1. Number Representation
ECE 645: Lecture 5
Number Representation
Part 2
Little-Endian vs. Big-Endian Representations
Floating Point Representations
Rounding
Representation of the Galois Field elements

2. Required Reading Endianness, from Wikipedia, the free encyclopedia
Behrooz Parhami,
Computer Arithmetic: Algorithms and Hardware Design
Chapter 17, Floating-Point Representations

3. Little-Endian vs. Big-Endian
Representation of
Integers

4. Little-Endian vs. Big-Endian Representation
A0 B1 C2 D3 E4 F
MSB
LSB
Little-Endian
Big-Endian
MSB = A0
LSB = 89 B1 67 C2 F5 D3 E4
address E4 D3 F5 C2 67 B1
LSB = 89
MSB = A0
MAX

5. Little-Endian vs. Big-Endian Camps
MSB
LSB
. . .
. . .
address
LSB
MSB
MAX
Little-Endian
Big-Endian
Motorola 68xx, 680x0
Bi-Endian
Intel
IBM
AMD
Motorola Power PC
Hewlett-Packard
DEC VAX
Silicon Graphics MIPS
Sun SuperSPARC
RS 232
Internet TCP/IP

6. 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

8. Little-Endian vs. Big-Endian
Advantages and Disadvantages
Big-Endian
Little-Endian
easier to determine a sign of
the number
easier to compare two numbers
easier to divide two numbers
easier to print
easier addition and multiplication
of multiprecision numbers

9. Pointers (1) Big-Endian Little-Endian MAX int * iptr; 89
int * iptr; 89
(* iptr) = 8967;
(* iptr) = 6789; 67 F5
address E4
iptr+1 D3 C2 B1 A0
MAX

10. Pointers (2) Big-Endian Little-Endian MAX long int * lptr; 89
long int * lptr; 89
(* lptr) = 8967F5E4;
(* lptr) = E4F56789; 67 F5
address E4 D3 C2
lptr + 1 B1 A0
MAX

11. Floating Point Representations

16. The ANSI/IEEE standard floating-point number representation formats
Originally IEEE
Superseded by IEEE Standard.

19. Table 17.1 Some features of the ANSI/IEEE standard floatingpoint number representation formats

20. Exponent encoding in 8 bits for the single/short (32-bit)
ANSI/IEEE format
Decimal code 1
126
127
128
254
255
Hex code 00 01 7E 7F 80 FE FF
Exponent value
–126 –1 +1
+127
f = 0: Representation of 0
f  0: Representation of denormals,
0.f  2–126
f = 0: Representation of 
f  0: Representation of NaNs
1.f  2e

21. Fig. 17.4 Denormals in the IEEE single-precision format.

25. New IEEE Standard
Basic Formats

26. Binary Interchange Formats
New IEEE Standard
Binary Interchange Formats

27. Requirements for Arithmetic
Results of the 4 basic arithmetic operations (+, -, , ) as well as square-rooting must match those obtained
if all intermediate computations were infinitely precise
That is, a floating-point arithmetic operation should introduce no more imprecision than the error attributable to the final rounding of a result that has no exact representation (this is the best possible)
Example:
( )  ( )
Exact result
Rounded result Error = ½ ulp

28. Rounding 101

29. Rounding Modes The IEEE 754-2008 standard includes five
Default:
Round to nearest, ties to even (rtne)
Optional:
Round to nearest, ties away from 0 (rtna)
Round toward zero (inward)
Round toward + (upward)
Round toward – (downward)

30. Required Reading Parhami, Chapter 17.5, Rounding schemes
Rounding Algorithms 101

31. Rounding
Rounding occurs when we want to approximate a more precise number (i.e. more fractional bits L) with a less precise number (i.e. fewer fractional bits L')
Example 1:
old: (K=6, L=8)
new: (K'=6, L'=2)
Example 2:
new: (K'=6, L'=0)
The following pages show rounding from L>0 fractional bits to L'=0 bits, but the mathematics hold true for any L' < L
Usually, keep the number of integral bits the same K'=K

32. Rounding Equation y = round(x)
Whole part
Fractional part
xk–1xk– x1x0 . x–1x– x–l yk–1yk– y1y0
Round
y = round(x)

33. Rounding Techniques There are different rounding techniques:
1) truncation
results in round towards zero in signed magnitude
results in round towards -∞ in two's complement
2) round to nearest number
3) round to nearest even number (or odd number)
4) round towards +∞
Other rounding techniques
5) jamming or von Neumann
6) ROM rounding
Each of these techniques will differ in their error depending on representation of numbers i.e. signed magnitude versus two's complement
Error = round(x) – x

