Complement in DLD

Subtraction of Negative Numbers, Types of complement, r's complement and r-1's complement.

1. Complements Digital Logic Design By Shehzad Ali

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. Types of Complement There are two types of complements: R’s complement e.g. 2’s complement for base 2 and 10’s complement for base 10 R-1’s complement e.g. 1’s complement for base 2 and 9’s complement for base 10

4. Complements of N umbers (r-1 )’s Complement Given a number N in base r having n digits, the (r- 1)’s complement of N is defined as (r n - 1) - N For decimal numbers the base or r = 10 and r - 1 = 9, so the 9’s complement of N is (10 n -1)-N 99999……. - N D i g i t n D i g i t n-1 N ext digit N ext digit F i rs t d i g i t 9 9 9 9 9 -

5. 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 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

6. 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. D i g i t n D i g i t n-1 N ext digit N ext digit F i rs t d i g i t 1 1 1 1 1 Bit n-1 Bit n-2 ……. Bit 1 Bit -

7. 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 2 4 = (10000) 2 and 2 4 - 1 = (1111) 2 . The l’s complement of N is (2 4 - 1) - N. = (1111) - N

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

9. r ’ s C o m p l e m e n t Given a number N in base r having n digits, the r’s complement of N is defined as r n - N. For decimal numbers the base or r = 10, so the 10’s complement of N is 10 n -N. 100000……. - N D i g i t n D i g i t n-1 N ext digit N ext digit F i rs t d i g i t - 1

10. 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 4 5 3 3 - 1 2 3 8 9 1 8 7 6 1 1 1

11. For binary numbers, r = 2, r’s complement is the 2’s complement. The 2’s complement of N is 2 n - N. 2’s complement D i g i t n D i g i t n-1 N ext digit N ext digit F i rs t d i g i t - 1

12. 2’s complement Example The 2’s complement of 1011001 is 0100111 The 2’s complement of 0001111 is 1110001 - - 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

13. 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

14. 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 1101 100 is 0010 100 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.

15. E x amples 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

16. 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 co m p l e m ent of N = + 86750 S u m = 1 59282 Discard end carry 10 5 = - 100000 Answer = 59282 no complement

17. 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 S u m = 40718 Take 10’s complement of Sum = 100000 -40718 The number is : 59282 Place negative sign in front of the number: -59282