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[Evil-Knievel]

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[david123]

yeuh but what are you doing? are you performing calculations on an
elliptic curve?

From what I gather we are brute-forcing searching Curve25519 for a specific number not disclosed to us

So with searching you mean finding a logarithm? This is a well studied problem believed to be hard in general,
and there are software packages doing this as good as possible (which is not very good. this is the whole point
of using elliptic curves for cryptography^^)

I really don't get the point of this thread. 30 BTC bounty for cracking ECC?? I offer 100 BTC for cracking
SHA! (But maybe I have to obsure the problem and rephrase it into semi-mathematical terms for it to be
taken seriously..)

[bitspill]

Maybe each addition can be replaced by an arbitrary number of doubles? Not sure about that, but possible.

Example:
modulo 11
four doubles: 1*2*2*2*2 = 5
two doubled one add: 1*2*2 + 1 = 5

mod: 11
Goal: 6
9 doubles:

Code:

1 *2 *2 *2 *2 *2 *2 *2 *2 *2 = 6
   2  4  8 16
            5 10 20
                  9 18
                     7 14
                        3  6 = 6

4 doubles 1 add

Code:

1 *2 *2 *2 *2 +1 = 6
   2  4  8 16
            5  6 = 6

2 doubles 2 adds

Code:

1 *2 *2 +1 +1 = 6
   2  4  5  6 = 6

[Evil-Knievel]

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[bitspill]

@bitspill: This is the point ... maybe each combinations of adds and doubles can be represented by a (way higher) number of doubles only.

That does not appear to be the case from super basic testing

http://pastebin.com/raw.php?i=C9KhyFry
https://github.com/bitspill/Doubling/blob/master/main.cpp

A value of -999 means it attempted doubling 100 million times and did not find the goal, so I'm treating that as if it wont ever find it.

It's likely possible to add checks such that when all possibilities have been exhausted it quits rather than an arbitrary 100 million but I have not


So it would seem in some cases you can't simply double your way there, an add is required occasionally.


Edit: actually it shouldnt be checking for the exhaustion of possibilities but rather the entrance of a loop. Which is possible to implement on tgese small scale test but a loop would be nearly impossible to detect at the scale of 2^256

[swapcoiner]

After fixing my computer,my first job would be to sweep the fund here.

[Jesse James]

What Evil hints at is correct.  

If you have a generator element b of an additive group of order N, and you know it takes x repeated doubling operations (squaring operations in the context of a multiplicative group) on this element in order to reach q then you have effectively solved [multiplicative group notation]:

   q = b^2x

for x.

What we really want to do is crack the discrete logarithm, which means finding x in the context of:

   q = b^x

So the question becomes: if we can solve the first equation can we solve the second?  The answer is yes.  The reason is that the exponent of b is itself an element of it's own multiplicative group ℤ_N^× of which 2 is a generator since N is prime (at least in the context of Curve25519 or secp256k1).

So if you find x that satisfies:

   q = b^2x

then you can use your solution to solve the discrete log easy peasy lemon squeezy.:

   log_b(q) = 2^x mod N

Having at this point successfully solved the discrete log, you can happily collect your 30 BTC which will shortly be worthless since the security of bitcoin relies on the discrete logarithm assumption.

[fox19891989]

What a huge bounty, does anyone solve it? I just check your btc address, the balance is 0. I guess it must be solved. What a pity, I am good at math very much.

[bitspill]

What a huge bounty, does anyone solve it? I just check your btc address, the balance is 0. I guess it must be solved. What a pity, I am good at math very much.

It is not solved. Funds were used by the wallet for another bounty but he does have enough to cover it, and actually raised it to 40 BTC

[LiQio]

...

Having at this point successfully solved the discrete log, you can happily collect your 30 BTC which will shortly be worthless since the security of bitcoin relies on the discrete logarithm assumption.

OP should switch the bounty to fiat (or crypto money not based on discrete logarithm assumption)

[michietn94]

What a huge bounty, does anyone solve it? I just check your btc address, the balance is 0. I guess it must be solved. What a pity, I am good at math very much.

It is not solved. Funds were used by the wallet for another bounty but he does have enough to cover it, and actually raised it to 40 BTC

I just look OP's receive red feedback because offering this bounty for his research's good.

Hope he will paid what he promise or OP can use escrow to hold the fund.

