Slide 8-1 Copyright © 2003 Pearson Education, Inc. Overzicht Informatica College 10 – Oktober 31 Computer Science an overview EDITION 7 / 8 J. Glenn Brookshear.

Slide 8-1 Copyright © 2003 Pearson Education, Inc.

Overzicht Informatica

College 10 – Oktober 31

Computer Sciencean overview

EDITION 7 / 8J. Glenn Brookshear


C H A P T E R 8 (was Chap. 7)

Data Structures

• Abstractions of the actual data organization in main memory

• Allow users to perceive data as ‘logical units’ (e.g.: arrangement in rows and columns)


Data Structure Basics: Pointers

• Pointers:– pointer = location in memory that contains the

address of another location in memory– so: pointer points to data positioned elsewhere in

memory

F02A

F02A

FF8C

FF8C

64B0

64B0


Static versus Dynamic Data Structures

• Static:– shape & size of structure does not change over time– example in C: int Table[2][9];

Table:

65

32• Dynamic:

– shape & size may change– example: Stack

48

97

17

Stack:


Arrays

• Example: to store 24 hourly temperature readings…• … a convenient storage structure is 1-D homogeneous

array of 24 elements (e.g. in C: float Readings[24] )• In main memory:

… x + 23

Address of Readings[i] = x + (i-1)


Two-dimensional Arrays

• Sometimes 2-D homogeneous arrays are more useful (e.g. in C: float Table[4][5] )

=> Address of Table[i][j] = x + (nr_of_columns × (i-1)) + (j - 1)=> Example: address of Table[3][4] = x + 13


Opdracht:

Suppose an array with 6 rows and 8 columns is stored starting at address 20. If each entry in the array requires only one memory cell, what is the address of the entry in the 3rd row and 4th column? What if each entry requires two cells?

• If 1 cell per entry:– address at [3, 4] : 20 + 8 × (3-1) + (4-1) = 39.

• If 2 cells per entry:– address at [3, 4] : 20 + 2 × (8 × (3-1) + (4-1)) = 58.


Opdracht:

Describe a method for storing 3-D homogeneous arrays. What addressing formula would be used to locate the entry in the i-th plane, the j-th row, and the k-th column?

• 3D: place each 2D plane consecutively in memory

• Addressing formula (with x as start address):– x + r × c × (i - 1) + c × (j - 1) + (k - 1)

• 2D:


Lists

• To store an ordered list of names we could use 2-D homogeneous array (in C: char Names[10][8])

• However:– addition & removal of names requires expensive data

movements!


Linked Lists

• Data movements can be avoided by using a ‘linked list’, including pointers to list entries


Deleting an Entry from a Linked List

• A list entry is removed by changing a single pointer:


Inserting an Entry into a Linked List

• A new entry is inserted by setting pointer of– (1) new entry to address of entry that is to follow– (2) preceding entry to address of new entry:


Previous

Opdracht:

Which of the following routines correctly inserts 'NewEntry' immediately after the entry called 'PreviousEntry' in a linked list?Routine 11. Copy pointer field of 'PreviousEntry' into the pointer field of 'NewEntry'.2. Change pointer field of 'PreviousEntry' to the address of 'NewEntry'.

Routine 21. Change pointer field of 'PreviousEntry' to the address of 'NewEntry'.2. Copy pointer field of 'PreviousEntry' into the pointer field of 'NewEntry'.

(1)

(2)

=> routine 1 is correct


Stacks

• Disadvantage of contiguous array structures:– insertion / removal requires costly data movements

• Still okay if insertion / removal restricted to end of array:– stack (with push & pop operations)


A Stack in Memory

• Here:– conceptual structure close to identical to actual

structure in memory

• If maximum stack-size unknown:– pointers can be used ( => conceptual = actual structure )


Queues

• List where insertions take place at one end, and deletions at the other: ‘queue’ – To serve ‘objects’ in the order of their arrival

(waiting-line)


Queue “crawling” (1)

• Problem with queue shown so far:– queue moves downward in memory, destroying any

other data in its path:


Queue “crawling” (2)

• Can be overcome by:– circular movement of insertions / deletions through

pre-designated area of memory:

Conceptual view of circular queue


Opdracht:

Describe a data structure suitable for representing a board configuration during a chess game

• Simplest:– 8×8 homogeneous array, where each entry contains one of the

values {empty, king, queen, bishop, knight, rook, pawn}

• Other:– 2×16 homogeneous array: 1st dimension used to distinguish

between black / white; 2nd to enumerate remaining pieces of one color, incl. board position. To save memory space this 2nd dimension could be implemented as a linked list.

