DLD03 Complement

Digital Logic and Design

(aka DLD: EEE241)

Dr. Aamir M. Khan

PhD Embedded Systems

Lecture 3: N’s Complement

2

How To Represent Signed Numbers

• Plus and minus sign used for decimal numbers:

25 (or +25), -16, etc.

• For computers, desirable to represent everything as bits.

• Three types of signed binary number representations:

signed magnitude, 1’s complement, 2’s complement.

• In each case: left-most bit indicates sign: positive (0) or

negative (1).

3

Consider signed magnitude:

000011002 = 1210

Sign bit Magnitude

100011002 = -1210

Sign bit Magnitude

One’s Complement Representation

• The one’s complement of a binary number involves

inverting all bits.

• 1’s complement of 00110011 is 11001100


• For an n bit number N the 1’s complement is (2n-1) – N.

• Called diminished radix complement by Mano since 1’s

complement for base (radix 2).

• To find negative of 1’s complement number, take the 1’s

complement.

4

000011002 = 1210 111100112 = -1210

Two’s Complement Representation

• The two’s complement of a binary number involves inverting

all bits and adding 1.

• 2’s comp of 00110011 is 11001101

• 2’s comp of 10101010 is 01010110

• For an n bit number N the 2’s complement is (2n) – N.

• Called radix complement by Mano since 2’s complement for

base (radix 2).

• To find negative of 2’s complement number, take the 2’s

complement.

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000011002 = 1210 111101002 = -1210

Two’s Complement Shortcuts

• Algorithm 1 – Simply complement each bit and then add 1 to the

result.

• Finding the 2’s complement of (01100101)2 and of its 2’s complement…

N = 01100101 [N] = 10011011

10011010 01100100

+ 1 + 1

--------------- ---------------

10011011 01100101

• Algorithm 2 – Starting with the least significant bit, copy all of the

bits up to and including the first ‘1’ bit and then complementing the

remaining bits.

N = 0 1 1 0 0 1 0 1

[N] = 1 0 0 1 1 0 1 1

6

Finite Number Representation

• Machines that use 2’s complement arithmetic can

represent integers in the range

-2n-1 ≤ N ≤ 2n-1 – 1

where n is the number of bits available for representing

N. Note that

2n-1 – 1 = (011…11)2 and -2n-1 = (100…00)2

• For 2’s complement more negative numbers than

positive.

• For 1’s complement two representations for zero.

• For an n bit number in base (radix) r there are r n

different unsigned values.

(0, 1, …r n-1)

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Signed Binary Numbers

8

1’s Complement Addition

• Using 1’s complement numbers, adding numbers is easy.

• For example, suppose we wish to add (12)10 + (1)10

• Binary code for 12 is (1100)2 and for 1 is (0001)2.

• But in signed number form, we have 0 at the MSB for +ve numbers.

• (12)10 = +(1100)2 = 011002 in 1’s comp.

• (1)10 = +(0001)2 = 000012 in 1’s comp.

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Step 1: Add binary numbers Step 2: Add carry to low-order bit

0 1 1 0 0

+ 0 0 0 0 1

--------------

0 0 1 1 0 1

0

--------------

0 1 1 0 1

Add carry

Final Result

Add

1’s Complement Subtraction

• Using 1’s complement numbers, subtracting numbers is also easy.

• For example, suppose we wish to compute (12)10 - (1)10.

• Subtraction will be like: 12 + (-1)

• (12)10 = +(01100)2 = 011002 in 1’s comp.

• (-1)10 = - (00001)2 = 111102 in 1’s comp.

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0 1 1 0 0

- 0 0 0 0 1

--------------

0 1 1 0 0

+ 1 1 1 1 0

--------------

1 0 1 0 1 0

1

--------------

0 1 0 1 1

Add carry

Final Result

Step 1: Take 1’s complement of 2nd operand Step 2: Add binary numbers Step 3: Add carry to low order bit

1’s comp

Add

2’s Complement Addition

• Using 2’s complement numbers, adding numbers is easy.

