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Fixed-Point Negative NumbersTwo Common Forms:
1. Signed-Magnitude Form2. Complement Forms
Signed-Magnitude Numbers
• First Digit is Sign Digit, Remaining n-1 are the Magnitude
• Convention (binary)
– 0 is a Positive Sign bit
– 1 is a Negative Sign bit
• Convention (non-binary)
– 0 is a Positive Sign digit
– -1 is a Negative Sign digit
• Only 2 • n-1 Digit Sequences are Utilized
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Signed-Magnitude Example
Largest Representable Value is:
2 0 1 2
2 10
10
10
1101.110 1 (1 2 1 2 1 2 1 2 )
1 11 4 1 5.75
2 4
7 4 1 3 3 2 10s d n k k m
2 1 0 1 2 3
2 10
10 10
0111.111 1 (1 2 1 2 1 2 1 2 1 2 1 2 )
1 1 1 71 4 2 1 7
2 4 8 8
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Signed-Magnitude Example (cont)7 4 1 3 3 2 10s d n k k m
2 1 0 1 2 310
10 10
0111.111 1 (1 2 1 2 1 2 1 2 1 2 1 2 )
1 1 1 71 4 2 1 7
2 4 8 8
3 11 1 2
8ulp
1 3
1
1 3, 2 8
11 ( )
8
k s
m
s
k
ulp
1
11 ( )
[ , ]
k MAX
k
MIN
MIN MAX
X ulp
X ulp
X X X
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Signed-Magnitude Ternary Example7 4 1 3 3 3 10s d n k k m
2 0 1 2
10
10 10
10
2102.120 1 (1 3 2 3 1 3 2 3 )
1 2 51 9 2 11
3 9 9(11.555555 )
1 1
1 min{ 0} ( ) 27
m
i sulp x
1
11 ( )
[ , ]
k
MAX
k
MIN
MIN MAX
X ulp
X ulp
X X X
{0,1,2}i x
Notice that fractional part is infinite in =10 but finite in =3
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Signed-Magnitude Ternary Bounds7 4 1 3 3 3 10s d n k k m
2 1 0 1 2 3
3 10
10
1
0( 1)( 1)( 1).( 1)( 1)( 1)
0222.222 1 (2 3 2 3 2 3 2 3 2 3 2 3 )
2 2 2 261 18 6 2 263 9 27 27
127
27
MAX
k
X
ulp
{0,1,2}i x
Positive Numbers:1
[ 0, ]k
ulp
Negative Numbers:1
[ ( ), 0]k
ulp
Range:
1 1[ ( ), ]
k k ulp ulp
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Signed-Magnitude Comments
• Two Representations for zero, +0 and – 0
• Addition of +K and – K is not zero
EXAMPLE
• Disadvantage since algorithm requires comparison
of signs and, if different, comparison of magnitudes
10001010.002
+00001010.002
10010100.002
-1010+1010 Yields a Sum of – 2010!!!!!
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Complement Representations
• Two Types of Complement Representations
1. Radix Complement (binary – 2’s-complement)
2. Diminished-Radix Complement (binary – 1’s-complement)
• Positive Values Represented Same Way as Signed
Magnitude for Both Types
• Negative Value, -Y , Represented as R-Y Where R is a Constant
• Obeys the Identity: ( ) ( )Y R R Y
R R Y
Y
• Advantage is No Decisions Needed Based on Operand
Sign Before Operations are Applied
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Complement Representation Example
• If |Y | > X , Then the Answer is R - (Y - X )
• If X > |Y|, Then the Answer Should be X - Y
– But X + ( R - Y ) = R + ( X - Y ),
Thus R Must be Discarded!
• Solution is to Choose the Value of R Carefully
( ) [ ( )]
[ ]
( )
X R Y X R Y
R Y X
R Y X
• X is Positive, Y is Negative, Compute X + Y
Using Complement Representation
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Requirements for Complementation Value, R
• Select R to Simplify (or Eliminate) Correction forthe X > |Y | Case
• Calculation of Complement of Y or ( R-Y ) Should be
Simple and Fast• Definition of Complement for Single Digit, xi
• Definition of Digit Complement for a Word, X
( 1)i i x x
1 2( )k k m X x x x
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Complementation Value, R
• Add Word and Complement Together:
1 2
1 2
( )
( )
( 1)( 1) ( 1)
0 0 1
1 0 0 0
k k m
k k m
X x x x
X x x x
ulp
Answer to
AdditionNow Add1 ulp • Therefore, we see that:
k
k
X X ulp
X X ulp
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Radix-Complement Form
• The Radix Complement Form is Defined When:
• Using k is Convenient Since Storing Result in Register of
Length n Causes MSD of 1 to be Discarded due to Finite
Register Length
• Therefore, it is Easy to Compute the Complement of X by:
1. Take the Digit Complement of X
2. Add 1ulp to Complement
k
k
X X ulp
X X ulp
k
k
R
R X X X ulp
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Radix-Complement Form (cont)
• No Correction is Needed When We have Positive X and
Negative Y Such That:
• Since R= k
( ) 0 X R Y
( ) ( )k
X R Y R X Y
X Y
• And k is discarded Due to Finite Register Length
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Radix-Complement Example
• Since n = m + k
m = 0• Therefore 1 ulp = 20 = 1
• Given X , the radix complement (2’s complement) is:
2 4k n
1k R X X X ulp X
• Range of Positive Numbers is [0000,0111]
• 2’s Complement of Largest, 0111:
2 10
1000; 1 1001 7 X X
• In Radix Complement, There is a Single Representation of Zero
(0000) and Each Positive Number has Corresponding Negative
Number With MSB=1
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Radix-Complement Example
2 4k n
• In Radix Complement, There is a Single Representation of Zero
(0000) and Each Positive Number has Corresponding Negative
Number With MSB=1
• Accounts for 1(zero)+7(pos.)+7(neg.), But Extra Bit Pattern Left
• One Additional Negative Number, 10002=-810, -810 X +710
10 10
10 10
10 10
0010 (2 ) 0111 (7 )
1001 ( 7 ) 1110 ( 2 )0 1011 ( 5 ) 1 0101 (5 )
1011 0100 0100 1 0101 ( )decode
Same to Encode
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Diminished-Radix Complement
2 4k n
• In Diminished Radix Complement, the Complementation
Process is Easier Since the Addition of 1 ulp is Avoided
k
k
R ulp
R X ulp X X
• Range of Positive Numbers is: [00002,01112]=[010,710]
• 1’s Complement of Largest is 10002= -710
• 1’s Complement of Zero is 11112
• Two Representations of Zero!
