I am familiar with two's complement when performing addition in 4 bits, then I am confused when I face the question below
** Find the sum of the 2 two's complement binary number 010111 and 110101 in 8 bit output**
Below is my attempt, but I am in a dilemma, should I
(1) discard the carry, then add two 0, so the answer is 00001100, which is 12 in decimal
Thank you !
(2) just add 1 in the beginning, so it the answer is 11001100, which is 204 in decimal
For the Twos Complement in 8 bit you have to invert ALL 8 bits of the number
As far as I know the ones complement and twos complement operate on the absolute value so:
The Binary Number 010111 is represented in 8 bits with 00010111
C_1(00010111) = 00010111 xor 11111111 = 11101000
C_2(00010111) = C_1 + 1 = 00010111 + 1 = 00011000
The Binary Number 110101 is represented in 8 bits -> 00110101
C_1(00110101) = 00110101 xor 11111111 = 11001010
C_2(00010111) = C_1 + 1 = 11001010 + 1 = 11001011
Now add the two twos-complements:
C_2(00010111) + C_2(00010111) = 00011000 + 11001011 = 11100011
Pls correct me if messed something up with sign bits (I just took the binary numbers as is in 8 bit)...
I am writing a library for RTC module in Arduino where the data is stored in BCD. I know how the Decimal number is converted into BCD but having some problem while writing it programmatically. After searching the internet I got two formulas which are as follows and working perfectly but cannot understand how it is calculating.
1. Formula1
DEC to BCD
(value / 10 * 16) + (value % 10)
Example
DEC -> 40 which converted to 01000000 in BCD which is again equal to 64.
So if I put into the formula I get the same result.
(40/10 * 16) + (40%10)
= (4*16) + 0
= 64
BCD to DEC
(value / 16 * 10) + (value % 16)
2. Formula2
DEC to BCD
val + 6 * (val / 10)
BCD to DEC
val - 6 * (val >> 4)
If anyone can explain it in details it will be helpful.
Thanks to all in advance.
The correct conversion functions are:
byte bcdToDec(byte val)
{
return( (val/16*10) + (val%16) );
}
byte decToBcd(byte val)
{
return( (val/10*16) + (val%10) );
}
Why does this work? Let's take a single digit 5. In binary, it's
0101 = 5
Now let's take that same digit, shift it four places to the left by adding four zeroes to the right:
0101 0000 = 50 BCD
That's how BCD works. Since it takes four binary digits to represent the decimal digits 0 through 9, each single decimal digit is represented by four bits. The key is that shifting four places in binary multiplies or divides by 16, so that's the reason for the 16 values in the formulas.
So let's take 96:
0110 = 6
1001 = 9
1001 0110 = 96 BCD
This question already has answers here:
What does the caret (^) operator do?
(5 answers)
Closed 2 years ago.
What mathematical operation does XOR perform?
XOR is a binary operation, it stands for "exclusive or", that is to say the resulting bit evaluates to one if only exactly one of the bits is set.
This is its function table:
a | b | a ^ b
--|---|------
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0
This operation is performed between every two corresponding bits of a number.
Example: 7 ^ 10
In binary: 0111 ^ 1010
0111
^ 1010
======
1101 = 13
Properties: The operation is commutative, associative and self-inverse.
It is also the same as addition modulo 2.
^ is the Python bitwise XOR operator. It is how you spell XOR in python:
>>> 0 ^ 0
0
>>> 0 ^ 1
1
>>> 1 ^ 0
1
>>> 1 ^ 1
0
XOR stands for exclusive OR. It is used in cryptography because it let's you 'flip' the bits using a mask in a reversable operation:
>>> 10 ^ 5
15
>>> 15 ^ 5
10
where 5 is the mask; (input XOR mask) XOR mask gives you the input again.
A little more information on XOR operation.
XOR a number with itself odd number of times the result is number
itself.
XOR a number even number of times with itself, the result is 0.
Also XOR with 0 is always the number itself.
One thing that other answers don't mention here is XOR with negative numbers -
a | b | a ^ b
----|-----|------
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0
While you could easily understand how XOR will work using the above functional table, it doesn't tell how it will work on negative numbers.
How XOR works with Negative Numbers :
Since this question is also tagged as python, I will be answering it with that in mind. The XOR ( ^ ) is an logical operator that will return 1 when the bits are different and 0 elsewhere.
A negative number is stored in binary as two's complement. In 2's complement, The leftmost bit position is reserved for the sign of the value (positive or negative) and doesn't contribute towards the value of number.
In, Python, negative numbers are written with a leading one instead of a leading
zero. So if you are using only 8 bits for your two's-complement
numbers, then you treat patterns from 00000000 to 01111111 as the
whole numbers from 0 to 127, and reserve 1xxxxxxx for writing negative
numbers.
