Bit Manipulation

& - and - both values are 1

| - or - one or the other is 1

^ - xor - if only one value is 1

~ - not - flips the bit

<< - left shift - shifts the bits left, same thing as multiplying by 2! COOL!
00010110 << 00000010 = 01011000
22 * 2 * 2 = 88

>> - right shift - shifts bits to the right, same thing as diving by 2, with rounding down! COOL!
00010110 >> 00000010 = 00000101
22 / 2 / 2 = 5

------ Bit masking ------
A mask is just a positional reference to the bit that needs to be set.

SETTING A BIT AT POSITION

x = 000000, 6 bits.
If I want to clear the second bit (or the 4th column from the right) to 1, I make this mask:
mask = 1 << 4 or mask = 1 << 000100
x = x | mask or x |= mask, Which is 000000 | 010000

CLEARING A BIT AT POSITION

x = 001011, 6 bits.
If I want to set the third bit (or the 3th column from the right) to 0, I make this mask:
mask = 1 << 3 or mask = 1 << 000011
x = x & ~mask or x &= ~mask
x =    001011
m =   001000
~m = 110111
x&~m = 001011 & 110111 = 000011 = 3

FLIPPING A BIT

x = 001011, 6 bits
If I want to flip the second bit (or 4th column from the right) to 1, I make this mask:
mask = 1 << 4 or mask = 1 << 000100
x = x ^ mask or x ^= mask.

TESTING IF A BIT IS SET OR NOT

x = 001011, 6 bits
Testing the second bit (4th column from right, 1st column from left), I make this mask:
mask = x >> 000100
shifted = mask & 1
Shifted will be 1 if set, 0 if not set

CHECK IF NUMBER IS EVEN OR ODD

x = 001011, 6 bits
x & 1 == 0 or 001011 & 000001 = 000001

CHECK IF POWER OF TWO

  x = 001011, 6 bits
(x & x-1 == 0) or (001011 & 001011 - 000001) or (001011 &  001010 = 001010 which is not 0)

HOW MANY BITS ARE DIFFERENT BETWEEN TWO NUMBERS?

```
x = 0b001011
y = 0b111001
def how_many_different(x, y):
  dif = 1
  z = x ^ y
  for i in range(6):
    if (z>>i)&1=1
      dif+=1
  return dif

(how_many_different(x, y) == 3) is True
```