
Back = [X1|Rest]
; Acc = F,
Back = []
).
To my surprise, this works as intended. I can "seed" the fold with an initial value at the head of a partial list, and keep on adding the next element as I consume the current head. The definition of fac_foldl/4 is almost identical to the definition of fac_b_1/4 above: the only difference is that the state is maintained differently. My assumption here is that this should use constant memory: is that assumption wrong?
I know this is silly, but it could however be useful for folding over a list that cannot be known when the fold starts. In the original question we had to find a connected region, given a list of x-y coordinates. It is not enough to fold over the list of x-y coordinates once (you can however do it in two passes; note that there is at least one better way to do it, referenced in the same Wikipedia article, but this also uses multiple passes; altogether, the multiple-pass algorithms assume constant-time access to neighboring pixels!).
My own solution to the original "regions" question looks something like this:
set_region_rest([A|As], Region, Rest) :-
sort([A|As], [B|Bs]),
open_set_closed_rest([B], Bs, Region0, Rest),
sort(Region0, Region).
open_set_closed_rest([], Rest, [], Rest).
open_set_closed_rest([X-Y|As], Set, [X-Y|Closed0], Rest) :-
X0 is X-1, X1 is X + 1,
Y0 is Y-1, Y1 is Y + 1,
ord_intersection([X0-Y,X-Y0,X-Y1,X1-Y], Set, New, Set0),
append(New, As, Open),
open_set_closed_rest(Open, Set0, Closed0, Rest).
Using the same "technique" as above, we can twist this into a fold:
set_region_rest_foldl([A|As], Region, Rest) :-
sort([A|As], [B|Bs]),
foldl(region_foldl, [B|Back],
closed_rest(Region0, Bs)-Back,
closed_rest([], Rest)-[]),
!,
sort(Region0, Region).
region_foldl(X-Y,
closed_rest([X-Y|Closed0], Set)-Back,
closed_rest(Closed0, Set0)-Back0) :-
X0 is X-1, X1 is X + 1,
Y0 is Y-1, Y1 is Y + 1,
ord_intersection([X0-Y,X-Y0,X-Y1,X1-Y], Set, New, Set0),
append(New, Back0, Back).
This also "works". The fold leaves behind a choice point, because I haven't articulated the end condition as in fac_foldl/4 above, so I need a cut right after it (ugly).
The Questions
Is there a clean way of closing the list and removing the cut? In the factorial example, we know when to stop because we have additional information; however, in the second example, how do we notice that the back of the list should be the empty list?
Is there a hidden problem I am missing?
This looks like its somehow similar to the Implicit State with DCGs, but I have to admit I never quite got how that works; are these connected?
You are touching on several extremely interesting aspects of Prolog, each well worth several separate questions on its own. I will provide a high-level answer to your actual questions, and hope that you post follow-up questions on the points that are most interesting to you.
First, I will trim down the fragment to its essence:
essence(N) :-
foldl(essence_(N), [2|Back], Back, _).
essence_(N, X0, Back, Rest) :-
( X0 #< N ->
X1 #= X0 + 1,
Back = [X1|Rest]
; Back = []
).
Note that this prevents the creation of extremely large integers, so that we can really study the memory behaviour of this pattern.
To your first question: Yes, this runs in O(1) space (assuming constant space for arising integers).
Why? Because although you continuously create lists in Back = [X1|Rest], these lists can all be readily garbage collected because you are not referencing them anywhere.
To test memory aspects of your program, consider for example the following query, and limit the global stack of your Prolog system so that you can quickly detect growing memory by running out of (global) stack:
?- length(_, E),
N #= 2^E,
portray_clause(N),
essence(N),
false.
This yields:
1.
2.
...
8388608.
16777216.
etc.
It would be completely different if you referenced the list somewhere. For example:
essence(N) :-
foldl(essence_(N), [2|Back], Back, _),
Back = [].
With this very small change, the above query yields:
?- length(_, E),
N #= 2^E,
portray_clause(N),
essence(N),
false.
1.
2.
...
1048576.
ERROR: Out of global stack
Thus, whether a term is referenced somewhere can significantly influence the memory requirements of your program. This sounds quite frightening, but really is hardly an issue in practice: You either need the term, in which case you need to represent it in memory anyway, or you don't need the term, in which case it is simply no longer referenced in your program and becomes amenable to garbage collection. In fact, the amazing thing is rather that GC works so well in Prolog also for quite complex programs that not much needs to be said about it in many situations.
On to your second question: Clearly, using (->)/2 is almost always highly problematic in that it limits you to a particular direction of use, destroying the generality we expect from logical relations.