• Many, many more possibilities


Chapter 8 - Data Structures: Conclusions

• Pointers:– basic aid in definition of dynamic data structures

• Often used data structures:– Arrays (multi-dimensional)– Lists (contiguous & linked)– Stacks– Queues (crawling & circular)– Trees (not discussed…)– …


‘C H A P T E R’ 9.5 (was chap. 8)

File Structures

• Abstractions of the actual data organization on mass storage (hard disks, tapes, cd’s…)

• Again: differences between conceptual and actual data organization


Files, Directories & the Operating System

• OS storage structure:– conceptual hierarchy of directories and files

directory tree

files


Files: Conceptual vs. Actual View

• View at OS-level is conceptual– actual storage may differ significantly!


Text Files

• Sequential file consisting of long string of encoded characters (e.g. ASCII-code)– But: character-string still interpreted by word processor!

Same file in “MS Word”File in “Notepad”


From actual storage to conceptual view

sequential view

Interpretation by Application Program

Assembly by Operating System

actual storage

conceptual view

Sequential buffer


Quick File Access

• Disadvantage of sequential files:– no quick access to particular file data

• Two techniques to overcome this problem:– (1) Indexing or (2) Hashing

keys

12N67 John Smith 23-Jul-71 17,000.00 New York …13C08 Andrew White 27-Jun-70 24,500.00 Boston …23G19 Mary Jackson 5-Mar-39 41,000.00 San Francisco …24X17 Eleanor Tracy 17-Sep-63 9,635.00 Fort Lauderdale …26X28 Michael Flanagan 1-Nov-44 18,800.00 Washington …32E76 Glenn White 29-Feb-68 17,000.00 Detroit …36Z05 Virginia Moore 27-Jun-70 32,000.00 San Francisco …

: : : : : …: : : : : …: : : : : …

• Indexing:Indexed File Index

12N67 location13C08 location23G19 location24X17 location26X28 location32E76 location36Z05 location

: :: :: :

loaded into mainmemory when opened


Hashing

• Disadvantage of indexing is… the index– requires extra space + includes 1 extra indirection

• Solution: ‘hashing’– finds position in file using a key value (as in indexing)…

– … simply by identifying location directly from the key

• How?– define set of ‘buckets’

& ‘hash function’ that converts keys to bucket numbers …

key value

bucket number

0 1 2 3 … N

hash function


Hash Function: Example

• If storage space divided into 40 buckets and hash function is division:– key values 14, 54, & 94 all map onto same bucket

(collision)

Key values


Key field value can be anything


Handling Bucket Overflow

• When bucket-sizes are fixed:– buckets can fill up and overflow

• One solution:– designate special overflow storage area

not fixed in size!


Opdracht:

If we use division as a hash function and have 23 buckets, in which bucket should we search to find the record whose key is interpreted as the integer value 101?

…

101

bucket number: 9

0 1 2 … 9 … 23

Division: 101 / 23 = 4, remainder 9

…


Opdracht:

a) What advantage does an indexed file have over a hash file?b) What advantage does a hash file have over an indexed file?

• a) When key unique: index directly points to required data, while hashing oftens require an additional (sequential) bucket search (incl. bucket overflow).

• b) No additional index file storage is required.


Chapter 9.5 - File Structures: Conclusions

• File Structures:– abstractions of actual data organization on mass

storage

• Changes of ‘view’:– actual storage -> sequential view by OS ->

conceptual view presented to user

• Quick access to particular file data by– (1) indexing– (2) hashing (requires no index, but requires bucket search!)


C H A P T E R 9

Database Structures

• (Large) integrated collections of data that can be accessed quickly

• Combination of data structures and file structures


Historical Perspective

• Originally: departments of large organizations stored all data separately in flat files

• Problems: redundancy & inconsistencies


Integrated Database System

• Better approach: integrate all data in a single system, to be accessed by all departments


Disadvantages of Data Integration

• Disadvantages:– Control of access to sensitive data?!

• Bijvoorbeeld: personeelszaken heeft niets te maken met persoonlijke gegevens opgeslagen door de bedrijfsarts!

– Misinterpretation of integrated data• Supermarkt-database zegt dat een klant veel medicijnen

koopt. Wat betekent dit? Wat als deze klant solliciteert op een baan bij de supermarkt-keten?

– What about the right to hold/collect/interpret data?• Heeft een credit card company het recht gegevens over

koopgedrag van personen te gebruiken/verkopen?