With that in mind, lets understand how XOR works on negative number with an example. Lets consider the expression - ( -5 ^ -3 ) .
The binary representation of -5 can be considered as 1000...101 and
Binary representation of -3 can be considered as 1000...011.
Here, ... denotes all 0s, the number of which depends on bits used for representation (32-bit, 64-bit, etc). The 1 at the MSB ( Most Significant Bit ) denotes that the number represented by the binary representation is negative. The XOR operation will be done on all bits as usual.
XOR Operation :
-5 : 10000101 |
^ |
-3 : 10000011 |
=================== |
Result : 00000110 = 6 |
________________________________|
∴ -5 ^ -3 = 6
Since, the MSB becomes 0 after the XOR operation, so the resultant number we get is a positive number. Similarly, for all negative numbers, we consider their representation in binary format using 2's complement (one of most commonly used) and do simple XOR on their binary representation.
The MSB bit of result will denote the sign and the rest of the bits will denote the value of the final result.
The following table could be useful in determining the sign of result.
a | b | a ^ b
------|-------|------
+ | + | +
+ | - | -
- | + | -
- | - | +
The basic rules of XOR remains same for negative XOR operations as well, but how the operation really works in negative numbers could be useful for someone someday 🙂.
Another application for XOR is in circuits. It is used to sum bits.
When you look at a truth table:
x | y | x^y
---|---|-----
0 | 0 | 0 // 0 plus 0 = 0
0 | 1 | 1 // 0 plus 1 = 1
1 | 0 | 1 // 1 plus 0 = 1
1 | 1 | 0 // 1 plus 1 = 0 ; binary math with 1 bit
You can notice that the result of XOR is x added with y, without keeping track of the carry bit, the carry bit is obtained from the AND between x and y.
x^y // is actually ~xy + ~yx
// Which is the (negated x ANDed with y) OR ( negated y ANDed with x ).
The (^) XOR operator generates 1 when it is applied on two different bits (0 and 1). It generates 0 when it is applied on two same bits (0 and 0 or 1 and 1).
I'm trying to understand the binary operators in C# or in general, in particular ^ - exclusive or.
For example:
Given an array of positive integers. All numbers occur even number of times except one number which occurs odd number of times. Find the number in O(n) time and constant space.
This can be done with ^ as follows: Do bitwise XOR of all the elements. Finally we get the number which has odd occurrences.
How does it work?
When I do:
int res = 2 ^ 3;
res = 1;
int res = 2 ^ 5;
res = 7;
int res = 2 ^ 10;
res = 8;
What's actually happening? What are the other bit magics? Any reference I can look up and learn more about them?
I know this is a rather old post but I wanted simplify the answer since I stumbled upon it while looking for something else.
XOR (eXclusive OR/either or), can be translated simply as toggle on/off.
Which will either exclude (if exists) or include (if nonexistent) the specified bits.
Using 4 bits (1111) we get 16 possible results from 0-15:
decimal | binary | bits (expanded)
0 | 0000 | 0
1 | 0001 | 1
2 | 0010 | 2
3 | 0011 | (1+2)
4 | 0100 | 4
5 | 0101 | (1+4)
6 | 0110 | (2+4)
7 | 0111 | (1+2+4)
8 | 1000 | 8
9 | 1001 | (1+8)
10 | 1010 | (2+8)
11 | 1011 | (1+2+8)
12 | 1100 | (4+8)
13 | 1101 | (1+4+8)
14 | 1110 | (2+4+8)
15 | 1111 | (1+2+4+8)
The decimal value to the left of the binary value, is the numeric value used in XOR and other bitwise operations, that represents the total value of associated bits. See Computer Number Format and Binary Number - Decimal for more details.
For example: 0011 are bits 1 and 2 as on, leaving bits 4 and 8 as off. Which is represented as the decimal value of 3 to signify the bits that are on, and displayed in an expanded form as 1+2.