[Evil-Knievel]

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[mriulian]

hmm.at school i didn't studied anything like that..i will give my best

[pythonpro1337]

I HAVE SOLVED THE ISSUE WITH THE MATHEMATICAL EQUATION! ITS FINALLY CLEAR NOW!!!!
here
var equation = equation.replace("x", "(x²-1)² / (4*x*(x²+a*x+1))");
should read
var equation = equation.replace(/x/gi, "(x²-1)² / (4*x*(x²+a*x+1))");
which I already tested for you
<script>
var equation= "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))";
var a=0;var b=1;var x=9;

document.write(equation+"</br>");
while(a<b)
{
var equation = equation.replace(/x/gi, "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))");

document.write(equation+"</br>");
a=a+1;
}
var x2=x*x;


var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));

document.write(answer+"</br>");
var answer=((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");

var x=9;

var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");
var x=answer;
var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");



var x=9;
var answer=((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))-1)*((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))-1) / (4*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))+a*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))+1))
document.write(answer+"</br>");
</script>
I also changed (x²-1)² / (4*x*(x²+a*x+1)) and used (x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1)) instead because javascript can directly evaluate it. Just for demonstation purposes I left
var x=9;

var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");
var x=answer;
var answer=(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1));
document.write(answer+"</br>");
which is evaluation by recursion. Find the answer plug it back into the equation etc. Compared with the created equation
var x=9;
var answer=((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))-1)*((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))-1) / (4*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*((x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))+a*(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))+1))
document.write(answer+"</br>");
both of which return the same answer so the code in the end is
<script>
var equation= "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))";var a=0;var b=6;var x=9;var c=0;document.write(eval(equation)+"</br>");
while(c<b){var equation = equation.replace(/x/gi, "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))");
document.write(equation+"</br></br>");document.write(eval(equation)+"</br>");c=c+1;}
</script>
This will give you just the answers minus the equations
<script>
var equation= "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))";var a=0;var b=6;var x=9;var c=0;document.write(eval(equation)+"</br>");
while(c<b){var equation = equation.replace(/x/gi, "(x*x-1)*(x*x-1) / (4*x*(x*x+a*x+1))");
document.write(eval(equation)+"</br>");c=c+1;}
</script>




Please send bounty to THIS BTC adress!!!
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN
14DFu1bW6DSPPuJYnZLn9EfgKeFYcLaFAN

[Evil-Knievel]

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[pythonpro1337]

WHAT IS WRONG WITH WHAT I POSTED?! WHAT IS NOT CORRECT?! YOU POSTED SOMETHING YOU NEEDED AN ANSWER FOR, I DID THE MATH AND THE ALGORITHM, IT IS NOT INCORRECT!!!!

[Evil-Knievel]

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[pythonpro1337]

can you please give me exactly what you are looking for in a CLEAR AND PRECISE PRivate Message? cause i mean you were unclear to begin with and every post its all jumping around to different maths and shit. like seriously you want me to solve thois shit you gotta be clear what you are looking for what types of maths are involved and what type of algorithm or mathematical formula you are requesting. i only went off of what your OP stated (which, was still very unclear)

[ncsupanda]

can you please give me exactly what you are looking for in a CLEAR AND PRECISE PRivate Message? cause i mean you were unclear to begin with and every post its all jumping around to different maths and shit. like seriously you want me to solve thois shit you gotta be clear what you are looking for what types of maths are involved and what type of algorithm or mathematical formula you are requesting. i only went off of what your OP stated (which, was still very unclear)

Yes Evil, please give him the answer so that he can earn his 30 BTC by doing mindless calculations and no real math to prove anything.

If the question tells you exactly how to get the answer, then it is a tutorial.

The answer is to THINK, not just complain.

[pythonpro1337]

I'm not entirely sure what you're asking here OP, but a couple of things popped into my mind while RE-reading:

1. Using the formula for new_x i presented, if we have to start with x = 9, then we have (x2 - 1)2 = 6400, which means the formula will never reach 10 for any p where 6400 = 0 (mod p), or for which (64002 - 1)2 = 1677721518080001 = 0 (mod p), etc.
 
2. Though you possibly qualified it by saying "at least in the context of...", I just thought I'd note that 2 is not necessarily a generator of ℤp× where p is prime. Consider, for example, p = 7.

Of course, I may be misunderstanding entirely what you're wanting to do.