I have been searching for the answer to this and was unable to find an exact answer help will be much appreciated.
echo $[ 2 ^ 2 ]
returns value 0
echo $[ 2 ^ 3 ]
returns 1
echo $[ 2 ^ 4 ]
returns 6
My question is what math operation is taking place when using the ^ in this context?
I expected to see a power of function. Would really appreciate any clarification, thanks in advance.
It's a bitwise XOR operation.
It compares the bits for the two numbers, and if for a given position, one of the bits is 1, the resulting bit will also be set to 1. In all other cases, the resulting bit will be 0.
So, for your examples:
2 010
2 010
--------
0 000
2 010
3 011
--------
1 001
2 010
4 100
--------
6 110
I would say, your commands are doing a bit-xor with the numbers.
Related
I am currently learning cryptography,
I got different tasks. The current one features stream ciphers based on LFSR which can be solved with the Gaussian algorithm.
The cipher is: 0001 0010 01101101
I know that the first 8 bits of plaintext are all 1.
The length of n is 4.
So if I XOR them I get:
11101101
But now I am stuck and don't know how to translate it into a correct matrix.
1 1 1 0 | ?
1 1 0 1 | ?
1 0 1 1 | ?
0 1 1 0 | ?
Is this the correct way or do there only need to be 3 variables in front of the line? And where do I get the solution of the question marks from?
I hope someone can help me understanding cryptography. Thanks.
I am looking for a method for calculating similarity score for list of numbers. Ideally the method should give result in fixed range. For example from 0 to 1 where 0 is not similar at all and 1 means all numbers are identical.
For clarity let me provide a few examples:
0 1 2 3 4 5 6 7 8 9 10 => the similarity should be 0 or close to zero as all numbers are different
1 1 1 1 1 1 1 => 1
10 9 11 10.5 => close to 1
1 1 1 1 1 1 1 1 1 1 100 => score should be still pretty high as only the last value is different
I have tried to calculate the similarity based on normalization and average, but that gives me really bad results when there is one 'bad number'.
Thank you.
Similarity tests are always incredibly subjective, and the right one to use depends heavily on what you're trying to use it for. We already have three typical measures of central tendency (mean, median, mode). It's hard to say what test will work for you because there are different ways of measuring that will do what you're asking, but have wildly different measures for other lists (like [1]*7 + [100] * 7). Here's one solution:
import statistics as stats
def tester(ell):
mode_measure = 1 - len(set(ell))/len(ell)
avg_measure = 1 - stats.stdev(ell)/stats.mean(ell)
return max(avg_measure, mode_measure)
What is the correct name of this error correction method?
It is quite similar to Hamming Code, but much more simple. I also cannot find it in the literature any more. The only internet sources, I'm now able to find, which describes the method, are this:
http://www.mathcs.emory.edu/~cheung/Courses/455/Syllabus/2-physical/errors-Hamming.html
And the german-language Wikipedia.
http://de.wikipedia.org/w/index.php?title=Fehlerkorrekturverfahren
In the Wikipedia article, the method is called Hamming-ECC method. But I'm not 100% sure, this is correct.
Here is an example, which describes the way the method works.
Payload: 10011010
Step 1: Determine parity bit positions. Bits, which are powers of 2 (1, 2, 4, 8, 16, etc.) are parity bits:
Position: 1 2 3 4 5 6 7 8 9 10 11 12
Data to be transmitted: ? ? 1 ? 0 0 1 ? 1 0 1 0
Step 2: Calculate parity bit values. Each bit position in the transmission is assigned to a position number. In this example, the position number is a 4-digit number, because we have 4 parity bits. Calculate XOR of the values of those positions (in 4-digit format), where the payload is a 1 bit in the transmission:
0011 Position 3
0111 Position 7
1001 Position 9
XOR 1011 Position 11
--------------------
0110 = parity bit values
Step 3: Insert parity bit values into the transmission:
Position: 1 2 3 4 5 6 7 8 9 10 11 12
Data to be transmitted: 0 1 1 1 0 0 1 0 1 0 1 0
Is is quite simple to verify, if a received message was transmitted correctly and single-bit errors can be corrected. Here is an example. The receiver calulates XOR of the calculated and received payload bits where the value is a 1 bit. Is the result is 0, there the transmission is error-free. Otherwise the result contains the position of the bit with the wrong value.