As for what's going on with the logic behind XOR here are some examples
From the original post
2^3 = 1
2 is a member of 1+2 (3) remove 2 = 1
2^5 = 7
2 is not a member of 1+4 (5) add 2 = 1+2+4 (7)
2^10 = 8
2 is a member of 2+8 (10) remove 2 = 8
Further examples
1^3 = 2
1 is a member of 1+2 (3) remove 1 = 2
4^5 = 1
4 is a member of 1+4 (5) remove 4 = 1
4^4 = 0
4 is a member of itself remove 4 = 0
1^2^3 = 0Logic: ((1^2)^(1+2))
(1^2) 1 is not a member of 2 add 2 = 1+2 (3)
(3^3) 1 and 2 are members of 1+2 (3) remove 1+2 (3) = 0
1^1^0^1 = 1 Logic: (((1^1)^0)^1)
(1^1) 1 is a member of 1 remove 1 = 0
(0^0) 0 is a member of 0 remove 0 = 0
(0^1) 0 is not a member of 1 add 1 = 1
1^8^4 = 13 Logic: ((1^8)^4)
(1^8) 1 is not a member of 8 add 1 = 1+8 (9)
(9^4) 1 and 8 are not members of 4 add 1+8 = 1+4+8 (13)
4^13^10 = 3 Logic: ((4^(1+4+8))^(2+8))
(4^13) 4 is a member of 1+4+8 (13) remove 4 = 1+8 (9)
(9^10) 8 is a member of 2+8 (10) remove 8 = 2
1 is not a member of 2+8 (10) add 1 = 1+2 (3)
4^10^13 = 3 Logic: ((4^(2+8))^(1+4+8))
(4^10) 4 is not a member of 2+8 (10) add 4 = 2+4+8 (14)
(14^13) 4 and 8 are members of 1+4+8 (13) remove 4+8 = 1
2 is not a member of 1+4+8 (13) add 2 = 1+2 (3)
To see how it works, first you need to write both operands in binary, because bitwise operations work on individual bits.
Then you can apply the truth table for your particular operator. It acts on each pair of bits having the same position in the two operands (the same place value). So the leftmost bit (MSB) of A is combined with the MSB of B to produce the MSB of the result.
Example: 2^10:
0010 2
XOR 1010 8 + 2
----
1 xor(0, 1)
0 xor(0, 0)
0 xor(1, 1)
0 xor(0, 0)
----
= 1000 8
And the result is 8.
The other way to show this is to use the algebra of XOR; you do not need to know anything about individual bits.
For any numbers x, y, z:
XOR is commutative: x ^ y == y ^ x
XOR is associative: x ^ (y ^ z) == (x ^ y) ^ z
The identity is 0: x ^ 0 == x
Every element is its own inverse: x ^ x == 0
Given this, it is easy to prove the result stated. Consider a sequence:
a ^ b ^ c ^ d ...
Since XOR is commutative and associative, the order does not matter. So sort the elements.
Now any adjacent identical elements x ^ x can be replaced with 0 (self-inverse property). And any 0 can be removed (because it is the identity).
Repeat as long as possible. Any number that appears an even number of times has an integral number of pairs, so they all become 0 and disappear.
Eventually you are left with just one element, which is the one appearing an odd number of times. Every time it appears twice, those two disappear. Eventually you are left with one occurrence.
[update]
Note that this proof only requires certain assumptions about the operation. Specifically, suppose a set S with an operator . has the following properties:
Assocativity: x . (y . z) = (x . y) . z for any x, y, and z in S.
Identity: There exists a single element e such that e . x = x . e = x for all x in S.
Closure: For any x and y in S, x . y is also in S.
Self-inverse: For any x in S, x . x = e
As it turns out, we need not assume commutativity; we can prove it:
(x . y) . (x . y) = e (by self-inverse)
x . (y . x) . y = e (by associativity)
x . x . (y . x) . y . y = x . e . y (multiply both sides by x on the left and y on the right)
y . x = x . y (because x . x = y . y = e and the e's go away)
Now, I said that "you do not need to know anything about individual bits". I was thinking that any group satisfying these properties would be enough, and that such a group need not necessarily be isomorphic to the integers under XOR.
But #Steve Jessup proved me wrong in the comments. If you define scalar multiplication by {0,1} as:
0 * x = 0
1 * x = x
...then this structure satisfies all of the axioms of a vector space over the integers mod 2.
Thus any such structure is isomorphic to a set of vectors of bits under component-wise XOR.
This is based on the simple fact that XOR of a number with itself results Zero.
and XOR of a number with 0 results the number itself.
So, if we have an array = {5,8,12,5,12}.
5 is occurring 2 times.
8 is occurring 1 times.
12 is occurring 2 times.
We have to find the number occurring odd number of times. Clearly, 8 is the number.
We start with res=0 and XOR with all the elements of the array.
int res=0;
for(int i:array)
res = res ^ i;
1st Iteration: res = 0^5 = 5
2nd Iteration: res = 5^8
3rd Iteration: res = 5^8^12
4th Iteration: res = 5^8^12^5 = 0^8^12 = 8^12
5th Iteration: res = 8^12^12 = 8^0 = 8
The bitwise operators treat the bits inside an integer value as a tiny array of bits. Each of those bits is like a tiny bool value. When you use the bitwise exclusive or operator, one interpretation of what the operator does is:
for each bit in the first value, toggle the bit if the corresponding bit in the second value is set
The net effect is that a single bit starts out false and if the total number of "toggles" is even, it will still be false at the end. If the total number of "toggles" is odd, it will be true at the end.
