COMPLEMENTS

1
COMPLEMENTS

• Complements are used in digital computers for
  simplifying the subtraction operations and for
  logical manipulation. There are two types of
  complements for each base-r system.
• (1) the rs complement and
• (2) the (r-1)s complement.
• When the value of the base is substituted, the
  two types receives the names 2s and 1s
  complement for binary numbers or 10s and 9s
  complement for decimal numbers.

2
The rs Complement

• Given a positive number N in base r with an
  integer part of n digits, the rs complement of N
  is defined as rn-N for N ? 0 and 0 for N0. The
  following numerical example will help clarify the
  definition.
• The 10s complement of (52520)10 is 105-52520
  47480
• The number of digits in the number is n5
• The 10s complement of (0.3267)10 is 1- 0.3267
  0.6733
• No integer part so 10n1001
• The 10s complement of (25.639)10 is
  102-25.63974.361
• The 2s complement of (101100)2 is (26)10
  (101100)2(1000000 - 101100)010100
• The 2s complement of (0.0110)2 is
  (1-0.0110)20.1010

3
The (r-1)s Complement

• Given a positive number N in base r with an
  integer part of n digits, the (r-1)s complement
  of N is defined rn-r-m-N. Some numerical examples
  follow
• The 9s complement of (52520)10 is
  (105-1-52520)99999-5252047479.
• No fraction party, 10-m1001
• The 9s complement of (0.3267)10 is
  (1-10-4-0.3267)0.9999-0.32670.6732
• No integer part, so 10n1001.
• The 9s complement of (25.639)10 is
  (102-10-3-25.639)99.999-25.63974.360
• The 1s complement of (101100)2 is (26-1)
  (101100)(111111-101100)010011.
• The 1s complement of (0.0110)2 is (1-2-4)10
  (0.0110) 2 (0.1111-0.0110)2 0.1001

4

• From the examples, we see that the 9s complement
  of a decimal number is formed simply by
  subtracting every digit from 9. The 1s
  complement of a binary number is even simpler to
  form. The 1s are changed to 0s and the 0s to
  1s. Since the (r-1)s complement is very easily
  obtained.
• It is sometimes convenient to use it when the rs
  complement is desired. From the definitions and a
  comparison of the results obtained in the
  examples, it follows that the rs complement can
  be obtained from the (r-1)s complement after the
  addition of r-m to the least significant digit.
  For example, the 2s complement of 10110100 is
  obtained from the 1s complement 01001011 by
  adding 1 to give 01001100.

5
Subtraction with rs complement

• The direct method of subtraction taught in
  elementary schools uses the borrow concept. In
  this method, we borrow a 1 from a higher
  significant position when the minuend digit is
  smaller than the corresponding subtrahend digit.
• This seems to be easiest when people perform
  subtraction with paper and pencil. When
  subtraction is implemented by means of digital
  components, this method is found to be less
  efficient than the method that uses complements
  and addition as stated below. The subtraction of
  two positive numbers (M-N), both of base r, may
  be done as follows

6

• Add the minuend M to the rs complement of the
  subtrahend N.
• inspect the result obtained in step 1 for an end
  carry
• if an end carry discard it.
• If an end carry does not occur take the rs
  complement of the number obtained in step 1 and
  place a negative sign in front.
• The following examples illustrate the
  procedures using 10s complement.

7

• Subtract 72532-3250
• M72532
  72532
• N03250
• 10s complement of N
  96750
• End carry 1 69282
• Answer 69282
• Subtract (3250-72533)10
• M03250
• N72532
• 03250
• 10s complement of N
  27468
• No
  carry 30718
• Answer -69282(10s complement of 30718)

8
Subtraction with (r-1)s complement

• The procedure for subtraction with the (r-1)s
  complement is exactly the same as oe variation,
  called, end-around carry as shown below. The
  subtraction of M-N, both positive numbers in base
  r, may be calculated in the following manner.
• Add the minuend M to the (r-1)s complement of
  the subtrahend N.
• inspect the result obtained in step 1 for an end
  carry
• if an end carry occurs, add 1 to the least
  significant digit (end-around carry)
• if an end does not occur, take the (r-1)s
  complement of the number obtained in step 1 and
  place a negative sign infront.

9

• Example
• M 72532
• N03250
• 72532
• 9s complement of N
  96749
• 
  69281
• End around carry
  1
• 69282
• Answer 69282
• M 03250
• N 72532
• 03250
• 9s complement of N 27467
• No carry
  30717
• Answer-69282 -(9s complement of 30717)

10

• Using 1s complement
• (a) M 1010100
• N 1000100
• 1s complement of N 0111011
• 1010100
• 0111011
• End-around carry 1 0001111
• 
  1
• 
  0010000
• Answer 10000
• M1000100
• N1010100
• 1s complement of N 0101011
• 1000100
• 0101011
• No carry 1101111
• Answer -10000 -(1s complement of 1101111)

11

• Use 2s complement to perform M-N with the given
  binary numbers.
• M1010100
• N1000100
• 
  1010100
• 2s complement of N
  0111100
• 
  End carry 1 0010000
• M 1000100
• N1010100
• 
  1000100
• 2s complement of N 0101100
• No carry 0010000
• Answer -10000 -(2s complement of 1110000)

12
Comparison between 1s and 2s Complements

• A comparison between 1s and 2s complements
  reveals the advantages and disadvantages of each.
• The 1s complement has the advantage of being
  easier to implement by digital components since
  the only thing that must be done is to change 0s
  to 1s and 1s to 0s.
• the implementation of the 2s complement may be
  obtained in two ways (1) by adding 1 to the
  least significant digit of the 1s complement,
  and
• (2) by leaving all leading 0s in the least
  significant positions and the first 1 unchanged,
  and only then changing all 1s into 0s and all
  0s into 1s.
• During subtraction of two numbers by complement,
  the 2s complement is advantageous in that only
  arithmetic addition operation is required.
• The 1s complement requires two arithmetic
  additions when an end-around carry occurs. The
  1s complement has the additional disadvantage of
  possessing two arithmetic zeros one with all
  0sand one with all 1s.

13

• To illustrate this fact, consider the subtraction
  of the two equal binary numbers 1100- 11000
• Using 1s complement
• 1100
• 0011
• 1111
• Complement again to obtain -0000.
• Using 2s complement
• 1100
• 0100
• 0000
• while the 2s complement has only one arithmetic
  zero, the 1s complement zero can be positive or
  negative, which may complicate matters