1. 1 Basic Dividers Lecture 9

2. Required Reading 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: http://www.ece.ucsb.edu/~parhami/text_comp_arit_1ed.htm#errors Behrooz Parhami, Computer Arithmetic: Algorithms and Hardware Design

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. 5 Notation z Dividend z 2k-1 z 2k-2... z 2 z 1 z 0 d Divisor d k-1 d k-2... d 1 d 0 q Quotient q k-1 q k-2... q 1 q 0 s Remainder s k-1 s k-2... s 1 s 0 (s = z - dq)

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

7. 7 Unsigned Integer Division Overflow z = z H 2 k + z L < d 2 k Condition for no overflow (i.e. q fits in k bits): z = q d + s < (2 k -1) d + d = d 2 k z H < d 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

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

9. 9 Sequential Integer Division Justification s (1) = 2 z - q k-1 (2 k d) s (2) = 2(2 z - q k-1 (2 k d)) - q k-2 (2 k d) s (3) = 2(2(2 z - q k-1 (2 k d)) - q k-2 (2 k d)) - q k-3 (2 k d)... s (k) = 2(... 2(2(2 z - q k-1 (2 k d)) - q k-2 (2 k d)) - q k-3 (2 k d)... - q 0 (2 k d) = = 2 k z - (2 k d) (q k-1 2 k-1 + q k-2 2 k-2 + q k-3 2 k-3 + … + q 0 2 0 ) = = 2 k z - (2 k d) q = 2 k (z - d q) = 2 k s

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

11. Fractional Division

12. 12 Unsigned Fractional Division z frac Dividend.z -1 z -2... z -(2k-1) z -2k d frac Divisor.d -1 d -2... d -(k-1) d -k q frac Quotient.q -1 q -2... q -(k-1) q -k s frac Remainder.000…0s -(k+1)... s -(2k-1) s -2k k bits

13. 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 ) z frac = q frac d frac + s frac For Fractions: where z frac = z 2 -2k d frac = d 2 -k q frac = q 2 -k s frac = s 2 -2k

14. 14 Unsigned Fractional Division Overflow Condition for no overflow: z frac < d frac

15. 15 Sequential Fractional Division Basic Equations s (0) = z frac s (j) = 2 s (j-1) - q -j d frac 2 k · s frac = s (k) s frac = 2 -k · s (k)

16. 16 Sequential Fractional Division Justification s (1) = 2 z frac - q -1 d frac s (2) = 2(2 z frac - q -1 d frac ) - q -2 d frac s (3) = 2(2(2 z frac - q -1 d frac ) - q -2 d frac ) - q -3 d frac... s (k) = 2(... 2(2(2 z frac - q -1 d frac ) - q -2 d frac ) - q -3 d frac... - q -k d frac = = 2 k z frac - d frac (q -1 2 k-1 + q -2 2 k-2 + q -3 2 k-3 + … + q -k 2 0 ) = = 2 k z frac - d frac 2 k (q -1 2 -1 + q -2 2 -2 + q -3 2 -3 + … + q -k 2 -k ) = = 2 k z frac - (2 k d frac ) q frac = 2 k (z frac - d frac q frac ) = 2 k s frac

17. Restoring Unsigned Integer Division

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

19. 19 Fig. 13.6 Example of restoring unsigned division.

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

21. Non-Restoring Unsigned Integer Division

22. s (1) = 2 z - 2 k d for j = 2 to k if s (j-1)  0 q k-(j-1) = 1 s (j) = 2 s (j-1) - 2 k d else q k-(j-1) = 0 s (j) = 2 s (j-1) + 2 k d end for if s (k)  0 q 0 = 1 else q 0 = 0 Correction step

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

24. 24 Fig. 13.7 Example of nonrestoring unsigned division.

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

26. Signed Integer Division

27. 27 s (j) = 2 s (j-1) s (j+1) = 2 s (j) - 2 k d = = 4 s (j-1) - 2 k d s (j) = 2 s (j-1) - 2 k d s (j+1) = 2 s (j) + 2 k d = = 2 (2 s (j-1) - 2 k d) + 2 k d = = 4 s (j-1) - 2 k d Restoring division Non-Restoring division Non-Restoring Unsigned Integer Division Justification s (j-1) ≥ 0 2 s (j-1) - 2 k d < 0 2 (2 s (j-1) ) - 2 k d ≥ 0

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

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

30. Non-Restoring Signed Integer Division

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

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

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

34. 34 BSD  2’s Complement Conversion q = (q k-1 q k-2... q 1 q 0 ) BSD = = (p k-1 p k-2... p 1 p 0 1) 2’s complement where pipi qiqi 0 1 1 Example: 1 -1 1 1 q BSD p q 2’scomp 1 0 1 1 0 0 1 1 1 = 0 1 1 1 no overflow if p k-2 = p k-1 (q k-1  q k-2 )

35. 35 Fig. 13.10 Shift-subtract sequential nonrestoring divider. NOTE: 1.q k-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