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• Prolog: using the sort/2 predicate

  - by Øyvind Hauge

  So I'm trying to get rid of the wrapper clause by using the sort library predicate directly inside split. What split does is just generating a list of numbers from a list that looks like this: [1:2,3:2,4:6] ---split-- [1,2,3,2,4,6]. But the generated list contains duplicates, and I don't want that, so I'm using the wrapper to combine split and sort, which then generates the desired result: [1,2,3,4,6]. I'd really like to get rid of the wrapper and just use sort within split, however I keep getting "ERROR: sort/2: Arguments are not sufficiently instantiated." Any ideas? Thanks :) split([],[]). split([H1:H2|T],[H1,H2|NT]) :- split(T,NT). wrapper(L,Processed) :- split(L,L2), sort(L2,Processed).

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• prolog: reduce then write the value of a predicate

  - by jreid9001

  This is some of the code I am writing assert(bar(foo)), assert(foo(bar-5)), I'm not sure if it works though. I'm trying to get it to reduce foo by 5. I need a way to write the value of foo, but haven't found a way too. write('foo is' + foo) would be the logical way to me, but doesn't seem to work.

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• Using recustion an append in prolog

  - by Adrian

  Lets say that I would like to construct a list (L2) by appending elements of another list (L) one by one. The result should be exactly the same as the input. This task is silly, but it'll help me understand how to recurse through a list and remove certain elements. I have put together the following code: create(L, L2) :- (\+ (L == []) -> L=[H|T], append([H], create(T, L2), L2);[]). calling it by create([1,2,3,4], L2) returns L2 = [1|create([2,3,4], **)\. which is not a desired result.

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• Prolog for beginners about logic and syntax

  - by lnotik

  Hello everybody. I have this question : I need to create a paradict "rightGuesses" which will get 3 arguments , each one of them is a list of letters : 1) The list of guessed letters 2) The word i have to guess 3) The letters that where guessed so far . for example : rightGuesses([n,o,p,q], [p,r,o,l,o,g], Ans). will give us Ans = [p, -, o, -, o, -]. i made: rightGuesses([],T2,[ANS]) rightGuesses([A|T1],T2,[ANS]):- (member(A,T2))=\=true , rightGuesses(T1,T2,[ _ |'-']). rightGuesses([A|T1],T2,[ANS]):- member(A,T2), rightGuesses(T1,T2,[ _ |A]). but i get : ERROR: c:/users/leonid/desktop/file3.pl:5:0: Syntax error: Operator expected Warning: c:/users/leonid/desktop/file3.pl:6: when i trying to compile it what is my problem , and is there is a better way to do it ? thanks in advance.

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• copy file into another file in prolog

  - by smile

  Good morning/evening how can I write something in a file and then copy its content into the current file? for example I consult file1.pro then I have rule write something in file2.pro , after this rule finish its job I want append the content of the file2.pro int file1.pro . when I tried to append into file1.pro directly , the data appear like undefined symbols ,I don't know why please hellp me thank you.

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• prolog recursion

  - by AhmadAssaf

  am making a function that will send me a list of all possible elemnts .. in each iteration its giving me the last answer .. but after the recursion am only getting the last answer back .. how can i make it give back every single answer .. thank you the problem is that am trying to find all possible distributions for a list into other lists .. the code test :- bp(3,12,[7, 3, 5, 4, 6, 4, 5, 2], Answer), format("Answer = ~w\n",[Answer]). bp(NB,C,OL,A):- addIn(C,OL,[[],[],[]],A); bp(NB,C,_,A). addIn(_,[],Result,Result). addIn(C,[Element|Rest],[F|R],Result):- member( Members , [F|R]), sumlist( Members, Sum), sumlist([Element],ElementLength), Cap is Sum + ElementLength, (Cap =< C, append([Element], Members,New), insert( Members, New, [F|R], PartialResult), addIn(C,Rest,PartialResult,Result)). by calling test .. am getting back all the list of possible answers .. now if i tried to do something that will fail like bp(3,11,[8,2,4,6,1,8,4],Answer). it will just enter a while loop .. more over if i changed the bp(NB,C,OL,A):- addIn(C,OL,[[],[],[]],A); bp(NB,C,_,A). to and instead of Or .. i get error : ERROR: is/2: Arguments are not sufficiently instantiated appreciate the help .. Thanks alot @hardmath

