Download week 11 - NUS Physics

Information Representation
in Computer
Lecture Nine
1
Outline
 Everything is a number (in computer, c.f.
Pythagoras)
 Binary number system
 Negative integers
 Floating point numbers (reals)
 Accuracy and limitation in “computer
numbers”
 ASCII code for characters
 Code for programs, graphics, music, etc
2
Bit, Byte, and Word
0
0 1
A bit is a size that can store 1 digit of a binary number, 0 or 1.
1 0 1 1
0 0
A byte is 8 bits, which can store eight 0’s
or 1’s.
cont’d
0 1
1 1 0 1
1 0 0 1
0 0
0 0 0 1
0 0
1 1 1 0
0 1 1 1
1 0 0 0
1 1
A word is either 32 or 64 bits, depending on
computers. Regular PC’s are 32-bit word in
size, higher-end workstations are 64-bit. Word
size is the size of the registers.
What do these bits mean is a matter of interpretation! All
information in a computer are represented in a uniform
format of bit patterns.
3
Binary Nonnegative Integers
Given a 32-bit pattern, e.g.,
0000 0000 … 0000 1101 1100,
a31 highest bit
a0 lowest bit
it can represent the integer (if you interpret it that way)
n  a0  a1  2  a2  22  a3  23 
ak 2k 
,
ak  0,1
Note that if we use 32-bit word, the smallest number is 0, and the
largest number is 11111111 11111111 11111111 11111111, which
is 232-1=4294967295. Numbers bigger than this cannot be
represented. If such things happen in a calculation, we say it
overflowed. Such interpretation is called unsigned int in the
programming language C.
4
Negative Integers
 To represent negative numbers, we’ll agree
that the highest bit a31 representing sign,
rather than magnitude, 0 for positive, 1 for
negative numbers.
 More precisely, all modern computers use
2’s complement representations.
 The rule to get a negative number
representation is: first write out the bit
pattern of the corresponding positive
number, complement all bits, then add 1.
5
Property of 2’s Complement
Numbers
 Two’s complement m of a number n is
such that adding it together you get
zero (m + n = 0, modulo word size)
 Thus m is interpreted as negative of
n.
 The key point is that computer has a
finite word length, the last carry is
thrown away.
6
2’s Complement, example
 1 in bits is 00000000 00000000 00000000 00000001
complementing all bits, we get
11111111 11111111 11111111 11111110
Add 1, we get representation for –1,
as 11111111 11111111 11111111 11111111
Decimal
00000000 00000000 00000000 00000001
1
+ 11111111 11111111 11111111 11111111
–1
1 00000000 00000000 00000000 00000000
0
Overflow bit not
kept by computer
7
2’s Complement, another e.g.
 125 in bits is
00000000 00000000 00000000 01111101
complement all bits, we get
11111111 11111111 11111111 10000010
Add 1, we get representation for –125, as
11111111 11111111 11111111 10000011
Decimal
00000000 00000000 00000000 01111111
127
+ 11111111 11111111 11111111 10000011
–125
1 00000000 00000000 00000000 00000010
+2
Overflow bit not
kept by computer
8
Dual Interpretation of Bit Patterns
Signed char
0000 0000 = 0
1000 0000 = 128
or
-128
0000 0001 = 1
1000 0001 = 129
or
-127
0000 0010 = 2
1000 0010 = 130
or
-126
0000 0011 = 3
1000 0011 = 131
or
-125
0000 0100 = 4
...
0000 0101 = 5
1111 1100 = 252
or
-4
0000 0110 = 6
1111 1101 = 253
or
-3
1111 1110 = 254
or
-2
1111 1111 = 255
or
-1
...
0111 1111 = 127
A byte (or char in C)
Unsigned char
9
Signed and Unsigned Int
 If we interpret the number as unsigned, an
unsigned integer takes the range 0 to 232-1
(that is 000…000 to 1111….1111)
 If we interpret the number as signed value,
the range is –231 to 231-1 (that is
1000….000 to 1111…111, to 0, to
0111….1111).
 Who decide what interpretation to take?
What if we need numbers bigger than 232?
