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)...
Related
For instance, given a word size of 4 bits:
0b1001 * 0b0111 = 0b1111 // -7 * 7 = -1
0b0111 * 0b0111 = 0b0001 // 7 * 7 = 1
0b0111 * 0b0110 = 0b1010 // 7 * 6 = -6
0b1001 * 0b0110 = 0b0110 // -7 * 6 = 6
There's undoubtedly some modular arithmetic going on here, but the way you take mod seems to be quite inconsistent. Is there a neat mathematical formulation of two's complement multiplication?
The nice thing about twos complement is that addition, subtraction, and multiplication of signed operands are exactly the same operations, bit-for-bit, as the ones for unsigned operands, so the computer doesn't need to care whether you think of them as signed or not.
In terms of modular arithmetic as well, the operations mean exactly the same thing. With 4 bit words, when you say:
r = a * b;
You get r = a * b mod 16.
The only difference between signed and unsigned is the value we assign in our heads to the residues mod 16. If we think of the words as unsigned then we have values 0-15. But 15 = -1 mod 16, 14 = -2 mod 16, etc, and if we think of the words as signed, then we just think of the values -8 to 7 instead of 0 to 15.
The reminder operator % that you get in C, java, etc, is annoying in the way it handles negative numbers. If you wanted to express your 4-bit multiply using that operator in larger words, then you could say:
a * b = ( (a * b % 16) + 24 ) % 16 - 8
If the remainder operator worked "properly" so that -1 % 16 == 15, then you could write a * b = (a * b + 8) % 16 - 8
Can someone help me with this question:
“Convert the decimal number 10/32 to the 32-bit IEEE 754 floating point and
express your answer in hexadecimal. (Reminder: the 32 bits are used as
follows: Bit 1: sign of mantissa, bits 2-9: 8-bits of exponent in excess 127, bits 10-32: 23 bits for magnitude of mantissa.)”
I understand how to convert a decimal number to IEE 754. But I am confused on how to answer this—it only gives me a quotient? I am not allowed to use a calculator, so I am unsure how to work this out. Should I convert them both to binary first and divide them?
10/32 = 5/16 = 5•2−4 = 1.25•2−2 = 1.012•2−2.
The sign is +, the exponent is −2, and the significand is 1.012.
A positive sign is encoded as 0.
Exponent −2 is encoded as −2 + 127 = 125 = 011111012.
Significand 1.012 is 1.010000000000000000000002, and it is encoded using the last 23 bits, 010000000000000000000002.
Putting these together, the IEEE-754 encoding is 0 01111101 01000000000000000000000. To convert to hexadecimal, first organize into groups of four bits: 0011 1110 1010 0000 0000 0000 0000 0000. Then the hexadecimal can be easily read: 3EA0000016.
I see it like this:
10/32 = // input
10/2^5 = // convert division by power of 2 to bitshift
1010b >> 5 =
.01010b // fractional result
--^-------------------------------------------------------------
|
first nonzero bit is the exponent position and start of mantissa
----------------------------------------------------------------
man = (1)010b // first one is implicit
exp = -2 + 127 = 125 // position from decimal point + bias
sign = 0 // non negative
----------------------------------------------------------------
0 01111101 01000000000000000000000 b
^ ^ ^
| | mantissa + zero padding
| exp
sign
----------------------------------------------------------------
0011 1110 1010 0000 0000 0000 0000 0000 b
3 E A 0 0 0 0 0 h
----------------------------------------------------------------
3EA00000h
Yes the answer of Eric Postpischil is the same approach (+1 btw) but I didn't like the formating as it was not clear from a first look what to do without proper reading the text.
Giving the conversion of 10/322 without a calculator as an exercise is pure sadism.
There is a a general method doable without tools, but it may be tedious.
N is the number to code. We assume n<1
exp=0
mantissa=0
repeat
n *= 2
exp ++
if n>1
n = n-1
mantissa = mantissa <<1 | 1
else
mantissa = mantissa <<1
until mantissa is a 1 followed by 23 bits
Then you just have to code mantissa and (23-exp) in IEEE format.
Note that frequently this kind of computations lead to loops. Whenever you find the same n, you know that the sequence will be repeated.
