Chapter 3 Complements

1
Chapter 3 Complements

• Dr. Bernard Chen Ph.D.
• University of Central Arkansas
• Spring 2009

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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.

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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 9s 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
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9s complement Examples
9
9
9
9
9
9
-
5
4
6
7
0
0
2- Find the 9s complement of 546700 and 12389
The 9s complement of 546700 is 999999 -
546700 453299 and the 9s 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
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ls complement

• For binary numbers, r 2 and r 1 1,
• r-1s complement is the ls complement.
• The ls 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
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ls complement
Find r-1 complement for binary number N with four
binary digits. r-1 complement for binary means
2-1 complement or 1s complement. n 4, we have
24 (10000)2 and 24 - 1 (1111)2. The ls
complement of N is (24 - 1) - N. (1111) - N
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ls complement
1
1
1
1
1
1
1
-
0
1
1
0
0
1
1
The complement 1s of 1011001 is 0100110
1
0
0
1
1
0
0
1
1
1
1
1
1
1
-
0
0
1
1
1
1
0
The 1s complement of 0001111 is 1110000
1
1
0
0
0
0
1
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rs Complement

• Given a number N in base r having n digits,
• the rs complement of N is defined as
• rn - N.

• For decimal numbers the base or r 10,
• so the 10s complement of N is 10n-N.
• 100000. - N

0
0
0
0
0
1
-
Digit n
Digit n-1
Next digit
Next digit
First digit
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10s complement Examples
Find the 10s complement of 546700 and 12389 The
10s complement of 546700 is 1000000 - 546700
453300 and the 10s complement of 12389 is
100000 - 12389 87611. Notice that it is the
same as 9s complement 1.
0
0
0
0
0
0
1
-
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
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2s complement
For binary numbers, r 2, rs complement is the
2s complement. The 2s complement of N is 2n -
N.
0
0
0
0
0
1
-
Digit n
Digit n-1
Next digit
Next digit
First digit
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2s complement Example
0
0
0
0
0
0
0
1
-
The 2s complement of 1011001 is 0100111
0
1
1
0
0
1
1
1
0
0
1
1
1
0
-
The 2s complement of 0001111 is 1110001
0
0
1
1
1
1
0
1
1
0
0
0
1
1
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Fast Methods for 2s Complement
Method 1 The 2s complement of binary number is
obtained by adding 1 to the ls complement
value. Example 1s complement of 101100 is
010011 (invert the 0s and 1s) 2s complement of
101100 is 010011 1 010100
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Fast Methods for 2s Complement
 Method 2 The 2s complement can be formed by
leaving all least significant 0s and the first 1
unchanged, and then replacing ls by 0s and 0s
by ls in all other higher significant bits.
Example The 2s complement of 1101100 is
0010100 Leave the two low-order 0s and the
first 1 unchanged, and then replacing 1s by 0s
and 0s by 1s in the four most significant bits.
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Examples

• Finding the 2s complement of (01100101)2
• Method 1 Simply complement each bit and then
  add 1 to the result.
• (01100101)2
• N 2s complement 1s 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

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Subtraction of Unsigned Numbers using rs
complement

• Subtract N from M M N
• rs complement N ? (rn N )
• add M to ( rn N ) Sum M ( r n N)
•  take rs complement (If M ? N, the negative
  sign will produce an end carry ? rn we need to
  take the rs complement again.)

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Subtraction of Unsigned Numbers using rs
complement

• (1) if M ? N, ignore the carry without taking
  complement of sum.
• (2) if M lt N, take the rs complement of sum and
  place negative sign in front of sum. The answer
  is negative.

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Example 1 (Decimal unsigned numbers),   perform
the subtraction 72532 - 13250 59282.   M gt N
Case 1 Do not take complement of sum and
discard carry The 10s complement of 13250 is
86750. Therefore M 72532 10s complement
of N 86750 Sum 159282 Discard end carry
105 - 100000 Answer 59282 no complement
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Example 2 Now consider an example with M ltN.
The subtraction 13250 - 72532 produces negative
59282. Using the procedure with complements, we
have   M 13250 10s complement of N
27468 Sum 40718
Take 10s complement of Sum
100000 -40718 The number is
59282 Place negative sign in front of the
number -59282