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• Prolog Backtracking

  - by AhmadAssaf

  I am trying to do a word calculator .. read words from a file .. translate them into numbers and then calculate the result .. i managed to do all of that but i think i have two bugs in my program .. I mainly have two functions ... extract(Words), calculate( Words,0). extract will read from the file .. and then return a list of Words .. ex: [one,plus,three] .. now calculate will translate the value for these words into numbers and calculate .. i managed to do that also .. now the bugs are : i must stop reading and terminate if i encounter stop in the file .. so if Words was [stop] End. i tried the following ... execute :- extract(Words), Words = [stop],nl,print('Terminating ...'),!. execute :- extract(Words), calculate( Words,0). it successfully terminates .. but it skips lines as i extract more than once .. i have tried to do .. execute :- extract(Words), Words \= [stop],execute(Words). execute(Words) :- calculate( Words,0). if the Words is not stop .. then go and calculate .. but its not working !! i appreciate the help .. Thank You

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• Understanding this prolog implementation of head

  - by dmindreader

  head([Y],Y):-!. head([X|XS],X). I understand that the head of the list is stored on the variable X. What does the first clause mean? Is it a cut? Why?

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• Prolog: find all numbers of unique digits that can be formed from a list of digits

  - by animo

  The best thing I could come up with so far is this function: numberFromList([X], X) :- digit(X), !. numberFromList(List, N) :- member(X, List), delete(List, X, LX), numberFromList(LX, NX), N is NX * 10 + X. where digit/1 is a function verifying if an atom is a decimal digit. The numberFromList(List, N) finds all the numbers that can be formed with all digits from List. E.g. [2, 3] -> 23, 32. but I want to get this result: [2, 3] -> 2, 3, 23, 32 I spent a lot of hours thinking about this and I suspect you might use something like append(L, _, List) at some point to get lists of lesser length. I would appreciate any contribution.

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• Understanding prolog [lists]

  - by wwrob

  I am to write a program that does this: ?- pLeap(2,5,X,Y). X = 2, Y = 3 ; X = 3, Y = 4 ; X = 4, Y = 5 ; X = 5, Y = 5 ; false. (gives all pairs X,X+1 between 2 and 5, plus the special case at the end). This is supposedly the solution. I don't really understand how it works, could anyone guide me through it? pLeap(X,X,X,X). pLeap(L,H,X,Y) :- L<H, X is L, Y is X+1. pLeap(L,H,X,Y) :- L=<H, L1 is L+1, pLeap(L1,H,X,Y). I'd do it simply like this: pLeap(L,H,X,Y) :- X >= L, X =< H, Y is X+1. Why doesn't it work (ignoring the special case at the end)?

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• prolog sets problem, stack overflow