As an example, assume we have to code 3/14
3/14 -> 6/14 e=1 m=0
6/14 -> 12/14 e=2 m=00
12/14 -> 24/14-14/14=10/14 e=3 m=001
10->14 -> 20/14-14/14=6/14 e=4 m=0011
6/14 -> 12/14 e=5 m=00110
Great we found a loop !
6/14->12/14->10/14->6/14.
So the mantissa will be 110 iterated as required 110110110...
If we fill the mantissa with 24 bits, we need 26 iterations and exponent is 23-26=-3 (another way to get it is to remark that n became >1 for the first time at iteration 3 and exponent is -3 as 1≤3/14*2^3<2).
And we can do the IEEE754 coding with exponent=127-3=124 and mantissa =1.1011011011011....
I am trying to fit 3 numbers inside 1 number.But numbers will be only between 0 and 11.So their (base) is 12.For example i have 7,5,2 numbers.I come up with something like this:
Three numbers into One number :
7x12=84
84x5=420
420+2=422
Now getting back Three numbers from One number :
422 MOD 12 = 2 (the third number)
422 - 2 = 420
420 / 12 = 35
And i understanded that 35 is multiplication of first and the second number (i.e 7 and 5)
And now i cant get that 7 and 5 anyone knows how could i ???
(I started typing this answer before the other one got posted, but this one is more specific to Arduino then the other one, so I'm leaving it)
The code
You can use bit shifting to get multiple small numbers into one big number, in code it would look like this:
int a, b, c;
//putting then together
int big = (a << 8) + (b << 4) + c;
//separating them again
a = (big >> 8) & 15;
b = (big >> 4) & 15;
c = big & 15;
This code only works when a, b and c are all in the range [0, 15] witch appears to be enough for you case.
How it works
The >> and << operators are the bitshift operators, in short a << n shifts every bit in a by n places to the left, this is equivalent to multiplying by 2^n. Similarly, a >> n shifts to to the right. An example:
11 << 3 == 120 //0000 1011 -> 0101 1000
The & operator performs a bitwise and on the two operands:
6 & 5 == 4 // 0110
// & 0101
//-> 0100
These two operators are combined to "pack" and "unpack" the three numbers. For the packing every small number is shifted a bit to the left and they are all added together. This is how the bits of big now look (there are 16 of them because ints in Arduino are 16 bits wide):
0000aaaabbbbcccc
When unpacking, the bits are shifted to the right again, and they are bitwise anded together with 15 to filter out any excess bits. This is what that last operation looks like to get b out again:
00000000aaaabbbb //big shifted 4 bits to the right
& 0000000000001111 //anded together with 15
-> 000000000000bbbb //gives the original number b
All is working exactly like in base 10 (or 16). Here after your corrected example.
Three numbers into One number :
7x12^2=1008
5*12^1=60
2*12^0=2
1008+60+2=1070
Now getting back Three numbers from One number :
1070 MOD 12 = 2 (the third number)
1070/12 = 89 (integer division) => 89 MOD 12 = 5
89 / 12 = 7
Note also that the maximum value will be 11*12*12+11*12+11=1727.
If this is really programming related, you will be using 16bits instead of 3*8 bits so sparing one byte. An easyer method not using base 12 would be fit each number into half a byte (better code efficiency and same transmission length):
7<<(4+4) + 5<<4 + 2 = 1874
1874 & 0x000F = 2
1874>>4 & 0x000F = 5
1874>>8 & 0x0F = 7
Because MOD(12) and division by 12 is much less efficient than working with powers of 2
you can use the principle of the positional notation to change from one or the other in any base
Treat yours numbers (n0,n1,...,nm) as a digit of a big number in the base B of your choosing so the new number is
N = n0*B^0 + n1*B^1 + ... + nm*B^m
to revert the process is also simple, while your number is greater than 0 find its modulo in respect to the base to get to get the first digit, then subtracts that digit and divide for the base, repeat until finish while saving each digit along the way
digit_list = []
while N > 0 do:
d = N mod B
N = (N - d) / B
digit_list.append( d )
then if N is N = n0*B^0 + n1*B^1 + ... + nm*B^m doing N mod B give you n0, then subtract it leaving you with n1*B^1 + ... + nm*B^m and divide by B to reduce the exponents of all B and that is the new N, N = n1*B^0 + ... + nm*B^(m-1) repetition of that give you all the digit you start with
here is a working example in python
def compact_num( num_list, base=12 ):
return sum( n*pow(base,i) for i,n in enumerate(num_list) )
def decompact_num( n, base=12):
if n==0:
return [0]
result = []
while n:
n,d = divmod(n,base)
result.append(d)
return result
example
>>> compact_num([2,5,7])
1070
>>> decompact_num(1070)
[2, 5, 7]
>>> compact_num([10,2],16)
42
>>> decompact_num(42,16)
[10, 2]
>>>
When one byte is represented by 8 bits in binary notation you have a sequence of 8 possible 1's and 0's. So 00101010 can be shortened to 2A using hexadecimal notation. My book says you can shorten that representation by using hexadecimal after the 4th place from the right. For example...
