PPT 2 - CA unit II 22 7-2020

13-11-2020 1Dr. K.K. THYAGHARAJAN, RMD ENGINEERING COLLEGE
Contact E-mail:
acdean@rmd.ac.in
kkthyagharajan@yahoo.com
kkthyagharajan@gmail.com
Dr. K.K. THYAGHARAJAN
Professor & Dean (Academic)
Department of Electronics and Communication Engineering
RMD ENGINEERING COLLEGE
Dr. K.K. THYAGHARAJAN, RMD ENGINEERING COLLEGE
Multiplication & Division Algorithms
Click on the links given below to view video
https://youtu.be/thC8B4B-PyY
https://youtu.be/m7JtcP5QmFA
https://youtu.be/NbfTKSm4ubM
https://youtu.be/RhiBtztCESI

PART - 1
Modified Booth’s Algorithm for
Multiplication

Multiplication using bit-pair recoding
• Bit-pair recoding of multiplier is one of the modified Booth’s algorithm
• This speeds up multiplication process
• Reduces the number of summands to half.
Bit-pair recoding table Multiplier bit-pair Multiplier bit
on the right
Bit-pair recoded
bit at position i
i+1 i i-1
0 0 0 0
0 O 1 +1
0 1 0 +1
0 1 1 +2
1 0 0 -2
1 0 1 -1
1 1 0 -1
1 1 1 0
Remember this table

B  Multiplicand ; Q  Multiplier
Bit-pair recoding is applied only to the multiplier
If multiplier is negative, take 2’s complement before recoding
Hint: Add one extra bit in addition to the number of bits required
to represent B or Q which ever is maximum.
Example: Recode the multiplier value 3 using bit-pair recoding.
Use 4 bits.
Convert to binary Add 0 near LSb Pair (3) bits Get code from
the table
Put the codes in proper places
Q  +3 

Multiplicand B = -6 ; Multiplier Q = 3
Any negative number is represented using 2’s complement. Use 4 bits to
represent each number.
6  0110
-6  1’s complement + 1
Example 1 : Multiply -6 X 3 using bit-pair recoding method
8 4 2 1
0 1 1 0
1 0 0 1
1
1 0 1 0 B = 1010 = -6
Q = +3 = 0011
8 4 2 1
0 0 1 1
+6
Recoded Q
0 0 1 1 0
-1+1

Example 1: Contd. . . .
B = -6 =
Recoded Multiplier-Q
1 0 1 0
+1 -1
0 0 0 0 0 1 1 0
Multiplying -6 by -1 gives +6
i.e. 0110. Since this is +ve no.
extend the left sign bits by 0s
1 1 1 0 1 0
Multiply B by +1 and
shift left by 2 bits.
Since this is negative
number, extend the
left sign bits by 1s
1 1 1 0 1 1 1 0

Since MSB is 1, the result is negative and the value is 2’s
complement of 1 1 1 0 1 1 1 0
0 0 0 1 0 0 0 1
1
0 0 0 1 0 0 1 0
1’s Complement
2’s Complement
128 64 32 16 8 4 2 1
0 0 0 1 0 0 1 0
Result is 16+2 = 18 i.e. -18

Write the binary code of multiplicand & multiplier
Use same number of bits for both the values such that it is
the maximum number of bits required either by
multiplicand or by multiplier and add one more at the left
(MSB) for sign.
If any value is negative, represent it using 2’s complement
Find the recoded (bit-pair) values of multiplier by adding a
‘0’ at the right side. Extend the MSB if necessary.
If the recoded value has -1 or -2 , find out the 2’s
complement of the multiplicand for further use.
What do you have to understand?