Received message: 0001101100101101
Position: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Received data: 0 0 0 1 1 0 1 1 0 0 1 0 1 1 0 1
Parity bits: X X X X X
00101 Position 5
00111 Position 7
01011 Position 11
01101 Position 13
XOR 01110 Position 14
--------------------
01010 Parity bits calculated
XOR 00111 Parity bits received
--------------------
01101 => Bit 13 ist defective!
I hope, anybody here knows the correct name of the method.
Thanks for any help.
This looks like a complicated implementation of the Hamming(15,11) encoding & decoding algorithm.
Interleaving the parity bits with the information bits does not change the behaviour (or performance) of the code. Your description only uses 8 information bits, where the Hamming(15,11) corrects all single bit errors even with 11 information bits being transmitted.
Your description does not explain how the transmitted 12-bit message gets extended to a 16-bit message on the receive side.
I have been searching for long but unable to find a solution for this.
My question is "Suppose you have n street lights(cannot be moved) and if you get any m from them then it should have atleast k working.Now in how many ways can this be done"
This seems to be a combination problem, but the problem here is "m" must be sequential.
Eg:
1 2 3 4 5 6 7 (Street lamps)
Let m=3
Then the valid sets are,
1 2 32 3 43 4 54 5 65 6 7Whereas,1 2 4 and so are invalid selections.
So every set must have atleast 2 working lights. I have figured how to find the minimum lamps required to satisfy the condition but how can I find the number of ways in it can be done ?
There should certainly some formula to do this but I am unable to find it.. :(
Should always be (n-m)+1.
E.g., 10 lights (n = 10), 5 in set (m = 5):
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
6 7 8 9 10
Gives (10-5)+1 = 6 sets.
The answer should always be m choose k for all values of n where n > m > k. I'll try to explain why;
Given, for example, the values m = 10, n = 4, k = 2, you can start by generating all possible permutations of 1s and 0s for sets of 4 lights, with exactly 2 lights on;
1100
0110
0011
1001
0101
1010
As you can see, there are 6 permutations, because 4 choose 2 = 6. You can choose any of these 6 permutations to be the first 4 lights. You then continue the sequence until you get n (in this case 10) lights, ensuring that you only ever add a zero if you must in order to keep the condition true of having 2 lights on for every 4. What you will find is that the sequence simply repeats; for example:
1100 -> next can be 1, so 11001
Next can still be 1 and meet the condition, so 110011.
The next must now be a zero, giving 1100110, and then again -> 11001100. This simply continues until the length is n : 1100110011. Given that the starting four can only be one of the above set, you will only get 6 different permutations.
Now, since the sequence will repeat exactly the same for any value of n, it means that the answer will always be m choose k.
For your example in your comment of 6,3,2, I can only find the following permutations:
011011
110110
101101
Which works, because 3 choose 2 = 3. If you can find more, then I guess I'm wrong and I've probably misunderstood again :D but from my understanding of this problem, I'm certain that the answer will always be m choose k.
So I was practicing my binary subtraction. It's been a long while since my first exam and I decided to create my own tricky binary subtraction and I came up with this one:
1100
-1101
Of course the "borrowing trick" does not work for this problem at least I could not get it to work. Is my only choice to flip the bits of the second binary number(the bottom one) and then add a one basically doing 2's complement so 1101 becomes 0011. Then add the primary binary number(1100) with the 2's complement representation(0011) which means it would look like this:
1100 (-4) assume 2's complement
+ 0011 (3) assume 2's complement
sum:1111 (-1) assume 2's complement
I just need confirmation on this problem since its been a long time since I did binary subtraction.
1100
-1101
0 - 1 = 1 (borrow 1)
1100
-1101
1
=====
1
0 - 0 - 1 = 1 (borrow 1)
1100
-1101
11
=====
11
1 - 1 - 1 = 1 (borrow 1)
1100
-1101
111
=====
111
1 - 1 - 1 = 1 (borrow 1)
1100
-1101
1111
=====
1111
The result is 1111 with 1 borrowed. In terms of unsigned arithmetic, this means that either the result underflowed or you need to borrow from the next significant digit. (In terms of signed arithmetic there is no overflow as you have also borrowed the second bit and the calculation corresponds to -4 - -3 = -1.)