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Hey everyone, pretty new to Prolog and just started programming in it,hence the "newbie"-questions.
I decided to make a small software that, this is basic I know - but I have trouble getting started with it, anyways here's the example;
I'm a carpenter, wanting to build different things for my house so I have a few "recipes" for different stuff and their required ingredients. At home I also have a few ingredients, maybe not enough though.
My ingredients at home;
at_home([ingredient(nails,2), ingredient(wood,10)]).
and the "recipes";
recipe(wooden_roof,[ingredient(nails, 4), ingredient(hammer,1), ingredient(wood, 5)]).
recipe(stairs, [ingredient(nails, 2), ingredient(hammer,1), ingredient(wood, 2)]).
Now I want to be able to write ;
build(X) and then the software should return all the things I can build.
And the second part of the example is that I want a function (I've some programming in C# so I'm a bit damaged from that, lol)
called "buy" where it returns what I need to buy in order to be able to build everything, i.e;
buy(X,L)
And then it returns the ingredients needed to be bought.
Is there anyone who can help me with that? I really need a kick in the butt to get started.
Or do ya have any recommended tutorials? |
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joel76 (Programmer) |
2 Nov 11 15:53 |
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This is what I have tried so far;
buy(X,L) :- recipe(X,Ls), compare3(L,Ls).
compare3([], []).
compare3(K, [L|Ls]):- compareAtHome(L, at_home(Ks),K), compare3(L, Ls).
compareAtHome(ingredient(F1,S1),[ingredient(F2,S2)|Ks], L):- (F1==F2 -> to_buy(S1,S2, K), add(ingredient(F1,K),L,L1), compareAtHome(ingredient(F1,S1), [Ks]);compareAtHome(ingredient(F1,S1),[Ks])).
to_buy(X,Y,Z):- K is X-Y, K >= 0, Z is K.
add(X,[],[X]).
add(X,[A|L],[A|L1]):- add(X,L,L1). |
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joel76 (Programmer) |
3 Nov 11 6:02 |
Ok, I give you a solution in the manner you used, but there is a probleme, when an ingredient is not needed, the TT is set to 0, I'll give you a tip where when an ingredient is not needed it is not in the result list. CODEbuy(X,L) :-
recipe(X,Ls),
compare3(L, Ls).
compare3([], []).
compare3([K | Ks], [L|Ls]):-
compareAtHome(K, L),
compare3(Ks, Ls).
% compareAtHome(ingredient(F,TT),ingredient(F,S))
% @arg1 : ingredent to buy (maybe)
% @arg2 : ingredient needed
compareAtHome(ingredient(F,TT),ingredient(F,S)):-
at_home(Lst),
( member(ingredient(F, Nb), Lst) ->
% the ingredient is present at home,
% we check if we have enough (required S)
( Nb >= S -> % not needed we set the number to 0
TT = 0
; TT is S - Nb
)
; % the ingredient is not at home
% we need to buy all
TT = S). Now, if you don't want to have 0 is the list, you must walk through the list with compare3(LstNeeded, CurrentLstToBuy, FinalLstToBuy). CODE% the list is finished, we unify the 2 lists Current and Final
compare3([], LstToBuy, LstToBuy).
% here we study the current ingredient
compare3([ingredient(F, S) | T], CurrentLstToBuy, FinalLstToBuy) :-
........ what have we to do ????
compare3(T, NewCurrentLstToBuy, FinalLstToBuy). Hope this help ! |
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What do you mean by "walk through the list with compare3(LstNeeded, CurrentLstToBuy, FinalLstToBuy)." could you show an example or write it in the code? |
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joel76 (Programmer) |
10 Nov 11 3:54 |
Walk through the list : example sum of a list with accumulator. CODE% list is finished
sum([], Acc, Acc).
% one step
sum([H|T], CurAcc, FinalAcc) :-
Acc1 is CurAcc + H,
% next step
sum(T, Acc1, FinalAcc). |
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yea so i put "compare3([H|T], CurAcc, FinalAcc) :- Acc1 is CurAcc + H, sum(T, Acc1, FinalAcc)." after "compare3([], LstToBuy, LstToBuy)." but i still don't get it to work, what am i doing wrong? |
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