1.
Complements
digital electronics  13IDP1413IDP14
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
(r1 )’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
n1
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,
r1’s complement is the l’s complement.
The l’s complement of N is (2n  1)  N.
Digit
n
Digit
n1
Next
digit
Next
digit
First
digit
1 1 1 1 1
Bit n1 Bit n2 ……. Bit 1 Bit 0

6.
l’s complement
Find r1 complement for binary number N with four binary digits.
r1 complement for binary means 21 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
n1
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
n1
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 loworder 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
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