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ELEC 5200-001/6200-001 Computer Architecture and Design Fall 2007 Computer Arithmetic (Chapter 3)

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ELEC 5200-001/6200-001Computer Architecture and DesignFall 2007 Computer Arithmetic (Chapter 3)

Vishwani D. Agrawal

James J. Danaher Professor

Department of Electrical and Computer Engineering

Auburn University, Auburn, AL 36849

http://www.eng.auburn.edu/~vagrawal

ELEC 5200-001/6200-001 Lecture 11

- Machine instr.: add $t1, $s1, $s2
- What it means to computer:
000000 10001 10010 01000 00000 100000

Arithmetic Logic Unit

(ALU)

Control unit

Flags

Registers

Registers

ELEC 5200-001/6200-001 Lecture 11

- Hardware can only deal with binary digits, 0 and 1.
- Must represent all numbers, integers or floating point, positive or negative, by binary digits, called bits.
- Devise electronic circuits to perform arithmetic operations, add, subtract, multiply and divide, on binary numbers.

ELEC 5200-001/6200-001 Lecture 11

- Decimal system: made of 10 digits, {0,1,2, . . . , 9}
41 = 4×101 + 1×100

255 = 2×102 + 5×101 + 5×100

- Binary system: made of two digits, {0,1}
00101001= 0×27 + 0×26 + 1×25 + 0×24 +1×23 +0×22 + 0×21 + 1×20

=32 + 8 +1 = 41

11111111= 255, largest number with 8 binary digits, 28-1

ELEC 5200-001/6200-001 Lecture 11

- For decimal system, 10 is called the base or radix.
- Decimal 41 is also written as 4110 or 41ten
- Base (radix) for binary system is 2.
- Thus,41ten= 1010012 or 101001two
- Also,111ten = 1101111two
and 111two = 7ten

ELEC 5200-001/6200-001 Lecture 11

- Use fixed length binary representation
- Use left-most bit (called most significant bit or MSB) for sign:
0 for positive

1 for negative

- Example: +18ten = 00010010two
- 18ten = 10010010two

ELEC 5200-001/6200-001 Lecture 11

- Sign and magnitude bits should be differently treated in arithmetic operations.
- Addition and subtraction require different logic circuits.
- Overflow is difficult to detect.
- “Zero” has two representations:
+0ten = 00000000two

- 0ten = 10000000two

- Signed-integers are not used in modern computers.

ELEC 5200-001/6200-001 Lecture 11

- Use fixed-length representation, but no sign bit
- 1’s complement: To form a negative number, complement each bit in the given number.
- 2’s complement: To form a negative number, start with the given number, subtract one, and then complement each bit, or
first complement each bit, and then add 1.

- 2’s complement is the preferred representation.

ELEC 5200-001/6200-001 Lecture 11

- To change the sign of a binary integer simply complement (invert) each bit.
- Example: 3 = 0011, – 3 = 1100
- n-bit representation: Negation is equivalent to subtraction from 2n – 1

Infinite

universe

-9-6-30369

03691215

Modulo-16

universe

036

-6-3-0

000000110110100111001111

ELEC 5200-001/6200-001 Lecture 11

- Add 1 to 1’s-complement representation.
- Some properties:
- Only one representation for 0
- Exactly as many positive numbers as negative numbers
- Slight asymmetry – there is one negative number with no positive counterpart

ELEC 5200-001/6200-001 Lecture 11

Sign-magnitude

000 = +0

001 = +1

010 = +2

011 = +3

100 = - 0

101 = - 1

110 = - 2

111 = - 3

1’s complement

000 = +0

001 = +1

010 = +2

011 = +3

100 = - 3

101 = - 2

110 = - 1

111 = - 0

2’s complement

000 = +0

001 = +1

010 = +2

011 = +3

100 = - 4

101 = - 3

110 = - 2

111 = - 1

(Preferred)

