1. Managing XML and Semistructured Data
Lecture 4: Path Expressions
Prof. Dan Suciu
Spring 2001

2. In this lecture Path expressions Regular path expressions
Evaluation techniques
Resources:
Data on the Web Abiteboul, Buneman, Suciu : section 4.1

3. Path Expressions Examples: Bib.paper Bib.book.publisher
Bib.paper.author.lastname
Given an OEM instance, the answer of a path expression p is a set of objects

4. Path Expressions Examples: DB = Bib.paper={&o12,&o29}
&96
&243
&206
&25
“Serge”
“Abiteboul”
1997
“Victor”
“Vianu”
122
133
paper
book
references
author
title
year
http
publisher
page
firstname
lastname
first
last
Bib
&o44
&o45
&o46
&o47
&o48
&o49
&o50
&o51
&o52
Examples:
DB =
Bib.paper={&o12,&o29}
Bib.book.publisher={&o51}
Bib.paper.author.lastname={&o71,&206}

5. Answer of a Path Expression
Simple evaluation algorithms for Answer(P,DB):
Runs in PTIME in size(P), size(db):
PTIME complexity
Answer(P, DB) = f(P, root(DB))
Where:
f(e, x) = {x}
f(L.P, x) = {f(P,y) | (x,L,y) edges(DB)}

6. Regular Path Expressions
R ::= label | _ | R.R | (R|R) | R* | R+ | R?
Examples:
Bib.(paper|book).author
Bib.book.author.lastname?
Bib.book.(references)*.author
Bib.(_)*.zip

7. Applications of Regular Path Expressions
Navigating uncertain structure:
Bib.book.author.lastname?
Syntactic substitute for inheritance:
Bib.(paper|book).author
Better: Bib.publication.author, but we don’t have inheritance

8. Applications of Regular Path Expressions
Computing transitive closure:
Bib.(_)*.zip
= everything accessible
Bib.book.(references)*.author
= everything accessible via references
Some regular expressions of doubtful practical use:
(references.references)*
= a path with an even number of references
(_._)* = paths of even length
(_._._.(_)?)* = paths of length (3m + 4n) for some m,n
But make great examples for illustration 

9. Answer of a Regular Path Expression
Recall:
Lang(R) = the set of words P generated by R
Answer of regular path expressions:
Answer(R,DB) = {Answer(P,DB) | P  Lang(R)}
Need an evaluation algorithm that copes with cycles

10. Regular Path Expressions
Recall: each regular expression  NDFA
Example: R = (a.a)*.a.b
A = a
states(A) = {s1,s2,s3,s4}
initial(A) = s1
terminal(A) = {s4} s1 s2 a a b s3 s4

11. Regular Path Expressions
Canonical Evaluation Algorithm
Answer(R,DB):
construct A from R
construct product automaton G = A x DB:
nodes(G) = states(A) x nodes(db)
edges(G) = {((s,x),L,(s’,x’) | (s,L,s’)  edges(A), (x,L,x’)  edges(DB)}
root(G) = (initial(A), root(DB))
compute Gacc = set of nodes accessible from root(G)
return {x | s  terminal(A) s.t. (s,x)  Gacc}

12. Regular Path Expressions
Example: R = _.(_._)*.a
A = DB = s1 s2 s3 _ a
&o1
&o2
&o3
&o4 a b
Answer of R on DB = { &o2, &o3}

13. Compute Product Automaton G_ a _
s1,&o1
s2,&o1
s3,&o1 a a a
s1,&o2
s2,&o2
s3,&o2 a a a a a a
s1,&o3
s1,&o4
s2,&o3
s2,&o4
s3,&o3
s3,&o4 b b b

14. Compute Accessible Part Gacc_ a _
s1,&o1
s2,&o1
s3,&o1 a a a
s1,&o2
s2,&o2
s3,&o2 a a a a a a
s1,&o3
s1,&o4
s2,&o3
s2,&o4
s3,&o3
s3,&o4 b b b
Answer(R,DB) = {&o2, &o3}

15. Complexity of Regular Path Expressions
The evaluation algorithm runs in PTIME in size(R), size(DB)
Even when there are cycles in DB