I'm a beginner at prolog and I'm having trouble getting started with the following problem:
Define the predicate partstr/3, where the first argument is a list, that generates a list A of length L that you find consecutive in the first list.
You should be able to present all answers with backtracking.
E.g.:
?- partstr([1, 2 , 3], L, A).
If L = 2 then A = [1,2] and [2,3],
or if L = 2 then F=[1,2] and [2,3].
and so on...
I feel like you would use recursion to solve it, but I'm not sure where to start. I would really appreciate some tips on how to solve this because I feel like I'm getting nowhere.
The core of this problem is that you need a way to pull all the sublist of length N from a list, correct?
So...
Consider that append/3 can concatenate two lists: append( [a,b,c], [1,2,3], L) returns L as [a,b,c,1,2,3]. But it can also decompose a list into a prefix and a suffix, so
append( Pfx, Sfx, [a,b,c])
will, on backtracking, successively yield:
Pfx
Sfx
[]
[a,b,c]
[a]
[b,c]
[a,b]
[c]
[a,b,c]
[]
...and... length/2 can not only tell you the length of a list, but
can generate lists of a specified length populated with unique,
unbound variables, so length(L,3) returns [V1,V2,V3].
You can combine those to get the behavior you want:
partstr( Xs, N, SL ) :- % To get all the contiguous sublists of length N from Xs...
append(_,Sfx,Xs) , % - repeatedly get all possible suffixes of Xs, and...
length(SL,N) , % - construct an empty, unbound list of the desired length (N), and...
append(SL,_,Sfx) % - pull that prefix off the suffix
. % Easy!
That's one approach. I imagine that this is coursework and that your instructor likely would like you to roll your own solution from scratch.
To do that, we first need a predicate that will yield the source list, and on backtracking remove the head of the list. Something like:
suffix( Xs , Xs ) .
suffix( [_|Xs] , Sfx ) :- suffix(Xs,Sfx).
Then we need a way to grab the 1st n elements from a list, something like this:
take( _ , 0 , [] ) :- ! .
take( [X|Xs] , N , [X|Sfx] ) :- N1 is N-1 , take(Xs,N1,Sfx) .
Given those two...
partstr( Xs, N , SL ) :-
suffix(Xs,Sfx),
take(Sfx,N, SL )
.
You can even dispense with the suffix/2 predicate, thus, rolling its functionality into partstr/3 itself:
partstr( Xs , N , SL ) :- take(Xs,N,SL).
partstr( [_|Xs] , N , SL ) :- partstr(Xs,N,SL).
And that, I think, is the sweet spot: it is hard to beat 4 lines of code —
partstr( Xs , N , SL ) :- take(Xs,N,SL) .
partstr( [_|Xs] , N , SL ) :- partstr(Xs,N,SL) .
take( _ , 0 , [] ) :- ! .
take( [X|Xs] , N , [X|Sfx] ) :- N > 0 , N1 is N-1 , take(Xs,N1,Sfx) .\
Related
This is the predicate:
partList(Len,L,R):-
length(L,LL),
length(R,RR),
RR is LL/Len,
append(R,L).
The query shows:
42 ?- partList(2,[t,t,t,f,f,t,f,f],R).
R = [[], [], [], [t, t, t, f, f, t, f, f]] .
But I want to partition into
[[t,t],[t,f],[f,t],[f,f]].
How do I fix this? Thanks!
The easy way is to look at the problem are repeatedly stripping off the 1st N items from the head of the list (until the list is exhausted).
partition( [] , [] ) . % if the source list is exhausted, we're done.
partition( [X] , [X] ) . % if the source list contains just one item, we're done.
partition( [X,Y|Z] , [[X,Y]|R] ) :- % if the source list contains 2 or more items, we take the 1st two, and ...
partition(Z,R) % - recursively partition the remainder.
. % Easy!.
To make it generic isn't much more complex.