• For example, suppose we wish to add (12)10 + (1)10.

• As both numbers are +ve, their binary codes will have 0 at

MSB.

• (12)10 = +(1100)2 = 011002 in 2’s comp.

• (1)10 = +(0001)2 = 000012 in 2’s comp.

11

Step 1: Add binary numbers Step 2: Ignore carry bit (if any)

0 1 1 0 0

+ 0 0 0 0 1

--------------

0 1 1 0 1

Final Result

Add

2’s Complement Subtraction

• Using 2’s complement numbers, follow steps for subtraction

• For example, suppose we wish to compute (12)10 - (1)10.

• Subtraction will be performed as: 12 + (-1)

• Let’s

• (12)10 = + (01100)2 = 011002 in 2’s comp.

• (-1)10 = - (00001)2 = 111112 in 2’s comp.

12

0 1 1 0 0

- 0 0 0 0 1

--------------

0 1 1 0 0

+ 1 1 1 1 1

--------------

1 0 1 0 1 1

Final Result

Step 1: Take 2’s complement of 2nd operand Step 2: Add binary numbers Step 3: Ignore carry bit

2’s comp

Add

Ignore Carry

2’s Complement Subtraction: Example #2

• Let’s compute (13)10 - (5)10.

• (13)10 = + (1101)2 = (01101)2

• (-5)10 = - (0101)2 = (11011)2

• Adding these two 5-bit codes…

• Discarding the carry bit, the sign bit is seen to be zero,

indicating a correct result. Indeed,

(01000)2 = +(1000)2 = +(8)10.

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0 1 1 0 1

+ 1 1 0 1 1

--------------

1 0 1 0 0 0

carry

2’s Complement Subtraction: Example #3

• Let’s compute (5)10 - (12)10.

• (-12)10 = -(1100)2 = (10100)2

• (5)10 = +(0101)2 = (00101)2

• Adding these two 5-bit codes…

• Here, there is no carry bit and the sign bit is 1. This

indicates a negative result, which is what we expect.

(11001)2 = -(7)10.

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0 0 1 0 1

+ 1 0 1 0 0

--------------

1 1 0 0 1

Formal Definition of Complement

• Complements are used in digital computers for

simplifying subtraction operation.

• There are two types of Complements

• r’s Complement

• (r-1)’s Complement

where ‘r’ is the base

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Complements

• In base-10:

• 10’s Complement


• In base-2



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Remember the rule: r’s complement is +1 (add 1) to the (r-1)’s complement.

10’s Complement

• Given a positive number ‘N’ in base ‘r’ with an integer

part of ‘n’ digits, then r’s complement is defined as

rn - N

• 10’s complement of (52520)₁₀ is

10⁵ - 52520

= 100000 – 52520 = 47480

• Similarly,

10’s complement of 134795 is 865205

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10’s Complement

• 10’s complement of (0.3267)₁₀ = 10⁰ – 0.3267

= 1 - 0.3267

= 0.6733

• 10’s complement of (25.639)₁₀ = 10² - 25.639

= 74.361

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0.3267 has no integer part => 100

2’s Complement

• 2’s Complement of (101100)₂ is (2⁶)₁₀ – (101100)₂

= (1000000 – 101100)₂

= 010100

• 2’s Complement of (0.0110)2 is (20)₁₀ – (0.0110)₂

= (1.0000 – 0.0110)₂

= 0.1010

• Rules for binary Subtraction

• 0 - 0 = 0

• 0 - 1 = 1, and borrow 1 from the next more significant bit

• 1 - 0 = 1

• 1 - 1 = 0

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(26)10 = (64)10 = (1000000)2

(20)10 = (1)10 = (1)2

(r-1)’s Complements

• Given a positive number ‘N’ in base ‘r’ with an integer

part of ‘n’ digits and a fraction part of ‘m’ digits then (r-

1)’s complement is defined as

rn - r-m - N

• Need to accommodate the fraction part through ‘r-m’

• 9’s complement of (52520)₁₀ is (10⁵ – 10⁰ – 52520)

= 100000 – 1 – 52520

= 47479

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Remember the rule: r’s complement is +1 unit (add 1) to the (r-1)’s complement.