  - by garm0nboz1a

  Hi. I'm gonna show some code and ask, what could be optimized and where am I sucked? sublist([], []). sublist([H | Tail1], [H | Tail2]) :- sublist(Tail1, Tail2). sublist(H, [_ | Tail]) :- sublist(H, Tail). less(X, X, _). less(X, Z, RelationList) :- member([X,Z], RelationList). less(X, Z, RelationList) :- member([X,Y], RelationList), less(Y, Z, RelationList), \+less(Z, X, RelationList). lessList(X, LessList, RelationList) :- findall(Y, less(X, Y, RelationList), List), list_to_set(List, L), sort(L, LessList), !. list_mltpl(List1, List2, List) :- findall(X, ( member(X, List1), member(X, List2)), List). chain([_], _). chain([H,T | Tail], RelationList) :- less(H, T, RelationList), chain([T|Tail], RelationList), !. have_inf(X1, X2, RelationList) :- lessList(X1, X1_cone, RelationList), lessList(X2, X2_cone, RelationList), list_mltpl(X1_cone, X2_cone, Cone), chain(Cone, RelationList), !. relations(List, E) :- findall([X1,X2], (member(X1, E), member(X2, E), X1 =\= X2), Relations), sublist(List, Relations). semilattice(List, E) :- forall( (member(X1, E), member(X2, E), X1 < X2), have_inf(X1, X2, List) ). main(E) :- relations(X, E), semilattice(X, E). I'm trying to model all possible graph sets of N elements. Predicate relations(List, E) connects list of possible graphs(List) and input set E. Then I'm describing semilattice predicate to check relations' List for some properties. So, what I have. 1) semilattice/2 is working fast and clear ?- semilattice([[1,3],[2,4],[3,5],[4,5]],[1,2,3,4,5]). true. ?- semilattice([[1,3],[1,4],[2,3],[2,4],[3,5],[4,5]],[1,2,3,4,5]). false. 2) relations/2 is working not well ?- findall(X, relations(X,[1,2,3,4]), List), length(List, Len), writeln(Len),fail. 4096 false. ?- findall(X, relations(X,[1,2,3,4,5]), List), length(List, Len), writeln(Len),fail. ERROR: Out of global stack ^ Exception: (11) setup_call_catcher_cleanup('$bags':'$new_findall_bag'(17852886), '$bags':fa_loop(_G263, user:relations(_G263, [1, 2, 3, 4|...]), 17852886, _G268, []), _G835, '$bags':'$destroy_findall_bag'(17852886)) ? abort % Execution Aborted 3) Mix of them to finding all possible semilattice does not work at all. ?- main([1,2]). ERROR: Out of local stack ^ Exception: (15) setup_call_catcher_cleanup('$bags':'$new_findall_bag'(17852886), '$bags':fa_loop(_G41, user:less(1, _G41, [[1, 2], [2, 1]]), 17852886, _G52, []), _G4767764, '$bags':'$destroy_findall_bag'(17852886)) ?

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• passing back answers in prolog

  - by AhmadAssaf

  i have this code than runs perfectly .. returns a true .. when tracing the values are ok .. but its not returning back the answer .. it acts strangely when it ends and always return empty list .. uninstantiated variable .. test :- extend(4,12,[4,3,1,2],[[1,5],[3,4],[6]],_ExtendedBins). %printing basic information about the extend(NumBins,Capacity,RemainingNumbers,BinsSoFar,_ExtendedBins) :- getNumberofBins(BinsSoFar,NumberOfBins), msort(RemainingNumbers,SortedRemaining),nl, format("Current Number of Bins is :~w\n",[NumberOfBins]), format("Allowed Capacity is :~w\n",[Capacity]), format("maximum limit in bin is :~w\n",[NumBins]), format("Trying to fit :~w\n\n",[SortedRemaining]), format("Possible Solutions :\n\n"), fitElements(NumBins,NumberOfBins, Capacity,SortedRemaining,BinsSoFar,[]). %this is were the creation for possibilities will start %will check first if the number of bins allowed is less than then %we create a new list with all the possible combinations %after that we start matching to other bins with capacity constraint fitElements(NumBins,NumberOfBins, Capacity,RemainingNumbers,Bins,ExtendedBins) :- ( NumberOfBins < NumBins -> print('Creating new set: '); print('Sorry, Cannot create New Sets')), createNewList(Capacity,RemainingNumbers,Bins,ExtendedBins). createNewList(Capacity,RemainingNumbers,Bins,ExtendedBins) :- createNewList(Capacity,RemainingNumbers,Bins,[],ExtendedBins), print(ExtendedBins). createNewList(0,Bins,Bins,ExtendedBins,ExtendedBins). createNewList(_,[],_,ExtendedBins,ExtendedBins). createNewList(Capacity,[Element|Rest],Bins,Temp,ExtendedBins) :- conjunct_to_list(Element,ListedElement), append(ListedElement,Temp,NewList), sumlist(NewList,Sum), (Sum =< Capacity, append(ListedElement,ExtendedBins,Result); Capacity = 0), createNewList(Capacity,Rest,Bins,NewList,Result). fit(0,[],ExtendedBins,ExtendedBins). fit(Capacity,[Element|Rest],Bin,ExtendedBins) :- conjunct_to_list(Element,Listed), append(Listed,Bin,NewBin), sumlist(NewBin,Sum), (Sum =< Capacity -> fit(Capacity,Rest,NewBin,ExtendedBins); Capacity = 0, append(NewBin,ExtendedBins,NewExtendedBins), print(NewExtendedBins), fit(0,[],NewBin,ExtendedBins)). %get the number of bins provided getNumberofBins(List,NumberOfBins) :- getNumberofBins(List,0,NumberOfBins). getNumberofBins([],NumberOfBins,NumberOfBins). getNumberofBins([_List|Rest],TempCount,NumberOfBins) :- NewCount is TempCount + 1, %calculate the count getNumberofBins(Rest,NewCount,NumberOfBins). %recursive call %Convert set of terms into a list - used when needed to append conjunct_to_list((A,B), L) :- !, conjunct_to_list(A, L0), conjunct_to_list(B, L1), append(L0, L1, L). conjunct_to_list(A, [A]). Greatly appreciate the help

