Recursion Recall:grandparent(Pers, Gpar) :- parent(Pers, Par), parent(Par, Gpar). How to define...

RecursionRecall: grandparent(Pers, Gpar) :-

parent(Pers, Par), parent(Par, Gpar).

How to define greatgrandparents? grandparent(Pers, GGpar) :- parent(Pers, Par), parent(Par, Gpar),

parrent(Gpar, GGPar).

How to define ancestor? ancestor(Pers, A) :-parent(Pers, A).

ancestor(Pers, A) :-

parent(Pers, X1), parent(X1, A).

...

ancestor(Pers, A) :-

parent(Pers, X1), parent(X1, X2),

…parent(Pers,X2, A).

...

Better: use Recursionancestor(P, Anc) :- parent(P, Anc).

ancestor(P, Anc) :- parent(P, Par), ancestor(Par, Anc).

father(charles,philip).

father(ana,george).

father(philip,tom).

mother(charles,ana).

Order of Clauses : Base Case first!!

ancestor(P, Anc) :- parent(P, Par), ancestor(Par, Anc).

ancestor(P, Anc) :- parent(P, Anc).

father(charles,philip).

father(ana,george).

father(philip,tom).

mother(charles,ana).

Paths in Graphs

path(Node1, Node2) :- edge(Node1, Node2).

path(Node1, Node2) :-

edge(Node1, Hop),

path(Hop, Node2).

?- path(a, X).

X=b, X=c, X=e, X=d.

path(Node1, Node2) :- edge(Node1, Node2).Path(Node1, Node2) :- edge(Node1, Hop), path(Hop, Node2).

Terms as Data StructuresFrom other programming languages you know

recordsstructuresarraysstrings …

Prolog knows only Terms!

Terms can (and must) be used to represent all other data structures

Reminder: Terms have the following forms:•a where a is an constant•X where X is a variable•f(T1, …, Tn) where f is a functor and T1…Tn are terms

There are no type declarations in Prolog!

Records

Reords can be nested:

person(Name, Age, Profession).profession(Type, Employer).

e.g.

person(john, 40,profession(pilot, qantas)).

Struct person { becomes person(Name, Age, Prof).char[] Name;int age; for examplechar[] profession; }

person(martin, 36, lawyer).

RecordsReords can be queried:

person(john, 40,profession(pilot, qantas)).

“What is John’s profession?”

?- person(john, _Age, Prof).

“Who is John’s employer?”

?- person(john, _Age, profession(_Type, Employer)).

“Who are the persons working as pilots?”

?- person(Name, _Age, profession(pilot, _Employer)).

Unification•In traditional programming we use assignments

X := Y means “X gets the value if Y”

•In logic programming there are no assignmentsX = Y means “try to make X and Y equal, but keep them as general as possible”

X=abc => X=abcabc=X => X=abcX=Y => no change,

but both variables identical (as if X=_1, Y=_1)

f(a)=f(X) => X=af(a,Y)=f(X,X) => X=Y and X=Y=af(X)=X => ??????? (not permitted)

Unification computes the substitution required to make the terms equal

Unification Algorithm (Idea)Unify(C): C is a set of equations, S a substitution

Initialize S to emptyWhile C is not empty

select equation c from Cif c is of the from X=X then remove c from Celseif c is of the form f(X1,…,Xn)=g(Y1,…,Ym) then

return false if not(f=g) or not(m=n)elseif c is of the form f(X1, …, Xn)=f(Y1, …, Yn) then

replace c in C with X1=Y1, …, Xn=Ynelseif c is of the form term=X and X is a variable then

replace c in C with X=termelseif c is of the form X=term and X is a variable then

if term contains X then return falseelse (1) remove c from C,

(2) replace X with t throughout C and throughout S(3) add c to S

endifendif

endwhilereturn S

Unification Exercise

•abc = def ?

•abc = Y ?

•abc = f(Y) ?

•f(abc) = f(Y) ?

•g(abc) = f(Y) ?

•g(abc, def, X) = g(A, Y, A) ?

•g(abc, def, X) = g(A, A, A) ?

•g(f(a), h(b)) = g(f(X), Y) ? <- cf. record example

•g(f(a), h(b)) = g(X, k(b)) ?

•X=g(X) ?

•g(X,Y)=g(g(Y),a) ?

Lists (sequential collections of elements)

The most important data structure in Prolog.

(Partially because of the strong emphasis on recursive programming)

Lists can be represented using terms

Example: the list <a, b, c, d> can be represented as

cons(a, cons(b, cons(c, cons(d, nil))))

ListsLists are important enough to have a special notation

[] nil[X | Y] cons(X, Y) “Y is a list”[X1, X2, …, Xn] cons(X1, cons(X2, …,

cons(Xn, nil)))

[X1, X2, …, Xn | Y] cons(X1, cons(X2, …, cons(Xn, Y)))

Note: [X1, …, Xn | Y] is a list only if Y is a list.Its length is n + the number of elements in Y.

