CSC 270 – Survey of
Programming Languages
Prolog Lecture 2 – Unification and
Proof Search
Unification
• From Knowledge Base 4:
woman(mia).
woman(jody).
woman(yolanda).
…
• Querying Knowledge Base 4:
1 ?- woman(X).
X = mia .
2 ?-
• Prolog unifies woman(X) with woman(mia) , thereby
instantiating the variable X to mia
Types in Prolog – A Quick Review
• Recall that there are three types of term:
– Constants. These can either be atoms (such as
vincent ) or numbers (such as 24 ).
– Variables. (Such as X , Z3 , and List .)
– Complex terms. These have the form:
– functor(term_1,...,term_n) .
Unification – A Definition
• Two terms unify:
– they are the same term or
– if they contain variables that can be uniformly
instantiated with terms in such a way that the
resulting terms are equal.
Unification – An Example
• This means that
– mia and mia unify, because they are the same
atom.
– 42 and 42 unify, because they are the same
number.
– X and X unify, because they are the same variable.
– woman(mia) and woman(mia) unify, because they
are the same complex term.
Unification – An Example (continued)
• This means that
– woman(mia) and woman(vincent) do not unify,
because they are not the same (and neither of them
contains a variable that could be instantiated to
make them the same).
Unification and Instantiation
• mia and X are not the same, but X can be instantiated
to mia which makes them equal. Therefore, mia and
X unify.
• woman(X) and woman(mia) unify, because they can
be made equal by instantiating X to mia .
• loves(vincent,X) and loves(X,mia) do not unify
because it is impossible to find an instantiation of X
that makes the two terms equal.
A Formal Definition of Unification
• If term1 and term2 are constants, then term1
and term2 unify if and only if they are the
same atom, or the same number.
A Formal Definition of Unification (continued)
• If term1 is a variable and term2 is any type of term,
then term1 and term2 unify, and term1 is instantiated
to term2 .
• If term2 is a variable and term1 is any type of term,
then term1 and term2 unify, and term2 is instantiated
to term1 .
• If they are both variables, they’re both instantiated to
each other, and we say that they share values.
A Formal Definition of Unification (continued)
• If term1 and term2 are complex terms, then
they unify if and only if:
– They have the same functor and arity, and
– all their corresponding arguments unify, and
– the variable instantiations are compatible.
A Formal Definition of Unification (continued)
• E. g., it’s impossible to instantiate variable X to
mia when unifying one pair of arguments, and
to instantiate X to vincent when unifying
another pair of arguments .)
• Two terms unify if and only if it follows from
the previous three clauses that they unify.
Unification – Some Examples
1 ?- =(mia, mia).
true.
2 ?- =(2, 2).
true.
3 ?- mia = mia.
true.
4 ?- 2 = 2.
true.
5 ?- mia = vincent.
false.
6 ?- 'mia' = mia.
true.
7 ?- '2' = 2.
false.
8 ?-
8 ?- mia = X.
X = mia.
9 ?- mia = X.
X = mia.
10 ?- X = Y.
X = Y.
11 ?- X = _5067.
X = _5067.
12 ?- Y = _5067.
Y = _5067.
13 ?- X = mia, X = vincent.
false.
Unification – Another Example
2 ?- k(s(g), Y) = k(X, t(k)) .
Y = t(k),
X = s(g).
3 ?-
• We use clause 3 because we are trying to unify two complex
terms. Do both complex terms have the same functor and
arity? Yes.
• So do the first arguments, s(g) and X , unify? By clause 2, yes.
• So do the second arguments, Y and t(k) , unify? By clause 2,
yes.
Unification – Other Examples
3 ?- k(s(g), t(k)) = k(X, t(Y)).
X = s(g),
Y = k.
4 ?- loves(X, X) = loves(marcellus, mia).
false.
5 ?-
• These terms do not unify because while they are both complex
terms and have the same functor and arity, all corresponding
arguments do not unify in this example
Prolog and Unification Algorithm
• Prolog does not use a standard unification
algorithm when it performs its version of
unification; it takes a shortcut. You need to
know about this shortcut.
• Consider the following query:
?-
father(X)
=
X.
Prolog and Unification Algorithm (continued)
• Do these terms unify or not? No
• Why is that? Pick any term and instantiate X to
the term you picked.
father(father(father(butch)))
= father(father(butch))
and so on.
Prolog and Unification Algorithm (continued)
• Prolog unification won’t spot the problem and
halt. Having instantiated X to father(X) ,
Prolog is committed to carrying out an
unending sequence of expansions and will run
until it runs out of memory.
Prolog and Unification Algorithm (continued)
• There are actually 3 responses to “does
father(X) unify with X ”.
– Standard unification algorithm (no)
– Older Prolog implementations (run amok until they
use up the available memory)
– Sophisticated Prolog implementations (yes)
• In short, there is no ‘right’ answer to this
question.
occurs check
• A standard algorithm given two terms to unify
carries out what is known as the occurs check.
• This means that if it is asked to unify a variable
with a term, it first checks whether the variable
occurs in the term.
• If it does, the standard algorithm declares that
unification is impossible, for clearly it is the
presence of the variable X in father(X) which
leads to problems.
• Only if the variable does not occur in the term do
standard algorithms attempt to carry out the
unification.
Prolog and Unification Algorithm (continued)
• Prolog comes with a built-in predicate that
carries out standard unification (that is,
unification with the occurs check). The
predicate is
unify_with_occurs_check/2.
• So if we posed the query
?- unify_with_occurs_check
(father(X),X).
we would get the response no.
Unification Proof Search
f(a).
f(b).
g(a).
g(b).
h(b).
k(X):- f(X),g(X),h(X).
• try to match f (_1), g(_1), h(_1)
• When f(a), then _1 = a.
• Does it find f(a), g(a),h(a) – nope
• When f(b) then _1 = b.
• Does it find f(b), g(b), h(b) – yes so it stops and says true.
• See with trace
• Stop trace with notrace
Crossword puzzle match
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word(astante,a,s,t,a,n,t,e).
word(astoria,a,s,t,o,r,i,a).
word(baratto,b,a,r,a,t,t,o).
word(cobalto,c,o,b,a,l,t,o).
word(pistola,p,i,s,t,o,l,a).
word(statale,s,t,a,t,a,l,e).
crossword(X,Y, Z, A, B, C) :word(X,_,X1,_,X2,_,X3,_),
word(Y,_,Y1,_,Y2,_,Y3,_),
word(Z,_,Z1,_,Z2,_,Z3,_),
word(C,_,X3,_,Y3,_,Z3,_),
word(B,_,X2,_,Y2,_,Z2,_),
word(A,_,X1,_,Y1,_,Z1,_).
Summary
• Prolog proves by instantiating variables and
then checking truth
• Stops at first complete matching truth
• Will continue finding match when you press ;
• Trace using trace / notrace
• Can go into infinite checking unless you call
with unify_with_occurs_check