1 Basic Dividers Lecture 9

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: Behrooz Parhami, Computer Arithmetic: Algorithms and Hardware Design

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

Notation and Basic Equations

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 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 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 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 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 ) = = 2 k z - (2 k d) q = 2 k (z - d q) = 2 k s

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

Fractional Division

12 Unsigned Fractional Division z frac Dividend.z -1 z z -(2k-1) z -2k d frac Divisor.d -1 d d -(k-1) d -k q frac Quotient.q -1 q q -(k-1) q -k s frac Remainder.000…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 ) 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 Unsigned Fractional Division Overflow Condition for no overflow: z frac < d frac

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 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 q q … + 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

Restoring Unsigned Integer Division

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 Fig Example of restoring unsigned division.

20 Fig Shift/subtract sequential restoring divider.

Non-Restoring Unsigned Integer Division

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 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 Fig Example of nonrestoring unsigned division.

25 Fig Partial remainder variations for restoring and nonrestoring division.

Signed Integer Division

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 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 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

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

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 Example: q BSD p q 2’scomp = no overflow if p k-2 = p k-1 (q k-1  q k-2 )

35 Fig 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