What is the result of:

[a,b] = [a|X]?

[a]=[a|X]?

[a]=[a, b | X]?

[] = [X]?

Lists + RecursionLists are important, because they fit naturally into the recursive schema. The idea is:

•If the list is empty (or has one element), the answer should be easy.•If the list is longer, break it up and check for the shorter lists.

Consider the test for membership in a list:Element X is a member of the list L if

•L starts with X or •the rest of X contains X

member(X, []) :- fail.

member(X, [U|V]) :- X=U.

member(X, [U|V]) :- member(X, V).

Membermember(X, []) :- fail.member(X, [U|V]) :- X=U.member(X, [U|V]) :- member(X, V).

Better: member(X, [X | _ ]).

member(X, [_ | V]) :- member(X, V).

Example: member(3, [1,2,3]) ->

member(3, [2,3]) ->

member(3, [3]) -> yes.

Analyze: member(0, [1,2,3]) and

member(X, [a,b,c]).

member/3

We can also define a predicate member(X,L,R) that checks whether an

element X is in the List L and unifies R with L after removal of X

member(X, [X | Rest], Rest).

member(X, [_ | Rest], Rest1) :- member(X, Rest, Rest1).

Analyze: member(X, [a,b], R).member(a, [a,b,a,c], R).member(c, [a,b], R).

This does not achieve the desired effect, because it looses elements in front of the item that we are looking for!

member/3: Corrected

member(X, [X | Rest], Rest).

member(X, [F | Rest], [F | Rest1]) :- member(X, Rest, Rest1).

Analyze: member(X, [a,b], R).member(a, [a,b,a,c], R).member(c, [a,b], R).

This no longer “discards” items that are not the search element,but keeps track of them in F.The base case remains unchanged.

append/3

A frequently needed operation is the concatenation of two lists:

append(L1, L2, L3) is defined such that L3 is the concatenation of L1 and L2

append([], L2, L2).append([X | Rest], L2, [X | Result]) :- append(Rest, L2, Result).

append/3

What happens with the following queries?

append([a,b],[c,d], Result). Yes!append(X, [b,c], [a,b,c]). Yes!append(X, Y, [a,b,c]). Yes, generates all possible solutions

append(L1, [], Result). No!because first and third argumentcan infinitely be extended

The recursion tries to shorten the first and third argument. This must eventually terminate to reach base case or to fail.

append([], L2, L2).append([X | Rest], L2, [X | Result]) :- append(Rest, L2, Result).

reverse/3

As defined here, reverse is hideously inefficient,Because append has to traverse the list.

We will see a better solution later.

Here is a predicate to reverse the order of the elements in a list:

reverse([], []).reverse([H | T ], L) :-

reverse(T, RT),append(RT, [H], L).

Accumulators(A better definition of reverse)

reverse(X, Y) :- fast_reverse(X, [], Y).

fast_reverse([], Y, Y).

fast_reverse([H|T], Acc, Y) :- fast_reverse(T, [H|Acc], Y).

Note: fast_reverse traverses the list only once.

The second argument Acc is called an accumulator

Analyze: reverse([a,b], Y).

Execution of fast_reverse

reverse([a,b,c], Y) :- fast_reverse([a,b,c], [], Y).

fast_reverse ([a,b,c], [], Y) :- fast_reverse([b,c], [a], Y).

by rule 2

fast_reverse ([b,c], [a], Y) :- fast_reverse([c], [b,a], Y).

by rule 2

fast_reverse ([c], [b,a], Y) :- fast_reverse([], [c,b,a], Y).

by rule 1

fast_reverse ([], [c,b,a], [c,b,a])

TreesTrees are another very important and often used data structure,

particularly binary trees (i.e. trees of degree two).

astruct node {

struct node *left;

struct node *right;

elemType key;

}

b c

d e f g

In Prolog we use terms of structure tree(Left, Key, Right)

To represent trees, for example

tree(tree(tree(null, d, null), b, tree(null, e, null)),

tree(tree(null, f, null), c, tree(null, g, null)))

Trees Search

Since a binary tree has two recursions in the data structure, a recursive predicate that traverses a tree will typically havetwo recursive calls.

The structure of the recursion follows the structure of the data

tree_member(X, Tree) checks whether an Element X is in the Tree T

tree_member(X, tree(_, X, _)).tree_member(X, tree(L, _, _)) :- tree_member(X, L).tree_member(X, tree(_, _, R)) :- tree_member(X, R).

Tree Traversal

There are three different ways to traverse a tree. They are distinguished by the order in which the nodes are visited:

•Inorder: left child -> node -> right child

•Preorder: node -> left child -> right child

•Postorder: left child -> right child -> node

a

b c

d e f g

Preorder(null, []).Preorder(tree(L,X,R), [X | Rest]) :-

preorder(L, Left),preorder(R, Right),append(Left, Right, Rest).

