ECE 645: Lecture 2
Number Representation
Part 2
Fixed-Radix Signed Representations
Floating Point Representations
Little-Endian vs. Big-Endian Representations
Galois Field Representations
Required Reading
Behrooz Parhami,
Computer Arithmetic: Algorithms and Hardware Design
Chapter 2, Representing Signed Numbers,
Chapter 17, Floating-Point Representations
J-P. Deschamps, G. Bioul, G. Sutter,
Synthesis of Arithmetic Circuits: FPGA, ASIC and
Embedded Systems,
Chapter 3.2, Integers
Chapter 3.3, Real Numbers
Recommended Reading
(to be covered at the next lecture)
Behrooz Parhami,
Computer Arithmetic: Algorithms and Hardware Design
Chapter 5, Basic Addition and Counting
J-P. Deschamps, G. Bioul, G. Sutter,
Synthesis of Arithmetic Circuits: FPGA, ASIC and
Embedded Systems,
Chapter 4.1.1 Basic Algorithm
Chapter 11.1 Basic Adder
Chapter 11.2 Carry-Chain Adder
Signed Number Representations
Representations of signed numbers
Signed-magnitude
Biased
Radix-complement
r=2
Two’s complement
Complement
Diminished-radix
complement
(Digit complement)
r=2
One’s complement
Signedmagnitude
7
6
5
4
3
2
1
0
-0
-1
-2
-3
-4
-5
-6
-7
-8
0111
0110
0101
0100
0011
0010
0001
0000
1000
1001
1010
1011
1100
1101
1110
1111
Two’s
complement
One’s
complement
1111
1110
1101
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000
0111
0110
0101
0100
0011
0010
0001
0000
1111
1110
1101
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000
1111
1110
1101
1100
1011
1010
1001
1000
Biased
Signed-magnitude representation of signed numbers
k-1 k-2
0
magnitude
sign
Advantages:
• conceptual simplicity
• symmetric range: -(2k-1-1) .. -(2k-1-1)
• simple negation
Disadvantages:
• addition of numbers with the same sign and with
a different sign handled differently
Biased (excess-B) representation of signed integers
B = 2k-1, k=4
R(X) = X + B
-2k-1 ≤ X ≤ 2k-1-1
X
-8 -7 -6 -5 -4 -3 -2 -1 0 1
2 3 4 5 6 7
R
R(X)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Biased representation
with radix 2
Signed number X
Representation
mapping
Unsigned Representation R(X)
Binary mapping
Bit vector (xk-1xk-2...x0.x-1...x-l)
k 1
R( X )   xi  2i
i  l
Complement
Signed Number Representations
Complement representations
with radix 2
Signed number X
Representation
mapping
Unsigned Representation R(X)
Binary mapping
Bit vector (xk-1xk-2...x0.x-1...x-l)
k 1
R( X )   xi  2i
i  l
Useful dependencies
1 – xi = xi
xi 1 – xi x i
0
1
1
0
1
0
X when X  0
|X| =
- X when X < 0
One’s complement transformation
For
k 1
A   Ai  2i  0
i  l
def
OC(A) = A = 2k – 2-l - A
k-1 k-2 ...