There are several solutions for this. If your CLP(FD) system supports zcompare/3 or a similar feature, you can write essence_/3 as follows:
essence_(N, X0, Back, Rest) :-
zcompare(C, X0, N),
closing(C, X0, Back, Rest).
closing(<, X0, [X1|Rest], Rest) :- X1 #= X0 + 1.
closing(=, _, [], _).
Another very nice meta-predicate called if_/3 was recently introduced in Indexing dif/2 by Ulrich Neumerkel and Stefan Kral. I leave implementing this with if_/3 as a very worthwhile and instructive exercise. Discussing this is well worth its own question!
On to the third question: How do states with DCGs relate to this? DCG notation is definitely useful if you want to pass around a global state to several predicates, where only a few of them need to access or modify the state, and most of them simply pass the state through. This is completely analogous to monads in Haskell.
The "normal" Prolog solution would be to extend each predicate with 2 arguments to describe the relation between the state before the call of the predicate, and the state after it. DCG notation lets you avoid this hassle.
Importantly, using DCG notation, you can copy imperative algorithms almost verbatim to Prolog, without the hassle of introducing many auxiliary arguments, even if you need global states. As an example for this, consider a fragment of Tarjan's strongly connected components algorithm in imperative terms:
function strongconnect(v)
// Set the depth index for v to the smallest unused index
v.index := index
v.lowlink := index
index := index + 1
S.push(v)
This clearly makes use of a global stack and index, which ordinarily would become new arguments that you need to pass around in all your predicates. Not so with DCG notation! For the moment, assume that the global entities are simply easily accessible, and so you can code the whole fragment in Prolog as:
scc_(V) -->
vindex_is_index(V),
vlowlink_is_index(V),
index_plus_one,
s_push(V),
This is a very good candidate for its own question, so consider this a teaser.
At last, I have a general remark: In my view, we are only at the beginning of finding a series of very powerful and general meta-predicates, and the solution space is still largely unexplored. call/N, maplist/[3,4], foldl/4 and other meta-predicates are definitely a good start. if_/3 has the potential to combine good performance with the generality we expect from Prolog predicates.
If your Prolog implementation supports freeze/2 or similar predicate (e.g. Swi-Prolog), then you can use following approach:
fac_list(L, N, Max) :-
(N >= Max, L = [Max], !)
;
freeze(L, (
L = [N|Rest],
N2 is N + 1,
fac_list(Rest, N2, Max)
)).
multiplication(X, Y, Z) :-
Z is Y * X.
factorial(N, Factorial) :-
fac_list(L, 1, N),
foldl(multiplication, L, 1, Factorial).
Example above first defines a predicate (fac_list) which creates a "lazy" list of increasing integer values starting from N up to maximum value (Max), where next list element is generated only after previous one was "accessed" (more on that below). Then, factorial just folds multiplication over lazy list, resulting in constant memory usage.
The key to understanding how this example works is remembering that Prolog lists are, in fact, just terms of arity 2 with name '.' (actually, in Swi-Prolog 7 the name was changed, but this is not important for this discussion), where first element represents list item and the second element represents tail (or terminating element - empty list, []). For example. [1, 2, 3] can be represented as:
.(1, .(2, .(3, [])))
Then, freeze is defined as follows:
freeze(+Var, :Goal)
Delay the execution of Goal until Var is bound
This means if we call:
freeze(L, L=[1|Tail]), L = [A|Rest].
then following steps will happen:
freeze(L, L=[1|Tail]) is called
Prolog "remembers" that when L will be unified with "anything", it needs to call L=[1|Tail]
L = [A|Rest] is called
Prolog unifies L with .(A, Rest)
This unification triggers execution of L=[1|Tail]
This, obviously, unifies L, which at this point is bound to .(A, Rest), with .(1, Tail)
As a result, A gets unified with 1.
We can extend this example as follows:
freeze(L1, L1=[1|L2]),
freeze(L2, L2=[2|L3]),
freeze(L3, L3=[3]),
L1 = [A|R2], % L1=[1|L2] is called at this point
R2 = [B|R3], % L2=[2|L3] is called at this point
R3 = [C]. % L3=[3] is called at this point
This works exactly like the previous example, except that it gradually generates 3 elements, instead of 1.
As per Boris's request, the second example implemented using freeze. Honestly, I'm not quite sure whether this answers the question, as the code (and, IMO, the problem) is rather contrived, but here it is. At least I hope this will give other people the idea what freeze might be useful for. For simplicity, I am using 1D problem instead of 2D, but changing the code to use 2 coordinates should be rather trivial.
The general idea is to have (1) function that generates new Open/Closed/Rest/etc. state based on previous one, (2) "infinite" list generator which can be told to "stop" generating new elements from the "outside", and (3) fold_step function which folds over "infinite" list, generating new state on each list item and, if that state is considered to be the last one, tells generator to halt.
