Ok, here's the deal:
- I've got two piles of shirts
- I want to take a random shirt from each pile and put them in a new pile
- Then get the new pile out
And here is the code:
mix([],[],_).
mix(P1,P2, Pile):-
takeshirt(P1,1,Taken1,Rem1), takeshirt(P2,1,Taken2,Rem2), #Take one
append(Pile,Taken1,New), append(New,Taken2,NewPile), #Put both of them
mix(Remain1,Remain2,NewPile).
This is what the result look like:
1 ?- mix([a,b],[c,d],NewPile).
NewPile = [] .
I want it to look like:
1 ?- mix([a,b],[c,d],NewPile).
NewPile = [b, d, a,开发者_如何学JAVA c] .
Or whatever the result is. I checked via the graphical debugger and found out that when the final call to mix happens, the bindings are:
P1 = Taken1 = [b]
P2 = Taken2 = [c]
Pile = [a, d]
Rem1 = Rem2 = []
New = [a, d, b]
NewPile = [a, d, b, c] #<--- Interresting
So the wanted value is in NewPile when the final call to:
mix([],[],_).
happens. After this is it collapses like a house of cards.
So the question is:
mix([],[],_).
I'd like to return the _ value from the base case, this rule mix is actually used from a higher instance where I send in two piles and get the new pile out.
Update:
To clarify some comments about the takeshirt rule, here it is:takeshirt(_,0,[],_).
takeshirt(List,Number,[Element|Taken],Remain) :- N > 0,
length(List,Len),
Index is random(Len) + 1,
removeshirt_at(Element,List,Index,Remain),
Number1 is Number - 1,
takeshirt(Remain,Number1,Taken,Remain).
Consider the following modifications to your code:
mix([], [], []) :- !.
mix(P1, P2, Pile) :-
takeshirt(P1, 1, Taken1, Rem1),
takeshirt(P2, 1, Taken2, Rem2),
append(Taken1, Taken2, Pile0),
mix(Rem1, Rem2, Pile1),
append(Pile0, Pile1, Pile).
It seems you need to accumulate the 'shirts' (as list atoms). Here, we are recursively appending them onto the third argument of mix/3
(Pile
), until the base case (the first clause) is hit when both input lists are empty lists (note that the cut !
is necessary here as the binding pattern for the second clause matches the first, so we want to exclude it). The behaviour of the second clause, which takes a shirt from each input list for every step, requires that they must have been of equal length to start with.
To test this, I used the following definition of takeshirt/4
:
takeshirt(Ps, _, [P], Rem) :-
select(P, Ps, Rem).
Note that the second argument here is unused, as select/3
is being used to take a single element (a shirt) from the list, and return the remainder. The absence of a cut (!
) after the select
allows this predicate to backtrack in selecting all other elements (shirts) from the list. If we now execute your example query with this definition, we can get:
1 ?- mix([a,b],[c,d],NewPile).
NewPile = [a, c, b, d] ;
NewPile = [a, d, b, c] ;
NewPile = [b, c, a, d] ;
NewPile = [b, d, a, c] ;
false.
...we can see that mix/3
enumerates all possible 'piles' on backtracking by taking a shirt from the first pile (first input list), then a shirt from the second pile (second input list), and so on, until both input lists are empty. If your definition of takeshirt/4
doesn't leave choice-points, (is non-backtracking), then you could only get one solution, if any.
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