assignment #2 due: feb. 22, 2010
Post on 30-Dec-2015
32 Views
Preview:
DESCRIPTION
TRANSCRIPT
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 1
Assignment #2
Due: Feb. 22, 2010
1. Apply the following algorithm to the B+-tree shown in Fig. 4 to store it in a data file. Trace the computation process.
Algorithm:
push(root, -1, -1);while (S is not empty) do{ x := pop( );
store x.data in file F;assume that the address of x in F is ad;if x.address-of-parent -1 then {
y := x.address-of-parent;z := x.position;write ad in page y at position z in F;
}let x1, …, xk be the children of v;for (i = k to 1) {push(xi, ad, i)};
}
641 3
53 5
Fig. 4
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 2
data address-of-parent
position
key values
S:
3, 5 -1 -1
3 50
B+-tree file
6 0 3
4 0 2
1, 3 0 1
S:
1 3 50
B+-tree file
6 0 3
4 0 2
S:
1 1 3
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 3
data address-of-parent
position
key values
S:
1 3 2 50
B+-tree file
6 0 3
S:
1 1 3
4 2
1 3 2 50
B+-tree file S:
1 1 3
4 2
6 3
3
empty
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 4
2. Write an algorithm to transform a simplified XPath expression (in which the subexpressions in any condition can be connected only with , e.g., /StarMovieData/Star[//City = “Malibu” and //Street = “123 Maple St.”]/Name) to a tree structure.
/StarMovieData/Star[//City = “Malibu”]/Name
/a/b[//c = “Malibu” and //d = 5]/e
Assume that an XPath is stored in a character array A:
/ a / b [ / / c “ M a l i b “ a n d …
When we scan A, we will meet the following different cases:‘/’ followed by a character,‘//’,‘[’Comparison symbols: ‘=’, ‘<’, ‘>’, ‘<>’, ‘≠’
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 5
Before we scan A, a virtual root r is created.
Algorithm tree-generation(A):
Scan array A;If A[i] is ‘/’ followed by a character, create a nodeP labeled with the string A[i+1..j] such that A[j+1]is ‘/’,‘[‘, or a comparison symbol. i := j + 1.Make p a /-child of r. r := p.
If A[i..i+1] = ‘//’, create a nodeP labeled with the string A[i+2..j] such that A[j+1]is ‘/’,‘[‘, or a comparison symbol. i := j + 1.Make p a //-child of r. r := p.
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 6
If A[i] is a comparison symbol, create a node p forthe value. Make p a child of r. Label the edge withthe symbol.
If A[i] = ‘[’, recognize the conditions C between the‘[‘ and the matching ‘]’ following A[i]. Assume that itis A[j]. Let C = path1 and path2 and … and pathk.Do the following:
for a = 1 to k do{ call tree-generation(patha);
assume that Ta be the subtree created by therecursive call; make it a subtree of r;
} i := j + 1.
If i = n + 1, remove the virtual node and return thetree.
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 7
3. (a) Represent the following graph as an XML document.
sw
book
title year
CarrieFisher street city street city
Maple H’wood Locust Malibu
MarkHamill Oak B’wood 1977
cf mh
root
author
name nameaddress
addressstreet
city
writing
authorOf
writingauthorOf
StarWar
ad1 ad2
author
The MITPress
publisher
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 8
<? Xml version = “1.0” encoding = “utf-8” standalone = “yes” ?><StarMovieData>
<Star starID = “cf” starredIn = “sw”><Name>Carrie Fishes</Name><Address>
<Street>123 Maple St.</Street><City>Hollywood</City></Address><Address>
<Street>5 Locust Ln.</Street><City>Malibu</City><Address>
</Star><Star starID = “mh” starredIn = “sw”>
<Name>Mark Hamill</Name><Street>456 Oak Rd.</Street><City>Brentwood</City>
</Star><Movie movieID = “sw” starOf = “cf mh”>
<Title>Star Wars</title><Year>1977</Year></Movie>
</StarMovieData>
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 9
(b) The following is a DTD for books. Please produce an XML document conforming to the DTD.
<!DOCTYPE AuthorBook [<!ELEMENT AuthorBook (Author*, Book*)><!ELEMENT Author (Name, Address+)>
<!ATTLIST AuthorAuthorId ID #REQUIREDwriting INREFS #IMPLIED
><!ELEMENT Name (#PCDATA)><!ELEMENT Address (Street, City)><!ELEMNT Street (#PCDATA)><!ELEMENT City (#PCDATA)><!ELEMENT Book (Title, Publisher, Year)>
<!ATTLIST BookBookIn ID #REQUIREDauthorOf IDREFS #REQUIRED
><!ELEMENT Title (#PCDATA)> <!ELEMENT Publisher (#PCDATA)><!ELEMENT Year (#PCDATA)>
]>
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 10
4. Define an XML-schema which is equivalent to the DTD shown in Question 3(b).
<? Xml version = “1.0” encoding = “utf-8” ?><xs: schema xmlns: xs = “http://www.w3.org/2001/XMLSchema”>
<xs:complexType name = “authorType”><xs: sequence>
<xs: attribute name = “AuthorID” type = “xs: string” /></xs: attribute name = “Writing” type = “xs: string” />
</xs: sequence></xs: complexType>
<xs:complexType name = “bookType”><xs: sequence>
<xs: attribute name = “BookID” type = “xs: string” /></xs: attribute name = “authorOf” type = “xs: string” />
</xs: sequence></xs: complexType>
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 11
<xs:complexType name = “addressType”><xs: sequence>
<xs: element name = “street” type = “xs: string” /></xs: element name = “city” type = “xs: string” />
</xs: sequence></xs: complexType>
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 12
<xs: element name = “AuthorBook”><xs: complexTyp>
<xs: sequence><xs: element name = “Author” type = “authorType”
minOccurs = “0” maxOcurs = “unbouned” /><xs: complexType>
<xs: sequence><xs: element name = “name” type = “xs: string” /></xs: element name = “Address” type = “addressType”minOccurs = “1” maxOcurs = “unbouned” />
</xs:sequence></xs: complexType>
</xs:element>
Analysis of Midterm-Examination
Jan. 2012 ACS-7102 Yangjun Chen 13
<xs: element name = “Book” type = “bookType”minOccurs = “0” maxOcurs = “unbouned” /><xs: complexType>
<xs: sequence><xs: element name = “title” type = “xs: string” /><xs: element name = “Publisher” type = “xs: string” /><xs: element name = “year” type = “xs: string” />
</xs:sequence></xs: complexType>
</xs:element></xs: sequence>
</xs: complexTyp></xs: element>
</xs: schema>
top related