Transcript Document 7741882

CS149D Elements of Computer
Science
Ayman Abdel-Hamid
Department of Computer Science
Old Dominion University
Lecture 3: 9/3/2002
Lecture 3: 9/3/2002
CS149D Fall 2002
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Outline
•Fractions in Binary
•Storing Integers
•Sign and magnitude (covered last lecture)
•Two’s complement (covered last lecture)
•Excess Notation
•Storing Fractions
•Representing Text
•Representing Images
•Data Compression
By now, we should have covered sections 1.4-1.8 from Brookshear Text
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Fractions In Binary1/3
•Use radix point just like decimal
•To the right of radix point positions are numbered –1, -2 , ….
210 -1-2-3
i
101.101
d
101.101 =
(1 * 20) + (0 * 21) + (1*22) +
(1*2-1) + (0* 2-2) + (1 * 2-3)
=
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5
1 + 4 + ½ + 1/8 = 5 8
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Fractions In Binary2/3
Express the following in binary notation
35
16
•Convert the integer part
•Convert the fraction part. Try to map the fraction as a sum of
power of 2 fractions using the given denominator as a guide
•Put a radix point in between
310 is 112
3 5 11.01012
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5  1  4  1  1  0.01012
16 16 16 16 4
Make sure that
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11.01012  3 5
16
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Fractions In Binary3/3
Addition of binary numbers with radix points
•Align radix point
•Apply binary addition process
10.011 + 100.11
10.011
+ 100.11
__________
111.001
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Storing Integers1/3
•Binary number system not used as is to actually store an integer
number
•Use a numeric storage technique
Sign and Magnitude
Problems: Two zeros (+0, and –0), adding +2 and –2 does not give zero.
Straightforward binary addition does not apply
Two’s complement
Advantages: deal with subtractions the same as additions, simpler
hardware to perform integer arithmetic operations
Excess Notation
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Storing Integers2/3
Excess Notation
•All positive numbers begin with 1
Excess four notation
Bit pattern
Value
Unsigned Binary
•All negative numbers begin with 0
•0 is represented as 100
111
3
7
110
2
6
101
1
5
100
0
4
011
-1
3
010
-2
2
001
-3
1
•Smallest negative number is 000
000
-4
0
•Largest positive number is 111
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•Unsigned binary value exceeds
excess notation by 4, hence the
name excess four
• Why 4?
(2#of bits –1) = 23-1 = 4
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Storing Integers3/3
What is 101 in excess four notation?
101 if interpreted unsigned is 5
101 in excess four notation is (5-4) = 1
What is 3 represented in excess four notation?
Excess four means we need (4-1) = 3 bits for representation
Remember unsigned value exceeds excess notation by 4
Then get unsigned value = 3+4 = 7
Represent 7 in 3-bit binary = 111
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3 in excess four is 111
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Storing Fractions1/6
Need to represent number and position of radix point
Use floating-point notation (Textbook uses one-byte as example)
- --- ---- 8 bits = 1-byte
1 bit: Sign bit (0 positive, 1 negative)
3 bits: Exponent (encodes position of radix point)
4 bits: Mantissa (encodes number)
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Storing Fractions2/6
Bit pattern 01101011 in floating-point notation is what in decimal?
•First bit is 0, then positive
•Exponent is 110 mantissa is 1011
•Extract mantissa and place a radix point on its left side 0.1011
•Extract exponent and interpret as excess four notation
•110 in excess four is +2 (make sure?)
•+2 exponent means move radix point to the right by two bits
(a negative exponent means move radix to left)
•0.1011 becomes 10.11
10.11 is
23
4
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01101011 is 2 4
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Storing Fractions3/6
Bit pattern 10111100 in floating point is what in decimal?
•First bit is 1, then negative
•Exponent is 011 mantissa is 1100
•Extract mantissa and place a radix point on its left side 0.1100
•Extract exponent and interpret as excess four notation
•011 in excess four is -1 (make sure?)
•-1 exponent means move radix point to the left by 1 bit
•0.1100 becomes 0.01100
0.01100 is
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8
3
10111100 is  8
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Storing Fractions4/6
What is the following number stored in floating-point?
11
8
•Express number in binary to obtain 1.001 (make sure?)
•Copy bit pattern into mantissa field from left to right starting with the leftmost 1 in
binary representation
•Mantissa is 1001
•Compute exponent to get 1.001 from .1001 (imagine mantissa with radix point at its
left)
•need to move radix point to right one bit
•Exponent is +1 expressed in excess four notation is 101 (How?)
