bigger(elephant, horse). bigger(horse, monkey). We can quer

1
Constraint Logic Programming

• Lesson 1

2
Info

• course number 211113
• course page http//www.cs.utwente.nl/etalle/lp/
• here comes the study material
• dont use teletop much
• schedule (important) WILL PROBABLY CHANGE
• teacher Sandro Etalle etalle_at_cs.utwente.nl
• Marcin Czenko assists the practical part

3
This course

• The slides and the exercises are being
  rewritten dont download the slides of the next
  lesson.
• slides material available via my homepage are
  sufficient

• Goal
• learn to PROGRAM in Prolog and Eclipse
• learn and be able to put into practice the
  concepts of
• backtracking search
• constraint propagation
• branch and bound search
• ...

4
Exam

• written
• possible questions could be
• write a program that ...
• tell what this program exactly does ...
• apply the XXX consistency algorithm to the
  following CSP...
• some philosophical question about search
  propagation, backtracking, unification, etc..
• but dont complain if I come up with a different
  one.

5
Structure

• part 1 PROLOG
• part 2 Constraint Satisfaction
• CSP and CLP
• each part 2 theory and 2 practice lessons
• practice is done using the ECLIPSE compiler (you
  can download it)

6
Info/Warning

• Exercises are not obligatory, and can be done at
  home and/or in group
• You dont have to turn in the result
• BUT1 unless you are a genius, there is no way
  youll pass the exam if you havent done the
  exercises
• BUT2 programming in Prolog is totally different
  than programming in another language. Dont think
  you can learn it in a couple of days

7
First Lesson PROLOG (part 1)

• tutorial by Ulle Endriss chapter 1 and 2.
• available via the course page
• differences Endriss speaks about matching
  instead of unification.

8
Syntax Terms

• 3 kinds of terms
• constants
• numbers
• atoms
• variable
• compound terms

9
Numbers

• 3 kinds of terms
• constants
• numbers
• atoms
• variable
• compound terms

• In PROLOG follow the usual standard
• 0
• 1
• -33.75
• 12345E1

10
Atoms

• 3 kinds of terms
• constants
• numbers
• atoms
• variable
• compound terms

• IN PROLOG
• 3 possibilities
• start with a lowercase letter (might contain _
  and numbers)
• a2, a_2, pippo
• not a2, a-2, -a, _a
• between single quotes
• _asdkfa-23
• contain only signs
• --

11
Variables

• 3 kinds of terms
• constants
• numbers
• atoms
• variable
• compound terms

• IN PROLOG
• 2 possibilities
• start with an Uppercase letter
• A2, A_2, Pippo
• start with _
• _pippo, _12344
• might contain _ and numbers, just like atoms

12
Compound terms

• 3 kinds of terms
• constants
• numbers
• atoms
• variable
• compound terms

• functorarguments
• functor is an atom
• arguments are terms
• number of arguments arity
• f(X), (X,a,f(X,z))

13
Compound terms 2 operators

• 3 kinds of terms
• constants
• numbers
• atoms
• variable
• compound terms

• to simplify writing one can declare operators
• infix
• prefix
• suffix
• ab (a,b) if is declared as infix
• -c -(c), if is declared as prefix
• _xx yy yy(_xx) if yy is declared as suffix

14
Test

• 3 kinds of terms
• constants
• numbers
• atoms
• variable
• compound terms

• jan
• Jan
• -jan
• _jan
• jan-willem
• jan_willem
• _jan(willem(jan))
• jan(willem(_jan))
• (_willem(_jan))
• _willem _jan

15
Test

• jan
• atom
• Jan
• variable
• -jan
• if - is declared as prefix is equal to (jan)
  (compound), otherwise is wrong
• _jan
• variable
• jan-willem
• f - is declared as infix is equal to
  (jan,willem), otherwise is wrong

• jan_willem
• atom
• _jan(willem(jan))
• WRONG (why?)
• jan(willem(_jan))
• compound
• (_willem(_jan))
• WRONG (why?)
• _willem _jan
• if is declared as infix is equal to
  (_willem,_jan), otherwise is wrong.

