Representing Integers

For our discussions, I am going to assume that we will use a computer with a word size of 4 bits.
- This gives us a total space of $2^4 = 16$ different things we can represent.
- In a real computer this is 32 ie 4,294,967,296 different tings.
- Or perhaps 64 ie 18,446,744,073,709,551,616
So with four bits, we could represent the integers $0000_{2} = 0$ through $1111_2 = 15$.
How do we represent negative numbers in real life?
- We use a system called signed-magnitude.
- The sign is either + or -
- A missing sign is interpreted as +
- magnitude frequently refers to distance.
- +3 or just 3 is three units to the right of 0 on the number line.
- In this case the sign is + and the magnitude is 3.
- -7 is seven units to the left of 0 on the number line.
- In this case the sign is - and the magnitude is 7.
We could do the same with binary.
- Use one bit as a sign.
  - 0 -> +
  - 1 -> -
- $0001_2 = +1$
- $1001_2 = -1$
- What is the biggest negative number?
- What is the biggest positive number?
- This system has two problems.
- First is -0
  - What is $1000_2$?
  - This causes problems with algorithms.
- Second tell me what operation you would be doing for each of these problems.
  - 3 + 2
  - 2 + -3
  - 3 - 2
  - 2 - 3
  - -5 - -7
  - -4 + 5
  - 4 + -5
  - -8 - -3
  - -8 + -3
- In general, for + if the signs are the same add and keep the sign.
- In general, for + if the signs are different subtract the smaller from the larger and keep the sign of the larger.
- And the opposite for subtraction.
There is another system.
- In the book she introduces 10's compliment, but you can ignore that.
- We will take a look at two's compliment.
Two's compliment.
- This is generally for a fixed number of bits.
- In our case, we will look at 4 bit two's compliment numbers.
- The size is generally set by the word size.
I don't really care about the background theory, you will not need it.
Positive 4-bit two's complement numbers
- For n bits, we can represent 0 through $2^{n-1} - 1$
- In our case that is 0 through $2^{4-1} -1$
- Or 0 through 7
- 0000 through 0111
- Yes, I think you should know this.
- This is not bad, there are n bits, the first must be a 0
- So there are n-1 bits that can be used for the magnitude.
- And one of those represents 0.
For negative numbers
- They always start with a 1.
- So 1000 through 1111 are the negative numbers.
- We don't loose one for 0, since we don't have a -0
- So we can represent $-2^{n-1}$ to -1

But how do we do this?

To represent a negative number, flip the bits and add 1.

-3:

3 is 0011
Flip the bits  1100
Add 1     1101
1101 is -3

-7

7 is 0111   
flip the bits   1000
Add 1     1001
1001 is negative 7.

We do the same to convert a negative number

What is 1001  
It starts with a 1 so it is negative.
flip the bits   0110
Add 1          0111
This is 7, so 1001 is -7

What is 1111?

It starts with a 1, so it is negative.
Flip the bits 0000
Add 1    0001 = 1
So 1111 is -1

1000 is a special case.

It starts with a 1 so it is negative.
Flip the bits  1111
Add 1          1000
It is the same thing?   This is defined to be -8.

But why do this?

What is 3 + -2

0011 is 3.
0010 is 2,
0010 -> 1101 -> 1110 
so 1110 is -2
   
         0011 
        +1110 
         ----
       1|0001

For now ignore the final carry the answer is 1.

Try -3 + 2

-3 0011 -> 1100 -> 1101
2: 0010

       1101
      +0010
      -----
     0|1111
But what is 1111?   It is negative
Flip the bits 0000
Add 1   0001
so 1111 is -1 (but we did that above)

Notice, everything can be done as addition with no special comparisons.

What happens if we add 4 + 5

The answer should be 9, but is actually -7.
- This is called numeric overflow.
- We just don't have enough bits to represent 9.
- We can detect this when the carry in to the last bit does not match the carry out.
- There is really nothing we can do about overflow.