Complements
-
Upload
sudheesh-s-madhav -
Category
Technology
-
view
1.184 -
download
1
description
Transcript of Complements
Complements
digital electronics - 13IDP1413IDP14
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.
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
-
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
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
-
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
The complement 1’s of1011001 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 of0001111 is 1110000
0
1
1
l’s complement
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
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
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
2’s complement Example
The 2’s complement of1011001 is 0100111
The 2’s complement of0001111 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
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
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.
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
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.)
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.
Example 1 (Decimal unsigned numbers), perform the subtraction 72532 - 13250 = 59282. M > N : “Case 1” “Do not take complement of sumand discard carry”The 10’s complement of 13250 is 86750. Therefore:
M = 7253210’s complement of N =+86750
Sum= 159282Discard end carry 105= - 100000Answer = 59282 no complement
Example 2;Now consider an example with M <N. The subtraction 13250 - 72532 produces negative 59282. Using the procedure with complements, we have
M = 1325010’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