Prolog Programming

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Prolog ProgrammingProlog Programming

DATA STRUCTURES IN PROLOG PROGRAMMING TECHNIQUES CONTROL IN PROLOG CUTS

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DATA STRUCTURES IN PROLOG Lists in Prolog

List notation is a way of writing terms Terms as Data

Term correspond with list

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Lists in Prolog

The simplest way of writing a list is to enumerate its elements.

The list consisting of the 3 atoms a, b and c can be written as[a, b, c]The list that doesn’t have elements called empty list denoted as [ ]

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Lists in Prolog

We can also specify an initial sequence of elements and a trailing list, separated by |

The list [a, b, c] can also be written as[a, b, c | [ ] ][a, b | [c] ][a | [b, c] ]

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Lists : Head & Tail

A special case of this notation is a list with head H and tail T, written as [H|T]

The head is the first element of a list, and The tail is the list consisting of the remaining

elements.

The list [a, b, c] can also be separated as• Head:The first element is a• Tail:The list of remaining elements = [b, c]

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Lists : Unification

Unification can be used to extract the components of a list, so explicit operators for extracting the head and tail are not needed. The solution of the query

Bind variable H to the head and variable T to the tail of list [a, b, c].

?- [H | T] = [a, b, c].H = aT = [b, c]

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Lists : Specified terms

The query (partially specified terms)

The term [ a | T ] is a partial specification of a list with head a and unknown tail denoted by variable T.

Similarly, [ H, b, c] is a partial specification of a list with unknown head H and tail [b, c].

These two specification to unify H = a, T =[b,c]

?- [a | T] = [H, b, c].T = [b, c]H = a

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Lists in Prolog

Example 2 The append relation on lists is defined by the following rules:

Append([ ], Y, Y).Append([H | X], Y, [H | Z]) :- append(X,Y,Z).

In words, The result of appending the empty list [ ] and a list Y is Y.If the result of appending X and Y is Z, then the result of appending [H | X] and Y is [H | Z]

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Lists : Compute Arguments

The rules for append can be used to compute any one of the arguments from the other two:

Inconsistent arguments are rejected

?- append([a, b], [c, d], Z). Z = [a, b, c, d]?- append([a, b], Y, [a, b, c, d]). Y = [c, d]?- append(X, [c, d], [a, b, c, d]). X = [a, b]

?- append(X, [d, c], [a, b, c, d]). no

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Terms as Data

The Dot operator or functor ‘.’ corresponds to make list with H and T.

[H | T ] is syntactic sugar for the term .(H,T) Lists are terms. The term for the list [a, b, c] is

.(H,T)

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

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Terms as Data

following terms can be drawn a tree

There is a one-to-one correspondence between trees and terms

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

∙

∙

∙

a

b

c []

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Terms : Binary Tree

Binary trees can be written as terms An atom leaf for a leaf A functor nonleaf with 2 arguments

leaf

nonleaf(leaf,leaf)

nonleaf(nonleaf(leaf,leaf), nonleaf(leaf,leaf))

nonleaf(nonleaf(leaf,leaf),leaf)

nonleaf(leaf,nonleaf(leaf,leaf))

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List : tree

Example 3 A binary search tree is either empty, or it consists of a node with two binary search trees as subtrees. Each node holds an integer. Smaller elements appear in the left subtree of

a node and larger elements appear in the right subtree.

Let a term node(K,S,T) represent a tree

K

S T

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Binary search trees

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2 16

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129

0

3

19

10

2 12

9

153

0 16

3

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Binary search trees

The rules define a relation member to test whether an integer appear at some node in a tree. The two arguments of member are an integer and a tree.

member(K,_,_).member(K, node(N,S,_)) :- K < N, member(K, S).member(K, node(N,_,T)) :- K > N, member(K, T).

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PROGRAMMING TECHNIQUES

The strengths of Prolog namely, backtracking and unification. Backtracking allows a solution to be found if

one exists Unification allows variables to be used as

placeholders for data to be filled in later. Careful use of the techniques in this section

can lead to efficient programs. The programs rely on left-to-right evaluation of subgoals.

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Guess and Verify

A guess-and-verify query has the form

Where guess(S) and verify(S) are subgoals. Prolog respond to a query by generating

solutions to guess(S) until a solution satisfying verify(S) is found. Such queries are also called generate-and-test queries.

Is there an S such that guess(S) and verify(S)?

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Guess and Verify

Similarly, a guess-and-verify rule has the following form:

Example

Conslusion(…) if guess(…,S,…) and verify(…,S,…)

overlap(X, Y) :- member(M, X), member(M, Y).

Two lists X and Y overlap if there is some M that is a member of both X and Y. The first goal member(M, X) guesses an M from list X, and the second goal member(M, Y) verifies that M also appears in list Y.

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The rules for member are

member(M, [M |_]).Member(M, [_ |T]) :- member(M, T).

The first rule says that M is a member of a list with head M. The second rule says that M is a member of a list if M is a member of its tail T.

