View source for Bitwise math

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== Introduction to Binary ==

Binary math and hexadecimal math have slight differences to decimal math but the same principles apply. For example, in the decimal number 1234, the ‘4’ is in the “ones” placeholder, the ‘3’ is in the “tens” placeholder, the ‘2’ in the “hundreds” placeholder and the ‘1’ in the “thousands” placeholder.

Example 1: Decimal number 1234

{| class="wikitable"
|+ Example 1: Decimal number 1234
|-
! Thousands (1x10^3)
! Hundreds (1x10^2)
! Tens (1x10^1)
! Ones (1x10^0)
|-
| 1
| 2
| 3
| 4
|-
|}

Binary operates a little bit differently, instead of having 1’s, 10’s, 100’s, etc, it has 1’s, 2’s, 4’s, 8’s, etc. So analysing this for a moment, the binary number 1010, has a ‘1’ in the “eights” placeholder and a ‘1’ in the “twos” placeholder. Add these together and 10 is obtained in decimal numbers.

{| class="wikitable"
|+ Example 2: Binary number 1010 = 8+2 = Decimal number 10
|-
! Eights (1x2^3)
! Fours (1x2^2)
! Twos (1x2^1)
! Ones (1x2^0)
|-
| 1
| 0
| 1
| 0
|-
|}

Another example is 1111:

{| class="wikitable"
|+ Example 3: Binary number 1111 = 8+4+2+1 = Decimal number 15
|-
! Eights (1x2^3)
! Fours (1x2^2)
! Twos (1x2^1)
! Ones (1x2^0)
|-
| 1
| 1
| 1
| 1
|-
|}


{| class="wikitable"
|+ Detailed Example
|-
|
! scope="col" | Eights (1x2^3)
! scope="col" | Fours (1x2^2)
! scope="col" | Twos (1x2^1)
! scope="col" | Ones (1x2^0)
|-
! scope="row" | Binary Values
| 1
| 1
| 1
| 1
|-
! scope="row" | Decimal Values
| 8
| 4
| 2
| 1
|}

Through the use of the binary table, the binary values can each be multiplied (1’s) by the above multipliers (1x23 1x22 1x21 1x100) or “8, 4, 2, 1”. This will then give us, 8+4+0+1=13.

=== Basic Addition ===

Much like the decimal system, binary numbers can be added together. For this example the binary numbers 0110 and 0010 are going to be used and added.

{| class="wikitable"
|
! scope="col" | Eights
! scope="col" | Fours
! scope="col" | Twos
! scope="col" | Ones
! scope="col" | Total
|-
! scope="row" | Binary
| 0
| 1
| 1
| 0
| '''6'''
|-
! scope="row" | Decimal
| 0
| 4
| 2
| 0
| '''6'''
|}


{| class="wikitable"
|
! scope="col" | Eights
! scope="col" | Fours
! scope="col" | Twos
! scope="col" | Ones
! scope="col" | Total
|-
! scope="row" | Binary
| 0
| 0
| 1
| 0
| '''2'''
|-
! scope="row" | Decimal
| 0
| 0
| 2
| 0
| '''2'''
|}


Thus by using decimal addition, 4+2+2=8, the value of the addition of two binary numbers can be determined as 8.

== Binary to Hexadecimal ==

The past exercises have featured working with 4 bits at once (4 values ranging from 0-1, e.g. 0001). This is known as a /nybble/ in hexadecimal. A byte is made of two nybbles as 8 bits make a byte. In hexadecimal, there is a 1’s placeholder and a 16’s placeholder. Hexadecimal is 0 through 9 and A through F. A nybble can hold 16 unique values but the highest value is 15 because one of the values is 0. A nybble is a single hex digit. So, A = 10, B = 11, so on and so forth, F = 15.

In hex, AF is obtained as a byte, AF = 175 in decimal because A is in the 16’s placeholder, A = 10, 10*16=160, plus F which is in the 1’s placeholder, 15*1=15. Therefore 160(A)+15(F) = 175.

