Formal Models of Computation Part II The Logic Model Lecture 5 – Lists

formal models of computation Lists Data structure for non-numeric programming. A sequence of any number of items. The items can be any Prolog term (including nested lists!). Useful when we don’t know in advance how many items we are going to have. Flexible and generic. In Prolog, a list is represented as a sequence of terms, separated by commas and within “[…]”: [123,hello,Var1,p(a,X),[1,2,3]] formal models of computation

Types in lists: Haskell vs Prolog In Haskell, for every type t there is a type [t] for lists whose elements are of type t In Haskell, every list must have type [t] (for some t), and hence all its elements must be of the same type So you can’t for instance have a list, some of whose elements have type int and some of whose elements have type [int], e.g. [1,[2,3],4] Prolog places no restrictions on what elements of lists can be Less restrictive, but you lose the advantages of type checking formal models of computation

formal models of computation Lists Lists are ordinary Prolog terms. The notation […] is just a shorthand: [1,2,3]  .(1,.(2,.(3,[]))) Lists are thus represented internally as trees: N.B.: the last element is always the empty list [] . 1 2 3 [] formal models of computation

formal models of computation Lists Lists are recursively defined as: The empty list []; or A term followed by a list. Lists can enter queries: . 1 2 3 [] ?- List = [1,2,3]. List = [1,2,3] yes ?- formal models of computation

formal models of computation Lists (Cont’d) Prolog provides a notation for manipulating lists. The notation is [H|T], a shortand for .(H,T) H stands for the first element of the list. T stands for the remaining elements of the list and must be a list. We can build and decompose lists with [H|T]: ?- List = [1,2,3], List = [A|B]. A = 1, B = [2,3], List = [1,2,3] ?- List = [A|B], A = 1, B = [2,3]. formal models of computation

formal models of computation Lists (Cont’d) The [H|T] notation allows for variations: [E1,E2|T] (a list with at least 2 elements) [1|[2,3|T]] (a list [1,2,3|T]) [H|T] is also a term, hence a tree: ?- List = [A|B], A = 1, B = [2,3]. A = 1, B = [2,3], List = [1,2,3] List A B . 1 [] 2 . 3 formal models of computation

Haskell vs Prolog list notations [1,2,3] [] x:xs [X|Xs] or .(X,Xs) 1:2:xs [1,2|Xs] formal models of computation

formal models of computation Operations on Lists We can access elements from the front of a list We can add elements to the front of a list: ?- L = [2,3], NewL = [1|L] L [] 2 . 3 1 . NewL formal models of computation

Programming with Lists Because of their recursive definition, lists are naturally manipulated with recursive programs. Let’s revisit our first loop program: Enable it to build a list with values typed in An argument is required to return the value of the list built loop(L):- % a loop to create a list L L = [X|Xs], % List L has form [X|Xs] read(X), % X is read from the standard Input (keyboard) X \= end, % if input different from “end” then loop(Xs). % build the tail recursively loop([]). % if input is “end”, then List is empty. ?- loop(MyList). |: a. |: b. |: end. MyList = [a,b] formal models of computation

Programming with Lists (Cont’d) Its execution: Xs1 X1 . loop(L):- L = [X|Xs], read(X), X \= end, loop(Xs). loop([]). loop(L1):- L1 = [X1|Xs1], read(X1), X1 \= end, loop(Xs1). L1 a Xs2 X2 . loop(L2):- L2 = [X2|Xs2], read(X2), X2 \= end, loop(Xs2). L2 ?- loop(MyList). |: a. |: b. ?- loop(MyList). ?- loop(MyList). |: a. ?- loop(MyList). |: a. |: b. |: end. b loop(L3):- L3 = [X3|Xs3], read(X3), X3 \= end, loop(Xs3). loop(L3):- L3 = [X3|Xs3], read(X3), X3 \= end, loop(Xs3). Xs3 X3 . L3 end formal models of computation

Programming with Lists (Cont’d) Its execution: Xs1 X1 . loop(L):- L = [X|Xs], read(X), X \= end, loop(Xs). loop([]). loop(L1):- L1 = [X1|Xs1], read(X1), X1 \= end, loop(Xs1). L1 a Xs2 X2 . loop(L2):- L2 = [X2|Xs2], read(X2), X2 \= end, loop(Xs2). L2 ?- loop(MyList). |: a. |: b. |: end. ?- loop(MyList). |: a. |: b. |: end. MyList = [a,b] b [] loop([]). formal models of computation

Programming with Lists (Cont’d) To check if an element appears in a list: Check if element is the head; Otherwise, check if element appears in tail This formulation is guaranteed to stop because lists are finite (!) and the process breaks list apart; In Prolog (first formulation): member(X,List):- % X is a member of List List = [Y|Ys], % if List is of form [Y|Ys] X = Y. % and X is equal to Y member(X,List):- % otherwise (if not as above) List = [Y|Ys], % break the list apart member(X,Ys). % and check if X appears in Ys formal models of computation

Programming with Lists (Cont’d) Second formulation: member(X,[Y|Ys]):- % X is a member of [Y|Ys] X = Y. % if X is equal to Y member(X,[Y|Ys]):- % otherwise (if not as above) member(X,Ys). % check if X appears in Ys formal models of computation

Programming with Lists (Cont’d) Third (final) formulation: General point: Uses of the built-in predicate = can almost always be replaced more elegantly with pattern-matching in the head of a clause. member(X,[X|_]). % X is member if it is head of list member(X,[_|Ys]):- % otherwise (if not as above) member(X,Ys). % check if X appears in Ys formal models of computation

formal models of computation Prolog vs Haskell Cf Haskell member :: a -> [a] -> Bool member x [] = False member x (y:ys) = (x==y) || (member x ys) Note: [] case member(X,[X|_]). member(X,[_|Ys]) :- member(X,Ys). formal models of computation