–
0
-1
-2 ... -l
Properties:
1 1 ... 1 . 1 1 ... 1
Ak-1 Ak-2 … A0 . A-1 A-2 ... A-l
0  OC(A)  2k – 2-l
Ak-1 Ak-2 … A0 . A-1 A-2 ... A-l
OC(OC(A)) = A
One’s Complement Representation
of Signed Numbers
For
–(2k-1 – 2-l)  X  2k-1 – 2-l
X
for X > 0
def
R(X) =
0 or OC(0) for X = 0
OC(|X|)
for X < 0
0  R(X)  2k – 2-l
One’s complement representation of signed integers
X>0
0
X
0,
2k-1
X<0
k=4
X+2k-1 = 2k-1 - |X|
-8 -7 -6 -5 -4 -3 -2 -1 0 1
2 3 4 5 6 7
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
One’s complement representation of signed numbers
Two’s complement transformation (1)
For
k 1
A   Ai  2i  0
i  l
def
TC(A) =
A + 2-l = 2k – A for A > 0
0
2k – A = 2k – A – 2-l + 2-l =
for A = 0
Properties:
0  TC(A)  2k – 2-l
= (2k – 2-l – A)+2-l = A + 2-l
TC(TC(A)) = A
Two’s complement transformation (2)
For
k 1
A   Ai  2i  0
i  l
def
TC(A) = A + 2-l mod 2k = 2k – A mod 2k
Two’s Complement Representation
of Signed Numbers
For
–2k-1  X  2k-1 – 2-l
def
X
for X  0
TC(|X|)
for X < 0
R(X) =
0  R(X)  2k – 2-l
Two’s complement representation of signed integers
X>0
0
X
0
X<0
k=4
X+2k = 2k - |X|
-8 -7 -6 -5 -4 -3 -2 -1 0 1
2 3 4 5 6 7
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Two’s complement representation of signed integers
Signed-magnitude representation of signed numbers
X>0
0
X
0,
2k-1
X<0
k=4
| X|+2k-1 = -X+2k-1
-8 -7 -6 -5 -4 -3 -2 -1 0 1
2 3 4 5 6 7
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Signed-magnitude representation of signed numbers
Biased representation of signed numbers
X>0
0
X<0
X+B
B
X+B
-8 -7 -6 -5 -4 -3 -2 -1 0 1
B = 2k-1, k=4
2 3 4 5 6 7
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Biased representation of signed numbers
Arithmetic Operations in
Signed Number Representations
Unsigned addition vs. signed addition
Programmer
Machine
128 64 32 16 8 4
weight
carry
Unsigned
mind
2 1
Signed
mind
1 1 1
0 0 0 1 0 0 1 1
1 0 0 0 0 1 0 1
1 0 0 1 1 0 0 0
X
+Y
=S
x6 y6
x7 y7
c8
FA
s7
c7
FA
s6
x5 y5
c6
FA
s5
x3 y3
x4 y4
c5
FA
s4
c4
FA
s3
x2 y2
c3
FA
s2
x1 y1
c2
FA
s1
x0 y0
c1
FA
s0
Out of range flags
Carry flag - C
C=1
0
out-of-range for unsigned numbers
if
result > MAX_UNSIGNED or
result < 0
otherwise
where MAX_UNSIGNED = 28-1 for 8-bit operands
216-1 for 16-bit operands
Overflow flag - V
V=1
0
out-of-range for signed numbers
if
result > MAX_SIGNED or
result < MIN_SIGNED
otherwise
where MAX_SIGNED = 27-1 for 8-bit operands
215-1 for 16-bit operands
MIN_SIGNED = -27 for 8-bit operands
-215 for 16-bit operands
Overflow for signed numbers
Indication of overflow
Positive
+ Positive
= Negative
Negative
+ Negative
= Positive
Formulas
Overflow2’s complement = xk-1 yk-1 sk-1 + xk-1 yk-1 sk-1 =
= ck  ck-1
Two’s complement representation of signed integers
Addition and subtraction
Two’s complement
Numbers of the same sign
-16 8 4 2 1
Numbers of the opposite sign
-16 8 4 2 1
11011
10110
-5
-10
00101
10110
5
-10
110001
-15
11011
-5
carry but not overflow
-16 8 4 2 1
00111
01010
7
10
10001
-15
no carry but overflow
-16 8 4 2 1
01010
11011
100101
10
-5
5
carry but not overflow
Two's Complement Adder/Subtractor
Architecture
Can replace this
with k xor gates
Arithmetic Shift
Two’s complement
Sh.L {001012 = +5} = 010102 = +10
Sh.L {110112 = -5} = 101102 = -10
Sh.L {010102 = +10} = 101002 = - 12
Shift left may cause overflow
overflow
Sh.R {001012 = +5} = 000102 = +2
rem 1
Sh.R {110112 = -5} = 111012 = -3
rem 1
Shift right requires sign extension
Addition and subtraction
One’s complement
Numbers of the same sign
Numbers of the opposite sign
-15 8 4 2 1
-15 8 4 2 1
11010
10101
+
-5
-10
101111
1
10000
end-arround carry
+
-15
01010
11010
100100
1
10
-5
00101
5
Disadvantage: Need another adder after
the addition is complete!