First, we need a way to partition the list into a prefix, containing N items (or fewer if the list isn't sufficiently long) and a suffix, containing whatever's left (which might be nothing):
take_prefix( _ , [] , [] , [] ) . % if the source list is empty, both prefix and suffix are empty, regardless of the value of N.
take_prefix( 0 , [X|Xs] , [] , [X|Xs] ) . % if N is 0, The prefix is the empty list and the suffix is the source list.
take_prefix( N , [X|Xs] , [X|P] , S ) :- % otherwise, add the head to the prefix,
N > 0 , % - assuming N > 0
N1 is N-1 , % - decrement N
take_prefix(N1,Xs,P,S) % - and recurse down.
. % Easy!
This is the crux of the matter. Once you have that, it's just a matter of repeatedly (and recursively) applying it until you get to the empty list:
partition( _ , [] , [] ) . % if the source list is empty, we're done.
partition( N , L , [P|R] ) :- % otherwise...
take_prefix(N,L,P,S) , % - break it up into a prefix and a suffix,
partition(N,S,R) % - and recurse down on the suffix.
. % Easy!
I want to remove the element in list1 when it is equal to element in list2.
The query and the expected output is:
filter( [[1,2,3],[1]] , [[1]] , X ).
X = [[1, 2, 3]] ;
filter( [[1,2,3],[1],[2,3,4],[2]] , [[1],[2]] , X ).
X = [[1, 2, 3],[2,3,4]] ;
What I have done right now is :
filter(_,[],_).
filter([A|B],[A|D],E):-
filter(B,D,E).
filter([A|B],[C|D],[A|E]):-
A\=C,
filter(B,D,E).
but it seems not right and gives the output like this:
11 ?- filter([[1,2,3],[1]],[[1]],X).
X = [[1, 2, 3]|_G488] ;
Can anyone help? Maybe I was near success.
Your program will not work correctly because you are removing one element from the second list each time. Also your base case (the first clause) should not be an uninstantiated variable (that is what is giving you the |_G488 in your output.
You have to iterate over the first list filtering the elements found in the second list, but not remove the elements of the second list.
For example:
filter([], _, []).
filter([Item|L1], L2, L3):-
(member(Item, L2) -> L4=L3;L3=[Item|L4]),
filter(L1, L2, L4).
The first clase is the base case of the recursion. It states that the output for an empty list would be an empty list.
The second clause checks to see if the first item of the input list is found on the second list. If it is found then it won't be added to the resulting list; otherwise it is added. Then it recursively calls itself with the rest of the input list.
But in no case it removes the elements from the second list.
Well, the lazy way would be to use the build-ins, findall/3 and member/2:
filter( Xs , Ys , Rs ) :-
findall( X , ( member(X,Xs), \+ member(X,Ys) ) , Rs ) .
which says to find all X such that X is a member of the list Xs and Xs is *not* a member of the listYs`.
Assuming your instructor wants you to come up with your own implementation, you probably first want to decompose your problem. You need to do two things:
Iterate over a list, removing items that are found in another list.
Given an item, determine whether it is contained in another list.
Both of these are simple. To determine if an item is contained in a list, you could say something like:
contained_in( X , [X|Ys] ) :- % if we find the item in the list,
! . % - we succeed and eliminate any alternatives.
contained_in( X , [_|Ys] ) :- % otherwise, we discard the head of the list,
contained_in(X,Ys) . % - and keep looking by recursing down.
The actual filtering is pretty simple, too:
filter( [] , _ , [] ) . % once the source list is exhausted, we're done.
filter( [X|Xs] , Ys , [X|R] ) :- % otherwise...
\+ contained_in(X,Ys) , % - if the current list item is not contained in the list of items to be removed,
! , % - cut off alternatives,
filter( Xs , Ys , R ) . % - add the current item to the result list and recurse down.
. %
filter( [_|Xs] , Ys , R ) :- % if the current list item IS contained in the list of items to be removed,
filter(Xs , Ys , R ) % - discard it and recurse down.
. % Easy!
