Compare lists and return if true.

[Pshycoone]

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?

 

[joel76]

What have you tried ?

 

[Pshycoone]

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).

 

[joel76]

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.

buy(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).

% 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 !

 

[qwertqwert7]

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?

 

[joel76]

Walk through the list : example sum of a list with accumulator.

% list is finished
sum([], Acc, Acc).

% one step
sum([H|T], CurAcc, FinalAcc) :-
   Acc1 is CurAcc + H,
   % next step
   sum(T, Acc1, FinalAcc).

 

[qwertqwert7]

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?