It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 11 years ago.
If someone can help me, can you help me identify this hash?
51a5a55a2a173a177a30a167a70a145a105a94a121a112a10a87
It looks like some data separated by "a".
The parts, after split by a:
51
5
55
2
173
177
30
167
70
145
105
94
121
112
10
87
It's not ASCII. Another operation must have been applied to it.
That thing is 52 hexadecimal digits, i.e. 26 bytes, i.e. 208 bits. I've never seen "standard" hashes of 208 bits, in general they are in powers of 2. It may as well be anything, probably it is just "normal" binary data.
Edit actually, as #Vamana noticed, it probably isn't hexadecimal as it may have looked at first glance, since the only non-numerical digit that appears is a, which instead is probably a separator of some kind.
Edit again are you sure that you didn't miss some leading zero? IMO those "A" started as 0A characters, which are newlines.
Perhaps this has no connection at all, but it is interesting that in this question somebody asked about a "hash" that is a base64 encoded version of a 26 byte hash. So perhaps there is some exotic 26 byte hash? Or perhaps some kind of 26 byte long data/ID?
it is
QUR1w0wEYA!!
hexxed, joined by a (in wierd positions)...
http://home2.paulschou.net/tools/xlate/ pwns
did i kill the fun ? :( sorry
Related
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Closed 10 years ago.
I have a given set of alphabets say 11 letters. I want to find all possible word combinations of specific length say 5. These are numbers are dynamic. It doesn't matter to me if the resulted words do have a meaning or not.
Can someone please provide me with this kind of code or algo to follow. Or guide me right direction.
Thanks
This algorithm is a permutation. A relatively easy way is to use recursion to solve it.
Read this SO Question:
Sequence combinations of varying length using two 'alphabets'
Read "UPDATE 2:" in this question there he post his working code
All the best
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Closed 10 years ago.
We're examining the RSA algorithm and would like to know how much time it would take an intel i-7 core (# 2.50 gHz) to factorise the RSA-public key.
we wrote a piece of java for this, and I don't know how effective it is
public static String factorise(long l)
{
double a = Math.floor(Math.sqrt(l));
while(l/a != Math.round(l/a))
{
a--;
}
return (long)a + ", " + (long)(l/a);
}
With a number around 2^45 it took the PC approximately 33 milliseconds. In theory, how long would it take to factorise a number around 2^1024?
Thanks in advance :)
Your algorithm is O(2^n), where n is the number of bits in the original number l. (that means that a single bit more will double the runtime, because twice as many numbers a must be checked - on average)
If 45 bits took 33 ms, then 1024 bits will take approx. 2^1024 / 2^45 * 33ms = 5.34654 * 10^285 years.
This of course assumes, that the 1024bit code is exactly as efficient as your code for long numbers (64bit?). Which is a bold statement, considering that 10^285 years is more than enough time to switch to the General number field sieve and scratch a few million years of that time...
In 2009 the 768 bit number rsa-768 was cracked using about 1000 cores and 2 years of calculations. Assuming they used the General number field sieve (a very fair assumption) it would take them 7481 years to crack a 1024 bit number using the same hardware.
Or using only your i7 with this algorithm: about 3 million years. Still a long time.... ;)
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Closed 11 years ago.
select round(product_qty * 100) - product_qty as test,
id, product_qty
from stock_move
where product_id=63
and state='done'
and id=45058;
test | id | product_qty
54.45 | 45058 | 0.55
(1 ligne)
select round(product_qty * 100) - (product_qty*100) as test,
id,
product_qty
from stock_move
where product_id=63
and state='done'
and id=45058;
test | id | product_qty
-7.105427357601e-15 | 45058 | 0.55
(1 ligne)
can somebody explain me these results?
Floating-point representations like double precision fields in pg are by their very nature subject to rounding errors. This paper gives a good mathematical introduction.
How you deal with these rounding errors depends on a bit on your application, and varies between:
just living with them eg by sort of ignoring them - format that result as %.2f and the error will become invisible. Just make sure you understand all the consequences of that choice!!
switching datatypes to something that by design is not subject to rounding issues - check out pg's money type and the exact numeric types.
The latter should be the preferred approach especially in contexts where your application handles money.
You're getting those results because product_qty*100 has different IEEE-754 binary representation than round(product_qty * 100):
without round : 100000001001011100000000000000000000000000000000000000000000001
with round : 100000001001011100000000000000000000000000000000000000000000000
It's easy to see that 55 should be represented as 1,71875 * 25 (as in with round), but due to floating point arithmetic product_qty*100 is slightly different.
If you want to still live with floating point number, then I recommend to read "What every computer scientist should know about floating-point arithmetic" paper, or just switch to numeric type with arbitrary precision.
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Closed 12 years ago.
I can calculate the expressions, and term but I don't understand what is tregonometric functions. Are they Expressions or Terms or Items? How to write the algorithm of calculation of Pi and tregonometric functions. Cos, sin....
Please if you know give me a hint.
Try a web search again, this time using the correct spelling "trigonometric functions". Algorithms for calculations like this date back at least to ancient Greece and you'll find them in a great many maths textbooks.
I suggest you try to do your own homework, then (and only then) come back to stack overflow if you have some particular issue with Scala syntax that prevents you from implementing your chosen algorithm. In the meantime, this isn't a proper or valid question.
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
matlab7.10.0(R2010a)
I=imread('input-small.jpg');
???Error using ==> jpeg_depth
JPEG parameter struct mismatch: library thinks size is 632, caller expects 616
Error in ==> readjpg at 13
[depth,msg]=jpeg_depth(filename);
Error in ==> imread at 441
[X,map]=feval(fmt_s.read,filename,extraArgs{:});
I just got this error with R2010b. There's no relation to the picture itself.
It was using the system's JRE. When I switched it back to the JRE shipped with Matlab it worked.