Isomorphic TreesTwo Binary Trees T1, T2 are called isomorphic

If T2 can be obtained from T1 by swapping left and right subtrees.

a

b c

d e f g

a

c b

g f d e

isoTree(null, null).

isoTree(tree(L1, X, R1), tree(L2, X, R2)) :-isoTree(L1, L2), isoTree(R1, R2).

isoTree(tree(L1, X, R1), tree(L2, X, R2)) :-isoTree(L1, R2), isoTree(R1, L2).

iso

Visualization of Data & Predicates

As described in: G.A. Ringwood, “Predicates and Pixels”,New Generation Computing 7 (1989): 59-80

...visualizes data (terms) and program (predicate) structurein one homogeneous form...

•as graphs•execution semantics defined as sub-graph substitution

The system in “Predicates and Pixels” is actually defined for a related logic programming language called FGDC (“Flat Guarded Definite Clauses”) defined in G.A. Ringwood, “A comparative Exploration of Concurrent Logic Languages”,Knowledge Engineering Review 4(1989):305-332.

For the purpose of our discussion we can ignore the differences.

Example: Quicksort (with Difference Lists)

we will illustrate “Predicates and Pixels” with a quicksort programthat uses difference lists.

Difference lists are a standard programming technique in logicprogramming that utilizes unification in a “tricky” form to avoid appending (because appending is a costly operation).

The meaning of a difference list written “Xs\Ys” is “the list Xs disregarding the tail Ys”, e.g.[a,b,c,d,e] \ [c,d,e] = [a,b]

Xs\Ys

Ys\Zs

Xs\Zs

= Xs

= Ys

= Zs

= L with append(Xs,Ys,L)

Standard Quicksortquicksort([X|Xs], Ys) :-

partition(Xs, X, pair(Ls, Gs)),quicksort(Ls, LsSorted),quicksort(Gs, GsSorted),append(LsSorted, [X | GsSorted], Ys).

quicksort([], []).

partition([X | Xs], P, pair([X | L1s], Gs) :- X <= P, partition(Xs, P, L1s, Bs).

partition([X | Xs], P, pair(Ls, [X | G1s]) :- X >= P, partition(Xs, P, pair(Ls, G1s)).

partition([], _Any, pair([], [])).

Note the inefficiency cause by using append/3.

Quicksort with Difference Lists

Note that append for difference lists can simply be declared as:

append_dl(Xs \ Ys, Ys \ Zs, Xs \ Zs).

This can obviously be executed in constant time, but only makes sense if the tail of the first DL can be unified with the head ofthe second DL.

?- append_dl([a,b,c | Xs] \ Xs, [1,2] \ [], Ys) => Ys=[a,b,c,1,2] \ []

quicksort(Xs, Ys) :- quicksort_dl(Xs, Ys \ []).

quicksort_dl([X | Xs], Y0s \ Y1s) :-partition(Xs, X, pair(Ls, Gs)) % see last slidequicksort_dl(Ls, Y0s \ [X | Y3s]),quicksort_dl(Gs, Y3s \ Y1s).

quicksort_dl([], Xs\Xs).

Quicksort with Difference Lists

quicksort(Xs, Ys) :- quicksort_dl(Xs, Ys \ []).

quicksort_dl([X | Xs], Y0s \ Y1s) :-partition(Xs, X, pair(Ls, Gs)) % see last slidequicksort_dl(Ls, Y0s \ [X | Y3s]),quicksort_dl(Gs, Y3s \ Y1s).

quicksort_dl([], Xs\Xs).

Y0s \ [X | Y3s ]

Y3s\Y1s

= Y0s

= Y3s

= Y1s

X

= [ X | Y3s ]

Y0s \ Y1s

Quicksort with Difference Lists (1)

in the following we use:list(X, Xs) for [X | Xs] and diff(Xs, Ys) for Xs \ Ys.

Domain-Specific Data Visualization in VPLs

Data Visualization only visualizes the terms used in a logic program as application specific / domain specific notations.

Such programs can easily be translated into standard textual logic programs.

Example: MPL (Matrix Programming Language) which incorporates a particular array notation. [6]

General Heterogeneous VPLs

A Generic Schema for Data Visualization: Arbitrary Visualizations for terms can be coded into a translation module which translates the visual notation to a textual standard term.

This makes domain-specific visual logic programming languages for arbitrary domains possible.

Example: Finite State Automatons

Exercise: Code the execution of state automatons in standard Prologand compare to this program.

Execution of Heterogeneous VPLsExecution is achieved in three steps:

(1) A preprocessor splits the program into pictures and textual program (with gaps for the pictures),

(2) A customized picture parser translates the visual terms into

their textual equivalents(3) The textual program and the translation of the visual

terms are recombined by a standard parser, resultingin a standard (Prolog) program.

(4) This program can be executed by a standard compiler.

For detailed information see: Heterogeneous Visual Languages. M. Erwig and B. MeyerInternational IEEE Workshop on Visual Languages 1995, Darmstadt, Germany.

Recursion Recall:grandparent(Pers, Gpar) :- parent(Pers, Par), parent(Par, Gpar). How to define...

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