10
Addition, Multiplication, and
Division in Binary
00010011
19+37=56
+ 00100101
00111000
0111
75=35
 0101
0111
+ 0111
1001
100011
1000 1001010
–1000
1010
–1 0 0 0
74  8 = 9 remainder 2
10
11
Floating Point Numbers (reals)
 To represent numbers like 0.5, 3.1415926,
etc, we need to do something else. First,
we need to represent them in binary, as
n
 am 2m
1
 a2 22  a1 2  a0  a1   a2  22  a3  23 
2
a k 2 k 
E.g. 11.00110 for 2+1+1/8+1/16=3.1875
 Next, we need to rewrite in scientific
notation, as 1.100110 21. That is, the
number will be written in the form:
1.xxxxxx…  2e
x = 0 or 1
12
Real Numbers on Computer

 d0  d1 
1
 d p 1
 ( p 1)

On modern
computers the
base β is
exclusively 2.
e
d0  0, 0  di   , emin  e  emax
Example for β=2, p=3, emin= -1, emax=2
ε
0 0.5
1
2
3
4
5
6
7
13
IEEE 754 standard
 The IEEE standard for floating numbers
specifies for 32-bit single precision (float in
C) as follows:
 The highest 31-th bit a31 is used for sign, 0
for positive, 1 for negative
 8 bits, from position 23 to 30, are used to
store an integer, for the biased exponent
e+127.
 The rest 23 bits are used for the significand.
bit 31
sign
exponent
significand
bit 0
14
IEEE Floating Point
a31 e7
sign
e1 e0 s-1 s-2 s-3
exponent
s-23
significand
bit 0
Bit 31 (a31) specifies the sign
Bit 23 to 30 (e7,e6,…,e1,e0), interpreted as an 8-bit
unsigned integer, specifies the biased exponent E = e +
127.
Bit 0 to 22 (s-1,s-2,…,s-22,s-23) specifies the fractional
part of number in scientific notation. The value of the
pattern is:
(1)a31  (1  s1 / 2  s2 / 4  s3 / 8 
s k 2 k
However, the number +0 is a string of all zeros.
 s23 223 )  2 E 127
15
Example of Floating Point Number
bit 31
0
sign
0 1 1 1 1 1 11
exponent
0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 00 0 0
significand
bit 0
The sign is positive, the biased exponent E=127=e+127,
so the original exponent e is 0. The significand f is 0.
Thus the pattern represents: +(1.0 + 0)20 = 1.0
bit 31
1
sign
0 1 1 1 1 1 10
exponent
1 0 1 0 0 0 00 0 0 0 0 0 0 00 0 0 0 00 0 0
significand
bit 0
The sign is negative, the biased exponent E=126=e+127,
so the original exponent e is –1. The significand f
=1/2+1/8. Thus the pattern represents: –(1.0 + f)2-1 =
–0.8125.
16
Encoding to IEEE Format
 What is the bit pattern of the real number 10.25 in
IEEE single precision (32-bit) format?
 Ans: first, we need to convert the number to binary.
We can consider separately the integer 10, and the
fractional part 0.25. For 10, we have
10 = 52 + 0
For 0.25, we multiply by 2
5 = 22 + 1
(repeatedly) and take the
2 = 12 + 0
integer part as the digit:
1 = 02 + 1
20.25 = 0.5 -> 0
20.5 = 1.0 -> 1
so 1010=10102
20.0 = 0.0 -> 0
Thus
10.2510=1010.01
=1.0100123
So the fractional part is
0.2510=0.0100002
17
10.25 in bits
 10.2510=1.01001223
 Thus sign is positive, f = .01001,
e=3, and E=e+127=130=100000102.
The bit pattern is
bit 31
0
sign
E=e+127
1 0 0 0 0 0 10
exponent
f
0 1 0 0 1 0 00 0 0 0 0 0 0 00 0 0 0 00 0 0
significand
bit 0
18
IEEE 754 (32-bit) Summary
The bit pattern
b-1 b-2
s
e
……
b-23
f = b-12-1 + b-22-2 + … + b-232-23
represents
If e = 0: (-1)s f 2-126
If 0<e<255: (-1)s (1+f) 2e-127
If e=255 and f = 0: + or - ∞
and f ≠ 0: NaN.
