Related
How does this bubble sort solution work in Prolog?
bubblesort([], []).
bubblesort([H], [H]).
bubblesort([H|D], R) :-
bubblesort(D, E),
[B|G] = E,
( (H =< B, R = [H|E])
; (H > B, bubblesort([B,H|G], R))
).
Here is an example trace: https://pastebin.com/T0DLsmAV
I understand the the line bubblesort(D,E), is responsible for sorting it down to one element, but I don't understand how this works. I understand the basics of lists in prolog, but still cannot work out how this solution operates.
The main difficulty with this code is that bad variable names were chosen and are making the logic harder to follow than it needs to be.
The first two cases are obviously base cases. The first says "the empty list is already sorted" and the second says "a singleton list is already sorted." This should make sense. The third case is where things get interesting.
Let's examine the first part.
bubblesort([H|D], R) :-
bubblesort(D, E),
All that's happened so far is we've named our result R and broken our inputs into a first element H and a tail D. From there, we have said, let's bubblesort the tail of our input and call that E. Maybe this would be a little easier to follow?
bubblesort([H|T], Result) :-
bubblesort(T, TSorted),
Next up,
[B|G] = E,
Again, bad names, but what the author is intending to do here is straightforward: take apart the result of sorting the tail so we can talk about whether the next item in the sorted tail is the right element for that position, or if it needs to switch places with the head of our input. Let's rename:
[HeadOfTSorted|RestOfTSorted] = TSorted,
Now we have a condition. Think of it in terms of prepending onto a sorted list. Say you have some element, like 3, and I hand you a sorted list. You want to determine if your 3 goes at the front or somewhere else. Well, suppose I gave you a sorted list that looked like [5,7,19,23,...]. You'd know that your 3 is right where it needs to be, and you'd hand back [3,5,7,19,23,...]. That's exactly the first case of the condition:
( (H =< HeadOfTSorted, Result = [H|TSorted])
Now consider another case, where I hand you a list that starts with [1,2,...]. You know you can't just put the three at the start and give me back [3,1,2,...]. But you don't really know where the 3 goes; it just doesn't go at the start. So what you have to do is resort the rest of the list with the 3 at the start, after the 1: [1 | resorted([3,2,...])]. That's effectively the other branch of the condition:
; (H > HeadOfTSorted, bubblesort([HeadOfTSorted,H|RestOfTSorted], R))
).
Hope this helps!
note: the key to recursive problem solving is exactly not to think about the minutiae of our code's operations. Imagine you already have the solution, then just use it to solve a smaller subproblem, thus arriving at the full problem's solution.
Your code, with more suggestive variable names so I could follow it, reads:
bubblesort([], []). % empty list is already sorted
bubblesort([H], [H]). % singleton list is already sorted
bubblesort([H|T], S) :- % `[H|T]` sorted is `S`, *if*
bubblesort(T, [M|R]), % `T` sorted is `[M|R]`, *and*
( % *either*,
H =< M, % in case `H` is not greater than `M`,
S = [H,M|R] % `S` is `[H,M|R]`,
; % *or*
H > M, % in case `H` is greater than `M`,
bubblesort([M,H|R], S) % `S` is `[M,H|R]` sorted by the same algorithm
).
(H is for "head", T is for "tail", S is "sorted", R "rest" and M is "minimum" -- see below for that).
We prove its correctness by structural induction. The induction hypothesis (IH) is that this definition is correct for shorter lists. We need to prove it is then also correct for a longer list. Indeed T is one-element shorter than [H|T]. Thus IH says [M|R] is sorted. This means M is the minimum element in T. It also means T is non-empty (sorting doesn't change the number of elements), so the clauses are indeed mutually-exclusive.
If H is not larger than the minimum element in T, [H,M|R] is obviously sorted.
Otherwise, we sort [M,H|R]. M is the minimum element and thus guaranteed to be the first in the result. What's actually sorted is [H|R], which is one element shorter, thus by IH sorting it works. QED.
If the last step sounds fishy to you, consider replacing the second alternative with the equivalent
; H > M, % in case `H` is greater then `M`,
bubblesort([H|R], S1), % `S1` is `[H|R]` sorted by the same algorithm
S = [M|S1]
).
where the applicability of the induction step is even clearer.