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• Prolog: declaring an operator

  - by B K

  I have defined ! (factorial) function and registered it as arithmetic function and an operator, so that I can execute: A is 6!. Now I'd like to define !! (factorial of odd numbers), but the same way - writing clauses, registering arithmetic_function and operator, calling A is 7!! - results in SyntaxError: Operator expected How should I, if possible, register !! operator ? Yes, I realize, ! is normally the cut.

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• Is Prolog the best language to solve this kind of problem?

  - by Milan Babuškov

  I have this problem containing some inequations and requirement to minimize a value. After doing some research on the Internet, I came to conclusion that using Prolog might be the easiest way to solve it. However, I never used Prolog before, and I would hate to waste my time learning it just to discover that it is not the right tool for this job. Please, if you know Prolog, take a look at this problem and tell me if Prolog is the right one. Or, if you know of some other language that is really suited for this. a + b + c >= 100 d + e + f >= 50 g + h >= 30 if (8b + 2e + 7h > 620) then y = 0.8 else y = 1.0 if (d > 35) then x = 0.9 else x = 1.0 5xa + 8yb + 5c + 3xd + 2ye + 2f + 6xg + 7yh = w. I need to find the values for a, b, c, d, e, f, g and h that minimize w. I'm not really asking for code, although I'd be grateful for some hint how to tackle this if Prolog is really good for it. Thanks.

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• Prolog Question

  - by Dan

  Assume the third parameter is the result. a( 1, [Hd | Tl], Hd ). a( N, [ | Tl], Elem ) :- N 1, N1 is N - 1, a( N1, Tl, Elem).

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• Get all sets of list in prolog

  - by garm0nboz1a

  How can I generate all the possible sets of the elements of a list with current length? get_set(X, [1,2,3]). X = [1,1,1], X = [1,1,2], X = [1,1,3], X = [1,2,1], X = [1,2,2], X = [1,2,3], X = [1,3,1], X = [1,3,2], X = [1,3,3], ..... X = [3,3,2], X = [3,3,3]. UPD: there is good answer given by Sharky. But maybe it's not the best. Here is another: get_set(X,L) :- get_set(X,L,L). get_set([],_). get_set([X|Xs],[_|T],L) :- member(X,L), get_set(Xs,T,L).

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• Why is Prolog associated with Natural Language Processing?

  - by kyphos

  I have recently started learning about NLP with python, and NLP seems to be based mostly on statistics/machine-learning. What does a logic programming language bring to the table with respect to NLP? Is the declarative nature of prolog used to define grammars? Is it used to define associations between words? That is, somehow mine logical relationships between words (this I imagine would be pretty hard to do)? Any examples of what prolog uniquely brings to NLP would be highly appreciated.

