Prolog programming

2
Prolog ProgrammingProlog Programming
 DATA STRUCTURES IN PROLOG
 PROGRAMMING TECHNIQUES
 CONTROL IN PROLOG
 CUTS

3
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 = a
T = [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
15
2 16
10
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 = a
T = q([b], [])
Y = b
U = 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
X
leave(X,S,T)
a
T
_3
b
Y
leave(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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 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
GOAL
Suffix([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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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

53
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

54
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.

56
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 Yes
X=2Yes
X=1
backtrack
a(X)
b e
!c d Yes
X=2
backtrack
c

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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(_).

Prolog programming

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