ELEC 5200-001/6200-001 Lecture 11

000

-1

+1

0

-1

111

001

+1

Positive numbers

010

+2

Negative numbers

-2

110

011

+3

-3

101

- 4

100

Overflow

Negation

ELEC 5200-001/6200-001 Lecture 11

- Range:–2n –1 through 2n –1 – 1
- Unique zero: 00000000 . . . . . 0
- Negation rule: see slide 9.
- Expansion of bit length: stretch the left-most bit all the way, e.g., 11111101 is still – 3.
- Overflow rule: If two numbers with the same sign bit (both positive or both negative) are added, the overflow occurs if and only if the result has the opposite sign.
- Subtraction rule: for A – B, add – B to A.

ELEC 5200-001/6200-001 Lecture 11

n-2

an-1an-2 . . . a1a0 = -2n-1an-1 + Σ 2i ai

i=0

8-bit conversion box

-128 64 32 16 8 4 2 1

-128 64 32 16 8 4 2 1

1 1 1 1 1 1 0 1

Example

-128+64+32+16+8+4+1 = -128 + 125 = -3

ELEC 5200-001/6200-001 Lecture 11

- Chapter 2 in D. E. Knuth, The Art of Computer Programming: Seminumerical Algorithms, Volume II, Second Edition, Addison-Wesley, 1981.
- A. al’Khwarizmi, Hisab al-jabr w’al-muqabala, 830.

Abu Abd-Allah ibn Musa al’Khwarizmi (~780 – 850)

Donald E. Knuth (1938 - )

ELEC 5200-001/6200-001 Lecture 11

- MIPS architecture uses 32-bit numbers. What is the range of integers (positive and negative) that can be represented?
Positive integers: 0 to 2,147,483,647

Negative integers: - 1 to - 2,147,483,648

- What are the binary representations of the extreme positive and negative integers?
0111 1111 1111 1111 1111 1111 1111 1111 = 231 - 1= 2,147,483,647

1000 0000 0000 0000 0000 0000 0000 0000 = - 231 = - 2,147,483,648

- What is the binary representation of zero?
0000 0000 0000 0000 0000 0000 0000 0000

ELEC 5200-001/6200-001 Lecture 11

- Adding bits:
- 0 + 0 = 0
- 0 + 1 = 1
- 1 + 0 = 1
- 1 + 1 = (1) 0

carry

1 1 0

0 0 0 . . . . . . 0 1 1 1 two = 7ten

+ 0 0 0 . . . . . . 0 1 1 0 two = 6ten

= 0 0 0 . . . . . . 1 (1)1 (1)0 (0)1 two = 13ten

ELEC 5200-001/6200-001 Lecture 11

- Direct subtraction
- Two’s complement subtraction

0 0 0 . . . . . . 0 1 1 1 two = 7ten

- 0 0 0 . . . . . . 0 1 1 0 two = 6ten

= 0 0 0 . . . . . . 0 0 0 1two = 1ten

1 1 1 . . . . . . 1 1 0

0 0 0 . . . . . . 0 1 1 1 two = 7ten

+ 1 1 1 . . . . . . 1 0 1 0 two = - 6ten

= 0 0 0 . . . . . . 0 (1) 0 (1) 0 (0)1 two = 1ten

ELEC 5200-001/6200-001 Lecture 11

- Examples: Addition of 3-bit integers (range - 4 to +3)
- -2-3 = -5110 = -2
+ 101 = -3

= 1011 = 3 (error)

- 3+2 = 5011 = 3
010 = 2

= 101 = -3 (error)

