Logic Programming
Some "declarative" slides on
logic programming and Prolog.
James Brucker
Introduction to Logic Programming

Declarative programming describes what is desired
from the program, not how it should be done

Declarative language: statements of facts and
propositions that must be satisfied by a solution to the
program
real(x).
proposition: x is a real number.
x > 0.
proposition: x is greater than 0.
Declarative Languages

what is a "declarative language"?

give another example (not Prolog) of a declarative
language.
SELECT * FROM COUNTRY WHERE CONTINENT = 'Asia';
Facts, Rules, ...

What is a proposition?

What are facts?

What are rules?

What is a predicate?

What is a compound term?
Facts:
fish(salmon).
likes(cat, tuna).
Predicates:
fish, likes
Compound terms:
likes(cat, X), fish(X)
Atoms:
cat, salmon, tuna
Rule:
eats(cat,X)  likes(cat,X), fish(X).
A Really Simple Directed Graph
(1) edge(a, b).
a
(2) edge(a, c).
(3) edge(c, d).
(4) path(X, X).
b
c
(5) path(X, Y) 
edge(X, N), path(N, Y).
d
Question: What are the...

atoms

facts

rules
Clausal Form

Problem: There are too many ways to express
propositions.
 difficult for a machine to parse or understand

Clausal form: standard form for expressing propositions
B1  B2  Bm  A1  A2  An
Consequent
Antecedent
Example:
path(X, Y)  edge(X, N)  path(N, Y).
Clausal Form Example
path(X, Y)  edge(X, N)  path(N, Y).
Meaning:
if there is an edge from X to N and there a path from N
to Y,
then there is a path from X to Y.
The above is also called a "headed Horn clause".
In Prolog this is written as a proposition or rule:
path(X, Y) :- edge(X, N) , path(N, Y).
Query
A query or goal is an input proposition that we want
Prolog to "prove" or disprove.
 A query may or may not require that Prolog give us a
value that satisfies the query (instantiation).

1 ?- edge(a,b).
Yes
2 ?- path(c,b).
No
3 ?- path(c,X).
X = c ;
X = d ;
No
Logical Operations on Propositions

What are the two operations that a logic programming
language performs on propositions to establish a
query?
That is, how does it satisfy a query, such as:
Unification
Unification is a process of finding values of variables
(instantiation) to match terms. Uses facts.
(1-3) edge(a,b).
edge(a,c).
edge(c,d).
(Facts)
(4) path(X,X).
(Rule)
(5) path(X,Y) := edge(X,N), path(N,Y).
(Rule)
?- path(a,d).
This is the query (goal).
Instantiate { X=a, Y=d }, and unify path(a,d) with Rule 5.
After doing this, Prolog must satisfy:
edge(a,N).
This is a subgoal.
path(N,d).
This is a subgoal.
Unification in plain English
Compare two atoms and see if there is a substitution
which will make them the same.
1. edge(a, b).
(Fact)
5. path(X, Y) :- edge(X, N) , path(N, Y).
6. path(a, Z).
How can we unify 6 with 5?
Let X := a
Let Y := Z
(Query)
Resolution
Resolution is an inference rule that allows propositions to
be combined.
 Idea: match the consequent (LHS) of one proposition
with the antecedent (RHS term) of another.
P1 ( X )  P2 ( X )
Q1 ( X )  Q2 ( X )
If P2 ( y )  Q1 ( y ) then P1 ( y )  Q2 ( y ).

Examples are in the textbook and tutorials.
Resolution Example
1. edge(a, b).
(Fact)
5. path(X, Y) :- edge(X, N) , path(N, Y).
6. path(a, Z).
How can we unify 6 with 5?
Let X := a
Let Y := Z
Resolution:
(Query)
Resolution
Resolution is an inference rule that allows propositions to
be combined.
 Idea: match the consequent (LHS) of one proposition
with the antecedent (RHS term) of another.
P1 ( X )  P2 ( X )
Q1 ( X )  Q2 ( X )
If P2 ( y )  Q1 ( y ) then P1 ( y )  Q2 ( y ).

