Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Fig. 13.2 Examples of sequential division with integer and fractional operands.Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Fig. 13.6 Example of restoring unsigned division.Fig. 13.5 Shift/subtract sequential restoring divider.Slide 21Slide 22Slide 23Fig. 13.7 Example of nonrestoring unsigned division.Fig. 13.8 Partial remainder variations for restoring and nonrestoring division.Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Fig. 13.9 Example of nonrestoring signed division.Slide 34Fig. 13.10 Shift-subtract sequential nonrestoring divider.1Basic DividersLecture 9Required ReadingChapter 13, Basic Division Schemes13.1, Shift/Subtract Division Algorithms13.3, Restoring Hardware Dividers13.4, Non-Restoring and Signed DivisionNote errata at:http://www.ece.ucsb.edu/~parhami/text_comp_arit_1ed.htm#errorsBehrooz Parhami, Computer Arithmetic: Algorithms and Hardware DesignRecommended ReadingJ-P. Deschamps, G. Bioul, G. Sutter, Synthesis of Arithmetic Circuits: FPGA, ASIC and Embedded Systems Chapter 6, Arithmetic Operations: Division6.1, Natural Numbers6.2.1, General Algorithm6.2.2, Restoring Division Algorithm6.2.3, Base-2 Non-Restoring Division AlgorithmNotation and Basic Equations5Notationz Dividend z2k-1z2k-2 . . . z2 z1 z0d Divisor dk-1dk-2 . . . d1 d0q Quotient qk-1qk-2 . . . q1 q0s Remainder sk-1sk-2 . . . s1 s0 (s = z - dq)6Basic Equations of Divisionz = d q + ssign(s) = sign(z)| s | < | d |z > 00  s < | d |z < 0- | d | < s  07Unsigned Integer Division Overflowz = zH 2k + zL < d 2kCondition for no overflow (i.e. q fits in k bits):z = q d + s < (2k-1) d + d = d 2kzH < 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 bits8Sequential Integer DivisionBasic Equationss(0) = zs(j) = 2 s(j-1) - qk-j (2k d)s(k) = 2k s9Sequential Integer DivisionJustifications(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 s10Fig. 13.2 Examples of sequential division with integer and fractional operands.Fractional Division12Unsigned Fractional Divisionzfrac Dividend .z-1z-2 . . . z-(2k-1)z-2kdfrac Divisor .d-1d-2 . . . d-(k-1) d-kqfrac Quotient .q-1q-2 . . . q-(k-1) q-ksfrac Remainder .000…0s-(k+1) . . . s-(2k-1) s-2kk bits13Integer vs. Fractional Division For Integers:z = q d + s  2-2kz 2-2k = (q 2-k) (d 2-k) + s (2-2k)zfrac = qfrac dfrac + sfracFor Fractions:wherezfrac = z 2-2kdfrac = d 2-kqfrac = q 2-ksfrac = s 2-2k14Unsigned Fractional Division OverflowCondition for no overflow:zfrac < dfrac15Sequential Fractional DivisionBasic Equationss(0) = zfracs(j) = 2 s(j-1) - q-j dfrac2k · sfrac = s(k)sfrac = 2-k · s(k)16Sequential Fractional DivisionJustifications(1) = 2 zfrac - q-1 dfracs(2) = 2(2 zfrac - q-1 dfrac) - q-2 dfracs(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 sfracRestoring Unsigned Integer Division18Restoring Unsigned Integer Divisions(0) = zfor 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)19Fig. 13.6 Example of restoring unsigned division.20Fig. 13.5 Shift/subtract sequential restoring divider.Non-Restoring Unsigned Integer DivisionNon-Restoring Unsigned Integer Divisions(1) = 2 z - 2k dfor 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 dend forif s(k)  0 q0 = 1else q0 = 0 Correction step23Non-Restoring Unsigned Integer DivisionCorrection stepz = q d + sz = (q-1) d + (s+d)z = q’ d + s’z, q, d ≥ 0 s<0 s = 2-k · s(k)24Fig. 13.7 Example of nonrestoring unsigned division.25Fig. 13.8 Partial remainder variations for restoring andnonrestoring division.Signed Integer Division27 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 Restoring divisionNon-Restoring divisionNon-Restoring Unsigned Integer DivisionJustification s(j-1) ≥ 0 2 s(j-1) - 2k d < 0 2 (2 s(j-1) ) - 2k d ≥ 028Signed Integer Divisionzd| z | | d | sign(z) sign(d)| q | | s |sign(s) = sign(z)sign(q) =+-Unsigneddivisionsign(z) = sign(d)sign(z)  sign(d)q s29Examples of division with signed operandsz = 5 d = 3  q = 1 s = 2z = 5 d = –3  q = –1 s = 2z = –5 d = 3  q = –1 s = –2z = –5 d = –3  q = 1 s = –2Magnitudes of q and s are unaffected by input signsSigns of q and s are derivable from signs of z and d Examples of Signed Integer DivisionNon-RestoringSigned Integer Division31Non-Restoring Signed Integer Divisions(0) = zfor 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_step32Non-Restoring Signed Integer DivisionCorrection stepz = q d + sz = (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)33Fig. 13.9 Example of nonrestoring signed division.========================z 0 0 1 0 0 0 0 124d 1 1 0 0 1 –24d 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),+24d 1 1 0 0 1 so set q3 = 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), +(–24d) 0 0 1 1 1 so set q2 = 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),+24d 1 1 0 0 1 so set q1 = 1 and