Exponentiation and Division

1. Lecture 9 Exponentiation and Division

2. Modular Multiplication Special Cases k bits a a a x = p = pH 2k + pL x x pL pH p a x mod 2k = pL a x mod 2k-1 = pL + pH + carry a x mod 2k+1 = pL - pH - borrow

3. Modular Multiplication Special Case (1) a x mod 2k-1 = (pH 2k + pL) mod (2k-1) = = (pH (2k mod 2k-1) + pL) mod (2k-1) = = pH + pL mod (2k-1) = = pH + pL if pH + pL < 2k - 1 pH + pL -(2k-1)if pH + pL 2k - 1 pL + pH + carry = carry = carry from addition pL + pH

4. Fig. 12.15 Design of a 4 ´ 4 modulo-15 multiplier.

5. Fig. 12.16 One way to design of a 4 ´ 4 modulo-13 multiplier.

6. Exponentiation: Y = XE mod N Right-to-left binary exponentiation Left-to-right binary exponentiation E = (eL-1, eL-2, …, e1, e0)2 Y = 1; S = X; for i=0 to L-1 { if (ei == 1) Y = Y  S mod N; S = S2 mod N; } Y = 1; for i=L-1 downto 0 { Y = Y2 mod N; if (ei == 1) Y = Y  X mod N; }

7. Exponentiation Example: Y = 712 mod 11 Right-to-left binary exponentiation Left-to-right binary exponentiation 12 = (1 1 0 0)2 i3 2 1 0 ei 1 1 0 0 Y 1 7 2 4 5 i 0 1 2 3 ei 0 0 1 1 Sbefore 7 5 3 9 Yafter 1 1 1 3 5 Safter 7 5 3 9 4 Sbefore - S before round i is computed Safter - S after round i is computed

8. Right-to-Left Binary Exponentiation in Hardware X 1 enable S Y E SQR MUL output

9. Left-to-Right Binary Exponentiation in Hardware 1 Y X Control Logic E MUL output

10. Notation z Dividend z2k-1z2k-2 . . . z2 z1 z0 d Divisor dk-1dk-2 . . . d1 d0 q Quotient qk-1qk-2 . . . q1 q0 s Remainder sk-1sk-2 . . . s1 s0 (s = z - dq)

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

12. Unsigned Integer Division Overflow Condition for no overflow: z = q d + s < (2k-1) d + d = d 2k z = zH 2k + zL < d 2k zH < d

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

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

15. 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(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

16. Unsigned Fractional Division zfrac Dividend .z-1z-2 . . . Z-(2k-1)z-2k dfrac Divisor .d-1d-2 . . . d-(k-1) d-k qfrac Quotient .q-1q-2 . . . q-(k-1) q-k sfrac Remainder .000…0s-(k+1) . . . s-(2k-1) s-2k k bits

17. 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

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

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

20. 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(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-1 2-1 + q-2 2-2 + q-3 2-3 + … + q-k2-k) = = 2k zfrac - (2k dfrac) qfrac = 2k (zfrac - dfrac qfrac) = 2k sfrac

21. 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) - qk-j (2k d) else qk-j = 0 s(j) = 2 s(j-1)

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

23. Fig. 13.6 Example of restoring unsigned division.

24. Restoring Signed Integer Division z 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

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

26. Non-Restoring Unsigned Integer Division Justification 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. Fig. 13.7 Example of nonrestoring unsigned division.

28. Fig. 13.8 Partial remainder variations for restoring andnonrestoring division.

29. Fig. 13.10 Shift-subtract sequential nonrestoring divider.

30. 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) Correction_step q = BSD_2’s_comp_conversion(q);

31. Fig. 13.9 Example of nonrestoring signed division.

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”

33. BSD  2’s Complement Conversion q = (qk-1 qk-2 . . . q1 q0)BSD = = (pk-1 pk-2 . . . p1 p0 1)2’s complement where Example: qBSD 1 -1 1 1 qi pi p -1 0 1 0 1 1 1 1 q2’scomp 0 0 1 1 1 = 0 1 1 1 no overflow if pk-2 = pk-1 (qk-1 qk-2)

34. Fig. 14.8 Block diagram of a radix-2 divider with partialremainder in stored-carry form.

35. 1 2 Using Carry-Save Adders with the Dividers sum = u = u1u0.u-1u-2u-3u-4….u-k carry = v = v1v0.v-1v-2v-3v-4….v-k t = u1u0.u-1u-2 + v1v0.v-1v-2 u + v - t = 00.00u-3u-4….u-k + 00.00v-3v-4….v-k < 0 

36. 1 1 1 1 1 2 2 2 2 2 Using Carry-Save Adders with the Dividers - 0 t - t < -  t < 0 t  0 -  u+v < u+v  0 u+v < 0 q-j = -1 q-j = 0 q-j = 1

37. Classification of Dividers Array Dividers Dividers by Convergence Sequential Radix-2 High-radix • Restoring • Non-restoring • regular • SRT • using carry save adders • SRT using carry save adders

38. Fig. 15.7 Restoring array divider composed of controlledsubtractor cells.

39. Fig. 15.8 Nonrestoring array divider built of controlledadd/subtract cells.

40. Fig. 15.9 Sequential radix-2 multiply/divide unit.