#import "@local/handout:0.1.0": *

= Integers

#definition()
A _bit string_ is a string of binary digits. \
In this handout, we'll denote bit strings with the prefix `0b`. \
#note[This prefix is only notation---it is _not_ part of the string itself.] \
For example, $1001$ is the number "one thousand and one," while $#text([`0b1001`])$ is the string of bits "1 0 0 1".

#v(2mm)
We will separate long bit strings with underscores for readability. \
Underscores have no meaning: $#text([`0b1111_0000`]) = #text([`0b11110000`])$.

#problem()
What is the value of the following bit strings, if we interpret them as integers in base 2?
- `0b0001_1010`
- `0b0110_0001`

#solution([
  - $#text([`0b0001_1010`]) = 2 + 8 + 16 = 26$
  - $#text([`0b0110_0001`]) = 1 + 32 + 64 = 95$
])

#v(1fr)

#definition()
We can interpret a bit string in any number of ways. \
One such interpretation is the _unsigned integer_, or `uint` for short. \
`uint`s allow us to represent positive (hence "unsigned") integers using 32-bit strings.

#v(2mm)

The value of a `uint` is simply its value as a binary number:
- $#text([`0b00000000_00000000_00000000_00000000`]) = 0$
- $#text([`0b00000000_00000000_00000000_00000011`]) = 3$
- $#text([`0b00000000_00000000_00000000_00100000`]) = 32$
- $#text([`0b00000000_00000000_00000000_10000010`]) = 130$

#problem()
What is the largest number we can represent with a 32-bit `uint`?

#solution([
  $#text([`0b11111111_11111111_11111111_11111111`]) = 2^(32)-1$
])

#v(1fr)
#pagebreak()

#problem()
Find the value of each of the following 32-bit unsigned integers:
- `0b00000000_00000000_00000101_00111001`
- `0b00000000_00000000_00000001_00101100`
- `0b00000000_00000000_00000100_10110000`
#hint([The third conversion is easy---look carefully at the second.])

#instructornote[
  Consider making a list of the powers of two $>= 1024$ on the board.
]

#solution([
  - $#text([`0b00000000_00000000_00000101_00111001`]) = 1337$
  - $#text([`0b00000000_00000000_00000001_00101100`]) = 300$
  - $#text([`0b00000000_00000000_00000010_01011000`]) = 1200$
  Notice that the third int is the second shifted left twice (i.e, multiplied by 4)
])

#v(1fr)


#definition()
In general, fast division of `uints` is difficult#footnote([One may use repeated subtraction, but this isn't efficient.]). \
Division by powers of two, however, is incredibly easy: \
To divide by two, all we need to do is shift the bits of our integer right.

#v(2mm)

For example, consider $#text[`0b0000_0110`] = 6$. \
If we insert a zero at the left end of this string and delete the zero at the right \
(thus "shifting" each bit right), we get `0b0000_0011`, which is 3. \

#v(2mm)

Of course, we lose the remainder when we right-shift an odd number: \
$9$ shifted right is $4$, since `0b0000_1001` shifted right is `0b0000_0100`.

#problem()
Right shifts are denoted by the `>>` symbol: \
$#text[`00110`] #text[`>>`] n$ means "shift `0b0110` right $n$ times." \
Find the value of the following:
- $12 #text[`>>`] 1$
- $27 #text[`>>`] 3$
- $16 #text[`>>`] 8$
#note[Naturally, you'll have to convert these integers to binary first.]

#solution[
  - $12 #text[`>>`] 1 = 6$
  - $27 #text[`>>`] 3 = 3$
  - $16 #text[`>>`] 8 = 0$
]

#v(1fr)
