1. Lecture 9
Basic Dividers

2. Computer Arithmetic: Algorithms and Hardware Design
Required Reading
Behrooz Parhami,
Computer Arithmetic: Algorithms and Hardware Design
Chapter 13, Basic Division Schemes
13.1, Shift/Subtract Division Algorithms
13.3, Restoring Hardware Dividers
13.4, Non-Restoring and Signed Division
Note errata at:

3. Recommended Reading
J-P. Deschamps, G. Bioul, G. Sutter,
Synthesis of Arithmetic Circuits: FPGA, ASIC and
Embedded Systems
Chapter 6, Arithmetic Operations: Division
6.1, Natural Numbers
6.2.1, General Algorithm
6.2.2, Restoring Division Algorithm
6.2.3, Base-2 Non-Restoring Division Algorithm

4. Notation and Basic Equations

5. Notation
z Dividend z2k-1z2k z2 z1 z0
d Divisor dk-1dk d1 d0
q Quotient qk-1qk q1 q0
s Remainder sk-1sk s1 s0
(s = z - dq)

6. Basic Equations of Division
z = d q + s
| s | < | d |
sign(s) = sign(z)
z > 0
0  s < | d |
z < 0
- | d | < s  0

7. Unsigned Integer Division Overflow
Must check overflow because obviously the quotient q can also be 2k bits.
For example, if the divisor d is 1, then the quotient q is the dividend z, which is 2k bits
Condition for no overflow (i.e. q fits in k bits):
z = q d + s < (2k-1) d + d = d 2k
z = zH 2k + zL < d 2k
zH < d

8. Sequential Integer Division
Basic Equations
s(0) = z
s(j) = 2 s(j-1) - qk-j (2k d)
s(k) = 2k s

9. Sequential Integer Division
Justification
s(1) = 2 z - qk-1 (2k d)
s(2) = 2(2 z - qk-1 (2k d)) - qk-2 (2k d)
s(3) = 2(2(2 z - qk-1 (2k d)) - qk-2 (2k d)) - qk-3 (2k d)
s(k) = 2( (2(2 z - qk-1 (2k d)) - qk-2 (2k d)) - qk-3 (2k d) . . .
- q0 (2k d) =
= 2k z - (2k d) (qk-1 2k-1 + qk-2 2k-2 + qk-3 2k-3 + … + q020) =
= 2k z - (2k d) q = 2k (z - d q) = 2k s

10. Fig. 13.2 Examples of sequential division with integer and fractional operands.

11. Fractional Division

12. Unsigned Fractional Division
zfrac Dividend z-1z z-(2k-1)z-2k
dfrac Divisor d-1d d-(k-1) d-k
qfrac Quotient q-1q q-(k-1) q-k
sfrac Remainder …0s-(k+1) s-(2k-1) s-2k
k bits

13. Integer vs. Fractional Division
For Integers:
z = q d + s
 2-2k
z 2-2k = (q 2-k) (d 2-k) + s (2-2k)
For Fractions:
zfrac = qfrac dfrac + sfrac
where
zfrac = z 2-2k
dfrac = d 2-k
qfrac = q 2-k
sfrac = s 2-2k

14. Unsigned Fractional Division Overflow
Condition for no overflow:
zfrac < dfrac

15. Sequential Fractional Division
Basic Equations
s(0) = zfrac
s(j) = 2 s(j-1) - q-j dfrac
2k · sfrac = s(k)
sfrac = 2-k · s(k)

16. Sequential Fractional Division
Justification
s(1) = 2 zfrac - q-1 dfrac
s(2) = 2(2 zfrac - q-1 dfrac) - q-2 dfrac
s(3) = 2(2(2 zfrac - q-1 dfrac) - q-2 dfrac) - q-3 dfrac
s(k) = 2( (2(2 zfrac - q-1 dfrac) - q-2 dfrac) - q-3 dfrac . . .
- q-k dfrac =
= 2k zfrac - dfrac (q-1 2k-1 + q-2 2k-2 + q-3 2k-3 + … + q-k20) =
= 2k zfrac - dfrac 2k (q q q … + q-k2-k) =
= 2k zfrac - (2k dfrac) qfrac = 2k (zfrac - dfrac qfrac) = 2k sfrac

17. Restoring Unsigned Integer Division

18. Restoring Unsigned Integer Division
s(0) = z
for j = 1 to k
if 2 s(j-1) - 2k d > 0
qk-j = 1
s(j) = 2 s(j-1) - 2k d
else
qk-j = 0
s(j) = 2 s(j-1)