- -2-3 = -5110 = -2

000

111

0

001

1

-1

010

– +

2

110

-2

3

-3

011

- 4

101

100

Overflow

ELEC 5200-001/6200-001 Lecture 11

- Adding two bits:
abhalf_sumcarry_out

0000

0110

1010

1101

- Half-adder circuit

a

half_sum

XOR

b

carry_out

AND

ELEC 5200-001/6200-001 Lecture 11

- One-bit full-adder truth table
abcihalf_sumcarry_outsum co

00 0000 0

00 1001 0

01 0101 0

01 1100 1

10 0101 0

10 1100 1

11 0010 1

11 1011 1

ELEC 5200-001/6200-001 Lecture 11

ci

FAi

XOR

sumi

ai

XOR

AND

bi

AND

OR

Ci+1

ELEC 5200-001/6200-001 Lecture 11

c0

a0

b0

sum0

FA0

sum1

a1

b1

FA1

sum2

a2

b2

FA2

sum31

FA31

a31

b31

ELEC 5200-001/6200-001 Lecture 11

- Longest delay path (critical path) runs from cin to sum31.
- Suppose delay of full-adder is 100ps.
- Critical path delay = 3,200ps
- Clock rate cannot be higher than 1/(3,200×1012) Hz = 312MHz.
- Must use more efficient ways to handle carry.

ELEC 5200-001/6200-001 Lecture 11

a0-a15

16-bit

ripple

carry

adder

b0-b15

cin

sum0-sum15

a16-a31

16-bit

ripple

carry

adder

0

b16-b31

0

sum16-sum31

Multiplexer

a16-a31

16-bit

ripple

carry

adder

1

b16-b31

This is a carry-select adder

1

ELEC 5200-001/6200-001 Lecture 11

- In general, any output of a 32-bit adder can be evaluated as a logic expression in terms of all 65 inputs.
- Number of levels of logic can be reduced to log2N for N-bit adder. Ripple-carry has N levels.
- More gates are needed, about log2N times that of ripple-carry design.
- Fastest design is known as carry lookahead adder.

ELEC 5200-001/6200-001 Lecture 11

Reference: J. L. Hennessy and D. A. Patterson, Computer Architecture: A Quantitative Approach, Second Edition, San Francisco, California, 1990, page A-46.

ELEC 5200-001/6200-001 Lecture 11

- Arithmetic: add, sub, addi, addu, subu, addiu, mfc0
- Data transfer: lw, sw, lhu, sh, lbu, sb, lui
- Logical: and, or, nor, andi, ori, sll, srl
- Conditional branch: beq, bne, slt, slti, sltu, sltiu
- Unconditional jump: j, jr, jal

ELEC 5200-001/6200-001 Lecture 11

- If an overflow is detected while executing add, addi or sub, then the address of that instruction is placed in a register called exception program counter (EPC).
- Instruction mfc0 can copy $epc to any other register, e.g., mfc0$s1, $epc
- Unsigned operations, addu, addiu and subu do not cause an exception or interrupt.

ELEC 5200-001/6200-001 Lecture 11

operation

c0

a0

b0

result0

ALU0

result1

a1

b1

ALU1

result2

operation

a2

b2

ALU2

ci

ai

bi

FAi

3

NOR

2

resulti

mux

OR

1

result31

ALU31

a31

b31

AND

0

ELEC 5200-001/6200-001 Lecture 11

“I’ve been waiting more than 30 years to say this: Dad, I always told you I’d come back and get my degree.”

ELEC 5200-001/6200-001 Lecture 11

“I’ve been waiting more than 30 years to say this: Dad, I always told you I’d come back and get my degree.”

ELEC 5200-001/6200-001 Lecture 11

1 0 0 0 two= 8tenmultiplicand

1 0 0 1 two= 9tenmultiplier

____________

1 0 0 0

0 0 0 0partial products

0 0 0 0

1 0 0 0

____________

1 0 0 1 0 0 0two= 72ten

Basic algorithm: n = 1, 32

If nth bit of multiplier is 1,

Add multiplicand × 2 n –1

to product

ELEC 5200-001/6200-001 Lecture 11

Start

Initialize product register to 0

Partial product number, n = 1

LSB

of multiplier

?

Add multiplicand to

product and place result

in product register

1

0

Left shift multiplicand register 1 bit

Right shift multiplier register 1 bit

n = 32

n < 32

i = ?