9’s Complement

• 9’s complement of (0.3267)₁₀ is (10⁰ – 10⁻⁴ – 0.3267)

= 0.9999 – 0.3267

= 0.6732

• 9’s complement of (25.639)₁₀ is (10² – 10⁻³ – 25.639)

= 99.999 – 25.639

= 74.360

• Find the 9’s and 10’s complements of 314700.

• 9’s complement = 685299

• 10’s complement = 685300

• To find the 10’s complement of a decimal number leave all leading

zeros unchanged. Then subtract the first non-zero digit from 10 and

all the remaining digits from 9’s.

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A version of Algorithm 2 on Slide 6, for binary numbers

1’s Complement


which can be obtained by replacing each one by a zero and each

zero by one.

• 1’s Complement of (101100)2 is (26 - 1)10 – (101100)2

= (1000000 – 1 – 101100)2

= 010011

• 1’s Complement of (0.0110)2 is (20 – 2– 4)10 – (0.0110)2

= (0.1111 – 0.0110)2

= 0.1001

22

Remind it: 20 – 2–4 = rn – n-m

20 – 2–4 = (1-0.0625) = (0.9375)10 = (0.1111)2

Use method of Example 1.3

Exercise

• Find the 1’s and 2’s complements of the binary number

1101001101

• 1’s complement is 0010110010


• Find the 1’s and 2’s complements of 100010100



23

Subtraction with r’s Complements

• The subtraction of two positive number (M-N), both of

base-r may be done as follows.

1. Add the minuend M to the r’s complement of the

subtrahend N.

2. Inspect the result obtained above for an end carry:

a) If an end carry occurs, discard it and write the answer as it is

b) If an end carry does not occur, then take r’s complement of the

number obtained in step 1, and place a negative sign in front.

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Exercise

• Using 10’s complement subtract 3250 from 72532

M = 72532

N = 03250

• 10’s complement of N = 96750

72532

+ 96750

= 169282

Answer = 69282

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Exercise

• Subtract (3250 – 72532) using 10’s Complement.

M = 03250

N = 72532


03250

+ 27468

= 030718

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Exercise


M = 1010100

N = 1000100


1010100

+ 0111100

= 10010000

Answer = 10000

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Exercise

• Subtract (1000100 – 1010100) using 2’s Complement

M = 1000100

N = 1010100


1000100

+ 0101100

= 01110000

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Subtraction with (r-1)’s Complements

• The subtraction of two positive number (M-N), both of

base-r may be done as follows.

1. Add the minuend M to the (r-1)’s complement of the

subtrahend N.

2. Inspect the result obtained above for an end carry:

a) If an end carry occurs, add 1 to the number.

b) If an end carry does not occur, then take (r-1)’s complement of

the number obtained in step 1, and place a negative sign in

front.

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Exercise


M = 72532

N = 03250


72532

+ 96749

= 169281

• Add this end carry to the above

i.e. 69281 +1

Answer = 69282

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Exercise

• Subtract (3250 – 72532) using 9’s Complement.

M = 03250

N = 72532


03250

+ 27467

= 030717

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Exercise


M = 1010100

N = 1000100


1010100

+ 0111011

= 10001111

• Add this end carry to the above

i.e. 0001111 + 1

Answer = 10000

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Exercise

• Subtract (1000100 – 1010100) using 1’s Complement

M = 1000100

N = 1010100


1000100

+ 0101011

= 01101111

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Sample Quiz

• Convert the following into Binary, Hexadecimal & Octal.

• 53.0286

• 106.274

• Write the 10’s Complement of 786.543 & 9’s

Complement of 12345

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DLD03 Complement

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