00101010 can be represented with a mix of hexadecimal notation and binary notation by taking the 4 digits on the left 0010 and and representing that sequence to equal 2 in hexadecimal. I understand because 0010 equals 32 and when you are using hexadecimal notation that has a base of 16 that equals to 2.
What I don't understand is how the right side of the sequence is represented. My book says 1010 can be represented by the letter A which equals to 10. 1010 in binary notation equals 8 + 2 = 10. Here is the issue I'm having. Applying the same concept to the right side as the left side of the 8 bit sequence shouldn't you divide the ride side 10 by 2 since binary notation is using the power of 2 like you divided the left side by 16 since you're using hexadecimal notation which has the power of 16? Am i thinking about it wrong?
Let's start with the complete 8-bit byte, writing the place value under each digit:
0 0 1 0 1 0 1 0
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
↓ ↓ ↓ ↓ 8 4 2 1
↓ ↓ ↓ 16
↓ ↓ 32
↓ 64
128
So, in base 10, this is 32 + 8 + 2 = 42.
If we split the 8-bit byte into two 4-bit nybbles, you have
0 0 1 0 1 0 1 0
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
8 4 2 1 8 4 2 1
You'll notice each 4-bit nybble can hold a value from 0 to 15. So the nybbles can represent two hexadecimal digits.
We treat the two nybbles the same way when we calculate their values. From left to right, the digits in each nybble have place values 8, 4, 2, 1. So the upper (left) nybble has a value of 2, and the lower (right) nybble has a value of 8 + 2 = 10. As you wrote, the decimal number 10 is written as A in hexadecimal, so the hexadecimal byte is written 2A.
Remember, though, that this is hexadecimal. So the places values are powers of 16:
2 A
↓ ↓
↓ 1
16
So, converting back to decimal, 2A = 2×16 + 10 = 32 + 10 = 42.
When converting binary to hexadecimal, the left 4 bits gets the same treatment as the right 4 bits. All that "binary to hex" does is replacing any sequence of 4 bits with a single hexadecimal digit -- at that stage you don't have to worry about conversion to a 'full' number.
Your example 00101010 can be split in two 4-bit sequences: 0010 and 1010. Converting each of these into hex is done by adding them up from right to left, multiplying each next bit by two. This is exactly what you ate used to in base-10; a value 532 represents "10^0 * 2 + 10^1 * 3 + 10^2 * 5" (where ^ is the usual shorthand for "power of"). So, for the first 4 bits you get
0*1 + 1*2 + 0*4 + 0*8 = 2
and the second set of 4 bits
0*1 + 1*2 + 0*4 + 1*8 = 10
In hexadecimal, each value from 0 to 15 is represented by a single 'digit', and we run out of single digits at 9. So starting at 10, we use A,B,C,D,E, and F to represent the decimal values 10, 11, 12, 13, 14, and 15.
Hence, the hex representation of 1010 is A; and your binary number translates to 2A.
Converting this, in turn, to decimal again also works the same as in decimal, only now each 'digit' represents a next power of 16. So this evaluates as
16 * 2 + 1 * A
or (decimal)
16 * 2 + 1 * 10 = 42
You can verify that this is the same decimal value as the starting binary 00101010 by adding up all of its binary numbers:
1 * 0
2 * 1
4 * 0
8 * 1
16 * 0
32 * 1
64 * 0
128 * 0
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.