Chapter 1 o level computer science with past paper

Number
System
CHAPTER 1.1
0101110?
AE73F9?

Why Binary System
• Any form of information needs to be converted into a binary format so that it can
be processed by a computer
On = 1 Off = 0
Motivation
• Computer contains millions and millions of tiny switches, which can be turned on
and off.
• Therefore, the binary system is chosen as the way for a computer to represent any
sort of data.

Eg. 365
102
=100
3 6 5
Denary System
(3x100) + (6x10) + (5x1) = 365
Explanation
101
=10 100
=1
Explanation: Multiply the digit value (eg.3) by the place value (eg. 100)

Explanation
Denary
Binary
Binary
Denary

1 1 1
(1x4)
Explanation
Denary
Binary
22
=4 21
=2 20
=1
+ (1x2) + (1x1) = 7 in denary

Eg. "1011"
0 1 1
1
Denary
Binary
22
=4 21
=2 20
=1
23
=8
Explanation
(1x8) + (0x4) + (1x2) = 11 in denary
+ (1x1)
Explanation: Multiply the digit value (eg.1) by the place value (eg. 8).
Then sum it all up!

DIY
1 1 0
1 0
What is the denary form of "11100"?
Denary
Binary
22
=4 21
=2 20
=1
23
=8
24
=16

DIY
1 1 0
1 0
ANSWER
Denary
Binary
22
=4 21
=2 20
=1
23
=8
24
=16
(1x16) + (1x8) + (1x4) + (0x2) + (0x1)
= 28 in denary

5
2
2
2
1
1
0
Denary Binary
Convert 5 to binary:
5
Explanation
remainder
remainder
2 0 remainder
1
Read the remainder
from bottom to top
Answer: 101

39
2
2
19
9
1
1
Denary Binary
39
Explanation
remainder
remainder
2 4 remainder
1 Read the remainder
from bottom to top
Answer: 100111
2 2 remainder
0
2 1 remainder
0
2 0 remainder
1

DIY
What is the binary form of 42?
Denary Binary
22
=4 21
=2 20
=1
23
=8
24
=16
25
=32

DIY
ANSWER
42
2
2
21
10
0
1
42
remainder
remainder
2 5 remainder
from bottom to top
2 2 remainder
1
2 1 remainder
0
2 0 remainder
1
Answer: 101010
Denary Binary

Denary System
7 6 5
102 101 100
(7x100) + (6x10)
+ (5x1) = 765
Ones
Tenth
Hundredth
RECAP

Denary System
7 6 5
102 101
100
Binary System
1 1 1
22 21
20
(7x100) + (6x10)
+ (5x1) = 765
(1x4) + (1x2)
+ (1x1) = 7
RECAP

DIY
1 0 1 0
What is the denary form of "1010"?
Denary
Binary
22
=4 21
=2 20
=1
23
=8

DIY
1 0 1 0
ANSWER
Denary
Binary
22
=4 21
=2 20
=1
23
=8
(1x8) + (1x2) = 10 in denary

DIY
What is the binary form of 38?
Denary Binary
22
=4 21
=2 20
=1
23
=8
24
=16
25
=32

DIY
ANSWER
38
2
2
19
9
0
1
38
remainder
remainder
2 4 remainder
from bottom to top
2 2 remainder
0
2 1 remainder
0
2 0 remainder
1
Answer: 100110
Denary Binary (Method 2)

Hexadecimal System
Motivation
• It is a base 16 system.
• It uses 16 digits to represent each value
Number System Digits used to represent each value
Denary
Binary
Hexadecimal
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
0, 1
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 , A, B, C, D, E, F

Hexadecimal System
Explanation
3 6 5
(3x100) + (6x10) + (5x1) = 365
1 0 1
(1x4) + (0x2) + (1x1) = 5
Denary System Binary System
Hexadecimal System
3 E 5
22
=4 21
=2 20
=1
102
=100 101
=10 100
=1
?2
=? ?1
=? ?0
=?