== NOT, AND, OR and XOR ==

Note: all examples in this section will be using hexadecimal and binary.

=== NOT ===

Not is an instruction that takes only ONE operand. The best way to describe it is that it inverts the binary value. 

Example:

  A = 1010 in binary or 10 in decimal. NOT A results in the inversion of 1010 which is 0101. Therefore NOT A = 5.

=== AND ===

The next instruction is AND, which returns “TRUE” per bits that are the same and takes two operands. True is 1 and false is 0. It operates bit by bit, just like NOT. 

Example:

  F AND A = A, F AND * = *

Any time F is AND’d with something, whatever it is AND’d with is the result.

Example: F AND A

                -----------
  F = 1111 so...> 1|1|1|1 < - All four bits are true.
                -----------
  A = 1010 so...> 1|0|1|0 < - The first and third bits are true.
                -----------
  1010          > 1|0|1|0 < - The first and third bits are the only true values in both F and A.
                -----------

Because of the bits in F and A that are the same, it generates A, however…

Example:

                -----------
  6 = 0110 so...> 0|1|1|0 < - The second and third bits are true.
                -----------
  5 = 0101 so...> 0|1|0|1 < - The second and fourth bits are true.
                -----------
  4 = 0100      > 0|1|0|0 < - The second bit is the only true value in both 6 and 5.
                -----------

AND compares each bit and if both bits per placeholder are true, then it returns a true for that placeholder, all else gets turned into 0. 

=== OR ===

OR will return true for each placeholder if ANY of the bits are true. 

Example: 5 OR C

  (0101 = 5) OR (1100 = C) = 1101 which equals D.
                -----------
  5 = 0101 so...> 0|1|0|1 < - The second and fourth bits are true.
                -----------
  C = 1100 so...> 1|1|0|0 < - The first and second bits are true.
                -----------
  D = 1101      > 1|1|0|1 < - The first, second and fourth bits are true in at least one instance.
                -----------
			  
=== XOR ===

XOR is kind of a strange command, if the placeholder bits are different, it returns a true bit. If they are the same, it returns a false bit. Thus, anything XOR’d with itself, results in 0.

General Examples: 

  1 xor 1 = 0, 0 xor 0 = 0, 1 xor 0 = 1.

Detailed Example: A xor F = 5

                -----------
  A = 1010 so...> 1|0|1|0 < - The first and third bits are true.
                -----------
  F = 1111 so...> 1|1|1|1 < - The first, second and third bits are true.
                -----------
  0101 = 5      > 0|1|0|1 < - The second and fourth bits are true in ONLY one instance as opposed
                              to two found in bits 1 and 3.
         
       -----------
{| class="wikitable" 
|-
!Operation
!Hexadecimal
!Binary
!comment
|-   <!--indicates new row, each "|" is a cell in that row-->
| 


<center>'''xor''' </center> 
|
A

F                     
|
1010

1111
|
The first and third bits are true.  

the first,second, and third bits are true.
|-
|<center>'''='''</center>
|5
|0101
|the second and fourth bits are true in ONLY one instance as opposed to two, found in bits 1 and 3
|-
|}


The 1’s placeholders are the same so they return 0, same with the 8’s placeholder. The 4’s and 2’s are different, therefore return true. In short, XOR determines which bits differ in two numbers.

== Shift and rotate ==

We're about to cover two more bitwise operations. One is called shift, the other, rotate. For those who already understand these concepts, /Logical/ shifts and Rotates /without carry/ (also known as /Circular shifts/) will be covered. Each of these operations can occur either to the right or to the left. Following the examination of nibbles, that is the size of binary that will be used in the examples (4 bits).

For logical shifts, there is the assumption of [[big endian] http://www.blackhatacademy.org/security101/Byte#Big-endianness].