16
Substitutions

• Substitution mapping from vars to terms
• ? X1/t1,...,Xn/tn
• Dom(?) X1,...,Xn and must be finite
• Ran(?) Var(t1,...,tn)
• Var(?) Dom(?) U Ran(?)
• e.g. ? X/f(Z)
• h(X,Z) ? h(f(Z),Z)
• renaming if t1,...,tn is a set of distinct
  variables
• ground, if t1,...,tn contain no variables

17
Unification

• Basic computing mechanism
• two terms t and s UNIFY if there exist a
  substitution ? such that t? s?.
• there exist an algorithm computing the MOST
  GENERAL UNIFIER
• ? is more general than ? if ? ? ? for some
  ?
• ? X/f(a),Z/a is a unifier of t(X) and
  t(f(Z)), but the most general unifier is ?
  X/f(Z)
• In the tutorial of Endriss, this is called
  matching

18
Unifying in practice

• In Prolog, t s is a unification test
• ?- f(X,a) f(b,Y).
• X b
• Y a
• Yes.
• ?- f(X,a) f(b,X).
• No.
• ?- f(X,a) f(b,Y). is a query. (?- might be
  omitted in Eclipse). the full stop at the end is
  important.

• X b Y a is the computed answer substitution
  (c.a.s.) i.e., the result of the query.
• a query can FAIL (second example).
• once a variable is instantiated to a value, it
  cannot be instantiated again
• ?- Xa, bX.
• No.
• In this case , gt and operaror

19
Test compute the c.a.s (if any)

• _x y.
• g(X) f(Y).
• f(Y,Y) f(a,Z).
• f(Y) f(a,Z).
• f(Y,Y) f(a,b).
• f(Z,Z) f(f(a,W),f(Y,b)).
• X g(Y).
• Y g(Y).

20
Test compute the c.a.s (if any)

• _x y.
• _x y is the cas.
• g(X) f(Y).
• fails two terms with a different functor are not
  unifiable
• f(Y,Y) f(a,Z).
• Y a, Z a
• f(Y) f(a,Z).
• fails two terms cannot be unified if they have
  different arity
• f(Y,Y) f(a,b).
• fails, a and b cannot unify

• f(Z,Z) f(f(a,W),f(Y,b)).
• Y a
• W b
• Z f(a,b)
• X g(Y)
• X g(Y)
• Y g(Y)
• this unification fails (at least in principle) Y
  cannot be unified with a term containing it (try
  to find the right unifier!). In practice, PROLOG
  interpreters dont carry out the so-called
  occur-check, and this would succeed with cas Y
  g(Y), i.e. Y g(g(g(g(g(g(...))))))

21
Facts

• Prolog programs contain FACTS and RULES
• Example of facts (well see the rules later)
• bigger(elephant, horse).
• bigger(horse, monkey).
• We can query them
• ?- bigger(elephant, horse).
• Yes
• ?- bigger(elephant, monkey).
• No
• The program doesnt know

22
Computed answer substitutions and multiple answers

• our program
• bigger(elephant, horse).
• bigger(horse, monkey).
• We can query them
• ?- bigger(X, horse).
• X elephant
• Yes

• even better
• ?- bigger(X, Y).
• X elephant
• Y horse
• More?
• X horse
• Y monkey
• peculiar of PROLOG
• More? yes, but the re might be other answers
• input from the user I want to see more
  answers (if any).

23
Complex Goals and Left-to-Right execution

• consider the program
• friend(a,b).
• friend(a,c).
• friend(c,d).
• friend(d,e).
• And the query
• ?- friend(X,Y), friend(Y,Z).
• What are going to be the results?