Hexadecimal System
3 6 5
(3x100) + (6x10) + (5x1) = 365
1 0 1
(1x4) + (0x2) + (1x1) = 5
Denary System Binary System
Hexadecimal System
3 E 5
Explanation
162
=256 161
=16 160
=1
22
=4 21
=2 20
=1
102
=100 101
=10 100
=1

Binary
Binary
Hexadecimal
Hexadecimal
Conversion

Binary Hexadecimal Explanation
• Since 16 = 24
this means that FOUR binary digits are equivalent to each hexadecimal
digit.

1 0 1 1 1 1 1 0 0 0 0 1
1011 111 0 00 01
1
E
B
ANSWER : BE1

1 0 0 0 0 1 1 1 1 1 1 1 0 1
10
D
F
1
ANSWER : 21FD
0001 1111 1101
00
2

DIY
What is the hexadecimal form of
0111010011100?
Binary Hexadecimal DIY

DIY
ANSWER
Binary Hexadecimal DIY
0111010011100
1100
1001
1110
0000
C
9
E
0

Binary
Hexadecimal Explanation
F 9 3 5
1111
1001
0011 0101
Answer: 1111 1001 0011 0101

DIY
What is the binary form of
BF08?
Binary
Hexadecimal DIY

B F 0 8
1111
0000 1000
Answer:
DIY
Binary
Hexadecimal DIY
1011
1011 1111 0000 1000

Denary
Denary
Hexadecimal
Hexadecimal
Conversion

Eg. "111"
1 1 1
(1x4) + (1x2) + (1x1) = 7 in denary
RECAP
Denary
Binary
22
=4 21
=2 20
=1

Denary
Eg. "45A"
4 5 A
(4x256) + (5x16) + (10x1) = 1114 in denary
Note: A=10
162
=256 161
=16 160
=1

Denary
Eg. "C8F"
C 8 F
(12x256) + (8x16) + (15x1) = 3215 in denary
Note: C=12, F=15
162
=256 161
=16 160
=1

DIY
What is the denary form of BF08?
Denary
Hexadecimal DIY
B F 0 8
162
=256 161
=16 160
=1
163
=4096

DIY
ANSWER
Denary
Hexadecimal DIY
B F 0 8
(11x4096) + (15x256) + (0x16) + (8x1) = 48904 in denary
162
=256 161
=16 160
=1
163
=4096

5
2
2
2
1
1
0
Denary Binary
5
RECAP
(Method 2)
remainder
remainder
2 0 remainder
1
Read the remainder
from bottom to top
Answer: 101

Denary Hexadecimal Explanation
Eg. "2004"
2004
16
16
125 remainder
2004/16 =
4
125/16 =
7 13
remainder
16 0 remainder
7
Answer: 7D4
Note: 13=D
125
remainder = 4
7
remainder = 13

DIY
What is the hexadecimal form of 3179?
Denary Hexadecimal DIY
3179
16
16
198 remainder
?
? ?
remainder
16 ? remainder
?
3179/16 = ?

DIY
16 6
What is the hexadecimal form of 3179?
Denary Hexadecimal DIY
3179
16 198 remainder
11
12 remainder
16 0 remainder
12
3179/16
198/16
Answer: C6B
= 198
remainder = 11
= 12
remainder = 6

Chapter 1.2
Use of hexadecimal
system

Discussion Time
Use of hexadecimal system
Binary Hexadecimal
1101011111101
00111001
1AFD39
Brainstorm time: Why is Hexadecimal used?

Explanation
Use of hexadecimal system
• One hex digit represents four binary digits
• The hex number is far easier for humans to
remember, copy and work with
• Four uses of the hexadecimal system

Explanation
Usage 1: Error Code
• Error codes are often shown as hexadecimal
values.
• These numbers refer to the memory location
of the error.
• They are generated by the computer.
• The programmer needs to know how to
interpret the hexadecimal error codes.

Explanation
Usage 1: Error Code

Explanation
Usage 2: MAC address
• Media Access Control (MAC) address refers to
a number which uniquely identifies a device on a network.
• The MAC address refers to the network interface card (NIC)
which is part of the device
• The MAC address is rarely changed so that a particular device
can always be identified no matter where it is.

