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ncsupanda

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Quote from: Josepht on January 29, 2015, 05:32:37 PM

Hey,

I am not much of a mathematician, but I'll give it a try.
I have two questions:
1. The a in the formula. Does it mean it can be any number?
2. You said the z value is always 1. What does that mean? I can't find a value z.

With z I believe he is referring to the standard deviation.

http://en.wikipedia.org/wiki/Standard_score

Josepht

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Hey,

I am not much of a mathematician, but I'll give it a try.
I have two questions:
1. The a in the formula. Does it mean it can be any number?
2. You said the z value is always 1. What does that mean? I can't find a value z.

Evil-Knievel

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ncsupanda

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I've done more calculations, and replaced the previous calculations with them.
http://www4.ncsu.edu/~ltwalz/bounty.txt

And the graphic visualization:
http://www4.ncsu.edu/~ltwalz/Algorithm.png

Still haven't been able to get the next x value to match the current (x=9). I could let it run for a lot longer, but the number drops off exponentially after the first few iterations.

Maybe I'm approaching this incorrectly, but just throwing it out there.

pythonpro1337

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THE FORMULA YOU REQUIRE IS CALLED pemdas

izanagi narukami

legendary

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Quote from: Watoshi-Dimobuto on January 29, 2015, 11:33:26 AM

What is a? What is z? Is both user defined?

a and z are constant of the Montgomery curve algorithm (http://en.wikipedia.org/wiki/Montgomery_curve)

Quote from: Evil-Knievel on January 28, 2015, 02:00:23 PM

the z value is always 1.

Watoshi-Dimobuto

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Yes!

Can someone simplify the question?

Quote

Problem Description:

Being a some constant, further assume that we
are in a factor ring (basically all operations modulo some sumber p). Note, that the division below is a multiplication by the modular inverse.
You always have to start with x=9.
Consider the following recursive formula:

Code:
new_x = (x²-1)² / (4*x*(x²+a*x+1))

How often do you have to perform this operation to get a specific x (basically getting the new_x and feeding it back into the formula to get another new_x, and so on)?
Note: You can start multiple such chains beginning at x=9, and add the resulting x values
using the addition algorithm from http://en.wikipedia.org/wiki/Montgomery_curve (Montgomery arithmetic section).
Note, that the x value, is the value you get at the end of such calculation-chain, and the z value is always 1.

I have to start with x = 9 and get new x(s) using the formula? and I have to find how often to do this to get a specific x?
Is this right?

What is a? What is z? Is both user defined?

gbianchi

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Quote from: Evil-Knievel on January 28, 2015, 02:00:23 PM

The winner:
The winner is the first person to post such formula in private. The formula must work in all cases, and be comutationally feasible (let us say calculatable in less than 24 hours).
If there are any wrong descriptions in this post at this time, they may be corrected or adjusted later on. Bounty ends on 02/15/2015.

calculate in less than 24 hours ... what is the magnitude of a, given x and p ?

izanagi narukami

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Quote from: david123 on January 29, 2015, 03:31:49 AM

Don't expect a reasonable answer in the question is still unclear.

My friend said that it still connect on ECDSA scheme ?
Is it true ?

Quote from: Evil-Knievel on January 29, 2015, 02:53:24 AM

Also, additionally to that, the division must me a multiplication with the modular inverse modulo p (just as a hint for future tries).

Thanks hope this help

david123

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Don't expect a reasonable answer in the question is still unclear.

Evil-Knievel

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redsn0w

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#Free market

Quote from: pythonpro1337 on January 29, 2015, 01:42:20 AM

""The winner is the first person to post such formula in private. The formula must work in all cases, and be computationally feasible (let us say calculatable in less than 24 hours)""

Well simply put, i did this, but with php and html lol the formula is written in a script in a private message in his inbox. i hope he gets on son cause the anticipation is killing me

Keep calm and wait him , if you have sent the solution (and it is valid) for first to him, you will be the winner.

pythonpro1337

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Activity: 99

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""The winner is the first person to post such formula in private. The formula must work in all cases, and be computationally feasible (let us say calculatable in less than 24 hours)""

Well simply put, i did this, but with php and html lol the formula is written in a script in a private message in his inbox. i hope he gets on son cause the anticipation is killing me

Vortex20000

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sucker got hacked and screwed --Toad

Silence in this thread...

ncsupanda

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Quote from: pythonpro1337 on January 29, 2015, 01:28:27 AM

its funny cause the script ive written and sent him like 5 hours ago wil display the correct answer using the algorithm that i used i like how everyone is trying to step on my fucking toes cause i KNOW the one i sent him is correct. OP please make the right decision here as to who gets the bounty cause ive sent multiple pm's thanks again

The guy is likely sleeping. Nobody is trying to "step on your toes" and you ought to relax. In fact, if you looked at what I provided, it was not an answer as much as it was a simulation. He'll choose based on the effort placed on the solution and how correct/helpful that solution is. You could send first and be completely correct, but it is his discrepancy and that's just how it is. Thanks.

pythonpro1337

member

Activity: 99

Merit: 10

its funny cause the script ive written and sent him like 5 hours ago wil display the correct answer using the algorithm that i used i like how everyone is trying to step on my fucking toes cause i KNOW the one i sent him is correct. OP please make the right decision here as to who gets the bounty cause ive sent multiple pm's thanks again

ncsupanda

legendary

Activity: 1628

Merit: 1012

I've written a script to solve the iterations and print outputs for them.

