unification

问题 plays(alice, leadguitar). plays(noah, drums). plays(mike, leadguitar). plays(mike, drums). plays(katie, baseguitar). plays(drew, leadguitar). plays(drew, baseguitar). duetwith(Person1,Person2):- plays(Person1,L), plays(Person2,L), Person1 \= Person2. I write a new rule called combo that determines whether or not three people can make a combo with a drummer, a base guitar, and a lead guitar. combo(Person1,Person2,Person3):- plays(Person1,X), plays(Person2,Y), plays(Person3,Z), X \= Y, Y \= Z,

问题 When I am using Instagram sharing then how can I get username and password from Instagram (current login user). $app = new UEApp(self::APP_SECRET, self::APP_SECRET); $user = new UEUser($this->currentUser->unification_userkey, $this->currentUser->unification_usersecret); $con = $access_token."@instagram.com/?username=rajneesh49&password=test&@instagram.com"; Now I am using static password and username then its working fine but How can I get current username and password(required filed by

问题 Has anyone an idea how the type inference problem E > hd (cons 1 nil) : α0 with the typing environment E={ hd : list(α1 ) → α1 , cons : α2 → list(α2 ) → list(α2 ), nil : list(α3 ), 1 : int } can be transferred in an unification problem? Any help would really be appreciated! 回答1: First, rename type variables so that none of the variables in your expression collide with variables in the typing environment. (In your example, this is already done since the expression references { a0 }, and the

问题 I'm trying to work my way through the exercises at the bottom of this page and I find myself utterly confused on number 3. We are given the following knowledge base of travel information: byCar(auckland, hamilton). byCar(hamilton, raglan). byCar(valmont, saarbruecken). byCar(valmont, metz). byTrain(metz, frankfurt). byTrain(saarbruecken, frankfurt). byTrain(metz, paris). byTrain(saarbruecken, paris). byPlane(frankfurt, bangkok). byPlane(frankfurt, singapore). byPlane(paris, losAngeles).

问题 I'm trying to make what one might call a decidable parser in Idris. At first I am just looking at parsing natural numbers, but have ran into an unexpected problem. A minimum example of the code that produces it is this: data Digit : Char -> Type where Zero : Digit '0' One : Digit '1' digitToNat : Digit a -> Nat digitToNat Zero = 0 digitToNat One = 1 natToChar : Nat -> Char natToChar Z = '0' natToChar (S Z) = '1' natToDigit : (n : Nat) -> Digit (natToChar n) natToDigit Z = Zero natToDigit (S Z

问题 I have a function to reconstruct a tree from 2 lists. I return a list on all branches, but I am getting an error that I don't understand. But I assume it has to do with the return types. The error is this: Can't unify ''a with ''a list (Type variable to be unified occurs in type) Found near recon ( ::( preoH, preoT), ::( inoH, ...)) Exception- Fail "Static errors (pass2)" raised The line the error occurs at is the headline of the function definition fun recon (preoH::preoT, inoH::inoT) = What

问题 I'm trying to understand why the type of fun g x = ys where ys = [x] ++ filter (curry g x) ys is ((a, a) -> Bool) -> a -> [a] . I understand that: filter :: (a -> Bool) -> [a] -> [a] and that curry :: ((a, b) -> c) -> a -> b -> c But I don't understand how to continue. 回答1: The approach below is not necessarily the easiest or fastest, but it's relatively systematic. Strictly speaking, you're looking for the type of \g -> (\ x -> let ys = (++) [x] (filter (curry g x) ys) in ys) ( let and where

问题 I'm trying to further my understanding of Prolog, and how it handles unification. In this case, how it handles unification with lists. This is my knowledgebase; member(X, [X|_]). member(X, [_|T]):- member(X, T). If I'm understanding the process correctly. If member(X, [X|_]) is not true, then it moves into the recursive rule, and if X is in list T , then [_|T] is unified with T . So what happens to the anonymous variable in my recursive predicate? Does it get discarded? I'm having difficulty

问题 I'm trying to get my head around how type inference is implemented. In particularly, I don't quite see where/why the heavy lifting of "unification" comes into play. I'll give an example in "pseudo C#" to help clarify: The naive way to do it would be something like this: Suppose you "parse" your program into an expression tree such that it can be executed with: interface IEnvironment { object lookup(string name); } interface IExpression { // Evaluate this program in this environment object

问题 likes(alice, sports). likes(alice, music). likes(carol, music). likes(david,animals). likes(david,X) :- likes(X,sports). likes(alice,X) :- likes(david,X). ?- likes(alice,X). I've been trying to learn prolog an few days now, and when I attempted this question, I realised that I don't completely understand when the variables are instantiated and used. The initial goal is : likes(alice , X) . After that, the next goal to prove is likes(david , X) ? Then is it likes(X, sports) . Then does X