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We consider procedures which amount to performing a complete match between two structures

**Matching like Patterns**

We consider procedures which amount to performing a
complete match between two structures

The match will be accomplished by comparing the two
structures and testing for equality among the corresponding parts

Pattern variables will be used for instantiations
of some parts subject to restrictions Matching Substrings

A basic function required in many match algorithms
is to determine if a substring S_{2} consisting of m characters occurs
somewhere in a string S_{1} of n characters, m ≤ n

A direct approach to this problem is to compare the
two strings character-by-character, starting with the first characters of both
S_{1} and S_{2}

If any two characters disagree, the process is
repeated, starting with the second character of S_{1} and matching
again against S_{2} character-by-character until a match is found or
disagreements occurs again

This process continues until a match occurs or S_{1}
has no more characters

Let i and j be position indices for string S1 and a
k position index for S_{2}

We can perform the substring match with the
following algorithm

:= 0

While i ≤ (n – m + 1) do

begin

i := i + 1;

j := i;

k :=1;

while S_{1}(j) = S_{2}(k) do

begin

if k = m

writeln(“success”)

else do

begin

j := j +1;

k := k + 1;

end

end

end

writeln(“fail”)

end

This algorithm requires m(n - m) comparisons in the
worst case

A more efficient algorithm will not repeat the same
comparisons over and over again

One such algorithm uses two indices I and j, where
I indexes the character positions in S_{1} and j is set to a “match
state” value ranging from 0 to m

The state 0 corresponds to no matched characters
between the strings, while state 1 corresponds to the first letter in S_{2}
matching character i in S_{2}

State 2 corresponds to the first two consecutive
letters in S_{2} matching letters i and i+1 in S_{1}
respectively, and so on, with state m corresponding to a successful match

Whenever consecutive letters fail to match, the
state index is reduced accordingly Matching Graphs

Two graphs G_{1} and G_{2} match if
they have the same labeled nodes and same labeled arcs and all node-to-node arcs
are the same

If G_{2} with m nodes is a subgraph of G_{1}
with n nodes, where n ≥ m

In a worst cas match, this will require n!/(n - m)!
node comparison and O(m^{2}) arc comparisons

Finding subgraph isomorphisms is also an important
matching problem

An isomorphism between the graphs G_{1} and
G_{2} with vertices V_{1}, V_{2} and edges E1, E2, that
is, (V1, E1) and (V2, E2) respectively, is a one-to-one mapping to f between V1
and V2, such that for all v1 € V1, f(v1) = v2, and for each arc e1 € E1
connecting v1 and v1^{’} , there is a corresponding arc e2 € E2
connecting f(v1) and f(v1^{’})

Matching Sets and Bags

An exact match of two sets having the same number
of elements requires that their intersection also have the number of elements

Partial matches of two sets can also be determined
by taking their intersections

If the two sets have the same number of elements
and all elements are of equal importance, the degree of match can be the
proportion of the total members which match

If the number of elements differ between the sets,
the proportion of matched elements to the minimum of the total number of
members can be used as a measure of likeness

When the elements are not of equal importance,
weighting factors can be used to score the matched elements

For example, a measure such as

s(S1,S2) = (∑ w_{i}
N(a_{i}))/m

could be used, where w_{i} =1 and N(a_{i})
= 1 if a_{i} is in the intersection; otherwise it is 0

An efficient way to find the intersection of two
sets of symbolic elements in LISP is to work through one set marking each
element on the elements property list and then saving all elements from the
other list that have been marked

The resultant list of saved elements is the
required intersection

Matching two bags is similar to matching two sets
except that counts of the number of occurrences of each element must also be
made

For this, a count of the number of occurrences can
be used as the property mark for elements found in one set. This count can then
be used to compare against a count of elements found in the second set

Matching to Unify Literals

One of the best examples of nontrivial pattern
matching is in the unification of two FOPL literals

For example, to unify P(f(a,x), y ,y) and P(x, b,
z) we first rename variables so that the two predicates have no variables in
common

This can be done by replacing the x in the second
predicate with u to give P(u, b, z)

Compare the two symbol-by-symbol from left to right
until a disagreement is found

Disagreements can be between two different
variables, a nonvariable term and a variable, or two nonvariable terms. If no
disagreements is found, the two are identical and we have succeeded

If disagreements is found and both are nonvariable
terms, unification is impossible; so we have failed

If both are variables, one is replaced throughout
by the other. Finally, if disagreement is a variable and a nonvariable term,
the variable is replaced by the entire term

In this last step, replacement is possible only if
the term does not contain the variable that is being replaced. This matching
process is repeated until the two are unified or until a failure occurs

For the two predicates P, above, a disagreement is
first found between the term f(a,x) and variable u. Since f(a,x) does not
contain the variable u, we replace u with f(a,x) everywhere it occurs in the
literal

This gives a substitution set of {f(a,x)/u} and the
partially matched predicates P(f(a,x),y ,y) and P(f(a,x), b, z)

Proceeding with the match, we find the next
disagreements pair y and b, a variable and term, respectively

Again we replace the variable y with the terms b
and update the substitution list to get {f(a,x)/u, b/y}

The final disagreements pair is two variables.
Replacing the variable in the second literal with the first we get the
substitution set {f(a,x)/u, b/y, y/z}or {f(a,x), b/y, b/z}

For example, a LISP program which uses both the
open and the segment pattern matching variables to find a match between a
pattern and a clause

(defun match (pattern clause)

(cond ((equal pattern clause) t ) ; return t if

((or (null pattern) (null clause)) nil) ; equal, nil

; if not

((or (equal (car pattern) (car clause)) ;not, ?x

; binds

(equal (car pattern) ‘?x)) ; to single

(match (cdr pattern) (cdr clause))) ; term, *y

; binds

((equal (car pattern) ‘*y) ; to several

(or (match (cdr pattern) (cdr clause)) ;contiguous

(match pattern (cdr clause)))))) ; terms

When a segment variable is encountered (the *y),
match is recursively executed on the cdrs of both pattern and clause or on the
cdr of clause and pattern as *y matches one or more than one item respectively

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