Done

n = n + 1

ELEC 5200-001/6200-001 Lecture 11

shift left

shift right

Multiplicand (expanded 64-bits)

32-bit multiplier

64

64

shift

Test LSB

32 times

add

64-bit ALU

LSB

= 1

LSB = 0

64

3 operations per bit:

shift right

shift left

add

Need 64-bit ALU

64-bit product register

write

ELEC 5200-001/6200-001 Lecture 11

Multiplicand

2 operations per bit:

shift right

add

32

32

add

1

Test LSB

32 times

32-bit ALU

LSB

1

32

write

64-bit product register

shift right

00000 . . . 00000 32-bit multiplier

Initialized prod. Reg.

ELEC 5200-001/6200-001 Lecture 11

0010two× 0011two = 0110two, i.e., 2ten×3ten = 6ten

ELEC 5200-001/6200-001 Lecture 11

- Convert numbers to magnitudes.
- Multiply the two magnitudes through 32 iterations.
- Negate the result if the signs of the multiplicand and multiplier differed.
- Alternatively, the previous algorithm will work with some modifications, listed next.

ELEC 5200-001/6200-001 Lecture 11

- Use one extra bit for multiplicand addition.
- Extend sign bit in right shift (Examples 1 and 2).
- If multiplier is negative, then the last addition is replaced by subtraction (Example 2). Why? See slide 14.
- See B. Parhami, Computer Architecture, New York: Oxford University Press, 2005, pp. 199-200.

ELEC 5200-001/6200-001 Lecture 11

1010two× 0011two = 101110two, i.e., -6ten×3ten = -18ten

ELEC 5200-001/6200-001 Lecture 11

1010two× 1011two = 011110two, i.e., -6ten×(-5ten) = 30ten

*Last iteration with a negative multiplier in 2’s complement.

ELEC 5200-001/6200-001 Lecture 11

- A. D. Booth, “A Signed Binary Multiplication Technique,” Quarterly Journal of Mechanics and Applied Math., vol. 4, pt. 2, pp. 236-240, 1951.
- Direct multiplication of positive and negative integers using two’s complement addition.

ELEC 5200-001/6200-001 Lecture 11

- Consider decimal multiplication:
4 5 7

9 9 9 0 1

4 5 7

4 1 1 3

4 1 1 3

4 1 1 3

4 5 6 5 4 7 5 7Three additions

- Operations for each digit of multiplier:
- Do nothing if the digit is 0
- Shift left, i.e., multiply by some power of 10
- Multiply by the digit, i.e., by a number between 1 and 9

ELEC 5200-001/6200-001 Lecture 11

- Examine multiplier:
99901 = 100000 – 100 + 1

- Multiply as follows:
457 × 100000=45700000

457 × ( - 100) = - 45700subtraction

45654300

457 × 1= 457addition

45654757

Reduced from three to two operations.

ELEC 5200-001/6200-001 Lecture 11

- Consider a multiplier, 00011110 (30)
- We can write, 30 = 32 – 2, or
00100000(32)= 25

+11111110(–2) =–21

00011110(0) 30

- We can write, 30 = 32 – 2, or

- kth bit is 1, (k-1)th bit is 0, multiplier contains -2k term
- kth bit is 0, (k-1)th bit is 1, multiplier contains 2kterm
- kth bit is 1, (k-1)th bit is 1, -2k is absent in multiplier
- kth bit is 0, (k-1)th bit is 0, 2k is absent in multiplier

ELEC 5200-001/6200-001 Lecture 11

- 7 × 3 = 21
0111multiplicand= 7

×0011(0)multiplier= 3

11111001bit-pair 10, add -7 in two’s com.

bit-pair 11, do nothing

000111bit-pair 01, add 7

bit-pair 00, do nothing

0001010121

ELEC 5200-001/6200-001 Lecture 11

- 7 × (-3) = -21
0111multiplicand= 7

×1101(0)multiplier= -3

11111001bit-pair 10, add -7 in two’s com.