Examples are in the textbook and tutorials.
How to handle failures

Prolog can work backwards towards the facts using
resolution, instantiation, and unification.

As it works, Prolog must try each of several choices.

These choices can be stored as a tree.
?- path(a,d).
The goal.
Unify:
unify path(a,d)with Rule 5 by instantiate { X=a,Y=d }
Subgoal:
edge(a,N).
Instantiate: N=b which is true by Fact 1.
Subgoal:
path(b,d).
Unify:
path(b,d)with Rule 5: path(b,d) :- edge(b,N),path(N,d)
Failure:
can't instantiate edge(b,N) using any propositions.
How to handle failures (2)

When a solution process fails, Prolog must undo some
of the decisions it has made.

This is called backtracking.


same as backtracking you use in recursion.
Marks a branch of the tree as failed.
How it Works (1)
There are 2 search/execution strategies that can be
used by declarative languages based on a database of
facts.

1.
Forward Chaining
2.
Backward Chaining
what are the meanings of these terms?
How it Works (2)
1.
Forward Chaining
2.
Backward Chaining

Which strategy does Prolog use?

Under what circumstances is one strategy more
effective than the other?
Consider two cases:

large number of rules, small number of facts

small number of rules, large number of facts
PROLOG: PROgramming in LOGic
The only "logic" programming language in common use.
3 Parts of a Prolog Program
1. A database contains two kinds of information.

What information is in a database?
2. A command to read or load the database.

in Scheme you can use load("filename")

in Prolog use consult('filename')
3. A query or goal to solve.
Ancestors
File: ancestors.pl
ancestor(X,Y) :- parent(X,Y).
ancestor(X,Y) :ancestor(X,Z), ancestor(Z,Y).
parent(X,Y) :- mother(X,Y).
parent(X,Y) :- father(X,Y).
father(bill, jill).
mother(jill, sam).
mother(jill, sally).
Query the Ancestors
?- consult('/pathname/ancestors.pl').
ancestor(bill,sam).
Yes
?- ancestor(bill,X).
X = jill ;
X = sam ;
ERROR: Out of local stack
?- ancestor(X,bob).
ERROR: Out of local stack
Understanding the Problem

You need to understand how Prolog finds a solution.
ancestor(X,Y) :- parent(X,Y).
Depth-first search
ancestor(X,Y) :ancestor(X,Z), ancestor(Z,Y). causes immediate
parent(X,Y) :- mother(X,Y).
parent(X,Y) :- father(X,Y).
father(bill,jill).
mother(jill,sam).
father(bob,sam).
recursion
Factorial
File: factorial1.pl
factorial(0,1).
factorial(N,N*M) :- factorial(N-1,M).

The factorial of 0 is 1.

The factorial of N is N*M if the the factorial of N-1 is M
?- consult('/path/factorial1.pl').
?- factorial(0,X).
X = 1
Yes
?- factorial(1,Y).
ERROR: Out of global stack
Query Factorial
?- consult('/path/factorial1.pl').
?- factorial(2,2).
No
?- factorial(1,X).
ERROR: Out of global stack
Problem: Arithmetic is not performed automatically.
?- 2*3 = 6.
No
?- 2*3 = 2*3.
Yes
?- 6 is 2*3.
is(6,2*3).
Yes
?- 2*3 is 6.
No
l-value = r-value ?
Arithmetic via Instantiation: is

"=" simply means comparison for identity.
factorial(N, 1) :- N=0.

"is" performs instantiation if the left side doesn't have a
value yet.
product(X,Y,Z) :- Z is X*Y.

this rule can answer the query:
product(3,4,N).
Answer: N = 12.

but it can't answer:
product(3,Y,12).
is does not mean assignment!

This always fails:
N is N - 1.
% sumto(N, Total): compute Total = 1 + 2 + ... + N.
sumto(N, 0) :- N =< 0.
sumto(N, Total) :=
Total is Subtotal + N,
N is N-1,
always fails
sumto(N, Subtotal).
?- sumto(0, Sum).
Sum = 0.
Yes
?- sumto(1, Sum).
No
is : how to fix?