19. Fig. 13.6 Example of restoring unsigned division.

20. Fig. 13.5 Shift/subtract sequential restoring divider.

21. Non-Restoring Unsigned Integer Division

22. Non-Restoring Unsigned Integer Division
s(1) = 2 z - 2k d
for j = 2 to k
if s(j-1)  0
qk-(j-1) = 1
s(j) = 2 s(j-1) - 2k d
else
qk-(j-1) = 0
s(j) = 2 s(j-1) + 2k d
end for
if s(k)  0
q0 = 1
q0 = 0
Correction step

23. Non-Restoring Unsigned Integer Division
Correction step
s = 2-k · s(k)
z = q d + s
z, q, d ≥ s<0
z = (q-1) d + (s+d)
z = q’ d + s’

24. Fig. 13.7 Example of nonrestoring unsigned division.

25. Fig. 13.8 Partial remainder variations for restoring and nonrestoring division.

26. Non-Restoring Unsigned Integer Division
Justification
s(j-1) ≥ 0
2 s(j-1) - 2k d < 0
2 (2 s(j-1) ) - 2k d ≥ 0
Restoring division
Non-Restoring division
s(j) = 2 s(j-1)
s(j+1) = 2 s(j) - 2k d =
= 4 s(j-1) - 2k d
s(j) = 2 s(j-1) - 2k d
s(j+1) = 2 s(j) + 2k d =
= 2 (2 s(j-1) - 2k d) + 2k d =
= 4 s(j-1) - 2k d

27. Signed Integer Division

28. Signed Integer Divisionz d
| z |
| d |
sign(z)
sign(d)
Unsigned
division
sign(s) = sign(z) +
sign(z) = sign(d)
| q |
| s |
sign(q) = -
sign(z)  sign(d) q s

29. Examples of Signed Integer Division
Examples of division with signed operands
z = 5 d = 3  q = 1 s = 2
z = 5 d = –3  q = –1 s = 2
z = –5 d = 3  q = –1 s = –2
z = –5 d = –3  q = 1 s = –2
Magnitudes of q and s are unaffected by input signs
Signs of q and s are derivable from signs of z and d

30. Non-Restoring Signed Integer Division

31. Non-Restoring Signed Integer Division
s(0) = z
for j = 1 to k
if sign(s(j-1)) == sign(d)
qk-j = 1
s(j) = 2 s(j-1) - 2k d = 2 s(j-1) - qk-j (2k d)
else
qk-j = -1
s(j) = 2 s(j-1) + 2k d = 2 s(j-1) - qk-j (2k d)
q = BSD_2’s_comp_conversion(q)
Correction_step

32. Non-Restoring Signed Integer Division
Correction step
s = 2-k · s(k)
z = q d + s
sign(s) = sign(z)
z = (q-1) d + (s+d)
z = q’ d + s’
z = (q+1) d + (s-d)
z = q” d + s”

33. Fig. 13.9 Example of nonrestoring signed division.
======================== z
24d
–24d
s(0)
2s(0) sign(s(0))  sign(d),
+24d so set q3 = -1 and add
––––––––––––––––––––––––
s(1)
2s(1) sign(s(1)) = sign(d),
+(–24d) so set q2 = 1 and subtract
s(2)
2s(2) sign(s(2))  sign(d),
+24d so set q1 = -1 and add
s(3)
2s(3) sign(s(3)) = sign(d),
+(–24d) so set q0 = 1 and subtract
s(4) sign(s(4))  sign(z),
+(–24d) so perform corrective subtraction
s(4) s
q ========================
p = Shift, compl MSB
Add 1 to correct
Check: 33/(-7) = -4

34. BSD  2’s Complement Conversion
q = (qk-1 qk q1 q0)BSD =
= (pk-1 pk p1 p0 1)2’s complement
where
Example:
qBSD qi pi -1 p 1 1
q2’scomp =
no overflow if pk-2 = pk (qk-1  qk-2)

35. Fig. 13.10 Shift-subtract sequential nonrestoring divider.
NOTE:
qk-j and add/subtract control determined by both divisor sign and sign of new partial remainder (via an XOR operation). Divisor sign is 0 for unsigned division.
2. note that cout is the complement of the sign of the new partial remainder, i.e. if new partial remainder = 0, then cout = 1; if new partial remainder = 1, then cout=0