=== Logical Shifts ===

On a binary level this stuff is really simple, given that only the nibble is being used. Using A or 1010 as an example.

  vvvvvvvv
  |-1010-|	Shifted left once, the 1 at the beginning gets shifted off.
  |-0100-|      Thus, the value becomes 0100. 
  ^^^^^^^^                
The end value "0" is /always/ used to replace the residual stuff. If A is shifted to the left twice, it goes from 1010 to 1000:

  vvvvvvvv
  |-1010-|	The first `10' is shifted off and the remaining two placeholders get zeroed out.
  |-1000-|      <-x2     
  ^^^^^^^^                

If 'A' is shifted to the right once it goes from 1010 to 0101

  vvvvvvvv
  |-1010-|	->x1
  |-0101-|         
  ^^^^^^^^ 

If 'A' is shifted to the right twice it goes from 1010 to 0010

  vvvvvvvv
  |-1010-|	->x2
  |-0010-|         
  ^^^^^^^^ 

At this point, understanding should be emerging into real hacking and cryptography. If understanding is not attained regarding this knowledge, further extension topics will prove difficult. Thus, it is considered a baseline knowledge.

=== Exercises: ===

1) Solve for B left shift one

  vvvvvvvv
  |-????-|	Value of B
  |-????-|    <-x1     
  ^^^^^^^^

2) Solve for F right shift three

  vvvvvvvv
  |-????-|	Value of F
  |-????-|    ->x3     
  ^^^^^^^^

Answers: 1) 0110 or 6. 2) 0001 or 1.

== Circular Shift or Bit Rotation ==

Circular shifts, or bit rotation occurs in processors which involves rotate /with carry/. Circular shifts are the same as rotate /without/ carry. So, continuing the usage of a nibble and A (1010) as the example...

  vvvvvvvv
  |-1010-|	If 1 is rotated to the left it becomes 0101 or 5 
  |-0101-|    <-x1     
  ^^^^^^^^                 

Instead of replacing a value with zero, like a normal shift, the value is taken and shifted off of one side and applied to the other side. So for 1100 (C)

  vvvvvvvv
  |-1100-|	Shift left 1, becomes 9 (1001) 
  |-1001-|    <-x1     
  ^^^^^^^^ 

  vvvvvvvv
  |-1100-|	Shift right 1, becomes 6 (0110) 
  |-0110-|    ->x1     
  ^^^^^^^^   

Based upon these examples, a circular shift is not the same thing as a logical shift. Proceeding further, two's complement and something small about binary will be explained. 

Remember in the previous lesson, a nibble (four bits) can hold 16 values (the uppermost being 15 because one of these values is 0). Four bits maximum value also = 2^4 - 1. 4 is taken from the number of bits and the -1 from the zero placeholder. If a full byte was going to be shifted, it would be 2^8 - 1. The maximum value of which is 255.

== Two's Complement ==

Two's complement is a way to represent negative numbers in binary,  the same thing as giving an integer a sign. Getting into the mechanics of two's compliment, though it ends up with a few ways to do it. Signing bits are used these days.

A two's complement is basically a NOT operation performed on the nibble. Any binary number is converted between positive and negative by computing its two's complement. Basically, the leftmost bit is called the sign bit. There are two ways. (taking nibbles as an example) -1 could be 1001. This is not useful, as now there are two ways to represent 0, 1000 for -0 and 0000 for ordinary 0. Also the addition and subtraction has to be changed for this. This is all not nice.
This is why people invented two's complement, two's complement is basically using 1111 for -1 and 1000 for -8. 

The first operation performed is a NOT operation on the nibble, -1 has to be done. To determine the two complements the positive representation of the number is required. If it's negative, complement it and add one. Thus, to determine the two complements, positive representation of the number is required. If the original value is negative, determine the compliment and do +1. If its positive, the easy way to do it, is to tack zeroes on in the right number of bits. If negative tack 1's instead.