Your filter is just subtract available with many Prolog systems (I've tested with B-Prolog, SWI-Prolog and ECLiPSe):
?- subtract( [[1,2,3],[1]] , [[1]] , X ).
X = [[1, 2, 3]].
?- subtract( [[1,2,3],[1],[2,3,4],[2]] , [[1],[2]] , X ).
X = [[1, 2, 3], [2, 3, 4]].
You can look at the sources of SWI-Prolog or ECLiPSe for implementation details.
So I am currently trying to compute a formula using Prolog. I currently have part of the formula done, but I am having trouble implementing the next part where I need to add the elements of the list and then square root the sum. Not sure how I would do that.
What I currently have:
formula([], [], []).
formula([], [H2|T2], [L|Ls]) :-
L = H2,
formula([], T2, Ls).
formula([H1|T1], [], [L|Ls]) :-
L = H1,
formula(T1, [], Ls).
formula([H1|T1], [H2|T2], [L|Ls]) :-
L is (H1 - H2)*(H1 - H2),
formula(T1, T2, Ls).
Your original formula
formula([], [], []).
formula([], [H2|T2], [L|Ls]) :-
L = H2,
formula([], T2, Ls).
formula([H1|T1], [], [L|Ls]) :-
L = H1,
formula(T1, [], Ls).
formula([H1|T1], [H2|T2], [L|Ls]) :-
L is (H1 - H2)*(H1 - H2),
formula(T1, T2, Ls).
can be simplified to make the pattern matching more explicit:
formula( [] , [] , [] ) .
formula( [] , [Y|Ys] , [Y|Zs] ) :- formula( [] , Ys , Zs ) .
formula( [X|Xs] , [] , [X|Zs] ) :- formula( Xs , [] , Zs ) .
formula( [X|Xs] , [Y|Ys] , [Z|Zs] ) :-
L is ( X - Y ) * ( X - Y ) ,
formula(Xs,Ys,Zs)
.
I assume your instructor wants you to roll your own here and learn about recursion rather than using a built-in predicate. So, ... You could sum the elements of a list like this (the naive implementation):
sum_of( [] , 0 ) . % the sum of the empty list is zero.
sum_of( [X|Xs] , S ) :- % the sum of an empty list is computed by
sum(Xs,T) , % - computing the sum of the tail of the list
S is T+X % - and adding that to the value of the head of the list.
. %
But that will fail with a stack overflow once the list gets sufficiently long as each recursive call pushes a new frame onto the stack. Prolog has an nifty optimization (tail recursion optimization) that effectively converts recursion into iteration by recognizing when it can reuse the stack frame. To do that, the recurive call must be the very last thing done.
This introduces a common pattern in prolog programming:
a public interface predicate (here, sum_of/2),
that invokes a "private" tail-recursize worker predicate (here, sum_of/3) that uses an accumulator argument to build up its result.
Using that pattern, we get this implementation:
sum_of(Xs,Sum) :- sum_of(Xs,0,Sum) .
sum_of( [] , S , S ) . % the sum of the empty list is 0.
sum_of( [X|Xs] , T , S ) :- % the sum of a non-empty list is computed by
T1 is T+X , % incrementing the accumulator by the value of the head of the list, and
sum_of( Xs , T1 , S ) % recursing down on the tail.
. % Easy!
This will work for lists of any length.
Using the SWI-Prolog library predicate sum_list/2:
list_summed_and_square_rooted(List, Value) :-
sum_list(List, Sum),
Value is sqrt(Sum).
You probably don't need to write a separate predicate for relating a list to the square root of the sum of its elements, unless you'll be needing to use that particular relation often. Your formula/3 makes one list out of two, but ultimately you seem to be after a numerical value, so you probably do want another predicate to describe the relation between two lists and the resultant numerical value.
lists_processed_in_some_way(List1, List2, Value) :-
formula(List1, List2, CombinedList),
sum_list(CombinedList, Sum),
Value is sqrt(Sum).