Explanation
00_1C_B3_4F_25_FE
00_1C_C3_4F_23_AE
Mac address uniquely identify a device on a Local Area Network
Message

Explanation
00-1C-B3-4F-25-FE
NN-NN-NN-DD-DD-DD
00:1C:B3:4F:25:FE
NN:NN:NN:DD:DD:DD
Form 1
Form 2 2
Mac Address comes with 2 forms

Explanation
00-1C-B3 4F-25-FE
Identity number of the
manufacturer
Serial number of
a device
Eg. 00 – 14 – 22 which identifies devices made by Dell
00 – a0 – c9 which identifies devices made by Intel

Explanation
Usage 3: Internet
Protocol Addresses
• Each device connected to a network is given an address know
as the Internet Protocol address
• An IPv4 address is a 32-bit number written in denary or
hexadecimal form e.g. 109.108.158.1 (or 77.76.9e.01 in hex)
• IPv4 has recently been improved upon by the adoption of IPv6
An IPv6 address is a 128-bit number broken down into 16-bit
chunks, represented by a hexadecimal number.
Eg. a8f b:7a88:fff0:0fff:3d21:2085:66f b:f0fa

Explanation
Usage 4: HyperText Markup
Language (HTML) colour code
• HyperText Mark-up Language (HTML) is used when writing and
developing web pages.
• It is not a programming language, but a markup language.
• A mark-up language is used in the processing, definition and
presentation of text.

Explanation
Overview
The 4 usages of Hexadecimals - EMIH
1. E - Error Codes
2. M - MAC Address
3. I - Internet Protocol Address
4. H - HTML Colour Code

Chapter 1.3
Addition of
binary number

Explanation
How do we perform add and carry in denary?
0 + 0 = 0
0 + 9 = 9
9 + 0 = 9
9 + 1 = 10
9
+1
0
1
1
Addition of binary number

Explanation
How do we perform add and carry in denary?
56
+79
6+9 = 15 (>9)
5
1
1+5+7 = 13 (>9)
3
1
1

Explanation
How do we perform add and carry in binary?
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10

Explanation
00100111
+01001010
1
1
0
0
1
0
1
1
1
1
0

DIY
Perform
01111110 + 00111110

Explanation
The overflow condition
01101110
+11011110
1
1 0
0
1
0
1
1
0
1
1
1
1
0
1

Explanation
The overflow condition 01101110
+11011110
1
1 0
0
1
0
1
1
0
1
1
1
1
0
1
• The maximum denary of an 8-bit binary number (11111111)
is (28
- 1 ) = 255
• The generation of a 9th bit is a clear indication that the
sum has exceeded this value.
• This is known as an overflow error. The sum is too big to be
stored using 8 bits.

Explanation
The overflow condition 01101110
+11011110
1
1 0
0
1
0
1
1
0
1
1
1
1
0
1
01101110 = 110
11011110 = 222
110 + 222 = 322
322 > 255 (overflow)
The sum is too big to be stored in a 8 bit binary.

Lesson Objectives
Last lesson on the binary system
BINARY SHIFTING
Multiplication
and division
of binary
numbers
TWO
COMPLEMENTS
Represent
negative
number in
binary
Chapter 1.4: Binary Shifting

BINARY SHIFTING
• Binary shift is a process that a CPU uses to
perform multiplication and division.

BINARY SHIFTING -
MULTIPLICATION
• For a CPU to multiply a binary number, the
number needs to be shifted to the left and will
fill the remaining gaps with zeros.

16
32 8 4 2 1
64
16
32 8 4 2 1
64
16
32 8 4 2 1
64
BINARY SHIFTING -
MULTIPLICATION
Examples: 111 (Binary)
0
0 0 1 1 1
0
0
0 0
1 1
1
0
0
0 0
1
1
1
0

16
32 8 4 2 1
64
BINARY SHIFTING -
MULTIPLICATION
0
0 0 1 1 1
0
Multiply by 2, shift 1 place to the left 111
0
Multiply by 4, shift 2 place to the left
Multiply by 8, shift 3 place to the left
1110
0
11100
0
Multiply by 2^n, shift n place to the left

BINARY SHIFTING - DIVISION
• For a CPU to multiply a binary number, the
number needs to be shifted to the right.