Just using one byte. It has 256 possible values
Doing something like:

  -42
  -42 + 256
  -42 + 255 + 1
  255 - 42  + 1
  213 + 1
  214

The binary representation of both numbers /without/ a signing bit is the same.

Computers use this stuff to represent signing amongst other things. When doing arithmetic stuff gets messy. This gets even more messy when doing arithmetic shifts and rotates with carry on a CPU, since that's what modern day CPU's use. 

Taking the example of -42. First, its positive representation is taken. So 42 in binary is: 00101010 (32+8+2). All of those placeholders have 1's and the rest have 0's.

Next step, as mentioned is to perform a NOT operation. So NOT(00101010) WILL BE 11010101. Replace 1 by 0 and vice-versa. Final step is to add 1 to the number obtained in above step.
Thus:

  11010101
  +
  00000001
  ----------
  = 11010110

What can also be done, is go at the bits one by one, starting at the right, and copy all the zeroes, until a 1 is obtained. Copy that 1 too, then flip all the remaining bits. So 42 is obtained again, which is 00101010. Then move from right-to-left, and get 11010110.

== Overflows ==

Now, overflow occurs in a "signed" integer when it increments too far. Using a nibble for an example,

  -8 4 2 1

The highest positive value is 7, 0111, the lowest negative value is -8, 1000. On a processor, the increment instruction(inc) changes this. If -8 is incremented, -7 is obtained, since it is incremented. Basically when moving to 7 and incremented, the next step in binary is 1000 and it lands on -8.

In an arithmetic shift, this is typically what processors do since signed numbers are worked with most of the times. Now of course, there are ways to use unsigned numbers, but this will be left for later. Most languages use an arithmetic shift, which is also called a 'signed shift'. This is because the number is "signed", that is, it has a + or - attatched to its value. Thus, when reading the hex, the way this works isn't going to change so terribly.

  vvvvvvvv
  |-0110-|	(6)
  |-1100-|	left shift 1 becomes 1100
  ^^^^^^^^

The difference here is if that first bit is -8 and not an 8. A positive number has just been turned into a negative number. 6 left shift 1 = -4

  -8 4 2 1
   1 1 0 0
  -8 + 4 = -4

It's now -4. 

If shifted to the right, a single shift will /always/ change the sign. If shifted to the left on any /unsigned/ binary number (shift by n), it is the same as multiplying it by 2^n. If shifted to the right on a two's complement signed number it is the same as dividing it by 2^n and always rounds down.

Example:

-8 right shift 1 or 1000 right shift 1 becomes 4. In other words, any signed number right shifted becomes positive immediately. Regardless of whether it was positive or not, it will always be positive once a right shift has been applied.

As stated before, x right shift n on a signed number is the same as x/(2^n). The problem that this could cause is that it always rounds down. So, if it always rounds down from 2^n (i.e. a fraction is obtained) it will round down until the fraction is gone.

The flow on problem from this is that if it rounds down and n is signed, the processor can run into a divide by zero error. 

== Rotate With Carry ==

Moving on now to rotate with carry. Rotate with carry occurs on a computer when a number is too large to fit into a single register. For example: On 32 bit systems the largest value an unsigned variable can hold is 2^32 - 1. An 8 bit system is being used just for sanity:

  2^8 - 1 = 255

That would be the largest value that can be held. If wanting to hold 256, it would need two pieces of processor memory to do this so rotate with carry. There's a bit on the CPU called CF or carry flag. It is just a little bit, 1 or 0. So on an 8 bit system a 16 bit number would want to be rotated. 

  1011 1000 1010 0011 (that's a 16 bit number)

Now, the registers only contain 8 bits, so register 1 = 1011 1000 and register 2 = 1010 0011 to connect these during the shift operation rather the rotate operation. There is a small bit, the carry flag, that will store the value as the bits are shifted between the two registers. It will take the first "1" from register 2 into the CF bit then move all of register2 1 bit to the left so on and so forth.