I'm not so sure it's a bubble sort though.
update: Indeed, measuring the empirical orders of growth, its number of inferences grows as ~ n3 (or slower), but true bubble sort clocked at ~ n2.1 (close enough to the theoretical ~ n2), where n is the list's length:
tbs([], []). % 'true' bubble sort
tbs([H],[H]).
tbs(L,S):- bubble(L,B),
( L==B -> S=L ; tbs(B,S) ).
bubble([],[]).
bubble([A],[A]).
bubble([A,B|C],R):-
( A =< B -> bubble([B|C],X), R=[A|X]
; bubble([A|C],X), R=[B|X] ).
I'm generating random coordinates and adding on my list, but first I need verify if that coordinate already exists. I'm trying to use member but when I was debugging I saw that isn't working:
My code is basically this:
% L is a list and Q is a count that define the number of coordinate
% X and Y are the coordinate members
% check if the coordniate already exists
% if exists, R is 0 and if not, R is 1
createCoordinates(L,Q) :-
random(1,10,X),
random(1,10,Y),
convertNumber(X,Z),
checkCoordinate([Z,Y],L,R),
(R is 0 -> print('member'), createCoordinates(L,Q); print('not member'),createCoordinates(L,Q-1).
checkCoordinate(C,L,R) :-
(member(C,L) -> R is 0; R is 1).
% transforms the number N in a letter L
convertNumber(N,L) :-
N is 1, L = 'A';
N is 2, L = 'B';
...
N is 10, L = 'J'.
%call createCoordinates
createCoordinates(L,20).
When I was debugging this was the output:
In this picture I'm in the firts interation and L is empty, so R should be 1 but always is 0, the coordinate always is part of the list.
I have the impression that the member clause is adding the coordinate at my list and does'nt make sense
First off, I would recommend breaking your problem down into smaller pieces. You should have a procedure for making a random coordinate:
random_coordinate([X,Y]) :-
random(1, 10, XN), convertNumber(XN, X),
random(1, 10, Y).
Second, your checkCoordinate/3 is converting Prolog's success/failure into an integer, which is just busy work for Prolog and not really improving life for you. memberchk/2 is completely sufficient to your task (member/2 would work too but is more powerful than necessary). The real problem here is not that member/2 didn't work, it's that you are trying to build up this list parameter on the way out, but you need it to exist on the way in to examine it.
We usually solve this kind of problem in Prolog by adding a third parameter and prepending values to the list on the way through. The base case then equates that list with the outbound list and we protect the whole thing with a lower-arity procedure. In other words, we do this:
random_coordinates(N, Coordinates) :- random_coordinates(N, [], Coordinates).
random_coordinates(0, Result, Result).
random_coordinates(N, CoordinatesSoFar, FinalResult) :- ...
Now that we have two things, memberchk/2 should work the way we need it to:
random_coordinates(N, CoordinatesSoFar, FinalResult) :-
N > 0, succ(N0, N), % count down, will need for recursive call
random_coordinate(Coord),
(memberchk(Coord, CoordinatesSoFar) ->
random_coordinates(N, CoordinatesSoFar, FinalResult)
;
random_coordinates(N0, [Coord|CoordinatesSoFar], FinalResult)
).
And this seems to do what we want:
?- random_coordinates(10, L), write(L), nl.
[[G,7],[G,3],[H,9],[H,8],[A,4],[G,1],[I,9],[H,6],[E,5],[G,8]]
?- random_coordinates(10, L), write(L), nl.
[[F,1],[I,8],[H,4],[I,1],[D,3],[I,6],[E,9],[D,1],[C,5],[F,8]]
Finally, I note you continue to use this syntax: N is 1, .... I caution you that this looks like an error to me because there is no distinction between this and N = 1, and your predicate could be stated somewhat tiresomely just with this:
convertNumber(1, 'A').
convertNumber(2, 'B').
...
My inclination would be to do it computationally with char_code/2 but this construction is actually probably better.
Another hint that you are doing something wrong is that the parameter L to createCoordinates/2 gets passed along in all cases and is not examined in any of them. In Prolog, we often have variables that appear to just be passed around meaninglessly, but they usually change positions or are used multiple times, as in random_coordinates(0, Result, Result); while nothing appears to be happening there, what's actually happening is plumbing: the built-up parameter becomes the result value. Nothing interesting is happening to the variable directly there, but it is being plumbed around. But nothing is happening at all to L in your code, except it is supposedly being checked for a new coordinate. But you're never actually appending anything to it, so there's no reason to expect that anything would wind up in L.