16
32 8 4 2 1
16
32 8 4 2 1
16
32 8 4 2 1
0
1 1 1 0 0
1
0 0
1 1
0
1
0 1
0
1
0

16
32 8 4 2 1
0
1 1 1 0 0
Divide by 2, shift 1 place to the right 10110
Divide by 4, shift 2 place to the right
Divide by 8, shift 3 place to the right
101
1
10
1
Divide by 2^n, shift n place to the right

BINARY SHIFTING WITH 8-BIT BINARY
NUMBERS
• Registers contained within the CPU often have 8-
bits limits on the amount of data they can hold at
any one time.
• The multiplying shifting process can cause bits to
be lost at one end of the register, and zeros added
at the opposite end.
• This process is known as losing the most significant
bit.

16
32 8 4 2 1
64
12
8
NUMBERS
Examples: 10110101 (181 in denary)
1
1 0 1 0 1
0
1
10110101 ->
01101010
106 in denary
The bit lost is called the most significant bit, and
when it is shifted beyond the furthest-column the
binary data that is stored loses precision due to

16
32 8 4 2 1
64
12
8
16
32 8 4 2 1
64
12
8
NUMBERS
The same process can happen when dividing an 8-
bit binary number.
1
1 1 1 0 1
0
1
Example: 10111101 (189 in denary)
Divide this number by 32 (move 5 places to the
right)
0
0 0 1 0 1
0
0
Least
Significant
bit
The division shift
produces the binary
number 101 = 5, not 5.9
1110
1

Lesson Objectives
Last lesson on the binary system
BINARY SHIFTING
Multiplication
and division
of binary
numbers
TWO
COMPLEMENTS
Represent
negative
number in
binary
Chapter 1.5: Two Complements

TWO
COMPLEMENTS
• A PROCESSOR CAN ALSO REPRESENT
NEGATIVE NUMBERS.
• ONE OF THE METHOD THAT A PROCESS
REPRESENT NEGATIVE NUMBERS IS CALLED
TWO'S COMPLEMENT.

TWO
COMPLEMENTS
• TO REPRESENT NEGATIVE NUMBERS, IT IS IMPORTANT TO
THINK ABOUT THE PLACE VALUE OF THE FURTHEST-LEFT
BIT IN A DIFFERENT WAY.
• PROCESSOR CAN BE SET UP TO SEE THE BIT IN THE
EIGHTH COLUMN AS A SIGN BIT.
0 = POSITIVE
1 = NEGATIVE

16
32 8 4 2
64
-128
CONVERT POSITIVE BINARY INTEGER TO A TWO'S COMPLEMENT 8-BIT
INTEGER
Examples:13
0
0 1 1 0
0
0
1
1
Step 2: Put the number into the place value
column
Step 3: Ensure that the the leftmost bit is 0 (+).
Step 1: Convert 13 into binary.
1101 in
binary

DI
Y
Convert 19 into a Two's complement
8-bit Integer

16
32 8 4 2
64
-128
CONVERT POSITIVE BINARY INTEGER TO A TWO'S COMPLEMENT 8-BIT
INTEGER
Examples:19
1
0 0 0 1
0
0
1
1
column
Step 3: Ensure that the the leftmost bit is 0 (+).
Step 1: Convert 19 into binary.
10011 in
binary
Answer:
00010011

16
32 8 4 2
64
-128
CONVERT TWO'S COMPLEMENT 8-BIT INTEGER TO A POSITIVE BINARY
INTEGER
Examples: Convert 00010011 (two's complement) to
denary
1
0 0 0 1
0
0
1
1
column
Step 2: This shows that it is a positive number, we
can just convert the binary into denary directly.
Step 3: Calculate the denary value.
(1x16) + (1x2) + (1x1) = 19