Multiplicand B = 13 ; Multiplier Q = -6
Any negative number is represented using 2’s complement. Use 5 bits (one for sign)
to represent each number because 13 requires 4 bits
Example 2 : Multiply 13 X -6 using bit-pair recoding method
16 8 4 2 1
0 0 1 1 0
1 1 0 0 1
1
1 1 0 1 0
Q = 11010 = -6
+6
Recoded Q
1 1 0 1 0
-2-1
6 = 00110
Q = -6 = (1S complement of +6 )+1
0
Sign bit
To bit-pair recode add a zero
near LSB , get table and get code

B = +13 = 01101
16 8 4 2 1
0 1 1 0 1
Since the recoded multiplier has negative values, find out the 2’s complement of the
multiplicand (B) for future use
Sign bit
0 1 1 0 1B = +13
2’s complement
1 0 0 1 0
1
1 0 0 1 1- 13 1 0 0 1 1
1’s complement

To multiply +13 by -2, Take the 2’s
complement of 13 shift left by one
bit and add 0 at the right
0 1 1 0 1
0 -1 -2
B= +13
Recoded Multiplier
1 1 1 1 1 0 0 1 1 0
To multiply +13 by -1, Take the 2’s
complement of 13. Put this two
positions left. Extend the sign bits
1 1 1 1 0 0 1 1
1 1 1 0 1 1 0 0 1 0
0 0 0 0 0 0multiply +13 by 0. Put this two
positions left
1
Since MSB is 1 the result is negative and the value is obtained by 2’s complement

1 1 1 0 1 1 0 0 1 0
Since MSB is 1 the result is negative and the value is obtained by 2’s complement of
0 0 0 1 0 0 1 1 0 1 1’s complement
1
0 0 0 1 0 0 1 1 1 0 Result in binary
0 0 0 1 0 0 1 1 1 0
128 64 32 16 8 4 2 1
Result in decimal value
Result : -(64+8+4+2) = -78
─

PART - 2
Division Algorithm

Binary Division
Manual Division
1001010
Dividend
/ 1000
Divisor
1000
0 0 0 1 0 0 1
1 0 0 1 0 1 0
1 0 0 0
1 0 1 0
1 0 0 0
0 0 1 0 Remainder
Quotient
Divisor has been shifted right
Quotient bit is shifted left and 1
or 0 is added in the right

Division-Hardware
Remainder
Quotient
Dividend
64-bit ALU
Shift Right
Shift Left
Write
Control
32 bits
64 bits
64 bits
64-bit Divisor register, 64-bit ALU, 64-bit Remainder register, and
32-bit Quotient register
Divisor
subtract

Division Algorithm
2b. Restore the original value by adding the
Divisor register to the Remainder register, &
place the sum in the Remainder register. Also
shift the Quotient register to the left, setting
the new least significant bit to 0.
Test
Remainder
Remainder < 0Remainder  0
Subtract the Divisor register from the
Remainder register, and place the result
in the Remainder register.
2a. Shift the Quotient
register to the left setting
the new rightmost bit to 1.
3. Shift the Divisor register right by 1 bit.
Done
Yes: n+1 repetitions )
Place Dividend in Remainder
n+1
repetition?
No: < n+1 repetitions
‘n’ is the number
of bits in the
quotient and
remainder

Division Algorithms
Dividend Divisor Quotient Remainder Sign of Quotient Sign of Remainder
+ + + + Quotient is positive if both
dividend & divisor have same
sign Remainder has
the same sign
as dividend
- - + -
+ - - + Quotient is negative if sign of
dividend is different from the
sign of the divisor- + - -
What do you have to remember?
 A-register is used to store remainder. It is initialized with zero and has one bit
more than the number of bits in the divisor.
 Initially dividend is stored in Q-register and after performing the division Q-
register holds Quotient.
 Divisor is stored in B-register. One extra ‘0’ is added as MSB
 If the operands are signed values, proceed assuming both as positive and at the
end assign signs as given in the above table.

Q0 = 0 ; A  A+B
Is
MSB of A = 1?
YesNo
A  0 ; Q  Dividend
B  Divisor ; SC  no_of_bits in the dividend
Q  1
SC  SC-1.
Stop
Yes
start
Is
SC<0?
No
Shift left A,Q
A  A-B
Restoring
Division
Algorithm