34. 1) Truncation
The simplest possible rounding scheme: chopping or truncation
xk–1xk– x1x0 . x–1x– x–l xk–1xk– x1x0
trunc
ulp
Truncation in signed-magnitude results in a number chop(x) that is always of smaller magnitude than x. This is called round towards zero or inward rounding
(3.5)10  011 (3)10
Error = -0.5
(-3.5)10  111 (-3)10
Error = +0.5
Truncation in two's complement results in a number chop(x) that is always smaller than x. This is called round towards -∞ or downward-directed rounding
(-3.5)10  100 (-4)10

35. Truncation Function Graph: chop(x)– 4 3 2 1
chop( x ) x – 4 3 2 1 – 4 3 2 1
Fig Truncation or chopping of a signed-magnitude number (same as round toward 0).
Fig Truncation or chopping of a 2’s-complement number (same as round to -∞).

36. Bias in two's complement truncation
X (binary)
X (decimal)
chop(x) (binary)
chop(x)
(decimal)
Error
011.00 3
011
011.01
3.25
-0.25
011.10
3.5
-0.5
011.11
3.75
-0.75
100.01
-3.75
100 -4
100.10
-3.5
100.11
-3.25
101.00 -3
101
Assuming all combinations of positive and negative values of x equally possible, average error is
In general, average error = -(2-L'-2-L )/2, where L' = new number of fractional bits

37. Implementation truncation in hardware
Easy, just ignore (i.e. truncate) the fractional digits from L to L'+1
xk-1 xk-2 .. x1 x0. x-1 x-2 .. x-L
= yk-1 yk-2 .. y1 y0.
ignore (i.e. truncate the rest)

38. 2) Round to nearest number
Rounding to nearest number what we normally think of when say round
(2.25)10  010 (2)10
Error = -0.25
(2.75)10  011 (3)10
Error = +0.25
(2.00)10  010 (2)10
Error = +0.00
(2.5)10  011 (3)10
Error = +0.5 [round-half-up (arithmetic rounding)]
(2.5)10  010 (2)10
Error = -0.5 [round-half-down]

39. Round-half-up: dealing with negative numbers
Rounding to nearest number what we normally think of when say round
(-2.25)10  110 (-2)10
Error = +0.25
(-2.75)10  101 (-3)10
Error = -0.25
(-2.00)10  110 (-2)10
Error = +0.00
(-2.5)10  110 (-2)10
Error = +0.5 [asymmetric implementation]
(-2.5)10  101 (-3)10
Error = -0.5 [symmetric implementation]

40. Round to Nearest Function Graph: rtn(x) Round-half-up version
Asymmetric implementation
Symmetric implementation

41. Bias in two's complement round to nearest Round-half-up asymmetric implementation
X (binary)
X (decimal)
rtn(x) (binary)
rtn(x)
(decimal)
Error
010.00 2
010
010.01
2.25
-0.25
010.10
2.5
011 3
+0.5
010.11
2.75
+0.25
101.01
-2.75
101 -3
101.10
-2.5
110 -2
101.11
-2.25
110.00
Assuming all combinations of positive and negative values of x equally possible, average error is
Smaller average error than truncation, but still not symmetric error
We have a problem with the midway value, i.e. exactly at 2.5 or -2.5 leads to positive error bias always
Also have the problem that you can get overflow if only allocate K' = K integral bits
Example: rtn(011.10)  overflow
This overflow only occurs on positive numbers near the maximum positive value, not on negative numbers

42. Implementing round to nearest (rtn) in hardware Round-half-up asymmetric implementation
Two methods
Method 1: Add '1' in position one digit right of new LSB (i.e. digit L'+1) and keep only L' fractional bits
xk-1 xk-2 .. x1 x0. x-1 x-2 .. x-L
+ 1
= yk-1 yk-2 .. y1 y0. y-1
Method 2: Add the value of the digit one position to right of new LSB (i.e. digit L'+1) into the new LSB digit (i.e. digit L) and keep only L' fractional bits
+ x-1
yk-1 yk-2 .. y1 y0.
ignore (i.e. truncate the rest)
ignore (i.e truncate the rest)

43. Round to Nearest Even Function Graph: rtne(x)
To solve the problem with the midway value we implement round to nearest-even number (or can round to nearest odd number)
Fig R* rounding or rounding to the nearest odd number.
Fig Rounding to the nearest even number.