Convert 01010011 (two's complement)
to denary
DI
Y

16
32 8 4 2
64
-128
CONVERT TWO'S COMPLEMENT 8-BIT INTEGER TO A POSITIVE BINARY
INTEGER
Examples: Convert 01010011 (two's complement) to
denary
1
0 0 0 1
1
0
1
1
column
Step 2: This shows that it is a positive number, we
can just convert the binary into denary directly.
Step 3: Calculate the denary value.
(1x64) + (1x16) + (1x2) + (1x1) = 83

16
32 8 4 2
64
-128
CONVERT NEGATIVE BINARY NUMBERS IN TWO'S COMPLEMENT FORMAT
AND CONVERT TO DENARY
Examples: 10010011
1
0 0 0 1
0
1
1
1
column
Step 3: Compute the denary value as usual.
Step 2: The left-most bit is 1, this means that it is a
negative number.
(1x -128) + (1x16) + (1x2) + (1x1)
= -128 + 16 + 2 + 1
= -109

Convert 10110011 (Two's Complement)
to denary
DI
Y

16
32 8 4 2
64
-128
CONVERT NEGATIVE BINARY NUMBERS IN TWO'S COMPLEMENT FORMAT
AND CONVERT TO DENARY
Examples: 10110011
1
1 0 0 1
0
1
1
1
column
Step 3: Compute the denary value as usual.
Step 2: The left-most bit is 1, this means that it is a
negative number.
(1x -128) (1x32)+ (1x16) + (1x2) + (1x1)
= -128 + 32 + 16 + 2 + 1
= -77

CONVERTING NEGATIVE DENARY NUMBERS INTO BINARY NUMBERS IN
TWO’S COMPLEMENT FORMAT
Examples: -67
Step 1: Convert the number to positive.
67
Step 2: Write the number in binary form (8 bits).
0100001
1
Step 3: Invert each binary value.
1011110
0

Step 4: Add 1 to the binary number.
1
1011110
1
+
1011110
0
Step 5: This gives us -67.
16
32 8 4 2
64
-128
1
1 1 1 0
0
1
1
1
-128 + 32 + 16 + 8 + 4 + 1 = -67

Convert -65 to 8 bit two's complement
binary number
DI
Y

CONVERTING NEGATIVE DENARY NUMBERS INTO BINARY NUMBERS IN
TWO’S COMPLEMENT FORMAT
Examples: -65
Step 1: Convert the number to positive.
65
Step 2: Write the number in binary form (8 bits).
0100000
1
Step 3: Invert each binary value.
1011111
0

Step 4: Add 1 to the binary number.
1
1011111
1
+
Step 5: This gives us -65.
16
32 8 4 2
64
-128
1
1 1 1 1
0
1
1
1
-128 + 32 + 16 + 8 + 4 + 2 + 1 = -65
1011111
0

Summary: Convert negative denary to two's
complement
Examples: -65
01000001
10111110
65
1
10111111
Convert to (+)
Convert to binary
Invert the digit
+1
Final result

The number of bits used to represent sound amplitude in digital
sound recording, as known as bit depth

Measurement of
Data Storage and
Calculation of file
size
Chapter 1.7

Measurement
of Data Storage
• A bit is the basic unit of all computing
memory storage terms and is either 1 or 0.
• The byte is the smallest unit of memory in a
computer.
8 bits = 1 byte
4 bits = 1 nibble

Memory
Size
System
Based on the SI (base 10) system of units where
1 kilo is equal to 1000.

Memory
Size
System
Based on the IEC (base 2) system of units where
1 kilo is equal to 1024 (2^10).
As memory size is actually measured in terms of powers of 2...

Memory
Size
System
Converting Bytes into KiB, MiB and GiB
68719476736 Bytes
68719476736 Bytes / 1024
= = 67108864 KiB
= 67108864 KiB / 1024 = 65536 MiB
= 65536 MiB / 1024 = 64 GiB

Memory
Size
System
Converting Gib, Mib, Kib into bytes
= 68719476736 Bytes
64 x 1024
= = 65536 MiB
= 65536 x 1024 = 67108864 KiB
= 67108864 x 1024
64 GiB

DIY
Convert the size of GTA-V to bytes
= 77309411328 Bytes
72 x 1024
= = 73728 MiB
= 65536 x 1024 = 75497472 KiB
= 75497472 x 1024
72 GiB

Calculation
of file
size
Ima
ge
Aud
io

Calculation
of file
size - Image
Image Resolution - The number of
pixels that make up an image.
The higher the image resolution, the
higher the quality of the image.

