1. Chapter 3 Complements Dr. Bernard Chen Ph.D. University of Central Arkansas Spring 2009

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 (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 Digit n Digit n-1 Next digit First digit 99999 -

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 - 0 999 999 453 299 1238 - 9 99999 87610

5. 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 First digit 11111 Bit n-1Bit n-2…….Bit 1Bit 0 -

6. 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

7. The complement 1’s of 1011001 is 0100110 0 11 0 0 - 1 111111 100110 00111 - 1 111111 110000 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 r n - N. For decimal numbers the base or r = 10, so the 10’s complement of N is 10 n -N. 100000……. - N Digit n Digit n-1 Next digit First digit 00000 - 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. 54 6 7 0 - 0 000 0 00 45 3 3 00 1238 - 9 100000 87611 1

10. 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 Digit n Digit n-1 Next digit First digit 00000 - 1

11. 2’s complement Example The 2’s complement of 1011001 is 0100111 The 2’s complement of 0001111 is 1110001 011 00 - 1 000000 100111 00111 - 1 110001 1 0 0 0 1 1 00000001

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 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.

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 10 5 = - 100000 Answer = 59282no 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