Sign:
0 (positive)
Exponent: 101
11  01011001
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Mantissa: 1001
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Storing Fractions5/6
What is the following number stored in floating-point?
1
4
•Express number in binary to obtain 0.01 (make sure?)
•Copy bit pattern into mantissa field from left to right starting with the leftmost 1
in binary representation
•Mantissa is 1000 (you append zeros to fill the 4-bit mantissa)
•Compute exponent to get 0.01 from 0.1000
•need to move radix point to left one bit
•Exponent is -1 expressed in excess four notation is 011 (How?)
Sign:
1 (negative)
Exponent: 011
 1  10111000
4
Mantissa: 1000
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Storing Fractions6/6
What is the following number stored in floating-point?
25
8
•Express number in binary to obtain 10.101 (make sure?)
•Copy bit pattern into mantissa field from left to right starting with the leftmost 1
in binary representation
•Mantissa is 1010 (we run out of bits!! Rightmost 1 is lost (equivalent to 1/8))
•Compute exponent to get 10.101 from 0.1010
•need to move radix point to right two bits
•Exponent is +2 expressed in excess four notation is 110
Sign:
0 (positive)
Exponent: 110
25
8
Mantissa: 1010
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Truncation error
Round off error
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 01101010
2
which is really
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Representing Text1/2
ASCII (adopted by American National Standards Institute ANSI)
•American Standard Code for Information Interchange
•8-bit to represent each symbol
•Upper and lower case letters of English alphabet, punctuation symbols, digits 0 to 9,
and other symbols
•Can represent 256 (28) different symbols
Unicode
•16-bit to represent each symbol
•Can represent 65,536 (216) different symbols
ISO (International Organization for Standardization)
•32-bit to represent each symbol
•Can represent more than 17 million symbols
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Representing Text2/2
ASCII Chart sample
•Upper case A is 6510
•Lower case a is 9710
•Difference between lower
case and upper case of a
letter is always 3210
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Decimal
Hex
ASCII
Binary
65
66
67
68
69
41
42
43
44
45
A
B
C
D
E
0100 0001
0100 0010
0100 0011
0100 0100
0100 0101
48
49
50
51
52
53
54
55
56
57
30
31
32
33
34
35
36
37
38
39
0
1
2
3
4
5
6
7
8
9
0011 0000
0011 0001
0011 0010
0011 0011
0011 0100
0011 0101
0011 0110
0011 0111
0011 1000
0011 1001
97
98
99
100
101
61
62
63
64
65
a
b
c
d
e
0110 0001
0110 0010
0110 0011
0110 0100
0110 0101
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Representing Images
Bitmap techniques
•Image is a collection of pixels (picture element)
•Each pixel can be represented as a number of bits (collection of bits is bitmap)
1 bit/pixel  B/W
8 bits/pixel  Gray Scale (different shades of gray from black to white)
24 bits/pixel  1-byte for each of the primary colors RGB
•Size? (need for compression)
Vector techniques
•Image represented as collection of lines and curves
•Fonts on printers  scalable Fonts (True Type)
•CAD (Computer Aided Design)
•Quality problem
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Data Compression1/3
Data compression techniques
•run-length encoding
replace repeating sequences with a code indicating the value being repeated and
the number of times it is repeated, e.g., aaaaa is replaced by a5
•relative encoding
record differences between consecutive data blocks, e.g., consecutive frames in a
motion picture
•Frequency-dependent encoding (variable length code)
number of bits to represent an item is inversely related to the frequency of the
item’s use, e.g., (e, t, a, and i) in English would be represented by short bit
patterns, whereas (z, q, and x) would be represented by longer patterns
•Adaptive dictionary encoding
•Example is LZ77 (Lempel-Ziv)
•abaabcb (5,4,a) when decompressed is abaabcbaabca (see Textbook page 61)
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Data Compression2/3
Image Compression
GIF (Graphic Interchange Format)
•8 bits/pixel (1 byte/pixel) instead of 24 bits/pixel (3 bytes/pixel)
•Each of 256 potential pixel values are associated with a RGB combination by means of table
known as the palette
JPEG (Joint Photographic Experts Group)
Lossless mode
•store difference between consecutive pixels rather than pixel intensities themselves
•Differences encoded using variable-length code
•Resulting bit maps not manageable with today’s technology
Base standard (lossy)
Each pixel represented by a brightness component and 2 color components
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Data Compression3/3
Audio and Video compression
MPEG (Motion Picture Experts Group)
•Start a picture sequence with an image similar to JPEG and then
represent rest of sequence using relative encoding techniques
•MP3 (MPEG-1 Audio Layer-3) audio compression ratios of 12 to 1
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