0000111bit-pair 01, add 7

111001bit-pair 10, add -7 in two’s com.

bit-pair 11, do nothing

11101011- 21

ELEC 5200-001/6200-001 Lecture 11

- -7 × 3 = -21
1001multiplicand= -7 in two’s com.

×0011(0)multiplier= 3

00000111bit-pair 10, add 7

bit-pair 11, do nothing

111001bit-pair 01, add -7

bit-pair 00, do nothing

11101011- 21

ELEC 5200-001/6200-001 Lecture 11

- -7 × (-3) = 21
1001multiplicand= -7 in two’s com.

×1101(0)multiplier= -3 in two’s com.

00000111bit-pair 10, add 7

1111001bit-pair 01, add -7 in two’s com.

000111bit-pair 10, add 7

bit-pair 11, do nothing

0001010121

ELEC 5200-001/6200-001 Lecture 11

Serial multiplication

Booth algorithm

00010100 20

×00011110 30

00000000

00010100

00010100

00010100

00010100

00000000

00000000

00000000________

000001001011000 600

00010100 20

×000111100 30

111111111101100

00000010100

__________________

0000001001011000 600

Four partial product additions Two partial product additions

ELEC 5200-001/6200-001 Lecture 11

- Using repeated additions, we need as many clocks as there are bits, say n, in multiplier.
- Multiplication can be done in one clock. Of course, the period of clock will have to be longer; but may not be as long as n times.

ELEC 5200-001/6200-001 Lecture 11

Mplier_b1·Mcand

Mplier_b0·Mcand

32b

32b

Mplier_b2·Mcand

33b

p0

32b

Mplier_b3·Mcand

33b

p1

Clock period

32b

33b

p2

Mplier_b31·Mcand

32b

33b

p30

p31…p63

ELEC 5200-001/6200-001 Lecture 11

y3y2y1y0Multiplicand

x3x2x1x0Multiplier

________________________

x0y3x0y2x0y1x0y0

carry←x1y3x1y2x1y1x1y0Partial

carry←x2y3x2y2x2y1x2y0Products

carry← x3y3x3y2x3y1x3y0

__________________________________________________

p7p6p5p4p3p2p1p0

Requires three 4-bit adders. Slow.

ELEC 5200-001/6200-001 Lecture 11

y3y2y1y0Multiplicand

x3x2x1x0Multiplier

________________________

x0y3x0y2x0y1x0y0

x1y3x1y2x1y1x1y0Partial

x2y3x2y2x2y1x2y0Products

x3y3x3y2x3y1x3y0

__________________________________________________

p7p6p5p4p3p2p1p0

Note: Carry is added to the next partial product. Adding the carry

from the final stage needs an extra stage. These additions are

faster but we need four stages.

ELEC 5200-001/6200-001 Lecture 11

- Two-input AND
- Full-adder

yi xk

kth

partial

product

carry bits

from (k-1)th

product

yi x0

Full

adder

p0i = x0yi

0th partial product

carry bits

to (k+1)th

product

(k+1)th partial

product

ELEC 5200-001/6200-001 Lecture 11

y3 y2 y1 y0

x0

ppk

yj

xi

0

x1

ci

0

0

0

0

0

FA

x2

co

ppk+1

0

x3

Critical path

0

0

FA

FA

FA

FA

p3

p2

p1

p0

p7

p6

p5

p4

ELEC 5200-001/6200-001 Lecture 11

- Baugh-Wooley Algorithm: Signed product by two’s complement addition or subtraction according to the MSB’s.
- Booth multiplier algorithm
- Tree multipliers
- Reference: N. H. E. Weste and D. Harris, CMOS VLSI Design, A Circuits and Systems Perspective, Third Edition, Boston: Addison-Wesley, 2005.

