1) Subtraction can be performed using addition of complements in digital computers more efficiently than conventional subtraction. 2) The r's complement of a number N is defined as rn - N, where r is the base and n is the number of digits. This includes the 1's complement for binary and 9's complement for decimal. 3) To subtract N from M, the r's complement of N is added to M. If the sum is greater than or equal to the base, the r's complement is taken. Otherwise the negative sign is placed in front of the sum.

1. Chapter 3 Complements
Dr. Bernard Chen Ph.D.
University of Central Arkansas
Spring 2009

2. Subtraction using addition
 Conventional addition (using carry) is easily
 implemented in digital computers.
 However; subtraction by borrowing is difficult
and inefficient for digital computers.
 Much more efficient to implement subtraction
using ADDITION OF the COMPLEMENTS of
numbers.

3. Complements of numbers
(r-1 )’s Complement
•Given a number N in base r having n digits,
•the (r- 1)’s complement of N is defined as
(rn - 1) - N
•For decimal numbers the
base or r = 10 and r- 1= 9,
•so the 9’s complement of N
is (10n-1)-N
•99999……. - N
Digit
n
Digit
n-1
Next
digit
Next
digit
First
digit
9 9 9 9 9
-

4. 2- Find the 9’s complement of 546700 and 12389
The 9’s complement of 546700 is 999999 - 546700=
453299
and the 9’s complement of 12389 is
99999- 12389 = 87610.
9’s complement Examples
5 4 6 7 0- 0
9 9 9 9 9 9
4 5 3 2 9 9
1 2 3 8- 9
9 9 9 9 9
8 7 6 1 0

5. l’s complement
 For binary numbers, r = 2 and r — 1 =
1,
 r-1’s complement is the l’s complement.
 The l’s complement of N is (2n - 1) - N.
Digit
n
Digit
n-1
Next
digit
Next
digit
First
digit
1 1 1 1 1
Bit n-1 Bit n-2 ……. Bit 1 Bit 0
-

6. l’s complement
Find r-1 complement for binary number N with four binary digits.
r-1 complement for binary means 2-1 complement or 1’s complement.
n = 4, we have 24 = (10000)2 and 24 - 1 = (1111)2.
The l’s complement of N is (24 - 1) - N. = (1111) - N

7. The complement 1’s of
1011001 is 0100110
0 1 1 0 0- 1
1 1 1 1 1 1
1 0 0 1 1 0
0 0 1 1 1- 1
1 1 1 1 1 1
1 1 0 0 0 0
1
1
0
The 1’s complement of
0001111 is 1110000
0
1
1
l’s complement

8. r’s Complement
•Given a number N in base r having n digits,
•the r’s complement of N is defined as
rn - N.
•For decimal numbers the
base or r = 10,
•so the 10’s complement of N
is 10n-N.
•100000……. - N
Digit
n
Digit
n-1
Next
digit
Next
digit
First
digit
0 0 0 0 0
-
1

9. 10’s complement Examples
Find the 10’s complement of
546700 and 12389
The 10’s complement of 546700
is 1000000 - 546700= 453300
and the 10’s complement of
12389 is
100000 - 12389 = 87611.
Notice that it is the same as 9’s
complement + 1.
5 4 6 7 0- 0
0 0 0 0 0 0
4 5 3 3 0 0
1 2 3 8- 9
1 0 0 0 0 0
8 7 6 1 1
1

10. For binary numbers, r = 2,
r’s complement is the 2’s complement.
The 2’s complement of N is 2n - N.
2’s complement
Digit
n
Digit
n-1
Next
digit
Next
digit
First
digit
0 0 0 0 0
-
1

11. 2’s complement Example
The 2’s complement of
1011001 is 0100111
The 2’s complement of
0001111 is 1110001
0 1 1 0 0- 1
0 0 0 0 0 0
1 0 0 1 1 1
0 0 1 1 1- 1
1 1 0 0 0 1
1
0
0
0
1
1
0 0 0 0 0 001

12. Fast Methods for 2’s
Complement
Method 1:
The 2’s complement of binary number is obtained by adding 1 to the
l’s complement value.
Example:
1’s complement of 101100 is 010011 (invert the 0’s and 1’s)
2’s complement of 101100 is 010011 + 1 = 010100

13. Fast Methods for 2’s
Complement
Method 2:
The 2’s complement can be formed by leaving all least significant 0’s
and the first 1 unchanged, and then replacing l’s by 0’s and 0’s by l’s
in all other higher significant bits.
Example:
The 2’s complement of 1101100 is
0010100
Leave the two low-order 0’s and the first 1 unchanged, and then
replacing 1’s by 0’s and 0’s by 1’s in the four most significant bits.

14. Examples
– Finding the 2’s complement of (01100101)2
• Method 1 – Simply complement each bit and then
add 1 to the result.
(01100101)2
[N] = 2’s complement = 1’s complement (10011010)2 +1
=(10011011)2
• Method 2 – Starting with the least significant bit,
copy all the bits up to and including the first 1 bit
and then complement the remaining bits.
N = 0 1 1 0 0 1 0 1
[N] = 1 0 0 1 1 0 1 1

15. Subtraction of Unsigned Numbers
using r’s complement
Subtract N from M : M – N
 r’s complement N  (rn – N )
 add M to ( rn – N ) : Sum = M + ( r n – N)
 take r’s complement (If M  N, the negative
sign will produce an end carry  rn we need
to take the r’s complement again.)

16. Subtraction of Unsigned Numbers
using r’s complement
 (1) if M  N, ignore the carry without
taking complement of sum.
 (2) if M < N, take the r’s complement
of sum and place negative sign in front
of sum. The answer is negative.

17. Example 1 (Decimal unsigned numbers),
perform the subtraction 72532 - 13250 = 59282.
M > N : “Case 1” “Do not take complement of sum
and discard carry”
The 10’s complement of 13250 is 86750.
Therefore:
M = 72532
10’s complement of N =+86750
Sum= 159282
Discard end carry 105= - 100000
Answer = 59282 no complement

18. Example 2;
Now consider an example with M <N.
The subtraction 13250 - 72532 produces negative 59282. Using
the procedure with complements, we have
M = 13250
10’s complement of N = +27468
Sum = 40718
Take 10’s complement of Sum = 100000
-40718
The number is : 59282
Place negative sign in front of the number: -59282