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Chapter 3 Complements

Chapter 3 Complements . Dr. Bernard Chen Ph.D. University of Central Arkansas Spring 2009. Subtraction using addition. Conventional addition (using carry) is easily implemented in digital computers. However; subtraction by borrowing is difficult and inefficient for digital computers.

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Chapter 3 Complements

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  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 • (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 9 9 9 9 9 - Digit n Digit n-1 Next digit Next digit First digit

  4. 9’s complementExamples 9 9 9 9 9 9 - 5 4 6 7 0 0 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. 4 5 3 2 9 9 9 9 9 9 9 - 1 2 3 8 9 8 7 6 1 0

  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. Bit n-1 Bit n-2 ……. Bit 1 Bit 0 1 1 1 1 1 - Digit n Digit n-1 Next digit Next digit First digit

  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 24= (10000)2 and 24 - 1 = (1111)2. The l’s complement of N is (24 - 1) - N. = (1111) - N

  7. l’s complement 1 1 1 1 1 1 1 - 1 0 1 1 0 0 1 The complement 1’s of 1011001 is 0100110 0 1 0 0 1 1 0 1 1 1 1 1 1 1 - 0 0 0 1 1 1 1 The 1’s complement of 0001111 is 1110000 1 1 1 0 0 0 0

  8. 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 1 0 0 0 0 0 - Digit n Digit n-1 Next digit Next digit First digit

  9. 10’s complementExamples 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. 1 0 0 0 0 0 0 - 5 4 6 7 0 0 4 5 3 3 0 0 1 0 0 0 0 0 - 1 2 3 8 9 8 7 6 1 1

  10. 2’s complement For binary numbers, r = 2, r’s complement is the 2’s complement. The 2’s complement of N is 2n - N. 1 0 0 0 0 0 - Digit n Digit n-1 Next digit Next digit First digit

  11. 1 0 0 0 0 0 0 0 2’s complement Example 1 0 0 0 0 0 0 0 - The 2’s complement of 1011001 is 0100111 1 0 1 1 0 0 1 0 1 0 0 1 1 1 - The 2’s complement of 0001111 is 1110001 0 0 0 1 1 1 1 1 1 1 0 0 0 1

  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 negativesign 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 105= - 100000 Answer = 59282 no 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

  19. Gray code هذه الشفرة تسمى "منعكسة" (كما فى المرآة) البدء بالرقمين صفر وواحد ثم عكسهما واحد وصفر وهكذا كما فى الجدول المرفق، وللإيضاح وضعت ألوان توضح صفر-واحد بالأصفر والعكس بلون آخر هذا فى خانة الآحاد.

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