7 7 X
• In All Cases MSB is Sign Bit
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Comparison of Two’s Complement, One’s
Complement and Signed-Magnitude
Sequence Two’sComplement
One’sComplement
Signed-Magnitude
011 3 3 3
010 2 2 2
001 1 1 1
000 0 0 0
111 -1 -0 -3
110 -2 -1 -2101 -3 -2 -1
100 -4 -3 -0
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Signed-Number Arithmetic
• Signed Magnitude – Only Use Magnitude Digits
10
10
10
0 1011 ( 11 )0 0110 ( 6 )
1 0001 ( 1 )
Carry-out
Overflow
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Radix-Complement Arithmetic
•Radix Complement; In this case 2’s Complement
10
10
10
01101 ( 13 )
11000 ( 8 )100101 ( 5 )
Carry-out
Does NOT Mean
Overflow
5 5 2n k 10 1013 8 ? X Y X Y
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2’s-Complement Overflow•If X , Y have opposite signs overflow never occurs
whether carry-out exists or not
10
10
10
00101 ( 5 )
10110 ( 10 )
11011 ( 5 )
10
10
10
01010 ( 10 )
11011 ( 5 )
1 00101 ( 5 )
•If X , Y have same sign and result sign differs,
overflow occurs
10
10
10
11001 ( 7 )
10110 ( 10 )
1 01111 ( 15 )
10
10
10
00111 ( 7 )
01010 ( 10 )
10001 ( 15 )
No
Carry-outCarry-out
Carry-out, Overflow
No
Carry-out,Overflow
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1’s-Complement Overflow• One’s complement – carry-out indicates a correction
is needed
; X Positive Value Y NegativeValue
1'k
Y R Y R Y ulp Y Y s Complement Definition
2 ( ) (2 )n n
X Y X ulp Y X Y ulp
• If X > Y , then answer should be X -Y however; register
contains X -Y -ulp since 2n is carry-out bit, therefore must
“correct” by adding 1 ulp
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Example of 1’s-Complement Overflow
10
10
10
10
01010 ( 10 )
11010 ( 5 )
1 00100 ( 4 )
1 ( )
00101 ( 5 )
ulp
So-called
“end-around”
carry
Need
Correction
Since
Overflow
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“End-Around” Carry Design
• This is “end-around” carry
– always add carry-out to LSD
Carry-out
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Other Number Systems
• Binary Number Systems are Most Common
• In terms of building “fast” systems, we should consider:
– Negative Radix
– Signed Digit
– Log (logarithm)
– Signed Log
– Complex Radix
– Mixed Radix
– Residue Number Systems
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Negative-Radix Fixed-Position Systems
1 1 1
{0,1, , 1}
( )
,
,
i
k k k i i
i i i i
i m i m i m
i
i i
r
x
X x w x x r
r i evenwr i odd
Nega-decimal example:
2 1 0
10
10
1 0
10
10
10
(192) 1 (10) 9 ( 10) 2 (10)
100 90 2 12
(12) 1 ( 10) 2 (10)
10 2 8
r
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Nega-Decimal Number System
Finite Register Length, n=3 digits:
max 09090909.0909 X
Largest Positive Value, X max:
Smallest Value, X min:
min 909090.9090 X
min 10 10 10
max 10 10 10
[090,909]
(090) 1 (090) 90(909) 1 (909) 909
X
X X
Asymmetric System!!!: 10 times more positive than negative
values represented
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Nega-Decimal Number System
Finite Register Length, n=4 digits:
10 10
[9090,0909]
9090 909
X
X
Nega-decimal System Characteristics:
Now more Negative Values than Positive
• Arithmetic Operations Same Regardless of Sign of Number
• No Signed Digit/Complement Representation Needed
• Sign of X Determined by Position of First Non-zero Digit
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Nega-Binary Number System
Negative Radix:; 2r r
How is this Addition Operation Performed?????
Example
2 2
10 10
4 (1010) (0101)
10 5
n X
X
3 2 1 0
103 2 1 0
10
3 2 1 0
10
8 4 2 1
0 ( 2) 1 ( 2) 0 ( 2) 1 ( 2) 50 1 0 1
1 ( 2) 1 ( 2) 0 ( 2) 1 ( 2) 31 1 0 1
0 ( 2) 1 ( 2) 1 ( 2) 0 ( 2) 20 1 1 0
iw value
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Nega-Binary Number System
3 2 1 02 2 2 2
8 4 2 1
0 1 0 11 1 0 1
1 1 0
0 0 0
1 1 0
0 0 1
1 0 0 1 1 0
wi Values
(5)10
(1+1=4-2)10
(0+0=0)10
(4+4=16-8)10
(0-8=-8)10
(5-3=2)-10
Carry-out
(-3)10