It is worth to note that list's elements are used for no other reason but to inform generator to stop. All calculation state is stored inside accumulator.
Boris, please clarify whether this gives a solution to your problem. More precisely, what kind of data you were trying to pass to fold step handler (Item, Accumulator, Next Accumulator)?
adjacent(X, Y) :-
succ(X, Y) ;
succ(Y, X).
state_seq(State, L) :-
(State == halt -> L = [], !)
;
freeze(L, (
L = [H|T],
freeze(H, state_seq(H, T))
)).
fold_step(Item, Acc, NewAcc) :-
next_state(Acc, NewAcc),
NewAcc = _:_:_:NewRest,
(var(NewRest) ->
Item = next ;
Item = halt
).
next_state(Open:Set:Region:_Rest, NewOpen:NewSet:NewRegion:NewRest) :-
Open = [],
NewOpen = Open,
NewSet = Set,
NewRegion = Region,
NewRest = Set.
next_state(Open:Set:Region:Rest, NewOpen:NewSet:NewRegion:NewRest) :-
Open = [H|T],
partition(adjacent(H), Set, Adjacent, NotAdjacent),
append(Adjacent, T, NewOpen),
NewSet = NotAdjacent,
NewRegion = [H|Region],
NewRest = Rest.
set_region_rest(Ns, Region, Rest) :-
Ns = [H|T],
state_seq(next, L),
foldl(fold_step, L, [H]:T:[]:_, _:_:Region:Rest).
One fine improvement to the code above would be making fold_step a higher order function, passing it next_state as the first argument.
I am new to prolog , I have a list in prolog like A=[1,2,3,4], and than I accessed nth element using nth(N,[_|T],R). Now I have Nth element in R, than I have done some calculation on R. Now what I want is to update that nth element in list.
Because of I am doing a lot of calculations with each element in list I can't make a new list each time.
I didn't find any method to update list.
With regard to our conversation, you can add two lists together, creating a third, by specifying that the two head elements of the source lists, added together, make the head element of the result list, and that this applies to the remainder of the lists.
There is also a need for a base case, that is, when the two source lists are empty, so should the result list.
addLists([X|A], [Y|B], [Z|C]) :- Z is X+Y, addLists(A, B, C).
addLists([], [], []).
Remember you are always aiming to specify the constraints of the answer, more than the method of answering it. Prolog is very different to other programming languages in that you do not tell it how to do something, you simply tell it conditions that are true for the answer and let it extrapolate it.
From the comments you exchanged with #Orbling seems that what you need is a kind of maplist/4
process_list(A, B, C) :-
maplist(process_elem, A, B, C).
process_elem(A, B, C) :- C is A + B. % or whatever needed
If you are using the index in process_elem then this is not appropriate. Then make a recursive visit of list, passing down the index
process_list(A, B, C) :-
process_list(1, A, B, C).
process_list(I, [A|As], [B|Bs], [C|Cs]) :-
C is A + B * I, % or whatever needed
J is I + 1,
!, process_list(J, As, Bs, Cs).
process_list(_, [], [], []).
edit Just to add to the various ways exposed in answers to the question #Orbling suggests, here a way using nth0/4
?- I = 6, nth0(I,"hello world",_,T), nth0(I,U,0'W,T), format('~s',[U]).
hello World
In SWI-Prolog, I have a list whose elements are pairs of the form Key-ValuesList. For instance, one such list may look like:
[1-[a,b],2-[],3-[c]]
I would like to transform this list into a nested list of pairs of the form Key-[Value], where Value is an element in ValuesList. The above example would be transformed into:
[[1-[a],2-[],3-[c]], [1-[b],2-[],3-[c]]]
My current solution is the following:
% all_pairs_lists(+InputList, -OutputLists).
all_pairs_lists([], [[]]).
all_pairs_lists([Key-[]|Values], CP) :-
!,
findall([Key-[]|R], (all_pairs_lists(Values,RCP), member(R,RCP)), CP).
all_pairs_lists([Key-Value|Values], CP) :-
findall([Key-[V]|R], (all_pairs_lists(Values,RCP), member(V,Value), member(R,RCP)), CP).
Using this predicate, a call of the form
all_pairs_lists([1-[a,b],2-[],3-[c]],OutputLists).
Binds the variable OutputLists to the desired result mentioned above. While it appears correct, this implementation causes an "Out of global stack" error when InputList has very long lists as values.
Is there a less stack consuming approach to doing this? It would seem like quite a common operation for this type of data structure.
Well, to sum it up, you're doing it wrong.