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Consider query

These query

The first goal in this query generates solutions and the second goal tests to see whether they are acceptable.

?- overlap([a,b,c,d],[1,2,c,d]).

yes

?- member(M,[a,b,c,d]),member(M,[1,2,c,d]).

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Consider query

The solutions generated by the first goal are

Test the second goal

?- member(M,[a,b,c,d]). M = a; M = b; M = c; M = d; no

?- member(a,[1,2,c,d]). no?- member(b,[1,2,c,d]). no?- member(c,[1,2,c,d]). yes

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Hint

Since computation in Prolog proceeds from left to right, the order of the subgoals in a guess-and-verify query can affect efficiency.

Choose the subgoal with fewer solutions as the guess goal.

Example of the effect of goal order

?- X = [1,2,3], member(a,X). no?- member(a,X), X = [1,2,3]). [infinite computation]

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Variables as Placeholders in Terms

Variables have been used in rules and queries but not in terms representing objects.

Terms containing varibales can be used to simulate modifiable data structures;

The variables serve as placeholders for subterms to be filled in later.

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Represent Binary Trees in Terms

The terms leaf and nonleaf(leaf,leaf) are completely specified.

leaf

nonleaf(leaf,leaf)

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Partially specified list

The example list [a, b | X] has Its first element : a Its second element : b Do not yet know what X represents

“Open list” if its ending in a variable, referred “end marker variable”

“Close list” if it is not open.

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How prolog know variable

Prolog used machine-generated variables, written with a leading underscore (“_”) followed by an integer.

?- L = [a, b | X].L = [a, |_G172]X = _G172 Yes

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Prolog generates fresh variables each time it responds to a query or applies a rule.

An open list can be modified by unifying its end marker

?- L = [a, b | X], X = [c,Y].L = [a,b,c |_G236]X = [c,_G236]Y = _G236 Yes

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Extending an open list by unifying its end marker.

a b

L X

_172

a b

L X

_236

c(a) Before X is bound. (b) After X = [c | Y].

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Unification of an end-marker variable is akin to an assignment to that variable.

List L changes from

[a, b | _172] [a, b, c | _236]

when _172 unifies with [c | _236] Advantage of working with open lists is that

the end of a list can be accessed quickly.

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Open list implement queues

when a queue is created, where L is an open list with end marker E

When element a enters queue Q, we get queue R.

When element a leaves queue Q, we get queue R.

q(L,E)

enter(a,Q,R)

leave(a,Q,R)

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Open list implement queue

?- setup(Q).

?- setup(Q), enter(a,Q,R).

?- setup(Q), enter(a,Q,R), leave(S,R,T).

?- setup(Q), enter(a,Q,R), enter(b,R,S), leave(X,S,T),leave(Y,T,U), wrapup(q([],[])).

setup(q(X,X)).enter(A, q(X,Y), q(X,Z)) :- Y = [A | Z].leave(A, q(X,Z), q(Y,Z)) :- Y = [A | Y].wrapup(q([],[])).

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Test queue

?-setup(Q),enter(a,Q,R),enter(b,R,S),leave(X,S,T),leave(Y,T,U),wrapup(U).

Q = q([a, b], [a, b])R = q([a, b], [b])S = q([a, b], [])X = aT = q([b], [])Y = bU = q([], [])

Yes?-

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Operations on a queue

Q

_1

R

_2

a

a

T

_3

b

Q

Q R

setup(Q)

enter(a,Q,R)

enter(b,R,S)

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Operations on a queue

a

T_3

b

Xleave(X,S,T)

a

T_3

b

Yleave(Y,T,U)

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Internal Prolog

A queue q(L,E) consists of open list L with end marker E.

The arrows from Q therefore go to the empty open list _1 with end marker _1.

setup(q(X,X)).

?-setup(Q). Q = q(_1,_1)yes

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Second goal

To enter A into a queue q(X,Y),

bind Y to a list [A|Z],

where Z is a fresh end marker,

and return q(X,Z).

enter(A,q(X,Y),q(X,Z)):- Y = [A|Z].

?-setup(Q),enter(a,Q,R).Q = q([a|_2], [a|_2])R = q([a|_2], _2)

Unifies _1 with [a|_2],where _2 is a fresh end marker

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When an element leaves a queue q(L,E), the resulting queue has the tail of L in place of L. Note in the diagram to the right of leave(X,S,T) that the open list for queue T is the tail of the open list for S.

The final goal wrapup(U) checks that the enter and leave operations leave U in an initial state q(L,E), where L is an empty openlist with end marker E.

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Difference Lists

Difference List are a technique for coping with such changes.

Difference List consists of a list and its suffix. We write this difference list as

dl(L,E).

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Contents of Difference List

The contents of the difference list consist of the elements that are in L but not in E.

Examples of difference lists with contents [a,b] are

dl([a,b],[]).Dl([a,b,c],[c]).Dl([a,b|E],E).Dl([a,b,c|F],[c|F]).