Edit Notice that #lambda.xy.x solves the problem in their answer by prepending the new coordinate in the head of the clause and examining the list only after the recursive call in the body, obviating the need for the second list parameter.
Edit 2 Also take a look at #lambda.xy.x's other solution as it has better time complexity as N approaches 100.
Since i had already written it, here is an alternative solution: The building block is gen_coord_notin/2 which guarantees a fresh solution C with regard to an exclusion list Excl.
gen_coord_notin(C, Excl) :-
random(1,10,X),
random(1,10,Y),
( memberchk(X-Y, Excl) ->
gen_coord_notin(C, Excl)
;
C = X-Y
).
The trick is that we only unify C with the new result, if it is fresh.
Then we only have to fold the generations into N iterations:
gen_coords([], 0).
gen_coords([X|Xs], N) :-
N > 0,
M is N - 1,
gen_coords(Xs, M),
gen_coord_notin(X, Xs).
Remark 1: since coordinates are always 2-tuples, a list representation invites unwanted errors (e.g. writing [X|Y] instead of [X,Y]). Traditionally, an infix operator like - is used to seperate tuples, but it's not any different than using coord(X,Y).
Remark 2: this predicate is inherently non-logical (i.e. calling gen_coords(X, 20) twice will result in different substitutions for X). You might use the meta-level predicates var/1, nonvar/1, ground/1, integer, etc. to guard against non-sensical calls like gen_coord(1-2, [1-1]).
Remark 3: it is also important that the conditional does not have multiple solutions (compare member(X,[A,B]) and memberchk(X,[A,B])). In general, this can be achieved by calling once/1 but there is a specialized predicate memberchk/2 which I used here.
I just realized that the performance of my other solutions is very bad for N close to 100. The reason is that with diminishing possible coordinates, the generate and test approach will take longer and longer. There's an alternative solution which generates all coordinates and picks N random ones:
all_pairs(Ls) :-
findall(X-Y, (between(1,10,X), between(1,10,Y)), Ls).
remove_index(X,[X|Xs],Xs,0).
remove_index(I,[X|Xs],[X|Rest],N) :-
N > 0,
M is N - 1,
remove_index(I,Xs,Rest,M).
n_from_pool(_Pool, [], 0).
n_from_pool(Pool, [C|Cs], N) :-
N > 0,
M is N - 1,
length(Pool, L),
random(0,L,R),
remove_index(C,Pool,NPool,R),
n_from_pool(NPool, Cs, M).
gen_coords2(Xs, N) :-
all_pairs(Pool),
n_from_pool(Pool, Xs, N).
Now the query
?- gen_coords2(Xs, 100).
Xs = [4-6, 5-6, 5-8, 9-6, 3-1, 1-3, 9-4, 6-1, ... - ...|...] ;
false.
succeeds as expected. The error message
?- gen_coords2(Xs, 101).
ERROR: random/1: Domain error: not_less_than_one' expected, found0'
when we try to generate more distinct elements than possible is not nice, but better than non-termination.
I picked up prolog a couple of days ago and I 'm kind of stuck to this question. I want to subtract a number recursively until that number becomes less than 0. In pseudocode that would be like:
N:=0
while(Y>=X)
{
Y := Y-X
N := N+1
Y := Y+2
}
So for example if I have Y=20 and X=10 then we would get N=2 and Y=4.
Any ideas? Thanks in advance. Any help appreciated. I'm using SWI Prolog.
EDIT 1
What I've accomplished so far is(although I'm not sure even if its correct):
sufficient(X, Y, M, N, F) :-
F is Y-X,
Y>=X,
plus(M, 1, N),
sufficient(X, F, N, N, F).
I have problem finding my base case, I'm confused on how to implement it. Also, in the sufficient I have implemented, obviously when Y<X it terminates returning false. Is there a way to get the N and F before terminating? I am feeling that I am not thinking the "prolog" way, since I am mostly used on C and that vagues my thinking. Thanks.
EDIT 2
I have found my base case and I can stop recursion however, I can't manage to ge the correct values. My code:
sufficient(X, Y, M, N, F) :- Y<X.
sufficient(X, Y, M, N, F) :-
F is Y-X,
plus(M, 1, N),
sufficient(X, F, N, D, E).