• Xa
• Yc
• Zd
• More?
• Xc
• Yd
• Ze
• Yes

24
Rules

• Recall
• bigger(elephant, horse).
• bigger(horse, monkey).
• ?- bigger(elephant, monkey).
• No
• The program doesnt know
• Consider now
• drinks(john,water).
• drinks(john, whyskey).
• drinks(anna,water).
• drinks(jan,X)- drinks(john,X).
• What could this mean?

• - IF
• drinks(jan,X) is the HEAD
• drinks(john,X) is the BODY
• BODIES could be more complex
• drinks(jan,X)-
• drinks(john,X), drinks(anna,X).
• What could this mean?
• HOMEWORK What happens to the queries
• ?- drinks(jan,X).
• ?- drinks(X,Y).
• ?- drinks(X,Y), drinks(Z,Y).

25
Execution Model

• given a query A1,...,An
• looks for a (renaming of a) clause H -
  B1,...,Bm. whose head unifies with A1
• ? mgu(A1,H).
• proceeds with the query (B1,...,Bm,A2,...,An) ?.
  (the RESOLVENT)
• if (B1,...,Bm,A2,...,An) ?. is empty, we are
  finished and the c.a.s. is ?
• Otherwise
• if (B1,...,Bm,A2,...,An) ? succeeds with c.a.s.
  ?, then A1,...,An succeeds with c.a.s. ??
• if (B1,...,Bm,A2,...,An) ? fails then the system
  goes back to () (last choice point) to see if it
  finds another clause G-C1,...,Ck whose head
  unifies with A1
• if so, it proceeds with (C1,...,Ck,A2,...,An)?,
  where ? mgu(A1,G). Etcetera, in a recursive
  way.
• if not, A1,...,An fails.

26
Example

• program
• a(1).
• a(X) - c(X).
• c(2).
• b(2).
• query
• a(Y), b(Y).

• a(Y),b(Y).
• resolve with (1), Y 1
• b(1).
• no clause whose head match this gt go back to
  the previous query
• a(Y),b(Y).
• resolve with (2) YZ
• c(Z),b(Z).
• resolve with (3) Z2
• b(2).
• succeeds
• c.a.s (YZ)(Z2) Y2.

27
Need for operators

• Think of the expressios 32 37
• In Prolog (((3,2),3),7).
• Fortunately we can define our own operators.
• See tutorial, chapter 4 to see how you can define
  extra operators.

28
Arithmetic expressions

• Try
• ?- 4 22.
• No.
• Why?
• 4 does not unify with (2,2).
• to evaluate an arithmetic expression, we use is

• ?- 4 is 22.
• Yes
• ?- X is 22.
• X 4
• Yes
• ?- 2 2 is 4.
• No
• is evaluates the rhs only of the expression
• Operators which perform expression evaluation
  before they are evaluated
• , gt, gt, lt, \

29
Lists

• nil the empty list
• a,b,c contains 3 elements
• other representations (equivalent)
• a.b.c.nil
• .(a, .(b, .(c,)))
• The first element is the HEAD
• the rest is the TAIL

30
The BAR

• may appear before a suffix of the list
• a,b,c
• ab,c
• a,bc
• a,b,c
• are the same list
• Used to split a list
• ?- a,b,c HT.
• H a
• T b,c
• Yes

• ?- a,b,c H1,H2T
• H1 a
• H2 b
• T c
• Yes
• ?- a,b H1,H2T
• H1 a
• H2 b
• T
• Yes
• ?- a H1,H2T
• No

31
a useful program

• m(El,El_).
• m(El,_T)-m(El,T).
• _ is the anonymous variable it behaves like a
  variable whose name is different than all other
  variables.

• What are answers to
• ?- m(a,a,b).
• ?- m(1,a,b).
• ?- m(X,a,b).
• ?- m(X,Y).

32
append

• append(Xs,Ys,XsYs) -
• XsYs is the result of concatenating the
• list Xs and Ys.
• append(,Ys,Ys).
• append(XXs,Ys,XZs) - append(Xs,Ys,Zs).