Calculation
of file
size - Image
Formula
image resolution (pixels) x colour depths
(bits)

Calculation
of file
size - Image
Example 1
00 01
10 11
2px
2px
Total pixels = 2 x 2 = 4 Colour depth =
2
Calculation = (2x2) x 2
= 8 bits = 1 byte

Calculation
of file
size - Image
Example 2
Formula : image resolution (pixels) x colour depths
(bits)
Question:
Image Resolution = 1024 x 1080
Colour depth = 32
Calculate the size of this image in Bytes.
Workings:
1024 x 1080 = 1105920 pixels
1105920 x 32 = 35389440 bits
Answer in byte: 35389440/8 = 4423680 bytes

Calculation
of file
size - Image
Example 2
Question:
Colour depth = 32
Calculate the size of this image in Bytes. How many
photograph of this size would fit onto a memory stick of
64Gib.
Each image = 4423680
bytes
First convert 64 Gib into
bytes:
64 x 1024 = 65536 MiB
65536 x 1024 = 67108864 KiB
67108864 x 1024 = 68719476736 bytes

Calculation
of file
size - Image
Example 2
Question:
Colour depth = 32
Calculate the size of this image in Bytes. How many
photograph of this size would fit onto a memory stick of
64Gib.
Each image = 4423680
bytes
First convert 64 Gib into bytes = 68719476736
bytes
68719476736/4423680 = 15534 photos.

DIY
Question:
Colour depth = 16

DANSWER
Question:
Colour depth = 16
Answer:
2048 x 2048 x 16 = 67108864 bits
= 67108864/8
= 8388608 bytes

DIY
Question:
Colour depth = 16
Calculate the size of this image in Bytes (Answer: 8388608
bytes).
What is the size of the image in MiB.

DANSWER
Question:
Colour depth = 16
Calculate the size of this image in Bytes (Answer: 8388608
bytes).
What is the size of the image in MiB.
8388608 / 1024 = 8192 KiB
8192 / 1024 = 8 MiB

Calculation
of file
size - Sound
Formula
Sample Rate (in Hz) x Sample Resolution (in bits) x length of sample (in
seconds)

Calculation
of file
size - Sound
Mono Sound vs Stereo
Sound
Comparison

Calculation
of file
size - Sound
Example 1 - Mono Sound
Question:
Sample Rate: 44100
Sample Resolution: 8 bits
Length of the music: 20 seconds
Calculate the size of the audio in KiB.
44100 x 8 x 20 = 7056000 bits
7056000/8 = 882000 bytes
882000 / 1024 = 861.328 KiB

Calculation
of file
size - Sound
Example 1 - Stereo Sound
An audio CD has a sample rate of 44100 and a sample resolution of 16
bits. The music being sampled uses two channels to allow for stereo
recording. Calculate the file size for a 60-minute recording.
44100 x 16 x 3600 = 2540160000 bits
2540160000 x 2 = 5080320000 bits
5080320000 / 8 = 635040000 bytes
635040000 / 1024 = 620156.25 KiB
620156.25 / 1024 = 605.62 MiB

DI
Y
recording. Calculate the file size for a 25-minute recording.

