Basic Dividers Lecture 10. Required Reading Chapter 13, Basic Division Schemes 13.1, Shift/Subtract...

Basic Dividers

Lecture 10

Required Reading

Chapter 13, Basic Division Schemes13.1, Shift/Subtract Division Algorithms13.3, Restoring Hardware Dividers13.4, Non-Restoring and Signed Division

Chapter 15 Variation in Dividers15.6, Combined Multiply/Divide Units

Behrooz Parhami, Computer Arithmetic: Algorithms and Hardware Design

Notation and

Basic Equations

4

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)

5

Basic Equations of Division

z = d q + s

sign(s) = sign(z)

| s | < | d |

z > 00 s < | d |

z < 0- | d | < s 0

6

Unsigned Integer Division Overflow

z = zH 2k + zL < d 2k

Condition for no overflow (i.e. q fits in k bits):

z = q d + s < (2k-1) d + d = d 2k

zH < 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

7

Sequential Integer DivisionBasic Equations

s(0) = z

s(j) = 2 s(j-1) - qk-j (2k d) for j=1..k

s(k) = 2k s

8

Sequential Integer DivisionJustification

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

9

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

Fractional Division

11

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-2kk bits

12

Integer vs. Fractional Division

For Integers:

z = q d + s 2-2k

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

zfrac = qfrac dfrac + sfrac

For Fractions:

wherezfrac = z 2-2k

dfrac = d 2-k

qfrac = q 2-k

sfrac = s 2-2k

13

Unsigned Fractional Division Overflow

Condition for no overflow:

zfrac < dfrac

14

Sequential Fractional DivisionBasic Equations

s(0) = zfrac

s(j) = 2 s(j-1) - q-j dfrac for j=1..k

2k · sfrac = s(k)

sfrac = 2-k · s(k)

15

Sequential Fractional DivisionJustification

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

Restoring Unsigned Integer Division

17

Restoring Unsigned Integer Division

s(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) end for

18

Fig. 13.6 Example of restoring unsigned division.

19

Fig. 13.5 Shift/subtract sequential restoring divider.

Non-Restoring Unsigned Integer Division

Non-Restoring Unsigned Integer Division

s(0) = zs(1) = 2 s(0) - 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 step

22

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)

23

Example of nonrestoring unsigned division

24

Partial remainder variations for restoring andnonrestoring division

25

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 division Non-Restoring division

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

May 2012 Computer Arithmetic, Division Slide 26

Convergence of the Partial Quotient to q

In restoring division, the partial quotient converges to q from below

Example

(0 1 1 1 0 1 0 1)two / (1 0 1 0)two

(117)ten/(10)ten = (11)ten = (1011)two

In nonrestoring division, the partial quotient may overshoot q, but converges to it after some oscillations 0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111Partial quotient

Iteration0 1 2 3 4

q

q(1) q(2)

q(3)

q(4)

q(2)

Restoring

Nonrestoring

q(0)

Signed Integer Division

28

Signed Integer Division

z d

| z | | d | sign(z) sign(d)

| q | | s |

sign(s) = sign(z)

sign(q) =+

-

Unsigneddivision

sign(z) = sign(d)

sign(z) sign(d)

q s

29

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 signsSigns of q and s are derivable from signs of z and d

Examples of Signed Integer Division

Non-RestoringSigned Integer Division

31

Non-Restoring Signed Integer Division

s(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) end forq = 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

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 add––––––––––––––––––––––––s(3) 1 1 0 1 1 1 2s(3) 1 0 1 1 1 sign(s(3)) = sign(d), +(–24d) 0 0 1 1 1 so set q0 = 1 and subtract––––––––––––––––––––––––s(4) 1 1 1 1 0 sign(s(4)) sign(z),+(–24d) 0 0 1 1 1 so perform corrective subtraction––––––––––––––––––––––––s(4) 0 0 1 0 1 s 0 1 0 1 q

1 11 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

BSD 2’s Complement Conversion

q = (qk-1 qk-2 . . . q1 q0)BSD =

= (pk-1 pk-2 . . . p1 p0 1)2’s complement

where

piqi

-1 011

Example:

1 -1 1 1qBSD

p

q2’scomp

1 0 1 1

0 0 1 1 1 = 0 1 1 1

no overflow if pk-2 = pk-1 (qk-1 qk-2)

May 2012 Computer Arithmetic, Division Slide 35

Nonrestoring Hardware Divider

Quotient

k

Partial Remainder

Divisor

add/sub

k-bit adder

k

cout cin

Complement

qk–j 2s (j–1)MSB of

Divisor Sign

Complement of Partial Remainder Sign

36

Multiply/DivideUnit

37

The control unit proceeds through necessary steps for multiplication or division (including using the appropriate shift direction)

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

Multiplier x or quotient q

Mux

Adder out c

0 1

Partial product p or partial remainder s

Multiplicand a or divisor d

Shift control

Shift

Enable

in c

q k–j

MSB of 2s (j–1)

k

k

k

j x

MSB of p (j+1)

Divisor sign

Multiply/ divide control

Select

Mul Div

The slight speed penalty owing to a more complex control unit is insignificant

Multiply-Divide Unit