How would you fix this problem?
% sumto(N, Total): compute Total = 1 + 2 + ... + N.
sumto(N, 0) :- N =< 0.
sumto(N, Total) :=
N1 is N-1,
sumto(N1, Subtotal),
Total is Subtotal + N.
?= sumto(5, X).
always fails
Factorial revised
File: factorial2.pl
factorial(0,1).
factorial(N,P) :- N1 is N-1,
factorial(N1,M), P is M*N.
Meaning:

The factorial of 0 is 1.

factorial of N is P
if N1 = N-1
and factorial of N1 is M
and P is M*N.
Query Revised Factorial
?- consult('/path/factorial2.pl').
?- factorial(2,2).
Yes
?- factorial(5,X).
X = 120
Yes
but still has some problems...
?- factorial(5,X).
request another solution
X = 120 ;
ERROR: Out of local stack
?- factorial(X,120).
Factorial revised again
File: factorial3.pl
Makes the rules mutually exclusive.
factorial(0,1).
factorial(N,P) :- not(N=0), N1 is N-1,
factorial(N1,M), P is M*N.
?- factorial(5,X).
X = 120 ;
No
?-
Readability: one clause per line
factorial(0,1).
factorial(N,P) :- not(N=0), N1 is N-1,
factorial(N1,M), P is M*N.
factorial(0,1).
factorial(N,P) :not(N=0),
N1 is N-1,
factorial(N1,M),
P is M*N.
Better
Finding a Path through a Graph
edge(a,
edge(b,
edge(b,
edge(d,
path(X,
path(X,
?Yes
?Yes
?Yes
?No
b).
c).
d).
e). edge(d, f).
X).
Y) :- edge(X, Z), path(Z, Y).
edge(a, b).
path(a, a).
path(a, e).
path(e, a).
a
b
c
d
e
f
How To Define an Undirected Graph?
edge(a,
edge(b,
edge(b,
edge(d,
edge(d,
edge(X,
path(X,
path(X,
b).
c).
d).
e).
f).
Y) := not(X=Y), edge(Y, X).
X).
Y) :- edge(X, Z), path(Z, Y).
?- edge(b, a).
Yes
?- path(a, b).
Yes
?- path(b, e).
No
a
b
c
d
e
f
Queries and Answers

When you issue a query in Prolog, what are the
possible responses from Prolog?
% Suppose "likes" is already in the database
:- likes(jomzaap, 219212).
% Programming Languages.
Yes.
:- likes(papon, 403111).
% Chemistry.
No.
:- likes(Who, 204219).
% Theory of Computing?
Who = pattarin

Does this mean Papon doesn't like Chemistry?
Closed World Assumption

What is the Closed World Assumption?

How does this affect the interpretation of results from
Prolog?
List Processing
[Head | Tail] works like "car" and "cdr" in Scheme.
 Example:

?-
[H | T ] = [a,b,c,d,e].
returns:
H = a
T = [b,c,d,e]
This can be used to build lists and decompose lists.
 Can use [H|T] on the left side to de/construct a list:

path(X, Y, [X|P]) :edge(X, Node),
path(Node, Y, P).
member Predicate

Test whether something is a member of a list
?- member(a, [b,c,d]).
No.

can be used to have Prolog try all values of a list as
values of a variable.
?- member(X, [a1,b2,c3,d4] ).
X = a1
X = b2
X = c3
member Predicate example

Use member to try all values of a list.

Useful for problems like
 Queen safety
 enumerating possible rows and columns in a game.
% dumb function to find square root of 9
squareroot9(N) :member(N,[1,2,3,4,5,5,6,7,8,9]),
9 is N*N.
appending Lists
?- append([a,b],[c,d,e],L).
L = [a,b,c,d,e]
?- append([],[a,b,c],L).
L = [a,b,c]

append can resolve other parameters, too:
?- append(X, [b,c,d], [a,b,c,d] ).
X = a
Defining your own 'append'
append([], List, List).
append([Head|Tail], X, [Head|NewTail]) :append(Tail, X, NewTail).
Type Determination
Prolog is a weakly typed language.
 It provides propositions for testing the type of a variable

