Discussed basic approach of binary division, then restoring division and it's drawbacks and finally discussed non restoring division algorithm

1. Dr. Prasenjit Dey

2. 
© Dr. Prasenjit Dey 2
Integer Division
Dividend = Quotient × Divisor + Remainder
Divisor
Quotient
Dividend
Remainder
174÷8 = 21(Q) , 6(R)
1<8
17>8
14>8
6<8
Divisor
Quotient Dividend
Remainder
(174)10 = (1010 1110)2
(8)10 = (1000)2
1010>1000
101<1000
1011>1000
111<1000-0
1110
-1000 1110>1000
110 110<1000
8 174
-0
17
-16
14
-8
6
0 2 1
1000 10101110
-1000
101
-0
1011
-1000
111
1 0 1 0 1

3. 
Implementation 1
© Dr. Prasenjit Dey 3
32bit ALU
Divisor
shift right
Remainder
write
Quotient
shift left
Control
32 bits
32 bits 32 bits

4. 
Restoring Division (Flowchart)
start
A = 0
M = divisor
Q = Dividend
cnt = n
Shift left A,Q
A = A-M
A<0?
Q0 = 1
Q0 = 0
A = A + M
cnt = cnt -1
Cnt=0 stop
YesNo
No Yes
© Dr. Prasenjit Dey 4
Restoration of A

5. 
Step-1:
First the registers are initialized with corresponding values (Q = Dividend, M = Divisor, A
= 0, n = number of bits in dividend)
Step-2:
Then the content of register A and Q is shifted left as a single unit
Step-3:
Then content of register M is subtracted from A and result is stored in A
Step-4:
Then the most significant bit of the A is checked if it is 0 the LSB of Q is set to 1 otherwise
if it is 1 the LSB of Q is set to 0 and value of register A is restored i.e the value of A before
the subtraction with M
Step-5:
Decrement the counter n
Step-6:
When n becomes zero, we get out of the loop otherwise we repeat from step 2
Step-7:
Finally, the register Q contain the quotient and register A contain remainder
Restoring Division (Algorithm)
© Dr. Prasenjit Dey 5

6. 
© Dr. Prasenjit Dey 6
Restoring Division (Example)
Action M A Q Count
initialize 00011 00000 1011 4
shift left AQ 00001 011_
A=A-M (00001 - 00011) = 11110 011_
Q[0]=0 and restore A (11110 + 00011) = 00001 0110 3
shift left AQ 00010 110_
A=A-M (00010 - 00011) = 11111 110_
Q[0]=0 and restore A (11111 + 00011) = 00010 1100 2
shift left AQ 00101 100_
A=A-M (00101 - 00011) = 00010 100_
Q[0]=1 00010 1001 1
shift left AQ 00101 001_
A=A-M (00101 - 00011) = 00010 001_
Q[0]=1 00010 0011 0
Dividend (Q) = (11)10 = (1011 )2
Divisor (M) = (3)10 = (00011)2

7. 
Drawback
 Division still takes:
 2 ALU cycles per bit position
 1 to check for divisibility (subtract)
 One to restore (if needed)
 Can reduce to 1 cycle per bit
 Called non-restoring division
 Avoids restore of remainder when test fails
© Dr. Prasenjit Dey

8. 
Non-Restoring Division (Flowchart)
start
A = 0
M = divisor
Q = Dividend
cnt = n
A<0?
Shift left A,Q
A = A-M
Shift left A,Q
A = A+M
cnt = cnt -1
Cnt=0 stop
YesNo
No Yes
A<0? A = A+M
No
Yes
A<0?
Q0 = 0
YesNo
Q0 = 1
© Dr. Prasenjit Dey 8

9. Step-1: First the registers are initialized with corresponding values (Q = Dividend,
M = Divisor, A = 0, n = number of bits in dividend)
Step-2: Check the sign bit of register A
 If it is 1 shift left content of AQ and perform A = A+M, otherwise shift left AQ and
perform A = A-M
Step-3: Again check the sign bit of register A
 If sign bit is 1 Q[0] become 0 otherwise Q[0] become 1
Step-4: Decrements value of cnt by 1
Step-5: If cnt not equal to zero go to Step 2 otherwise go to next step
Step-6: If sign bit of A is 1 then perform A = A+M
Step-7: Register Q contain quotient and register A contain remainder
Non-Restoring Division (Algorithm)
© Dr. Prasenjit Dey 9

10. 
© Dr. Prasenjit Dey 10
Non-Restoring Division (Example)
Dividend (Q) = (11)10 = (1011 )2
Divisor (M) = (3)10 = (00011)2
Action M A Q count
Start 00011 00000 1011 4
Left shift AQ 00001 011_
A=A-M (00001 - 00011) = 11110 011_
Q[0]=0 11110 0110 3
Left shift AQ 11100 110_
A=A+M (11100 + 00011) = 11111 110_
Q[0]=0 11111 1100 2
Left Shift AQ 11111 100_
A=A+M (11111 + 00011) = 00010 100_
Q[0]=1 00010 1001 1
Left Shift AQ 00101 001_
A=A-M (00101 - 00011) = 00010 001_
Q[0]=1 00010 0011 0