ELEC 5200-001/6200-001 Lecture 11

- Separate 32-bit registers, Hi and Lo, to hold the 64-bit product.
- Multiply signed:
mult$s1, $s2# $s1 × $s2 = Hi, Lo

- Multiply unsigned:
multu$s1, $s2# $s1 × $s2 = Hi, Lo

- Product is copied into registers by
- mfhi$s1# copies Hi into $s1
- mflo$s1# copies Lo into $s1

ELEC 5200-001/6200-001 Lecture 11

1 3 Quotient

1 1 / 1 4 7Divisor / Dividend

1 1

3 7 Partial remainder

3 3

4 Remainder

0 0 0 0 1 1 0 1

1 0 1 1 / 1 0 0 1 0 0 1 1

1 0 1 1

0 0 1 1 1 0

1 0 1 1

0 0 1 1 1 1

1 0 1 1

1 0 0

ELEC 5200-001/6200-001 Lecture 11

0 0 0 1

0 0 0 0 1 1 0

1 1 0 0

1 1 0 0negative →quotient bit 0

0 1 0 0→restore remainder

0 0 0 0 1 1 0

1 1 0 0

1 1 0 1negative →quotient bit 0

0 1 0 0→restore remainder

0 0 0 1 1 0

1 1 0 0

1 1 1 1negative →quotient bit 0

0 1 0 0→restore remainder

0 0 1 1 0

1 1 0 0

00 1 0positive →quotient bit 1

- Dividend: 6 = 0110
- Divisor: 4 = 0100
- 4 = 1100

Iteration 4 Iteration 3 Iteration 2 Iteration 1

ELEC 5200-001/6200-001 Lecture 11

Start

$R=0, $M=Divisor, $Q=Dividend, count=n

$R (33 b) | $Q (32 b)

Shift 1-bit left $R, $Q

$R and $M have

one extra sign bit

beyond 32 bits.

$R ← $R - $M

No

Yes

$R < 0?

$Q0=0

$R←$R+$M

$Q0=1

Restore $R

(remainder)

count = count - 1

Done

$Q=Quotient

$R= Remainder

No

count = 0?

Yes

ELEC 5200-001/6200-001 Lecture 11

Remainder | Quotient

ELEC 5200-001/6200-001 Lecture 11

33-bit $M (Divisor)

Step 2:Subtract $R← $R - $M

33

33

33-bit ALU

Step 1: 1- bit left shift $R and $Q

32 times

33

33-bit $R (Remainder)

32-bit $Q (Dividend)

Step 3:If sign-bit ($R)=0, set Q0=1

If sign-bit ($R)=1, set Q0=0 and restore $R

Initialize

$R←0

V. C. Hamacher, Z. G. Vranesic and S. G. Zaky, Computer Organization, Fourth Edition,

New York: McGraw-Hill, 1996.

ELEC 5200-001/6200-001 Lecture 11

Initialize$R = 0 0 0 0 0$Q = 1 0 0 0$M = 0 0 0 1 1

Step 1, L-shift$R,Q = 0 0 0 0 1$Q =0 0 0 0

Step 2, Add- $M = 1 1 1 0 1

$R = 1 1 1 1 0

Step 3, Set Q0$Q =0 0 0 0

Restore+ $M = 0 0 0 1 1

$R = 0 0 0 0 1

Step 1, L-shift$R,Q = 0 0 0 1 0$Q =0 0 0 0$M =0 0 0 1 1

Step 2, Add - $M = 1 1 1 0 1

$R = 1 1 1 1 1

Step 3, Set Q0$Q =0 0 00

Restore + $M = 0 0 0 1 1

$R = 0 0 0 1 0

Iteration 1

Iteration 2

ELEC 5200-001/6200-001 Lecture 11

$R = 0 0 0 1 0$Q =0 0 0 0$M =0 0 0 1 1

Step 1, L-shift$R ,Q = 0 0 1 0 0$Q =0 0 0 0$M =0 0 0 1 1

Step 2, Add- $M = 1 1 1 0 1

$R = 0 0 0 0 1

Step 3, Set Q0$Q =0 0 0 1

Step 1, L-shift$R,Q = 0 0 0 1 0$Q =00 1 0$M =0 0 0 1 1

Step 2, Add- $M = 1 1 1 0 1

$R = 1 1 1 1 1

Step 3, Set Q0$Q =00 1 0Final quotient

Restore+ $M = 0 0 0 1 1

$R = 0 0 0 1 0

Iteration 3

Iteration 4

Remainder

Note “Restore $R” in Steps 1, 2 and 4. This method is known as

the RESTORING DIVISION.