In Prolog, when we want to express a relation instead of a function (several results possible instead of one), we don't use findall/3 and member/2 directly. We rather state what the relation is and then maybe once it's done if we need a list of results we use findall/3.
Here what it means is that we want to express the following relation:
Take a list of Key-Values and return a list of Key-[Value] where Value is a member of the Values list.
We could do so as follows:
% The base case: handle the empty list
a_pair_list([], []).
% The case where the Values list is empty, then the resulting [Value] is []
a_pair_list([Key-[]|List], [Key-[]|Result]) :-
a_pair_list(List, Result).
% The case where the Values list is not empty, then Value is a member of Values.
a_pair_list([Key-[Not|Empty]|List], [Key-[Value]|Result]) :-
member(Value, [Not|Empty]),
a_pair_list(List, Result).
Once this relation is expressed, we can already obtain all the info we wish:
?- a_pair_list([1-[a, b], 2-[], 3-[c]], Result).
Result = [1-[a], 2-[], 3-[c]] ;
Result = [1-[b], 2-[], 3-[c]] ;
false.
The desired list is now just a fairly straight-forward findall/3 call away:
all_pairs_lists(Input, Output) :-
findall(Result, a_pair_list(Input, Result), Output).
The important thing to remember is that it's way better to stay away from extra logical stuff: !/0, findall/3, etc... because it's often leading to less general programs and/or less correct ones. Here since we can express the relation stated above in a pure and clean way, we should. This way we can limit the annoying use of findall/3 at the strict minimum.
As #Mog already explained clearly what the problem could be, here a version (ab)using of the basic 'functional' builtin for list handling:
all_pairs_lists(I, O) :-
findall(U, maplist(pairs_lists, I, U), O).
pairs_lists(K-[], K-[]) :- !.
pairs_lists(K-L, K-[R]) :- member(R, L).
test:
?- all_pairs_lists([1-[a,b],2-[],3-[c]],OutputLists).
OutputLists = [[1-[a], 2-[], 3-[c]], [1-[b], 2-[], 3-[c]]].
I starting to study for my upcoming exam and I'm stuck on a trivial prolog practice question which is not a good sign lol.
It should be really easy, but for some reason I cant figure it out right now.
The task is to simply count the number of odd numbers in a list of Int in prolog.
I did it easily in haskell, but my prolog is terrible. Could someone show me an easy way to do this, and briefly explain what you did?
So far I have:
odd(X):- 1 is X mod 2.
countOdds([],0).
countOdds(X|Xs],Y):-
?????
Your definition of odd/1 is fine.
The fact for the empty list is also fine.
IN the recursive clause you need to distinguish between odd numbers and even numbers. If the number is odd, the counter should be increased:
countOdds([X|Xs],Y1) :- odd(X), countOdds(Xs,Y), Y1 is Y+1.
If the number is not odd (=even) the counter should not be increased.
countOdds([X|Xs],Y) :- \+ odd(X), countOdds(Xs,Y).
where \+ denotes negation as failure.
Alternatively, you can use ! in the first recursive clause and drop the condition in the second one:
countOdds([X|Xs],Y1) :- odd(X), !, countOdds(Xs,Y), Y1 is Y+1.
countOdds([X|Xs],Y) :- countOdds(Xs,Y).
In Prolog you use recursion to inspect elements of recursive data structs, as lists are.
Pattern matching allows selecting the right rule to apply.
The trivial way to do your task:
You have a list = [X|Xs], for each each element X, if is odd(X) return countOdds(Xs)+1 else return countOdds(Xs).
countOdds([], 0).
countOdds([X|Xs], C) :-
odd(X),
!, % this cut is required, as rightly evidenced by Alexander Serebrenik
countOdds(Xs, Cs),
C is Cs + 1.
countOdds([_|Xs], Cs) :-
countOdds(Xs, Cs).
Note the if, is handled with a different rule with same pattern: when Prolog find a non odd element, it backtracks to the last rule.
ISO Prolog has syntax sugar for If Then Else, with that you can write
countOdds([], 0).
countOdds([X|Xs], C) :-
countOdds(Xs, Cs),
( odd(X)
-> C is Cs + 1
; C is Cs
).
In the first version, the recursive call follows the test odd(X), to avoid an useless visit of list'tail that should be repeated on backtracking.
edit Without the cut, we get multiple execution path, and so possibly incorrect results under 'all solution' predicates (findall, setof, etc...)
This last version put in evidence that the procedure isn't tail recursive. To get a tail recursive procedure add an accumulator:
countOdds(L, C) :- countOdds(L, 0, C).
countOdds([], A, A).
countOdds([X|Xs], A, Cs) :-
( odd(X)
-> A1 is A + 1
; A1 is A
),
countOdds(Xs, A1, Cs).