Arithmetic Shift
One’s complement
Sh.L {001012 = +5} = 010102 = +10
Sh.L {110102 = -5} = 101012 = -10
Sh.L {010102 = +10} = 101002 = - 11
Shift left may cause overflow
overflow
Sh.R {001012 = +5} = 000102 = +2
rem 1
Sh.R {110112 = -5} = 111012 = -2
rem -1
Shift right requires sign extension
Addition and subtraction
Signed-magnitude
Numbers of the same sign
sign bit
+
0
0
Numbers of the opposite sign
magnitude
1011
0110
11
6
0 10001
17
carry = overflow
sign bit
+
1
0
magnitude
1011
0110
-11
6
11 > 6
–
1
1011
0110
11
6
0101
5
Signed Number Representations
Summary
Representing k-bit signed binary numbers
Representation
Representation Representation
for X>0
for 0
Signedmagnitude
X
0,
2k-1
Biased
X+B
B
Representation
for X<0
2k-1+|X|
X+B
typical B=2k-1 or 2k-1-ulp
Complement
X
Two’s
complement
X
One’s
complement
X
0, M mod 2k
0
0, 2k-ulp
M-|X|=M+X
2k-|X|= X  ulp
2k-ulp-|X|= X
Value of a number in the signed representations
Representation
Signedmagnitude
Value of
(xk-1 xk-2 … x1 x0.x-1 … x-l)
X  ( 1)
x k 1
k 2
i
x

2
i
i  l
k 1
Biased
Two’s
complement
X   xi  2i  B
i  l
X   xk 1 2
k 1
k 2
  xi  2i
i  l
k 2
One’s
complement
X   xk 1 (2k 1  ulp )   xi  2i
i  l
Extending the number of bits of a signed number
xk-1 xk-2 … x1 x0 . x-1 x-2 … x-l
X
Y
yk’-1 yk’-2 … yk yk-1 yk-2 … y1 y0 . y-1 y-2 … y-l y-(l+1) … y-l’
signed-magnitude
xk-1 0 0 0 0 0 0 0 xk-2 … x1 x0 . x-1 x-2 … x-l 0 0 0 0 0 0
two’s complement
xk-1 xk-1 xk-1 . . .xk-1 xk-2 … x1 x0 . x-1 x-2 … x-l 0 0 0 0 0 0
one’s complement
xk-1 xk-1 xk-1 . . .xk-1 xk-2 … x1 x0 . x-1 x-2 … x-l xk-1 . . . .xk-1
biased
xk-1 x k 1 . . . x k 1 xk-2 … x1 x0 . x-1 x-2 … x-l 0 0 0 0 0 0
Generalized Complement Representation
Generalized complement representation
of signed integers
Generalized complement representation
of signed integers
Floating Point Representations
The ANSI/IEEE standard floatingpoint number representation formats
Short (32-bit) format
8 bits,
bias = 127,
–126 to 127
Sign Exponent
11 bits,
bias = 1023,
–1022 to 1023
23 bits for fractional part
(plus hidden 1 in integer part)
IEEE 754 Standard
(now being revised
to yield IEEE 754R)
Significand
52 bits for fractional part
(plus hidden 1 in integer part)
Long (64-bit) format
Table 17.1 Some features of the ANSI/IEEE standard
floatingpoint number representation formats
Exponent Encoding
Exponent encoding in 8 bits for the single/short (32-bit)
ANSI/IEEE format
Decimal code
Hex code
Exponent value
0
00
1
01
–126
126 127 128
7E 7F 80
–1
0
+1
254 255
FE FF
+127
1.f  2e
f = 0: Representation of 0
f  0: Representation of denormals,
0.f  2–126
f = 0: Representation of 
f  0: Representation of NaNs
Fig. 17.4 Denormals in the IEEE single-precision format.
Little-Endian vs. Big-Endian
Representation of
Integers
Little-Endian vs. Big-Endian Representation
A0 B1 C2 D3 E4 F5 67 8916
MSB
LSB
Little-Endian
Big-Endian
0
MSB = A0
B1
C2
D3
E4
F5
67
LSB = 89
address
MAX
LSB = 89
67
F5
E4
D3
C2
B1
MSB = A0
Little-Endian vs. Big-Endian Camps
0
MSB
LSB
...
...
