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  • Designs For Emerald Cut Rings

    Step Cut type of the emerald cut stones appears to be like steps engraved into the jewel. Particularly, emerald cutting was rectangular in shape with little cropped corners. At first, this cut was bu... [Author: Scheygen Smith - Computers and Internet - March 21, 2010]

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  • Cut and paste out of/ into text fields in a separate window doesn't work

    - by Brian Postow
    I have a Macintosh Mozilla plugin which puts up a separate window for login information. It seems to work fine, it gets keyboard events like typing and hitting return to hit the default button. HOWEVER, it doesn't seem to get cut and paste events. When I hit Cmd-v, the edit menu flashes, but nothing happnes. Is this a problem with my responder chain? Do I have to specially tell Mozilla that I want these events? or am I likely to have some other problem that I haven't even thought of? thanks.

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  • Prolog : Simple question

    - by abduls85
    I want to add any strings user entered into a list run :- write('How many students you have: '),read(x),nl. enterNameOfStudents(x). enterNameOfStudents(x) :- for(A, 1, x, 1),write('Please enter the names of students'),read(A),??????,nl,fail. What do i put in the ?????? portion to ensure that whatever the user enter will go into a user-defined list which will be used for further processing later ? Please help. I have tried numerous stuff like append and other but it does not work :(

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  • assert fact into file in prolog

    - by smile
    Hello everyone, How can I assert a fact into a file without deleting the previous fact? In the following line, when I execute it twice, the second fact overwrites the first fact: tell('animal.txt'),write(Animal),nl,told. But when I use assert or assertz it will do nothing. Help me please. Thank you :)

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  • Prolog Program for a recordings database

    - by RP
    I have three types of facts: album(code, artist, title, date). songs(code, songlist). musicians(code, list). Example: album(123, 'Rolling Stones', 'Beggars Banquet', 1968). songs(123, ['Sympathy for the Devil', 'Street Fighting Man']). musicians(123, [[vocals, 'Mick Jagger'], [guitar, 'Keith Richards', 'Brian Jones']]. I need to create these 4 rules: together(X,Y) This succeeds if X and Y have played on the same album. artistchain(X,Y) This succeeds if a chain of albums exists from X to Y; two musicians are linked in the chain by 'together'. role(X,Y) This succeeds if X had role Y (e.g. guitar) ever. song(X,Y) This succeeds if artist X recorded song Y. Any help?

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  • Result in an argument isn't correct

    - by Paulo Nunes
    So I have this piece of prolog code: my_avalia(A,R) :- A=="Koza" -koza(R,0,0,e,89). koza(R,_,_,_,87):-!,write(R). koza(R,X,Y,V,C):-movex(V,X,X1),movey(V,Y,Y1),confirma(X1,Y1,Z),Z==1->(append(R,[emFrente],U),L is (C-1),koza(U,X1,Y1,V,L)). The matter is that when I write the "R" at koza(), it has the correct values, however it ends up with a empty list in my_avalia when I call it like this: my_avalia("Koza",R). My recursion might be incorrect but I don't really know what's wrong with it. Thanks in advance.

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  • Prolog: Sentence Parser Problem

    - by Devon
    Hey guys, Been sat here for hours now just staring at this code and have no idea what I'm doing wrong. I know what's happening from tracing the code through (it is going on an eternal loop when it hits verbPhrase). Any tips are more then welcome. Thank you. % Knowledge-base det(the). det(a). adjective(quick). adjective(brown). adjective(orange). adjective(sweet). noun(cat). noun(mat). noun(fox). noun(cucumber). noun(saw). noun(mother). noun(father). noun(family). noun(depression). prep(on). prep(with). verb(sat). verb(nibbled). verb(ran). verb(looked). verb(is). verb(has). % Sentece Structures sentence(Phrase) :- append(NounPhrase, VerbPhrase, Phrase), nounPhrase(NounPhrase), verbPhrase(VerbPhrase). sentence(Phrase) :- verbPhrase(Phrase). nounPhrase([]). nounPhrase([Head | Tail]) :- det(Head), nounPhrase2(Tail). nounPhrase(Phrase) :- nounPhrase2(Phrase). nounPhrase(Phrase) :- append(NP, PP, Phrase), nounPhrase(NP), prepPhrase(PP). nounPhrase2([]). nounPhrase2(Word) :- noun(Word). nounPhrase2([Head | Tail]) :- adjective(Head), nounPhrase2(Tail). prepPhrase([]). prepPhrase([Head | Tail]) :- prep(Head), nounPhrase(Tail). verbPhrase([]). verbPhrase(Word) :- verb(Word). verbPhrase([Head | Tail]) :- verb(Head), nounPhrase(Tail). verbPhrase(Phrase) :- append(VP, PP, Phrase), verbPhrase(VP), prepPhrase(PP).

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  • Prolog adding and removing list element if non present in second list

    - by logically
    I don't know what I'm missing here. I wan't to add an element if it is in arg1 but not in arg2 and want to remove an element if it is in arg1 but not in arg2. I'm using an if condition with includes function that return true if the element is in the arg2 list, false otherwise. Then use built it predicates append and select to add or remove. I'm getting false to all my objectives searches. I comment and uncomment depending on what predicate I want, add or remove. includes([],_). includes([P|Z],S) :- memberchk(P,S), includes(Z,S). addop([],list,res). addop([P|R],list,res) :- includes(P,s0) - addop(R,list,res) ; append(P,list,res), addop(R,list,res). rem([],list,res). rem([P|R],list,res) :- includes(P,list) - rem(R,list,res) ; select(P,list,res),rem(R,list,res). Thanks for help.

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  • Using recursion 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: Error Handling and Find Unique

    - by anotherstat
    Given: fruitid('Apple', 'Granny Smith', 1). fruitid('Apple', 'Cox', 2). fruitid('Pear', 'Bartlett', 3). How would I go about finding only unique items for instance: is_unique(FruitName):- In the example clauses the answer would be Pear. I'm also trying to add error handling to my code, so in this instance if an input is: is_unique(pineapple) How could I catch this and output an error message? Thanks, AS

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  • Prolog term concatenation

    - by d0pe
    Hi, I'm trying to format a result from a program but getting an hard time. I wanted to give something like this as result: Res = do(paint(x) do(clean(a), do(repair(b) , initialState))) basically, I want to concatenate successive terms to initialState atom but, it doesn't work with atom_concat since the other terms to concatenate aren't atoms and also I wanted to add the ) everytime I pass through the "do" function. So it would be something like: Res = initialState. When do function was called, I would have a function like concatenateTerm(Pred, Res, Res). Pred beeing repair(b) for instance and obtain the result: res = do(repair(b), initialState). Is this possible to be done? Thanks

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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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  • 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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  • 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: 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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  • 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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  • 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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  • 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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