1. Data Representation and Arithmetic Algorithms
Module II
Data Representation and Arithmetic Algorithms

2. Non-Restoring Division

3. END
A  A + M
START
A  0
M Divisor
Q Dividend
Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1

4. 10 / 3 A Q Cnt 0000 1010 4 M 0011 START Initial values A  0 M Divisor
END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1
10 / 3 A Q
Cnt
0000
1010 4
Initial values M
0011

5. 10 / 3 A Q Cnt 0000 1010 4 0001 0100 M 0011 START Initial values A  0
END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1
10 / 3 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift M
0011

6. 10 / 3 A Q Cnt 0000 1010 4 0001 0100 1110 M 0011 START Initial values
END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1
10 / 3 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift
1110
A - M M
0011

7. END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift
1110
A – M 3
Q0 = 0 M
0011
10 / 3

8. END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift
1110
A – M 3
Q0 = 0
1100
1000
1111
A + M M
0011
10 / 3

9. END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift
1110
A – M 3
Q0 = 0
1100
1000
1111
A + M 2 M
0011
10 / 3

10. END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift
1110
A – M 3
Q0 = 0
1100
1000
1111
A + M 2
0010 M
0011
10 / 3

11. END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift
1110
A – M 3
Q0 = 0
1100
1000
1111
A + M 2
0010 1
Q0 = 1 M
0011
10 / 3

12. END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift
1110
A – M 3
Q0 = 0
1100
1000
1111
A + M 2
0010 1
Q0 = 1
A - M M
0011
10 / 3

13. END
A  A + M
START
A  0
M Divisor
Q Dividend Count n
A < 0?
Yes No
Shift Left A,Q
A  A - M
A  A + M
Q0 = 0
Count =0?
Count  Count -1
Q0 = 1 A Q
Cnt
0000
1010 4
Initial values
0001
0100
Shift
1110
A – M 3
Q0 = 0
1100
1000
1111
A + M 2
0010 1
Q0 = 1
0011 M
0011
10 / 3

14. Floating Point Arithmetic

15. Addition and Subtraction
Phase 1: Zero check.
The process begins by changing the sign of the subtrahend if it is a subtract operation.
If either operand is 0, the other is reported as result.
Phase 2: Significand alignment.
This phase manipulate the numbers so that the two exponents are equal.

16. Addition and Subtraction
Phase 3: Addition
The two significands are added together, taking into account their signs.
If there is significand overflow by 1 digit,the result is shifted right and the exponent is incremented.
If exponent overflow occurs, this will be reported.
Phase 4: Normalization.
It consists of shifting significand digits left until the most significant digit is nonzero.

17. Example : Floating Point Adder Unit

18. Floating Point Adder Unit
The inputs to adder are two normalized floating point numbers of the form
A = a x 10p
B = b x 10q
where a and b are two fractions and p and q are their exponents.

19. Floating Point Adder Unit
Our purpose is to compute the sum
C = A + B = c x 10r = d x 10s
where r = max(p,q) and 0.1 ≤ d < 1
For example:
A= x 103
B= x 102
a = b=
p=3 & q =2

20. Floating Point Adder Unit
Operations performed are :
Compare p and q and choose the largest exponent, r = max(p,q)and compute t = |p – q|
Example:
r = max(p , q) = 3
t = |p-q| = |3-2|= 1

21. Floating Point Adder Unit
Shift right the fraction associated with the smaller exponent by t units to equalize the two exponents before fraction addition.
Example:
Smaller exponent, b=
Shift right b by 1 unit is 0.082

22. Floating Point Adder Unit
Perform fixed-point addition of two fractions to produce the intermediate sum fraction c
Example :
a = b= 0.082
c = a + b = =

23. Floating Point Adder Unit
Count the number of leading zeros (u) in fraction c and shift left c by u units to produce the normalized fraction sum d = c x 10u, with a leading bit 1. Update the large exponent s by subtracting s = r – u to produce the output exponent.
Example:
c = , u = -1  right shift
d = , s= r – u = 3-(-1) = 4
C = x 104

24. Floating Point Adder Unit
The above 4 steps can all be implemented with combinational logic circuits and the 4 stages are:
Comparator / Subtractor
Shifter
Fixed Point Adder
Normalizer (leading zero counter and shifter)

25. Example for floating-point adder
Exponents
Segment 1:
Segment 2:
Segment 3:
Segment 4: R
Adjust
exponent
Normalize
result
Add
mantissas
Align mantissas
Choose exponent
Compare
exponents
by subtraction
Difference=3-2=1
Mantissas b a A B
For example:
X=0.9504*103
Y=0.8200*102
0.082 3
S= =1.0324 -1

26. 4-STAGE FLOATING POINT ADDER
A = a x 2 p
B = b x 2 q a b A B
Exponent
subtractor
Fraction
selector
Fraction with min(p,q)
Right shifter
Other
fraction
t = |p - q|
r = max(p,q)
adder
Leading zero
counter r c
Left shifter s d
Stages: S1 S2 S3 S4
C= X + Y = d x 2s

27. Problem
Using Non-Restoring division algorithm perform
1011/0101