You can view the current values, including the previous x value and the new_x value here:
http://www4.ncsu.edu/~ltwalz/bounty.txt

And a graph of the comparison of the x value calculated compared to the previous value:
http://www4.ncsu.edu/~ltwalz/Algorithm.png

I'm happy to do more on request - usually seeing the numbers help people to visualize the problem they are approaching. Let me know if the links cause any problems (I'm hosting them through my university). Also, I did my modeling and calculations in MATLAB, so if anyone would like that as well, feel free to PM me.

ticoti

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Quote from: pythonpro1337 on January 28, 2015, 11:15:47 PM

anyone know where the original poster is at?!

relax sir, he must be sleeping he won't answer until tomorrow

pythonpro1337

member

Activity: 99

Merit: 10

anyone know where the original poster is at?!

pythonpro1337

member

Activity: 99

Merit: 10

Quote from: GLBrim on January 28, 2015, 10:43:54 PM

i tried and came up with this.. hope i can get some xD

=((x-1)^2 (x+1)^2)\/(4 x (a x+x^2+1))

=(x^4-2 x^2+1)\/(4 a x^2+4 x^3+4 x)

x = 1, a - 2=/0
x = 1, a + 2=/0

={x element R : (a<=-2 and x<0) or (a<=-2 and x>0 and sqrt(a^2-4)+a+2 x<0)
or (a<=-2 and sqrt(a^2-4)+a+2 x>0 and sqrt(a^2-4)>a+2 x)
or (a<=-2 and a+2 x>sqrt(a^2-4))
or (-2or (a>=2 and sqrt(a^2-4)>a+2 x and sqrt(a^2-4)+a+2 x>0)
or (a>=2 and sqrt(a^2-4)or (a>=2 and x>0)
or (a>=2 and sqrt(a^2-4)+a+2 x<0)}

1\/(4 x)-a\/4+1\/4 (a^2-3) x+(a-a^3\/4) x^2+1\/4 (a^4-5 a^2+4) x^3+O(x^4)\n(Laurent series)

x\/4-a\/4+(a^2-3)\/(4 x)+(a-a^3\/4)\/x^2+O((1\/x)^3)\n

(d)\/(dx)((x^2-1)^2\/(4 x (x^2+a x+1))) = ((x^2-1) (2 a (x^3+x)+x^4+6 x^2+1))\/(4 x^2 (a x+x^2+1)^2)

integral (-1+x^2)^2\/(4 x (1+a x+x^2)) dx = 1\/8 ((a^2-4) log(a x+x^2+1)+2 a sqrt(4-a^2) tan^(-1)((a+2 x)\/sqrt(4-a^2))-2 a x+x^2+2 log(x))+constant

BTC: 199hWWjMZdZ59dfyKpi7AG7wKv9LnqJSij

Quote from: GLBrim on January 28, 2015, 10:43:54 PM

i tried and came up with this.. hope i can get some xD

=((x-1)^2 (x+1)^2)\/(4 x (a x+x^2+1))

=(x^4-2 x^2+1)\/(4 a x^2+4 x^3+4 x)

x = 1, a - 2=/0
x = 1, a + 2=/0

={x element R : (a<=-2 and x<0) or (a<=-2 and x>0 and sqrt(a^2-4)+a+2 x<0)
or (a<=-2 and sqrt(a^2-4)+a+2 x>0 and sqrt(a^2-4)>a+2 x)
or (a<=-2 and a+2 x>sqrt(a^2-4))
or (-2or (a>=2 and sqrt(a^2-4)>a+2 x and sqrt(a^2-4)+a+2 x>0)
or (a>=2 and sqrt(a^2-4)or (a>=2 and x>0)
or (a>=2 and sqrt(a^2-4)+a+2 x<0)}

1\/(4 x)-a\/4+1\/4 (a^2-3) x+(a-a^3\/4) x^2+1\/4 (a^4-5 a^2+4) x^3+O(x^4)\n(Laurent series)

x\/4-a\/4+(a^2-3)\/(4 x)+(a-a^3\/4)\/x^2+O((1\/x)^3)\n

(d)\/(dx)((x^2-1)^2\/(4 x (x^2+a x+1))) = ((x^2-1) (2 a (x^3+x)+x^4+6 x^2+1))\/(4 x^2 (a x+x^2+1)^2)

integral (-1+x^2)^2\/(4 x (1+a x+x^2)) dx = 1\/8 ((a^2-4) log(a x+x^2+1)+2 a sqrt(4-a^2) tan^(-1)((a+2 x)\/sqrt(4-a^2))-2 a x+x^2+2 log(x))+constant

BTC: 199hWWjMZdZ59dfyKpi7AG7wKv9LnqJSij

can check the times of his post my posts on the forums and messages in your inbox i messaged you first money please and thank you

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