44. Bias in two's complement round to nearest even (rtne)
X (binary)
X (decimal)
rtne(x) (binary)
rtne(x)
(decimal)
Error
000.10
0.5
000
-0.5
001.10
1.5
010 2
+0.5
010.10
2.5
011.10
3.5
0100 (overfl) 4
100.10
-3.5
100 -4
101.10
-2.5 -2
110.10
-1.5
111.10
average error is now 0 (ignoring the overflow)
cost: more hardware

45. 4) Rounding towards infinity
We may need computation errors to be in a known direction
Example: in computing upper bounds, larger results are acceptable, but results that are smaller than correct values could invalidate upper bound
Use upward-directed rounding (round toward +∞)
up(x) always larger than or equal to x
Similarly for lower bounds, use downward-directed rounding (round toward -∞)
down(x) always smaller than or equal to x
We have already seen that round toward -∞ in two's complement can be implemented by truncation

46. Rounding Toward Infinity Function Graph: up(x) and down(x)
down(x) can be implemented by chop(x) in two's complement

47. Two's Complement Round to Zero
Two's complement round to zero (inward rounding) also exists
inward( x ) – 4 3 2 1

48. Other Methods
Note that in two's complement round to nearest (rtn) involves an addition which may have a carry propagation from LSB to MSB
Rounding may take as long as an adder takes
Can break the adder chain using the following two techniques:
Jamming or von Neumann
ROM-based

49. 5) Jamming or von Neumann
Chop and force the LSB of the result to 1
Simplicity of chopping, with the near-symmetry or ordinary rounding
Max error is comparable to chopping (double that of rounding)

50. 6) ROM Rounding
Fig ROM rounding with an 8  2 table.
Example: Rounding with a
32  4 table
Rounding result is the same as that of the round to nearest scheme in 31 of the 32 possible cases, but a larger error is introduced when
x3 = x2 = x1 = x0 = x–1 = 1
xk– x4x3x2x1x0 . x–1x– x–l xk– x4y3y2y1y0
ROM
ROM data
ROM address

51. Representation
of the Galois Field
elements

52. Evariste Galois ( )

53. Evariste Galois ( )
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.

54. Evariste Galois ( )
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.

55. Evariste Galois (1811-1832) May 1832 Galois provoked into a duel
The night before the duel he wrote a letter to his friend
containing the summary of his discoveries.
The letter ended with a plea:
“Eventually there will be, I hope, some people who
will find it profitable to decipher this mess.”
May 30, Galois was grievously wounded in the duel and died
in the hospital the following day.
Galois manuscript rediscovered by Joseph Liouville
1846 Galois manuscript published for
the first time in a mathematical journal.

57. 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.

58. 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 }

59. Finite Fields = Galois Fields
p – prime
pm – number of
elements in the field
GF(pm)
GF(p)
GF(2m)
Most significant
special cases
Arithmetic
operations
present
in many libraries
Polynomial basis
representation
Normal basis
representation
Fast in hardware
Fast squaring

60. Quotient and remainder
Given integers a and n, n>0
! q, r  Z such that
a = q n + r and  r < n a
q – quotient
r – remainder
(of a divided by n)
q =
= a div n n a
r = a - q n = a –
 n = n
= a mod n

61. mod 5 =
-32 mod 5 =

62. Two integers a and b are congruent modulo n
Integers coungruent modulo n
Two integers a and b are congruent modulo n
(equivalent modulo n)
written a  b
iff
a mod n = b mod n or
a = b + kn, k  Z
n | a - b

63. Laws of modular arithmetic

64. Rules of addition, subtraction and multiplication
modulo n
a + b mod n = ((a mod n) + (b mod n)) mod n
a - b mod n = ((a mod n) - (b mod n)) mod n
a  b mod n = ((a mod n)  (b mod n)) mod n

65. 9 · 13 mod 5 =
25 · 25 mod 26 =

66. Laws of modular arithmetic
Regular addition
Modular addition
a+b = a+c
iff
b=c
a+b  a+c (mod n)
iff
b  c (mod n)
Regular multiplication
Modular multiplication
If a  b  a  c (mod n)
and gcd (a, n) = 1
then
b  c (mod n)
If a  b = a  c
and a  0
then
b = c

67. Modular Multiplication: Example x
6  x mod 8 x
5  x mod 8

68. Finite Fields = Galois Fields
p – prime
pm – number of
elements in the field
GF(pm)
GF(p)
GF(2m)
Most significant
special cases
Arithmetic
operations
present
in many libraries
Polynomial basis
representation
Normal basis
representation
Fast in hardware
Fast squaring

69. 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)

70. 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}

71. Addition in the Galois Field GF(2m)
A  A(x)
B  B(x)
C  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

72. Multiplication in the Galois Field GF(2m)
A  A(x)
B  B(x)
C  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