By the way, you can simplify your formula/3 because you don't need L = H2:
formula([], [H2|T2], [H2|Ls]) :-
formula([], T2, Ls).
Also, it's generally good practice to name your predicates carefully, with something descriptive. It will help you reason about what your predicates do and help you communicate your programs to others.
I am trying to write a program which make the following:
?- g([2,3, [22,[3],9] ,4,[5],99],X).
X= [2,3,22,[3],9 ,4,5,99]
So it searches for lists in the given list and replace it by their elements without brackets [].
So I wrote this program:
The first block just searches for the first element in the list which is list
If there is no such element it returns [there_is_no_list].
first_list_in_the_list([],[there_is_no_list]):-!.
first_list_in_the_list([H|_],X):-is_list(H),X=H,!.
first_list_in_the_list([_|T],X):-first_list_in_the_list(T,X).
The first block works in prolog perfectly.
The second block just search in the list for an element X and then split the list into a two lists one is the list of all elements before X and the second is the elements after X.
splite_when_find_element([H|T],H,[],T):-!.
splite_when_find_element([H|T],X,F,G):-
splite_when_find_element(T,X,F1,G),append([H],F1,F).
it also works fine in Prolog.
and the third block is append, and it joins two list together in a new list.
append([],L,L).
append([H|T],L,[H|U1]):- append(T,L,U1).
and the last part is:
gg(L,L):-first_list_in_the_list(L,[there_is_no_list]),!.
gg(L,U):-first_list_in_the_list(L,X),
splite_when_find_element(L,X,F,G),gg(G,R),append(F,X,E),
append(E,R,U).
When I give a query [2,[3],5] I get also [2,[3],5] and I really don't understand why it does this.
A simple recursive solution will also work. Recursion is done by the head of the input list. In the non-trivial case, when the head is a list itself, we just append the rest of the flattened list to it. In the code below, it has not flattened Rest yet in append(H, Rest, Out), but it will be, after the recursive call of g(In, Rest). Cut after the append call ensures that backtracking won't consider the last case, where the head will appear in the output as-is.
% Base case, empty list.
g([], []).
% First recursive case: head is list.
% Append remaining elements to it.
g([H|In], Out):-
append(H, Rest, Out), !,
g(In, Rest).
% Second recursive case: head is not list.
% Will appear as-is in the output.
g([H|In], [H|Out]):-
g(In, Out).
also a DCG can do
lev, X --> [X], {is_list(X)}, lev.
lev, [X] --> [X], lev.
lev --> [].
test:
?- phrase(lev,[a,[b,c,[d]],e],L).
L = [a, b, c, [d], e] .
To flatten 1 level of a nested list, try something like this:
flatten1( Xs , Ys ) :- % to flatten a list
flatten1( Xs , [] , Ys ) , % - invoke the worker predicate
. %
flatten1( [] , T , R ) :- % if we get to to the empty list
reverse(T,R) % - just reverse the accumulator and we're done.
. %
flatten1( [X|Xs] , T , R ) :- % if the head of the list is unbound
var(X) , % - check for being a var
! , % - cut (to eliminate problems on backtracking
T1 = [X|T] , % - prepend the head of the list to the accumulator
flatten( Xs , T1 , R ) % - and recurse down
. %
flatten1( [[]|Xs] , T , R ) :- % if head of the list is an empty list, skip it
flatten1( Xs , T , R ) % - ignore it and recurse down
. %
flatten1( [[X|Ns]|Xs] , T , R ) :- % if head of the list is a non-empty list
X1 = [Ns|Xs] , % - prepend the tail of the sublist to the list
T1 = [X|T] , % - prepend the head of the sublist to the accumulator
flatten( X1 , T1 , R ) % - and recurse down
. %
flatten( [X|Xs] , T , R ) :- % if the head of the list is something else (except an unbound variable)
T1 = [X|T] , % - prepend the list head to the accumulator and
flatten( Xs , T1 , R ) % - recurse down
. %
Having trouble understanding how Prolog works. I'm tryig to write a rule that takes three lists of integers as input (representing sets) and puts the integers that belong to both the first and second list in the third list.