DI
Y
recording. Calculate the file size for a 25-minute recording in MiB.
44100 x 8 x 1500 = 529200000 bits
529200000 x 2 = 1058400000 bits
1058400000 / 8 = 132300000 bytes
132300000 / 1024 = 129199.218 KiB
129199.218 / 1024 = 126.17 MiB

• FILE SIZE OF IMAGES AND SOUND CAN BE VERY
LARGE.
• THEREFORE, DATA COMPRESSION IS NEEDED TO
REDUCE THE SIZE OF A FILE.
DATA
COMPRESSI
ON

WHAT ARE SOME BENEFITS OF
REDUCING THE FILE SIZE?
DATA
COMPRESSI
ON

Benefits of
Data
Compressio
n
SAVE
STORAGE
SPACE
REDUCE
STREAMIN
G TIME
REDUCE
TIME
TAKEN TO
UPLOAD
AND
DOWNLOA
D MEDIA
REDUCE
COST

LOSSY
FILE
COMPRESS
ION
LOSSLESS
FILE
COMPRESS
ION
DATA
COMPRESSI
ON

LOSSY FILE
COMPRESSI
ON
• FILE COMPRESSION ALGORITHM ELIMINATES
UNNECESSARY DATA FROM THE FILE.
• ORIGINAL FILE CANNOT BE RECONSTRUCTED
ONCE IT HAS BEEN COMPRESSED.
• IMAGE - REDUCE THE RESOLUTION // COLOUR
DEPTH
• SOUND - REDUCE THE SAMPLING RATE //
SAMPLING RESOLUTION
• SOME LOSSY FILE COMPRESSION ALGORITHMS
ARE:
MPEG-3 MPEG-4 JPEG

LOSSY FILE
COMPRESSIO
N
MPEG-3
• A COMPRESSION TECHNOLOGY
THAT REDUCES THE SIZE OF A
NORMAL MUSIC FILE BY ABOUT
90%.
SECRET
• REMOVE
SOUNDS
OUTSIDE THE
HUMAN EAR
RANGE
• ELIMINATE
THE SOFTER
SOUND -
PERCEPTUAL

LOSSY FILE
COMPRESSIO
N
MPEG-4
ALLOWS STORAGE OF MULTIMEDIA
FILES RATHER THAN JUST SOUND.
SECRET
• MOVIES CAN
BE STREAMED
USING THE
MP4 FORMAT
WITHOUT
LOSING ANY
REAL
DISCERNIBLE
QUALITY

LOSSY FILE
COMPRESSI
ON
JPEG
A LOSSY COMPRESSION ALGORITHM
USED FOR BITMAP IMAGES. ORIGINAL
FILE CAN NO LONGER BE
CONSTRUCTED
SECRET
• REMOVE
COLOUR
SHADES
(HUMAN
CAN'T NOTICE
THEM)

DATA
COMPRESSIO
N
LOSSLESS
FILE
COMPRESS
ION
• ALL THE DATA FROM THE ORIGINAL
UNCOMPRESSED FILE CAN BE
RECONSTRUCTED
• THIS IS IMPORTANT FOR FILES WHERE
LOSS OF DATA WOULD BE
DISASTROUS (COMPLEX
SPREADSHEET).
• LOSSLESS FILE COMPRESSION IS
DESIGNED SO THAT NONE OF THE
ORIGiNAL DETAIL FROM THE FILE IS
LOST.

LOSSLESS
FILE
COMPRESSIO
N
RUN-
LENGTH
ENCODIN
G
1. CAN BE USED FOR LOSSLESS COMPRESSION OF A NUMBER
OF DIFFERENT FILE FORMATS.
2. IT REDUCES THE SIZE OF A STRING OF ADJACENT, IDENTICAL
DATA.
3. A REPEATED STRING IS ENCODED INTO TWO VALUES.
NUMBER OF
IDENTICAL
DATA
DATA ITEM

LOSSLESS
FILE
COMPRESSIO
N
RUN-
LENGTH
ENCODIN
G
IN
ACTION
16 bytes

RUN-
LENGTH
ENCODING
IN
ACTION
Each digit - 1 byte
Total size = 8 bytes (50% reduction in size)
Does not work well when no repeated data!
LOSSLESS
FILE
COMPRESSIO
N

3,2,5,2,
4
2,4,3,4,
3
1.6.1.6.
2
0,
15,1
TO BE
CONTINUE
LOSSLESS
FILE
COMPRESSIO
N

DIY
REDUCE
COLOUR
DEPTH
REDUCE
IMAGE
RESOLUTIO
N

Chapter 1 o level computer science with past paper

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