# Digital Design – Complements

In the previous lesson, we learned about number systems – binary, decimal, octal and hexadecimal number systems in detail. Complements are very helpful in performing subtraction and save computing time and lead to simple circuits.

Before we move to subtraction, let’s understand what complement is all about. In simple words, if there is group with have R items, and you took S items from the group, then you need the complement R-S items to complete the group.

These R-S items are called the complements.

In a number system, there is a fixed range of numbers and when you pick any number from the system, you have a complement for that number in the number system.

### Types of Complements

There are two types of complements.

1. Diminished Radix Complement or r-1’s complement.
2. Radix Complement or r’s complement.

### Diminished Radix Complement [r-1’s complement]

Suppose a number N is given

N = Number

r = base of the number

n = Number of digits in the number.

The r-1’s complement of the number is given by the formula.

(rn – 1) – N

To find radix complement of a number N where

N = Number

r. = base of the number

n = Number of digits in the number.

The r’s complement of the number is given as

rn – N

or

r-1’s complement + 1 => [( rn -1) – N] + 1

### Examples – Diminished Radix Complements

Problem1:

Find 9’s complement of 258.

Solution:

N = 258

r = 10

n = 3

We use the formula for r-1’s complement.

(rn – 1) – N

(103 – 1) – 258

999 – 258

The 9’s complement of 258 is 741.

Problem2:

Find 1’s complement for 1101.

Solution:

In the given problem,

N = 1101

r = 2

n = 4

The formula for r-1’s complement where r = 2 is (rn – 1) – N.

(24 – 1) – 1101

Note: Binary equivalent for 24 = 10000.

(10000 – 1)1101

1111 – 1101

The 1’s complement for 1101 is 0010.

Method2

The second method to find the r-1’s complement is to subtract each digit in the number by r-1.

In our case, r = 2 so r – 1 = 1

Given that the 1’s complement of binary number is

1 – 1 = 0

1 – 1 = 0

1 – 0 = 1

1 – 1 = 0

Problem3:

Find 7’s complement of octal number 234.

Solution:

The range of digits for octal number is 0-to-7. Subtract each digit with 7 will give r-1’s complement for octal.

7 – 2 = 5

7 – 3 = 4

7 – 4 = 3

The r-1’s complement is 543.

There are two methods to find the r’s complement of a number.

Method 1: rn – N

Method 2: r-1’s complement + 1

Problem4:

Find 10’s complement for 432.

Solution:

Given that

N = 432

r = 10

n = 3

Method 1:

First we will find the 10’s complement using rn – N.

rn – N = 1000 – 432 = 568

Therefore,

10’s complement of 432 is 568.

Method 2:

By using method 2, you need to find the 9’s complement and then add 1 to the result to get 10’s complement.

Step1: Subtract each digit by 9 to get 9’s complement.

9 – 4 = 5

9 – 3 = 6

9 – 2 = 7

Step2: Add 1 to the result.

567 + 1

=568

568 is the 10’s complement of 432.

Problem5:

Find 2’s complement of the binary number 1001.

Solution:

Method 1:

To find 2’s complement using method 1 use the formula rn – N.

Given that

N = 1001

r = 2

n = 4

rn – N = 10000 – 1001

10000

– 1001

0111

The 2’s complement for 1001 is 0111.

Method 2:

To use second method, find the 1’s complement for 1001 and then add 1 to the result.

Step1: find the 1’s complement of 1001.

1001 -> 0110 (1’s complement)

0110 + 1 => 0111

Therefore,

0111 is the 2’s complement of 0110.

### Bibliography

John.F.Wakerly. 2008. Digital Design: Principles And Practices, 4/E. Pearson Education, India.

Mano, M. Morris. 1984. Digital Design. Pearson.