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• Removing duplicate solutions

  - by Enoon

  My code merges two lists of lists, item by item, in the following way: mergeL([[a,b],[c,d]], [[1,2],[3,4]], Result). Result = [[a,b,1,2],[c,d,3,4]] And this is the code i use: mergeL([],[],[]). mergeL(List, [], List). mergeL([], List, List). mergeL([X|Rest],[Y|Rest2], [XY|Res2]) :- mergeL(Rest, Rest2, Res2), append(X,Y,XY). This seems to work but if i call it with two lists of the same size i get three repeated results. Example (both list contain only one element): ?- mergeL([[a,b]],[[1,2,3]],Q). Q = [[a, b, 1, 2, 3]] ; Q = [[a, b, 1, 2, 3]] ; Q = [[a, b, 1, 2, 3]]. Is there a clean way to make this output only one solution?

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• sumList predicate needs to check to see if their Sum is equal to the sum of all the elements in the list

  - by nick bonnet

  I am implementing a method that when given Sum and a List. It will check to see that if you add the elements in the list, their sum is equal to the Sum given. Here is what I am trying to do thus far, but I am pretty sure it is wrong... I'm not really sure how to think about it. sumList([],0). sumList([X|Xrest], Sum) :- sumList[Xrest, Sum1), Sum is X + Sum1. Could you give me a point in the right direction or at least let me know how to try to think about the problem?

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• Why (not) logic programming?

  - by Anto

  I have not yet heard about any uses of a logical programming language (such as Prolog) in the software industry, nor do I know of usage of it in hobby programming or open source projects. It (Prolog) is used as an academic language to some extent, though (why is it used in academia?). This makes me wonder, why should you use logic programming, and why not? Why is it not getting any detectable industry usage?

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• swi prolog- IDE - Eclipse- how to use it in Linux?

  - by francis gan

  I read that Swi Prolog IDE - Eclipse - in Linux(Ubuntu) is possible. However I couldn't get the installed Eclipse to 'use' swi Prolog. Please help

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• Find the order among tasks in a company by using prolog?

  - by Cem

  First of all,I wish a happy new year for everyone.I searched more and worked a lot but I could not solve this question.I am quite a new in prolog and I must do this homework. In my homework,the question is like this: Write a prolog program that determines a valid order for the tasks to be carried out in a company. The prolog program will consist of a set of "before" predicates which denotes the order between task pairs. Here is an example; before(a,b). before(a,e). before(d,c). before(b,c). before(c,e). Here, task a should be carried before tasks b and e, d before c and so on. Hence a valid ordering of the tasks would be [a, b, d, c, e]. The order predicate in your program will be queried as follows. ?- order([a,b,c,d,e],X). X = [a, b, d, c, e] ; X = [a, d, b, c, e] ; X = [d, a, b, c, e] ; false. Hint: Try to generate different orders for the tasks (permutation) and then check if the order is consistent with the "before" relationships given. Even if you can generate a single valid order, you will get reasonable partial credits.

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• How can I insert an additional argument after the last argument in a prolog procedure?

  - by Meko

  I am new to learning prolog, and I want to know, if we have some procedure like father("Nic","Adam"). and I want to write a function that it will add new value to this father("Nic","Adam","something"..) how can I do this? Using list? Or what?

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• What does [a|b|c] evaluate to in Prolog?

  - by Ambrose

  The pipe operator in prolog returns one or more atomic Heads and a Tail list. ?- [a,b,c] = [a,b|[c]]. true. Nesting multiple pipes in a single match can be done similar to this: ?- [a,b,c] = [a|[b|[c]]]. true. What does the statement [a|b|c] infer about a, b and c?

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• prolog. recursive function returning multiple values.

  - by obtur

  I am new in Prolog. I need to declare a function which looks at a list like [[1, 0, 0], [1, 0, 0], [1, 0, 0]] and if the value is 0 returns its address(by considering it as a double array). I wrote a function basicly in the format: function(..., X) :- function(called by other values). How can I write a function, that returns a value when each times it called(recursively). May I get them (in above question) as alternative X's???

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