Programming with Lists (Cont’d) Let’s run the previous program: member(X,[X|_]). member(X,[_|Ys]):- member(X,Ys). The query member(3,[1,2,3])does not match the 1st clause!! Prolog matches member(3,[1,2,3])with head of 2nd clause: member(X1,[_|Ys1]):- member(X1,Ys1) We thus get the values for the variables: {X1/3, Ys1/[2,3]} Prolog now tries to prove member(X1,Ys1) that is member(3,[2,3]) ?- member(3,[1,2,3]). yes ?- ?- member(3,[1,2,3]). The goal member(3,[2,3])does not match the 1st clause!! Prolog matches member(3,[2,3])with head of 2nd clause: member(X2,[_|Ys2]):- member(X2,Ys2) We thus get the values for the variables: {X2/3, Ys2/[3]} Prolog now tries to prove member(X2,Ys2) that is member(3,[3]) The query member(3,[3])matches the 1st clause!! formal models of computation

Programming with Lists (Cont’d) Member program can be run in two ways: To check if an element occurs in a list; To obtain the elements of a list, one at a time It depends on the query – the program is exactly the same!! To obtain the elements of a list, we query: member(Elem,[1,2,3]) We then get in Elem, the different elements ?- member(Elem,[1,2,3]). Elem = 1 ? ; Elem = 2 ? ; Elem = 3 ? ?- member(Elem,[1,2,3]). Elem = 1 ? ; Elem = 2 ? ; Elem = 3 ? ; no ?- ?- member(Elem,[1,2,3]). Elem = 1 ? ; Elem = 2 ? ?- member(Elem,[1,2,3]). Elem = 1 ? ?- member(Elem,[1,2,3]). Can you trace this query? formal models of computation

Programming with Lists (Cont’d) To concatenate two lists Xs and Ys: Create a 3rd list Zs by copying all elements of Xs, When we get to the end of Xs, put Ys as the tail of Zs Diagrammatically: Ys [] a . b Xs 1 2 Zs . . 1 2 formal models of computation

Programming with Lists (Cont’d) We need a predicate with 3 arguments: 1st argument is the list Xs 2nd argument is the list Ys 3rd argument is the concatenation Xs.Ys (in this order) The first list Xs will control the loop As the first list Xs is traversed, its elements are copied onto the 3rd argument (the concatenation). When we get an empty list, we stop and make Ys the value of the 3rd argument formal models of computation

Programming with Lists (Cont’d) First attempt: A more compact format: append([],Ys,Zs):- % if Xs is empty, Zs = Ys. % then Zs is Ys append([X|Xs],Ys,[Z|Zs]):- % otherwise X = Z, % copy X onto 3rd argument append(Xs,Ys,Zs). % carry on recursively append([],Ys,Ys). % if Xs empty, then Zs is Ys append([X|Xs],Ys,[X|Zs]):- % else copy X onto 3rd arg. append(Xs,Ys,Zs). % carry on recursively formal models of computation

Programming with Lists (Cont’d) Execution: aliasing: Res/[X1|Zs1] append([],Ys,Ys). append([X|Xs],Ys,[X|Zs]):- append(Xs,Ys,Zs). append([X1|Xs1],Ys1,[X1|Zs1]):- append(Xs1,Ys1,Zs1). {X1/1,Xs1/[2],Ys1/[a,b]} append([X1|Xs1],Ys1,[X1|Zs1]):- append(Xs1,Ys1,Zs1). append([X1|Xs1],Ys1,[X1|Zs1]):- append(Xs1,Ys1,Zs1). {X1/1,Xs1/[2],Ys1/[a,b]} aliasing: Zs1/[X2|Zs2] ?- append([1,2],[a,b],Res). Res = [1,2,a,b] ? yes ?- ?- append([1,2],[a,b],Res). append([X2|Xs2],Ys2,[X2|Zs2]):- append(Xs2,Ys2,Zs2). append([X2|Xs2],Ys2,[X2|Zs2]):- append(Xs2,Ys2,Zs2). {X2/2,Xs2/[],Ys2/[a,b]} append([X2|Xs2],Ys2,[X2|Zs2]):- append(Xs2,Ys2,Zs2). {X2/2,Xs2/[],Ys2/[a,b]} aliasing: Zs2/Ys3 append([],Ys3,Ys3). {Ys3/[a,b]} append([],Ys3,Ys3). formal models of computation

Programming with Lists (Cont’d) 1 Res Zs1 X1 . 2 Zs2 X2 a b [] Execution: append([],Ys,Ys). append([X|Xs],Ys,[X|Zs]):- append(Xs,Ys,Zs). append([X1|Xs1],Ys1,[X1|Zs1]):- append(Xs1,Ys1,Zs1). {X1/1,Xs1/[2],Ys1/[a,b]} ?- append([1,2],[a,b],Res). Res = [1,2,a,b] ? yes ?- append([X2|Xs2],Ys2,[X2|Zs2]):- append(Xs2,Ys2,Zs2). {X2/2,Xs2/[],Ys2/[a,b]} append([],Ys3,Ys3). {Ys3/[a,b]} formal models of computation

“Reversible definitions” in Prolog As with “member”, “append” can be used in different ways, e.g. ?- append([a,b,c],[d],X). ?- append(X,[a,b,l,e],[e,a,t,a,b,l,e]). ?- append([e,a,t],X,[e,a,t,a,b,l,e]). ?- append(X,Y,[e,a,t,a,b,l,e]). Each of these would require a different function/method definition in Haskell/Java… formal models of computation