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CONTROL IN PROLOG

In the informal equation

“Logic” refers to the rules and queries in a logic program and

“control” refers to how a language computes a response to a query.

algorithm = logic + control

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CONTROL IN PROLOG

Control in Prolog is characterized by two decisions Goal order : Choose the leftmost subgoal. Rule order : Select the first applicable rule.

The response to a query is affected both by goal order within the query and by rule order with in the database of facts and rules.

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CONTROL IN PROLOG

start with a query as the current goal;while the current goal is nonempty do

choose the leftmost subgoal; if a rule applies to the subgoal then

select the first applicable rule; form a new current goal

else backtrack

end if

end while;succeed

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Example

A sublist S of Z can be specified in the following seemingly equivalent ways: preffix X of Z and suffix S of X. suffix S of X and prefix X of Z.

appen1([],Y,Y).appen1([H|X],Y,[H|Z]):- appen1(X,Y,Z).Prefix(X,Z) :- appen1(X,Y,Z).Suffix(Y,Z) :- appen1(X,Y,Z).appen2([H|X],Y,[H|Z]):- appen2(X,Y,Z).appen2([],Y,Y).

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Queries

The corresponding queries usually produce the same responses.

Rule order can also make a difference.

?-prefix(X,[a,b,c]),suffix([e],X).no?-suffix([e],X),prefix(X,[a,b,c]).[infinite computation]

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Queries

?- appen1(X,[c],Z).X = []Z = [c] ;

X = [_G230]Z = [_G230, c] ;

X = [_G230, _G236]Z = [_G230, _G236, c] ;?- appen2(X,[c],Z).

New Solutions are produced on demand for

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Unification an Substitutions

Unification is central to control in Prolog Substitution is a function from variables to

terms

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Applying a Rule to a Goal

A rule applies to a subgoal G if its head A unifies with G

Variables in the rule are renamed before unification to keep them distinct from variables in the subgoal.

A :- B1, B2, …, Bn

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A computation that succeeds without backtracking

GOALSuffix([a],L),prefix(L,[a,b,c]).

suffix([a],L) if append(_1,[a],L).Append(_1,[a],L),prefix(L,[a,b,c]).

{_1[],L[a]} append([],[a],[a]).Prefix([a],[a,b,c]).

prefix([a],[a,b,c]) if append([a],_2,[a,b,c])append([a],_2,[a,b,c]).

prefix([a],[a,b,c]) if append([],_2,[b,c])Append([],_2,[b,c]).

{_2[b,c]} append([],[b,c],[b,c])yes

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Prolog Search Trees

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Goal Order Changes Solutions

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Cuts

A cut prunes or “cuts out” and unexplored part of a Prolog search tree.

Cuts can therefore be used to make a computation more efficient by eliminating futile searching and backtracking.

Cuts can also be used to implement a form of negation

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Cuts

A cut, written as !, appears as a condition within a rule. When rule

is applied, the cut tells control to backtrack past Cj-1,…,C1,B, without considering any remaining rules for them.

B :- C1,…, Cj-1, !,Cj+1,…,Ck

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A cut as the First Condition

Consider rules of the form

If the goal C fails, then control backtracks past B without considering any remaining rules for B. Thus the cut has the effect of making B fail if C fails.

B :- !, C.

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Example

b :- c.b :- d.b :- e.

b,G

c,G e,G

X

d,G!,d,G

d,G

b :- c.b :- !,d.b :- e.

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Example

?-a(X).

a(1) :- b;a(2) :- e;b :- c.b :- d.

a(1) :- b;a(2) :- e;b :- !,c.b :- d.

a(X)

b e

c d YesX=2Yes

X=1backtrack

a(X)

b e

!c d YesX=2

backtrackc

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The Effect of Cut

As mentioned earlier, when a rule

is applied during a computation The cut tells control to backtrack past Cj-

1,..C1,B without considering any remaining rules for them.

The effect of inserting a cut in the middle of a guess-and-verify rule.

B :- C1,…, Cj-1, !,Cj+1,…,Ck

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The Effect of Cut

The right side of a guess-and-verify rule has the form guess(S), verify(S), where guess(S) generates potential solutions until one satisfying verify(S) is found.

The effect of insering a cut between them, as

is to eliminate all but the first guess.

Conclusion(S) :- guess(S), !, verify(S)

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a(X) :- b(X).a(X) :- f(X).b(X) :- g(X),v(X).b(X) :- X = 4, v(X).g(1).g(2).g(3).v(X).f(5)

a(X) :- b(X).a(X) :- f(X).b(X) :- g(X),!,v(X).b(X) :- X = 4, v(X).g(1).g(2).g(3).v(X).f(5)

(a) (b)

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a(Z)

b(Z) f(5)

g(Z),v(Z) v(4)

v(1) v(2) v(3)

a(Z)

b(Z) f(5)

g(Z),!,v(Z) v(4)

!v(X)

v(1)

v(2) v(3)

(a) (b)

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Negation as Failure

The not operator in Prolog is implemented by the rules

not(X) :- X, !, fail.not(_).