Example:
?-inter([10,20,30,40],[10,50,40,60], List3 )
List3 = [10, 40]
So far I have this, that can recognize if a list contains a certain letter:
mymember(X,[X|T]).
mymember(X,[H|T]) :- mymember(X,T).
There's actually an inbuilt library to sort that all out for you, known as ordsets.
inter(X, Y, Z) :-
list_to_ord_set(X, L1),
list_to_ord_set(Y, L2),
ord_intersection(L1, L2, Z).
Using your example input you get the following
| ?- inter([10,20,30,40],[10,50,40,60],X).
X = [10,40] ? ;
no
inter(Xs, Ys, Zs) will be true when each element in Zs also is in Xs and in Ys.
But Zs are unknown, then a more constructive approach is required.
Here it is: iterate on Xs and store in Zs each element that is in Ys.
An example of iteration is mymember/2, you can see that it requires a recursive predicate.
The other idiomatic part of the above statement is store in Zs, Prolog has a peculiar way to do such things, using pattern matching.
inter([X|Xs], Ys, [X|Zs]) :-
mymember(X, Ys), inter(Xs, Ys, Zs).
You will need to complete inter/3 with other 2 clauses: base recursion, i.e. when all Xs elements have been processed, and the case where X is not a member of Ys.
Try something like this, using the builtins member/2 and setof\3:
set_intersection( As , Bs , Xs ) :-
set_of( X , ( member(X,As) , member(X,Bs) ) , Xs )
.
One should note that this will fail if the lists As and Bs have no elements in common. An alternative would be use findall/3 rather than set_of/3. findall/3 will hand back and empty list rather than failure if the goal is not satisfied:
set_intersection( As , Bs , Xs ) :-
findall( X , ( member(X,As) , member(X,Bs) ) , Xs )
.
However findall/3 returns a bag (duplicates are allowed) rather than a set (no duplicates allowed), so if your two source lists aren't sets, you won't get a set out.
member/2 is a builtin predicate that unifies its first argument with an element of the list — the equivalent of
member(X,[X|_).
member(X,[_|Xs) :- member(X,Xs) .
And, finally, as #chac noted in his answer, you can recursively traverse the list.
set_intersection( [] , _ , [] ) . % the intersection of the empty set with anything is the empty set.
set_intersection( [A|As] , Bs , [A|Xs] ) :- % if the list is non-empty,
member(A,Bs) , % - and A is a member of the 2nd set
! , % - we cut off alternatives at this point (deterministic)
set_intersection( As , Bs , Xs ) % - and recurse down on the tail of the list.
.
set_intersection( [_|As] , Bs , Xs ) :- % if the list is non-empty, and A is NOT a embmer of the 2nd set
set_intersection( As , Bs , Xs ) % we just recurse down on the tail of the list.
.
#chac's technique builds the result list as he goes, something like:
[a|X]
[a,b|X]
[a,b,c|X]
The final unification, the special case of the empty list unifies the unbound tail of the list with [] making the list complete, so the final [a,b,c|X] becomes
[a,b,c]
A little prolog magic. An alternative that might be easier to understand is to use a worker predicate with an accumulator:
%
% set_intersection/3: the public interface predicate
%
set_intersection( As , Bs , Xs ) :-
set_intersection( As , Bc , [] , T ) % we seed our accumulator with the empty list here
.
%
% set_intersection/4: the private worker bee predicate
%
set_intersection( [] , _ , T , Xs ) :- % since our accumulator is essentially a stack
reverse(T,Xs) % we need to reverse the accumulator to
. % put things in the expected sequence
set_intersection( [A|As] , Bs , T , Xs ) :-
member( A, Bs ) ,
! ,
T1 = [A|T] ,
set_intersection( As , Bs , T1 , Xs )
.
set_intersection( [_|As] , Bs , T , Xs ) :-
set_intersection( As , Bs , T , Xs )
.