assignment #2 due: feb. 22, 2010

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



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



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



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: ‘=’, ‘<’, ‘>’, ‘<>’, ‘≠’



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.



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.



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



<? 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>



(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)>

]>



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:complexType name = “addressType”><xs: sequence>

<xs: element name = “street” type = “xs: string” /></xs: element name = “city” type = “xs: string” />




<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>



<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>

assignment #2 due: feb. 22, 2010

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