PREDICATE
var(X)
nonvar(X)
atom(A)
integer(K)
real(R)
number(N)
atomic(A)
functor(T,F,A)
T =..L
clause(H,T)
SATISFIED (TRUE) IF
X is a variable
X is not a variable
A is an atom
K is an integer
R is a floating point number
N is an integer or real
A is an atom or a number or a string
T is a term with functor F, arity A
T is a term, L is a list.
H :- T is a rule in the program
Tracing the Solution
?- trace.
[trace] ?- path(a,d).
Call: (8) path(a, d) ? creep
Call: (9) edge(a, _L169) ? creep
Exit: (9) edge(a, b) ? creep
Call: (9) path(b, d) ? creep
Call: (10) edge(b, _L204) ? creep
Exit: (10) edge(b, c) ? creep
Call: (10) path(c, d) ? creep
Call: (11) edge(c, _L239) ? creep
^ Call: (12) not(c=_G471) ? creep
^ Fail: (12) not(c=_G471) ? creep
Fail: (11) edge(c, _L239) ? creep
Fail: (10) path(c, d) ? creep
Redo: (10) edge(b, _L204) ? creep
Solution Process
(1) path(X,X).
(2) path(X,Y) := edge(X,Z), path(Z,Y).
?- path(a,c).
Stack
[path(a,c), path(X,X)]
[path(a,c), path(a,a)]
Undo.
[path(a,c), path(X,X)]
[path(a,c), path(c,c)]
Undo.
Substitution (Instantiation)
X=a
X=c
Solution Process (2)
(1) path(X,X).
(2) path(X,Y) := edge(X,Z), path(Z,Y).
?- path(a,c).
Stack
[path(a,c), path(X,Y)]
[path(a,c), path(a,Y)]
edge(a,Z), path(Z,c)
edge(a,b), path(b,c)
path(b,c)
Substitution (Instantiation)
(Rule 2)
X=a
X = a, Y = c
new subgoals
X = a, Y = c, Z = b
edge(a,b) is a fact - pop it.
What does this do?
% what does this do?
sub([], List).
sub([H|T], List) :member(H, List),
sub(T, List).
What does this do?
Underscore (_) means "don't care".
% what does this do?
It accepts any value.
foo([], _, []).
foo([H|T], List, [H|P]) :member(H, List),
foo(T, List, P).
foo([H|T], List, P) :not( member(H, List) ),
foo(T, List, P).
Max Function

Write a Prolog program to find the max of a list of
numbers:
 max( List, X).
 max( [3, 5, 8, -4, 6], X).
X = 8.

Strategy:
 use recursion
 divide the list into a Head and Tail.
 compare X to Head and Tail. Two cases:



Head = max( Tail ). in this case answer is X is Head.
X = max( Tail ) and Head < X.
what is the base case?
Max Function
% max(List, X) : X is max of List members
max([X], X).
base case
max([H|Tail], H) :-
1st element is max
max(Tail, X),
H >= X.
max([H|Tail], X) :complete this
case.
1st element not max
Towers of Hanoi
% Move one disk
move(1,From,To,_) :write('Move top disk from '),
write(From),
write(' to '),
See tutorials at:
write(To),
www.csupomona.edu and
nl.
www.cse.ucsc.edu
% Move more than one disk.
move(N,From,To,Other) :N>1,
M is N-1,
move(M,From,Other,To),
move(1,From,To,_),
move(M,Other,To,From).
Learning Prolog

The Textbook - good explanation of concepts

Tutorials:

http://www.thefreecountry.com/documentation/onlineprolog.sht
ml has annotated links to tutorials.

http://www.cse.ucsc.edu/classes/cmps112/Spring03/languages
/prolog/PrologIntro.pdf last section explains how Prolog
resolves queries using a stack and list of substitutions.

http://cs.wwc.edu/~cs_dept/KU/PR/Prolog.html explains
Prolog syntax and semantics.

http://www.csupomona.edu/~jrfisher/www/prolog_tutorial/conte
nts.html has many examples