Non-restoring division is a binary division algorithm used to divide one binary number by another. Unlike traditional division, which is based on restoring the dividend, non-restoring division does not restore the dividend to its original value. Instead, it uses a non-restoring method to find the quotient.

1. Computer Organization
and Architecture
Topic: Division Non-Restoring
Method
by
Upendra Mishra
(M. Tech., Ph.D.*)
KIET Group of Institutions

2. Prerequisites
 Shifting of Bits(Left or Right)
 Addition
 Subtraction
 2’s Complement

3. Division of unsigned binary integers
 Paper Pencil method
 Example:
 (Dividend)10/ (Divisor) 4 =(Quotient)2; (Remainder)2
Dividend
Divisor Quotient
Remainder
4 10 2
8
2

4.  Division of unsigned binary integers is
instructive method. There are two methods:
 Restore Method
 Non - Restore Method
Division of unsigned binary integers

6. Flowchart for Unsigned
Binary Division-
Non-Restore Method
start
nno of bits
M Divisor
A 0
Q Dividend
A < 0
Shift left AQ
A= A + M
Shift left AQ
A= A - M
A < 0
Q[0] = 0 Q[0] = 1
n=n-1
n=0
A < 0
Stop
A=A+M
Yes
No
Yes
No
No
No
Yes
Yes

7. Flowchart for Unsigned
Binary Division-
Non-Restore Method
start
nno of bits
M Divisor
A 0
Q Dividend
A < 0
Shift left AQ
A= A + M
Shift left AQ
A= A - M
A < 0
Q[0] = 0 Q[0] = 1
n=n-1
n=0
A < 0
Stop
A=A+M
Yes
No
Yes
No
No
No
Yes
Yes

8. Division by Non-Restore Method
11(Dividend)/3(Divisor) 3(Quotient) and 2(Remainder)
N M A Q ACTION
4 00011 00000 1011 Start
00001 011_ Left shift AQ
11110 011_ A=A-M
3 11110 0110 Q[0]=0
11100 110_ Left shift AQ
11111 110_ A=A+M
2 11111 1100 Q[0]=0
11111 100_ Left Shift AQ
00010 100_ A=A+M
1 00010 1001 Q[0]=1
00101 001_ Left Shift AQ
00010 001_ A=A-M
0 00010 0011 Q[0]=1
start
nno of bits
M Divisor
A 0
Q Dividend
A < 0
Shift left AQ
A= A + M
Shift left AQ
A= A - M
A < 0
Q[0] = 0 Q[0] = 1
n=n-1
n=0
A < 0
Stop
A=A+M
Yes
No
Yes
No
No
No
Yes
Yes
M = 00011
M’ = 11100
M’+1 = 11101
-M = 11101
-M = 11101

9. Division by Non-Restore Method
11(Dividend)/3(Divisor) 3(Quotient) and 2(Remainder)
N M A Q ACTION
4 00011 00000 1011 Start
00001 011_ Left shift AQ
11110 011_ A=A-M
3 11110 0110 Q[0]=0
11100 110_ Left shift AQ
11111 110_ A=A+M
2 11111 1100 Q[0]=0
11111 100_ Left Shift AQ
00010 100_ A=A+M
1 00010 1001 Q[0]=1
00101 001_ Left Shift AQ
00010 001_ A=A-M
0 00010(2)(R) 0011(3)(Q) Q[0]=1
start
nno of bits
M Divisor
A 0
Q Dividend
A < 0
Shift left AQ
A= A + M
Shift left AQ
A= A - M
A < 0
Q[0] = 0 Q[0] = 1
n=n-1
n=0
A < 0
Stop
A=A+M
Yes
No
Yes
No
No
No
Yes
Yes

10. 10
Question
• Try dividing 13/4 and 15/3

11. 11
Divide Overflow
Condition:
1. The number of bits in dividend is twice as the number of bits in the divisor. AND
2. The higher order half bits of the dividend have a value more than or equal to the value of the
total bits of the divisor.
OR
1. The divide overflow also checks that whether the divisor is zero or not.
• All these conditions are checked before the division process starts.
• Let dividend(Q) = 1010 1100 (length 8 bit)
Divisor(M) is 0111 (length 4 bit)
Here, the number of bits in Q is twice the number of bits in M
Higher order half bits of Q is 1010 value 10
total bits value of M is 0111 value 7 so Q>M
OR check whether M is 0000 or not
if yes also indicate Divide Overflow

12. 12
Reference
1.Computer System Architecture - Morris Mano
2.William Stallings, Computer Organization and Architecture-Designing for Performance,
Pearson Education, Seventh edition, 2006.