Just think "tiny array of boolean values" and it will start to make sense.
The definition of the XOR (exclusive OR) operator, over bits, is that:
0 XOR 0 = 0
0 XOR 1 = 1
1 XOR 0 = 1
1 XOR 1 = 0
One of the ways to imagine it, is to say that the "1" on the right side changes the bit from the left side, and 0 on the right side doesn't change the bit on the left side. However, XOR is commutative, so the same is true if the sides are reversed.
As any number can be represented in binary form, any two numbers can be XOR-ed together.
To prove it being commutative, you can simply look at its definition, and see that for every combination of bits on either side, the result is the same if the sides are changed. To prove it being associative, you can simply run through all possible combinations of having 3 bits being XOR-ed to each other, and the result will stay the same no matter what the order is.
Now, as we proved the above, let's see what happens if we XOR the same number at itself. Since the operation works on individual bits, we can test it on just two numbers: 0 and 1.
0 XOR 0 = 0
1 XOR 1 = 0
So, if you XOR a number onto itself, you always get 0 (believe it or not, but that property of XOR has been used by compilers, when a 0 needs to be loaded into a CPU register. It's faster to perform a bit operation than to explicitly push 0 into a register. The compiler will just produce assembly code to XOR a register onto itself).
Now, if X XOR X is 0, and XOR is associative, and you need to find out what number hasn't repeated in a sequence of numbers where all other numbers have been repeated two (or any other odd number of times). If we had the repeating numbers together, they will XOR to 0. Anything that is XOR-ed with 0 will remain itself. So, out of XOR-ing such a sequence, you will end up being left with a number that doesn't repeat (or repeats an even number of times).
This has a lot of samples of various functionalities done by bit fiddling. Some of can be quite complex so beware.
What you need to do to understand the bit operations is, at least, this:
the input data, in binary form
a truth table that tells you how to "mix" the inputs to form the result
For XOR, the truth table is simple:
1^1 = 0
1^0 = 1
0^1 = 1
0^0 = 0
To obtain bit n in the result you apply the rule to bits n in the first and second inputs.
If you try to calculate 1^1^0^1 or any other combination, you will discover that the result is 1 if there is an odd number of 1's and 0 otherwise. You will also discover that any number XOR'ed with itself is 0 and that is doesn't matter in what order you do the calculations, e.g. 1^1^(0^1) = 1^(1^0)^1.
This means that when you XOR all the numbers in your list, the ones which are duplicates (or present an even number of times) will XOR to 0 and you will be left with just the one which is present an odd number of times.
As it is obvious from the name(bitwise), it operates between bits.
Let's see how it works,
for example, we have two numbers a=3 and b=4,
the binary representation of 3 is 011 and of 4 is 100, so basically xor of the same bits is 0 and for opposite bits, it is 1.
In the given example 3^4, where "^" is a xor symbol, will give us 111 whose decimal value will be 7.
for another example, if you've given an array in which every element occurs twice except one element & you've to find that element.
How can you do that? simple xor of the same numbers will always be 0 and the number which occur exactly once will be your output. because the output of any one number with 0 will be the same name number because the number will have set bits which zero don't have.
This is about how to do number conversion between binary to octal, octal to hexadecimal, binary to hexadecimal.. ( in all these decimal is no where there, either in source or destination )
Whenever decimal is involved either in source or destination i have a general methodology as,
if decimal is source, do the mod operation of that number with the base of destination. ( and continue the same way, and get the result ),
Example: convert decimal 20 to octal
20 mod 8 = 4
2 = 2
So it is 24 in Octal. This way you can do for anything ( binary, hex ) from Decimal.
if decimal is destination, do the multiplication operation with the source's base from 0 to N which starts from left to right. And add up to get the decimal.
Example: convert binary 1010 to decimal
0 : 0 x 2**0 = 0
1 : 1 x 2**1 = 2
0 : 0 x 2**2 = 0
1: 1 x 2**3 = 8
The sum is 10 in Decimal. This way you can do for anything ( octal, hex ) to decimal.
Questions
Is there any similar general logic, which i can use for conversion between octal, hex and binary ?
Hope the above is clear.. Else let me know.
The conversion between binary and either octal or hex is even easier than decimal. Just break the binary digits up into groups of 3 or 4 (depending on oct or hex), then convert each group to a digit. (If you don't have a multiple of 3 or 4 binary digits, just left-pad the number with zeros.)
Example:
111101111010 binary
1111 0111 1010
F 7 A
F7A hex
111101111010
111 101 111 010
7 5 7 2
7572 oct
Converting back is just the opposite. Each digit in octal or hex gets converted directly to either 3 or 4 binary digits.