address
MSB
LSB
MAX
Big-Endian
Motorola 68xx, 680x0
Bi-Endian
IBM
Hewlett-Packard
Sun SuperSPARC
Internet TCP/IP
Motorola Power PC
Silicon Graphics MIPS
Little-Endian
Intel
AMD
DEC VAX
RS 232
Little-Endian vs. Big-Endian
Origin of the terms
Jonathan Swift, Gulliver’s Travels
• A law requiring all citizens of Lilliput to break their soft-eggs
at the little ends only
• A civil war breaking between the Little Endians and
the Big-Endians, resulting in the Big Endians taking refuge on
a nearby island, the kingdom of Blefuscu
• Satire over holy wars between Protestant Church of England
and the Catholic Church of France
Little-Endian vs. Big-Endian
Advantages and Disadvantages
Big-Endian
• easier to determine a sign of
the number
• easier to compare two numbers
• easier to divide two numbers
• easier to print
Little-Endian
• easier addition and multiplication
of multiprecision numbers
Pointers (1)
Big-Endian
0
address
MAX
Little-Endian
int * iptr;
89
67
F5
E4
D3
C2
B1
A0
(* iptr) = 8967;
iptr+1
(* iptr) = 6789;
Pointers (2)
Big-Endian
0
address
MAX
Little-Endian
long int * lptr;
89
67
F5
E4
D3
C2
B1
A0
(* lptr) = 8967F5E4;
lptr + 1
(* lptr) = E4F56789;
Polynomial Representation
of the Galois Field
elements
Evariste Galois (1811-1832)
Evariste Galois (1811-1832)
Studied the problem of finding algebraic solutions for the general
equations of the degree  5, e.g.,
f(x) = a5x5+ a4x4+ a3x3+ a2x2+ a1x+ a0 = 0
Answered definitely the question which specific equations of
a given degree have algebraic solutions
On the way, he developed group theory,
one of the most important branches of modern mathematics.
Evariste Galois (1811-1832)
1829
Galois submits his results for the first time to
the French Academy of Sciences
Reviewer 1
Augustin-Luis Cauchy forgot or lost the communication
1830
Galois submits the revised version of his manuscript,
hoping to enter the competition for the Grand Prize
in mathematics
Reviewer 2
Joseph Fourier – died shortly after receiving the manuscript
1831 Third submission to the French Academy of Sciences
Reviewer 3
Simeon-Denis Poisson – did not understand the manuscript
and rejected it.
Evariste Galois (1811-1832)
May 1832
Galois provoked into a duel
The night before the duel he writes a letter to his friend
containing the summary of his discoveries.
The letter ends with a plea:
“Eventually there will be, I hope, some people who
will find it profitable to decipher this mess.”
May 30, 1832 Galois is grievously wounded in the duel and dies
in the hospital the following day.
1843
Galois manuscript rediscovered by Joseph Liouville
1846
Galois manuscript published for
the first time in a mathematical journal
Field
Set F, and two operations typically denoted by
(but not necessarily equivalent to)
+ and *
Set F, and definitions of these two operations must
fulfill special conditions.
Examples of fields
Infinite fields
{
R= set of real numbers,
+ addition of real numbers
* multiplication of real numbers
}
Finite fields
{
set Zp={0, 1, 2, … , p-1},
+ (mod p): addition modulo p,
* (mod p): multiplication modulo p
}
Finite Fields = Galois Fields
GF(pm)
GF(p)
Arithmetic
operations
present
in many libraries
p – prime
pm – number of
elements in the field
GF(2m)
Polynomial basis
representation
Most significant
special cases
Normal basis
representation
Fast in hardware
Fast squaring
Elements of the Galois Field GF(2m)
Binary representation
(used for storing and processing in computer systems):
A = (am-1, am-2, …, a2, a1, a0)
ai  {0, 1}
Polynomial representation
(used for the definition of basic arithmetic operations):
m-1
A(x) =  aixi = am-1xm-1 + am-2xm-2 + …+ a2x2 + a1x+a0
i=0
 multiplication
+ addition modulo 2 (XOR)
Addition and Multiplication
in the Galois Field GF(2m)
Inputs
A = (am-1, am-2, …, a2, a1, a0)
B = (bm-1, bm-2, …, b2, b1, b0)
ai , bi  {0, 1}
Output
C = (cm-1, cm-2, …, c2, c1, c0)
ci  {0, 1}
Addition in the Galois Field GF(2m)
Addition
A
B
C



A(x)
B(x)
C(x) = A(x) + B(x) =
= (am-1+bm-1)xm-1 + (am-2+bm-2)xm-2+ …+
+ (a2+b2)x2 + (a1+b1)x + (a0+b0) =
= cm-1xm-1 + cm-2xm-2 + …+ c2x2 + c1x+c0
 multiplication
+ addition modulo 2 (XOR)
ci = ai + bi = ai XOR bi
C = A XOR B
Multiplication in the Galois Field GF(2m)
Multiplication
A
B
C



A(x)
B(x)
C(x) = A(x)  B(x) mod P(X)
= cm-1xm-1 + cm-2xm-2 + …+ c2x2 + c1x+c0
P(x) - irreducible polynomial of the degree m
P(x) = pmxm + pm-1xm-1 + …+ p2x2 + p1x+p0