13. Thank
You

14. Thank
You

15. Division by Restore Method
11(Dividend)/3(Divisor) 3(Quotient) and 2(Remainder)
Start
nno of bits
M Divisor
A 0
Q Dividend
Shift left AQ
A= A - M
MSB
of A
Q[0] = 1
Q[0] = 0
Restore A
n = n-1
Is n = 0
Quotient in Q
Remainder in A
Stop
=1
=0
Yes
No
n M A Q Operation
4 00011 (3) 00000 (0) 1011 (11) initialize
00011 00001 011_ shift left AQ
00011 11110 011_ A=A-M
00011 00001 0110
Q[0]=0 And
restore A
3 00011 00010 110_ shift left AQ
00011 11111 110_ A=A-M
00011 00010 1100
Q[0]=0And
restore A
2 00011 00101 100_ shift left AQ
00011 00010 100_ A=A-M
00011 00010 1001 Q[0]=1
1 00011 00101 001_ shift left AQ
00011 00010 001_ A=A-M
00011 00010 (2) 0011 (3) Q[0]=1
M = 00011
M’ = 11100
M’+1 = 11101
-M = 11101
-M = 11101

16. Division by Restore Method
11(Dividend)/3(Divisor) 3(Quotient) and 2(Remainder)
n M A Q Operation
4 00011 (3) 00000 (0) 1011 (11) initialize
00011 00001 011_ shift left AQ
00011 11110 011_ A=A-M
00011 00001 0110
Q[0]=0 And
restore A
3 00011 00010 110_ shift left AQ
00011 11111 110_ A=A-M
00011 00010 1100
Q[0]=0And
restore A
2 00011 00101 100_ shift left AQ
00011 00010 100_ A=A-M
00011 00010 1001 Q[0]=1
1 00011 00101 001_ shift left AQ
00011 00010 001_ A=A-M
00011 00010 (2) 0011 (3) Q[0]=1

17. Division by Restore Method
11(Dividend)/3(Divisor) 3(Quotient) and 2(Remainder)
Start
nno of bits
M Divisor
A 0
Q Dividend
Shift left AQ
A= A - M
MSB
of A
Q[0] = 1
Q[0] = 0
Restore A
n = n-1
Is n = 0
Quotient in Q
Remainder in A
Stop
=1
=0
Yes
No
n M A Q Operation
4 00011 (3) 00000 (0) 1011 (11) initialize
00011 00001 011_ shift left AQ
00011 11110 011_ A=A-M
00011 00001 0110
Q[0]=0 And
restore A
3 00011 00010 110_ shift left AQ
00011 11111 110_ A=A-M
00011 00010 1100
Q[0]=0And
restore A
2 00011 00101 100_ shift left AQ
00011 00010 100_ A=A-M
00011 00010 1001 Q[0]=1
1 00011 00101 001_ shift left AQ
00011 00010 001_ A=A-M
00011 00010 (2) 0011 (3) Q[0]=1
M = 00011
M’ = 11100
M’+1 = 11101
-M = 11101
-M = 11101

18. Division by Restore Method
11(Dividend)/3(Divisor) 3(Quotient) and 2(Remainder)
n M A Q Operation
4 00011 (3) 00000 (0) 1011 (11) initialize
00011 00001 011_ shift left AQ
00011 11110 011_ A=A-M
00011 00001 0110
Q[0]=0 And
restore A
3 00011 00010 110_ shift left AQ
00011 11111 110_ A=A-M
00011 00010 1100
Q[0]=0And
restore A
2 00011 00101 100_ shift left AQ
00011 00010 100_ A=A-M
00011 00010 1001 Q[0]=1
1 00011 00101 001_ shift left AQ
00011 00010 001_ A=A-M
00011 00010 (2) 0011 (3) Q[0]=1

19. Division by Restore Method
11(Dividend)/3(Divisor) 3(Quotient) and 2(Remainder)
Start
nno of bits
M Divisor
A 0
Q Dividend
Shift left AQ
A= A - M
MSB
of A
Q[0] = 1
Q[0] = 0
Restore A
n = n-1
Is n = 0
Quotient in Q
Remainder in A
Stop
=1
=0
Yes
No
n M A Q Operation
4 00011 (3) 00000 (0) 1011 (11) initialize
00011 00001 011_ shift left AQ
00011 11110 011_ A=A-M
00011 00001 0110
Q[0]=0 And
restore A
3 00011 00010 110_ shift left AQ
00011 11111 110_ A=A-M
00011 00010 1100
Q[0]=0And
restore A
2 00011 00101 100_ shift left AQ
00011 00010 100_ A=A-M
00011 00010 1001 Q[0]=1
1 00011 00101 001_ shift left AQ
00011 00010 001_ A=A-M
00011 00010 (2) 0011 (3) Q[0]=1