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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.

    (r^n - 1) - N

    Radix Complement [r’s complement]

    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

    r^n - N

    or

    r-1’s \hspace{3px}complement + 1 => [( r^n -1) – N] + 1

    Examples – Diminished Radix Complements

    Here are some example problems for diminished radix complements.

    Problem1:

    Find 9’s complement of 258.

    Solution:

    \begin{aligned}
&N = 258\\
&r = 10\\
&n=3\\\\
&We \hspace{3px} use \hspace{3px}the \hspace{3px}formula\hspace{3px} for \hspace{3px}r-1’s \hspace{3px}complement.\\\\
&(r^n – 1) – N\\
&(10^3 – 1) – 258\\
&999 – 258\\
&The \hspace{3px} 9’s \hspace{3px}complement \hspace{3px}of \hspace{3px}258 \hspace{3px}is \hspace{3px}741.
\end{aligned}

    Problem2:

    Find 1’s complement for 1101.

    Solution:

    In the given problem,

    \begin{aligned}
&N = 1101\\
&r = 2\\
&n = 4\\\\
&The \hspace{3px}formula \hspace{3px}for \hspace{3px}r-1’s \hspace{3px}complement \hspace{3px}where r = 2\hspace{3px} is \hspace{3px}(r^n - 1) – N.\\\\
&(2^4 - 1) - 1101\\\\
&Note:\hspace{3px} Binary \hspace{3px}equivalent for \hspace{3px}2^4  = 10000.\\\\
&(10000 - 1)1101\\
&1111 - 1101\\
&The \hspace{3px} 1’s \hspace{3px}complement \hspace{3px}for \hspace{3px}1101 \hspace{3px}is \hspace{3px}0010.
\end{aligned}

    Method 2

    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

    \begin{aligned}
&1 – 1 = 0\\
&1 – 1 = 0\\
&1 – 0 = 1\\
&1 – 1 = 0\\\\
&Answer: 0010
\end{aligned}

    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.

    \begin{aligned}
&7 - 2 = 5\\
&7 - 3 = 4\\
&7 - 4 = 3\\\\
&The \hspace{3px} r-1's \hspace{3px} complement \hspace{3px}is \hspace{3px} 543.
\end{aligned}

    Examples – Radix Complements

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

    \begin{aligned}
&Method \hspace{3px}1: r^n – N\\
&Method \hspace{3px}2: r-1’s \hspace{3px}complement + 1
\end{aligned}

    Problem4:

    Find 10’s complement for 432.

    Solution:

    Given that

    \begin{aligned}
&N = 432\\
&r = 10\\
&n = 3\\
&
\end{aligned}

    Method 1:

    \begin{aligned}
&First \hspace{3px}we \hspace{3px}will \hspace{3px}find \hspace{3px}the \hspace{3px} 10’s \hspace{3px}complement \hspace{3px}using \hspace{3px}r^n – N.\\
&r^n – N = 1000 – 432 = 568\\
&Therefore,\\
&10’s \hspace{3px}complement \hspace{3px}of\hspace{3px} 432 \hspace{3px}is \hspace{3px}568.
\end{aligned}

    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.

    \begin{aligned}
&9 - 4 = 5\\
&9 - 3 = 6\\
&9 - 2 = 7\\

\end{aligned}

    Step2: Add 1 to the result.

    \begin{aligned}
&567 + 1\\
&=568\\
&568 \hspace{3px}is \hspace{3px}the \hspace{3px}10’s \hspace{3px}complement \hspace{3px}of \hspace{3px}432.
\end{aligned}

    Problem5:

    Find 2’s complement of the binary number 1001.

    Solution:

    Method 1:

    We can obtain the solution in using two methods.

    \begin{aligned}
&To \hspace{3px} find \hspace{3px} 2’s \hspace{3px}complement \hspace{3px}using \hspace{3px}method 1 \hspace{3px}use \hspace{3px}the \hspace{3px}formula r^n – N.\\
&Given \hspace{3px} that\\
&N = 1001\\
&r = 2\\
&n = 4\\
&r^n - N = 10000 – 1001\\\\
&\hspace{8px}\sout10000\\
&-1001\\
&\hspace{8px}----\\
&\hspace{11px}0111\\\\
&The \hspace{3px}2’s \hspace{3px}complement \hspace{3px}for \hspace{3px}1001 \hspace{3px}is \hspace{3px}0111.
\end{aligned}

    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.

    \begin{aligned}
1001 -> 0110 (1’s complement)
\end{aligned}

    Step2: Add 1 to result.

    \begin{aligned}
&0110 + 1 => 0111\\
&Therefore,\\
&0111 \hspace{3px}is \hspace{3px} the \hspace{3px}2’s \hspace{3px}complement \hspace{3px}of \hspace{3px}0110.
\end{aligned}

    References

    1. John.F.Wakerly. 2008. Digital Design: Principles And Practices, 4/E. Pearson Education, India.
    2. Mano, M. Morris. 1984. Digital Design. Pearson.
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