ELEC 5200-001/6200-001 Lecture 11

Start

$R=0, $M=Divisor, $Q=Dividend, count=n

Shift 1-bit left $R, $Q

$R (33 b)|$Q (32 b)

$R and $M have

one extra sign bit

beyond 32 bits.

Reexamine the

restoring algorithm

from slide 61

$R ← $R - $M

No

Yes

$R < 0?

$Q0=0

$R←$R+$M

$Q0=1

Restore $R

(remainder)

Cycle contains

2 additions

count = count - 1

Done

$Q=Quotient

$R= Remainder

No

count = 0?

Yes

ELEC 5200-001/6200-001 Lecture 11

Start

$R=0, $M=Divisor, $Q=Dividend, count=n

Shift 1-bit left $R, $Q

$R (33 b)|$Q (32 b)

$R ← $R - $M

No

$R < 0?

Yes

$Q0=1

$Q0=0

No

Yes

Restore $R

(remainder)

$R < 0?

Rearrange

flowchart

$R←$R+$M

count = count - 1

Done

$Q=Quotient

$R= Remainder

No

count = 0?

Yes

ELEC 5200-001/6200-001 Lecture 11

Start

$R=0, $M=Divisor, $Q=Dividend, count=n

Shift 1-bit left $R, $Q

$R (33 b)|$Q (32 b)

$R ← $R - $M

No

$R < 0?

Yes

$Q0=1

$Q0=0

Rearrange

flowchart

count = count - 1

No

No

count = 0?

Yes

No

$R < 0?

$R < 0?

Yes

Yes

Restore $R

(remainder)

Restore $R

(remainder)

Done

$Q=Quotient

$R= Remainder

$R←$R+$M

$R←$R+$M

ELEC 5200-001/6200-001 Lecture 11

Start

$R=0, $M=Divisor, $Q=Dividend, count=n

$R,$Q = a

$M = b

Yes

$R < 0?

$R←$R+$M

Rearrange

flowchart

Restore $R

(remainder)

No

Shift 1-bit left $R, $Q

$R (33 b)|$Q (32 b)

$R ← $R - $M

(a + b)2 – b

= 2a + b

2a – b

No

$R < 0?

Yes

$Q0=1

$Q0=0

Restore $R

(remainder)

count = count - 1

No

Yes

No

count = 0?

$R < 0?

Yes

Restore $R

(remainder)

Done

$Q=Quotient

$R= Remainder

$R←$R+$M

ELEC 5200-001/6200-001 Lecture 11

Start

$R=0, $M=Divisor, $Q=Dividend, count=n

$R (33 b)|$Q (32 b)

$R,#Q = a

$M = b

No

Yes

$R < 0?

Shift 1-bit left $R, $Q

Shift 1-bit left $R, $Q

Functional

equivalence

$R ← $R - $M

$R←$R+$M

2a – b

2a + b

No

$R < 0?

Yes

$Q0=1

$Q0=0

count = count - 1

No

No

Yes

count = 0?

$R < 0?

Yes

Restore $R

(remainder)

Done

$Q=Quotient

$R= Remainder

$R←$R+$M

ELEC 5200-001/6200-001 Lecture 11

Start

$R=0, $M=Divisor, $Q=Dividend, count=n

$R (33 b)|$Q (32 b)

No

Yes

$R < 0?

Shift 1-bit left $R, $Q

Shift 1-bit left $R, $Q

$R ← $R - $M

$R←$R+$M

No

$R < 0?