Conceptual Database Layers

OperatingSystem

Actual datastorage

Data seen interms of asequential view

• Compare:


The Relational Model

• Relational Model– shows data as being stored in rectangular tables,

called relations, e.g.:

– row in a relation is called ‘tuple’– column in a relation is called ‘attribute’


Issues of Relational Design

• So, relations make up a relational database… • … but this is not so straightforward:

• Problem: more than one concept combined in single relation


Redesign by extraction of 3 concepts

Any information obtained

by combining information

from multiple relations


Example:

• Finding all departments in which employee 23Y34 has worked:


Relational Operations

• Extracting information from a relational database by way of relational operations– Most important ones:

• (1) extract tuples (rows) : SELECT

• (2) extract attributes (columns) : PROJECT

• (3) combine relations : JOIN

• Such operations on relations produce other relations– so: they can be used in combination, to create

complex database requests (or ‘queries’)


The SELECT operation


The PROJECT operation


The JOIN operation


Opdracht:

• RESULT := PROJECT W from X

X relationU V W

A Z 5B D 3C Q 5

Y relation R S

3 J 4 K

RESULTX.U X.V X.W Y.R Y.S

A Z 5 3 J A Z 5 4 K C Q 5 3 J C Q 5 4 K

SELECT from X where W=5PROJECT S from Y JOIN X and Y where X.W > Y.R


Opdracht:

PART relation

PartName Weight

Bolt 2X 1 Bolt 2Z 1.5 Nut V5 0.5

• a) Which companies make Bolt 2Z?

– NEW := SELECT from MANUFACTURER where PartName = Bolt2Z

– RESULT := PROJECT CompanyName from NEW

MANUFACTURER relation

CompanyName PartName Cost

Company X Bolt 2Z .03 Company X Nut V5 .01 Company Y Bolt 2X .02 Company Y Nut V5 .01 Company Y Bolt 2Z .04 Company Z Nut V5 .01


Opdracht:

PART relation

PartName Weight


• b) Obtain a list of the parts (+cost) made by Company X?

– NEW := SELECT from MANU’ER where CompanyName=CompanyX

– RESULT := PROJECT PartName, Cost from NEW





Opdracht:

PART relation

PartName Weight


• c) Which companies make a part with weight 1?

– NEW1 := JOIN MANUCTURER and PART where MANUFACTURER.PartName = PART.PartName

– NEW2 := SELECT from NEW1 where PART.Weight = 1

– RESULT := PROJECT MANU’ER.CompanyName from NEW2





Opdracht:

PART relation

PartName Weight





• c) Which companies make a part with weight 1?

– NEW1 := SELECT from PART where Weight = 1

– NEW2 := JOIN MANUCTURER and NEW1 where MANUFACTURER.PartName = NEW1.PartName

– RESULT := PROJECT MANU’ER.CompanyName from NEW2


Chapter 9 - Database Structures: Conclusions

• Database Structures:– (large) integrated collections of data that can be

accessed quickly

• Database Management System– provides high-level view of actual data storage

(database model)

• Relational Model most often used– relational operations: SELECT, PROJECT, JOIN, …

– high-level language for database access: SQL


Overzicht Informatica – Tentamen (1)

• Most important sections (editie 8) & keywords:

– Ch. 0 - 1, 3, 4: abstractie / algoritme

– Ch. 1 - 1, 2, 4, 5, 6, 7: bits / data opslag & representatie (ASCII, etc) / Boolse operaties / flipflops / geheugen-vormen en -karakteristieken / getalstelsels (binair, hexadecimaal, etc…) / overflow & truncation errors

– Ch. 2 - 1, 2, 3, 4, 6: cpu architectuur / machine language & instructions / programma executie / machine cycle / alternatieve architecturen

– Ch. 3 - 1, 2, 3, 4: operating systems / batch processing / time-sharing / multitasking / OS componenten / process vs. programma / competition

– Ch. 4 - 1, 2, 3, 4: network topologies / bridges / routers / client-server / the internet / world wide web / network protocols / the grid




– Ch. 5 - 1, 2, 4, 5, 6: algoritme (formeel) / primitiven / pseudo-code / syntax / semantiek / iteratie / loop control / recursie / efficientie

– Ch. 6 - 1, 2, 3, 4, 5: generaties: 1e, 2e, 3e / assembly language / compilers / machine independence / paradigma’s / imperatief / object-georienteerd / programming concepts / procedures / parameters / call by value/reference translation/compilation process

– Ch. 7 - 1, 2, 3: software life cycle / ontwikkelings-fase / modulariteit / koppeling / cohesie / documentatie / complexiteits-maat voor software

– Ch. 8 - 1, 2: datastructuren / abstractie / statisch vs. dynamisch / pointers / (arrays, lists, stacks, queues, etc…)




– Ch. 9 - 1, 2, 5: databases vs. ‘platte’ files / relaties / tuples / attributen / relationele operaties: SELECT, PROJECT, JOIN / files / sequential / tekst / indexed / hashing

– Ch. 10 – 1, 3, 4: intelligent agents / Turing-test / production systems / search trees / heuristics / artificial neural networks / training vs test

– Ch. 11 – 1, 2, 4: computability / Turing Machines / the ‘halting’ problem

Veel succes!

Slide 8-1 Copyright © 2003 Pearson Education, Inc. Overzicht Informatica College 10 – Oktober 31 Computer Science an overview EDITION 7 / 8 J. Glenn Brookshear.

Documents

line slide

correct slide

pearson education

actual structure slide

glenn brookshear slide

list entry

memory f02a ff8c 64b0

new entry