19
Limitations in 32-bit Integer and
Floating Point Numbers
 Limited range of values (e.g. integers only
from –231 to 231–1)
 Limited resolution for real numbers. E.g., if
x is a machine representable value, the
next value is x + ε (for some small ε).
There is not value in between. This causes
“floating point errors” in calculation. The
accuracy of a single precision floating point
number is about 6 decimal places.
20
Limitations of Single Precision
Numbers
 Given the representation of the single
precision floating point number
format, what is the largest magnitude
possible? What is the smallest
number possible?
 With floating point number, it can
happen that 1 + ε = 1. What is that
largest ε?
21
Double Precision
 An IEEE double precision takes twice
the number of the bits of the single
precision number, 64 bits.
 1 bit for the sign
 11-bit for exponent
 52-bit for significand
 The double precision has about 15
decimal places of accuracy
22
Floating Point Rounding Error
 Consider 4-bit mantissa floating point
addition:
 1.01022 + 1.1012-1
1.010000  22
+ 0.001101  22
Shift exponent to that of the
large number
1.011101  22
1.100
 22
Round to 4 bits
In decimal, it means 5.0 + 0.8125  6.0
23
Representation of Characters, the
ASCII code
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
HT
LF
VT
FF
CR
SO
SI
1 DLE DC1 DC2 DC3 DC4 NAK SYN ETB CAN EM
SUB ESC FS
GS
RS
US
2 SP
!
"
#
$
%
&
'
(
)
*
+
,
-
.
/
3 0
1
2
3
4
5
6
7
8
9
:
;
<
=
>
?
4 @
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
5 P
Q
R
S
T
U
V
W
X
Y
Z
[
\
]
^
_
6 `
a
b
c
d
e
f
g
h
I
j
k
l
m
n
o
7 p
q
r
s
t
u
v
w
x
y
z
{
|
}
~
DEL
0 NUL SOH STX ETX EOT ENQ ACK BEL BS
How to read the table: the top line specifies the last digit
in hexadecimal, the leftmost column specifies the higher
value digit. E.g., at location 4116 (=0100 00012=6510) is 24
the letter ‘A’.
Base-16 Number, or Hexadecimal
binary
hexadecimal
decimal
binary
hexadecimal
decimal
0000
0
0
1000
8
8
0001
1
1
1001
9
9
0010
2
2
1010
A
10
0011
3
3
1011
B
11
0100
4
4
1100
C
12
0101
5
5
1101
D
13
0110
6
6
1110
E
14
0111
7
7
1111
F
15

Instead of writing out strings of 0’s and 1’s, it is easier
to read if we group them in group of 4 bits. A four bit
numbers can vary from 0 to 15, we denote them by
0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F.
25
Program as Numbers
 High level programming language
C = A + B;
 Assembly language
add $5, $10, $3
 Machine code (MIPS computer)
bit 31
0 0
0
0 0 0
Not used
0
$10
1 0 1
0
source 1
$3
0 0 0
1 1
source 2
0
$5
0 1
0 1
0 0
0 0
0
destination not used
add
1 0 0
bit 0
0 0
0
operation
26
Graphics as Numbers
A picture like this is also
represented on computer
by bits. If you magnify the
picture greatly, you’ll see
square “pixels” which is
either black or white.
These can be represented
as binary 1’s and 0’s.
Color can also be presented
with numbers, if we allow
more bits per pixel.
27
Music as Numbers – MP3 format
 CD music is sampled 44,100 times
per second (44.1 kHertz), each
sample is 2 bytes long
 The digital music signals are
compressed, at a rate of 10 to 1 or
more, into MP3 format
 The compact MP3 file can be played
back on computer or MP3 players
28
Summary
 All information in computer are
represented by bits. These bits
encode information. It’s meaning has
to be interpreted in a specific way.
 We’ve learned how to represent
unsigned integer, negative integer,
floating pointer number, as well as
ASCII characters.
29