SC
A Q
Steps Explanation
A4 A3 A2 A1 A0 Q3 Q2 Q1 Q0
4 0 0 0 0 0 1 0 1 0 Initial
3 0 0 0 0 1 0 1 0 SHL A,Q A-B = A+(-B)= A + 2’s complement of B
A  00001 + 11101 ; A = 111101 1 1 1 0 0 1 0 A A-B
0 0 0 0 1 0 1 0 0 MSB = 1; So A A+B ; Q0 =0 A  11110 + 00011 ; A= 1 00001
2 0 0 0 1 0 1 0 0 SHL A,Q A-B = A+(-B)= A + 2’s complement of B
A  00010 + 11101 ; A = 111111 1 1 1 1 1 0 0 A A-B
0 0 0 1 0 1 0 0 0 MSB = 1; So A A+B ; Q0 =0 A  11111 + 00011 ; A= 1 00010
1 0 0 1 0 1 0 0 0 0 SHL A,Q
0 0 0 1 0 0 0 0 A A-B A  00101 + 11101 ; A= 1 00010
0 0 0 1 0 0 0 0 1 MSB = 0; So Q0 =1
0 0 0 1 0 0 0 0 1 SHL A,Q
0 0 0 0 1 0 0 1 A A-B A  00100 + 11101 ; A= 1 00001
0 0 0 0 1 0 0 1 1 MSB = 0; So Q0 =1
Restoring Division Example 1: 1010 / 0011  10/3
A= 00000 ; Q=1010 (Dividend) ; B= 00011 (Divisor with MS bit) ; -B = 2’s Complement of B = 1’s complement of B +1 = 11100 +1 = 11101
Remainder = 1 Quotient = 3

A  A+B
Is
MSB of A = 1?
YesNo
A  0 ; Q  Dividend
B  Divisor ; SC  no_of_bits in the dividend
Q0 = 1
SC  SC-1.
Stop
No
start
Is
SC<0?
No
Shift left A,Q
Non-Restoring
Division
Algorithm
Is
MSB of A = 1?
YesNo
A  A─B
Q0 = 0
Yes
Is
MSB of A = 1?
A  A+B
Yes

SC
A Q
Steps Explanation
A4 A3 A2 A1 A0 Q3 Q2 Q1 Q0
4 0 0 0 0 0 1 0 1 0 Initial
3 0 0 0 0 1 0 1 0 SHL A,Q A-B = A+(-B)= A+ 2’s complement of B
A  00001 + 11101 ; A = 111101 1 1 1 0 0 1 0 0 MSB = 0; A A-B
Now MSB =1, so Q0 =0
2 1 1 1 0 0 1 0 0 SHL A,Q A  11100 + 00011 ; A= 11111
1 1 1 1 1 1 0 0 0 MSB =1 so A A+B
1 1 1 1 1 1 0 0 0 SHL A,Q
0 0 0 1 0 0 0 0 1 MSB =1 so A A+B
A  11111 + 00011
A= 1 00010
0 0 0 1 0 0 0 0 1 SHL A,Q
0 0 0 0 1 0 0 1 1 MSB = 0; A A-B
A  00100 + 11101 ; A= 1 00001
Non-Restoring Division Example 1: 1010 / 0011  10/3
Remainder = 1 Quotient = 3

SC
A Q
Steps Explanation
A4 A3 A2 A1 A0 Q3 Q2 Q1 Q0
4 0 0 0 0 0 1 1 0 0 Initial
3 0 0 0 0 1 1 0 0 SHL A,Q A-B = A+(-B)= A+ 2’s complement of B
A  00001 + 11101 ; A = 111101 1 1 1 0 1 0 0 0 MSB = 0; A A-B
2 1 1 1 0 1 0 0 0 SHL A,Q A  11101 + 00011 ; A= 1 00000
0 0 0 0 0 0 0 0 1 MSB =1 so A A+B
1 0 0 0 0 0 0 0 1 SHL A,Q
1 1 1 0 1 0 0 1 0 MSB =0 so A A-B
A  00000+ 11101
A= 11101
0 1 1 0 1 0 0 1 0 SHL A,Q
1 1 1 0 1 0 1 0 0 MSB = 1; A A+B
A  11010 + 00011 ; A= 11101
1 1 1 0 1 0 1 0 0 MSB = 1; A A+B A  11101 + 00011; A=1 00000
Non-Restoring Division Example 2: -12 / +3  1100 / 0011 (don’t add any sign bit)
Remainder = 00000Quotient

Answer:
Remainder = 00000
Quotient = 0100
In this problem since the dividend is negative the
remainder will be negative (ref slide 18) i.e. -0 = 0
Since dividend & divisor have different signs, the
quotient is negative i.e. -4

PPT 2 - CA unit II 22 7-2020

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PPT 2 - CA unit II 22 7-2020