Yes

$Q0=1

$Q0=0

Cycle

contains

1 addition

count = count - 1

No

No

Yes

count = 0?

$R < 0?

Yes

Restore $R

(remainder)

Done

$Q=Quotient

$R= Remainder

$R←$R+$M

ELEC 5200-001/6200-001 Lecture 11

- Avoids the addition in the restore operation – does exactly one add or subtract per cycle.
- Non-restoring division algorithm:
- Step 1:Repeat 32 times
- if sign bit of $R is 0
Left shift $R,Q one-bit and subtract, $R ← $R - $M

else (sign bit of $R is 1)

Left shift $R,Q one-bit and add, $R ← $R + $M

- if sign bit of resulting $R is 0
Set Q0 = 1

else (sign bit of resulting $R is 1)

Set Q0 = 0

- if sign bit of $R is 0
- Step 2: (after 32 Step 1 iterations) if sign bit of $R is 1, add $R ← $R + $M

- Step 1:Repeat 32 times

ELEC 5200-001/6200-001 Lecture 11

Initialize$R = 0 0 0 0 0$Q = 1 0 0 0$M = 0 0 0 1 1

Step 1, L-shift$R,Q = 0 0 0 0 1$Q =0 0 0 ?$M =0 0 0 1 1

Subtract- $M =1 1 1 0 1

$R =1 1 1 1 0$Q =0 0 0 0

Step 1, L-shift$R,Q =1 1 1 0 0$Q =0 0 0?$M =0 0 0 1 1

Add+$M =0 0 0 1 1

$R =1 1 1 1 1$Q =0 0 0 0

Step 1, L-shift$R,Q =1 1 1 1 0$Q =0 0 0 ?$M =0 0 0 1 1

Add+$M =0 0 0 1 1

$R =0 0 0 0 1$Q =0 0 01

Step 1, L-shift$R,Q =0 0 0 1 0$Q =0 0 1 ?$M =0 0 0 1 1

Subtract -$M =1 1 1 0 1

$R =1 1 1 1 1$Q =0 0 10Final quotient

Step 2, Add $R ← $R + $M = 11111+00011 = 00010 (Final remainder)

Initialize$R = 0 0 0 0 0$Q = 1 0 0 0$M = 0 0 0 1 1

Step 1, L-shift$R,Q = 0 0 0 0 1$Q =0 0 0 ?$M =0 0 0 1 1

Subtract- $M =1 1 1 0 1

$R =1 1 1 1 0$Q =0 0 0 0

Step 1, L-shift$R,Q =1 1 1 0 0$Q =0 0 0 ?$M =0 0 0 1 1

Add+$M =0 0 0 1 1

$R =1 1 1 1 1$Q =0 0 00

Step 1, L-shift$R,Q =1 1 1 1 0$Q =0 0 0?$M =0 0 0 1 1

Add+$M =0 0 0 1 1

$R =0 0 0 0 1$Q =0 0 01

Step 1, L-shift$R,Q =0 0 0 1 0$Q =0 0 1?$M =0 0 0 1 1

Subtract -$M =1 1 1 0 1

$R =1 1 1 1 1$Q =0 0 10Final quotient

Step 2, Add $R ← $R + $M = 11111+00011 = 00010 (Final remainder)

Iteration 1

Iteration 2

Iteration 3

Iteration 4

ELEC 5200-001/6200-001 Lecture 11

- Remember the signs and divide magnitudes.
- Negate the quotient if the signs of divisor and dividend disagree.
- There is no other direct division method for signed division.

ELEC 5200-001/6200-001 Lecture 11

- div$s2, $s3# Lo = quotient
# Hi = remainder

- divu$s2, $s3# Lo = quotient
# Hi = remainder

- mflo and mfhi retrieve registers Lo and Hi
mflordst# move from Lo to rdst

mfhirdst# move from Hi to rdst

- Hardware ignores overflow, so software should determine if the quotient is too large.
- Software should also check if divisor = 0.

ELEC 5200-001/6200-001 Lecture 11