Thing is after the first recursion, if for example I call sufficient as sufficient(10,21,0,N,F). from the swi prolog command prompt, I 'll get N=1 and F=11. That happens because I make 2 new variables D and E. If I don't make those 2 new variables(D and E), and at the 3rd sufficient in the code I call N and F instead of D and E at the line F is Y-X, I get a false, because F is 11 and Y-X is 1. Do I have to set the a subtraction function myself, since F is Y-X is not exactly a subtraction? Any ideas on how to do it?
All recursive functions need at least one base case. In what circumstance should your function say, OK, I have the answer, no need to recurse?
It would be the case in which your pseudocode loop is done, right?
Usually we write it in this format:
factorial(0,1). % The factorial of 0 is 1.
factorial(N,Factorial) :-
N>0, % You may need to test applicability
% of this recursive clause
NMinus1 is N-1, % maybe some setup
factorial(NMinus1,FactorialOfNMinus1), %recursive call
Factorial is N*FactorialOfNMinus1). %and maybe some code after
I wouldn't want to do your homework for you, but this should get you going:
sufficient(X,Y,M,N,F) :- %whatever condition means you're done,
% and results = whatever they should
sufficient(X,Y,M,N,F) :- %whatever condition means you aren't done
% and setting up results w/ a recursive call
One more hint: looks like M is a temporary variable and need not be a parameter.
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'm trying to make a function that has a list of lists, it multiplies the sum of the inner list with the outer list.
So far i can sum a list, i've made a function sumlist([1..n],X) that will return X = (result). But i cannot get another function to usefully work with that function, i've tried both is and = to no avail.
Is this what you mean?
prodsumlist([], 1).
prodsumlist([Head | Tail], Result) :-
sumlist(Head, Sum_Of_Head),
prodsumlist(Tail, ProdSum_Of_Tail),
Result is Sum_Of_Head * ProdSum_Of_Tail.
where sumlist/2 is a SWI-Prolog built-in.
Usage example:
?- prodsumlist([[1, 2], [3], [-4]], Result).
Result = -36.
The part "it multiplies the sum of the inner list with the outer list" isn't really clear, but I believe you mean that, given a list [L1,...,Ln] of lists of numbers, you want to calculate S1*..*Sn where Si is the sum of the elements in Li (for each i).
I assume the existence of plus and mult with their obvious meaning (e.g. plus(N,M,R) holds precisely when R is equal to N+M). First we need predicate sum such that sum(L,S) holds if, and only if, S is the sum of the elements of L. If L is empty, S obviously must be 0:
sum([],0).
If L is not empty but of the form [N|L2], then we have that S must be N plus the sum S2 of the elements in L2. In other words, we must have both sum(L2,S2) (to get S2 to be the sum of the elements of L2) and plus(N,S2,S). That is:
sum([N|L2],S) :- sum(L2,S2), plus(N,S2,S).
In the same way you can figure out the predicate p you are looking for. We want that p(L,R) holds if, and only if, R is the product of S1 through Sn where L=[L1,...,Ln] and sum(Li,Si) for all i. If L is empty, R must be 1:
p([],1).
If L is not empty but of the form [LL|L2], then we have that R must be the product of 'S', the sum of the elements of LL, and 'P', the product of the sums of the lists in L2. For S we have already have sum(LL,S), so this gives us the following.
p([LL|L2],R) :- sum(LL,S), p(L2,P), mult(S,P,R).
One thing I would like to add is that it is probably not such a good idea to see these predicates as functions you might be used to from imperative or functional programming. It is not the case that sumlist([1,..,n],X) returns X = (result); (result) is a value for X such that sumlist([1,...,n],X) is true. This requires a somewhat different mindset. Instead of thinking "How can I calculate X such that p(X) holds?" you must think "When does P(X) hold?" and use the answer ("Well, if q(X) or r(X)!") to make the clauses (p(X) :- q(X) and p(X) :- r(X)).
Here is a rewrite of Kaarel's answer (that's the intention anyway!) but tail-recursive.
prodsumlist(List, Result) :-
xprodsumlist(List,1,Result).
xprodsumlist([],R,R).
xprodsumlist([Head|Rest],Sofar,Result) :-
sumlist(Head, Sum_Of_Head),
NewSofar is Sofar * Sum_Of_Head,
xprodsumlist(Rest, NewSofar, Result).