Ok , what is the k value ?
I didn't actually guess it, I was just being sarcastic sorry
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I didn't actually guess it, I was just being sarcastic sorry
if you solve this and you know how you did solve it (not just a guess) then you don't need to get paid here because you could have all Bitcoins you want 
Well What if I just guessed and solved it:P
Ok , what is the k value ?
if you solve this and you know how you did solve it (not just a guess) then you don't need to get paid here because you could have all Bitcoins you want
Well What if I just guessed and solved it:P
Question: If someone were to solve this, how would they know they would get paid?
if you solve this and you know how you did solve it (not just a guess) then you don't need to get paid here because you could have all Bitcoins you want
Question: If someone were to solve this, how would they know they would get paid?
Anyone form the user have found the solution 
?
?if anyone can find the solution then Bitcoin is simply doomed.
hehe indeed you are right.
not really
if you find the solution without a logical explanation then only that address is doomed!
for all BTC to be doomed you need to explain how you found out the value of k
if you simply replace k with a number and then you realize that you got the correct k you will only get half of the bounty!
I know, but that number (k) ranges from 1 to 2 ^ 256 xD
it's impossible with a bruteforce. Maybe with all the computers in the world ... it would take you 1000 years to find this number !
Anyone form the user have found the solution ?
?if anyone can find the solution then Bitcoin is simply doomed.
hehe indeed you are right.
not really
if you find the solution without a logical explanation then only that address is doomed!
for all BTC to be doomed you need to explain how you found out the value of k
if you simply replace k with a number and then you realize that you got the correct k you will only get half of the bounty!
Anyone form the user have found the solution ?
?if anyone can find the solution then Bitcoin is simply doomed.
hehe indeed you are right.
Anyone form the user have found the solution ?
?if anyone can find the solution then Bitcoin is simply doomed.
Anyone form the user have found the solution ?
? I will think about some ways and make them public here. I will probably have to drink some wine to be more creative.
The first idea would be:
If someone manages to get a mathematical closed formula for the x-value, he could just paste that thing into wolfram alpha and get k out of it right away.
Meaning:
You have two points (x,y) - one is called "other" and one is called "self"
The point addition of those points (x3,y3) can be calculated as follows.
Code:
l = ( ( other.__y - self.__y ) * inverse_mod( other.__x - self.__x, p ) ) % p
x3 = ( l * l - self.__x - other.__x ) % p
y3 = ( l * ( self.__x - x3 ) - self.__y ) % p
After this step, other.x is set to x3 and other.y is set to y3.
We have all we need (initial state):
Code:
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2FL (i guess)
other.x = 0x4641b45737ee8e11ae39899060160507d61a30928b0d3e37b6aede29b4ed807bL
other.y = 0xb61b706b81dbb5512c556dfd16815cced84e2fa12b5c8b6440057355f0df2a12L
self.x = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798L
self.y = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8L
All you gotta know is: How often do you have to perform the above "point addition" so that x3=self.x
The inversemod of this value (to the basis of p) is the k.
So does that mean how many times the point addition is run (lets say its count) so that x3 = self.x and do the inverse_mod(count, p) = k?
2 billion that is.
Of course it is arguable on how many computers to add each year, and about how fast they are going to get.
But even if you let is grow in a few orders of size, the point of my explanation stays the same. It is just too hard to do.
Of course it is arguable on how many computers to add each year, and about how fast they are going to get.
But even if you let is grow in a few orders of size, the point of my explanation stays the same. It is just too hard to do.
Yeah, the sheer magnitude of the number of values that need to be checked means that a few orders of magnitude isn't going to change much/anything really in the grand scheme of things. I suppose the people who are bothering with this are hoping to get lucky, but I'm pretty sure their chances are lower than winning the lottery 10 times in a row just to give a little perspective.
Ah, but one crucial fact you neg nellies are missing is that it depends how many computers you throw at it. If it would take a 1000 years on one PC, then it would take 1 year on a 1000. Especailly if using the earlier distributed idea where ranges are assigned. Yes, it may not be economically feasible to pay for that electricity versus just buying bitcoins, but it would make solving the problem quicker.
I already included all the computers existing today, see:
...
Simple crude calculation:
2160 bitcoin addresses
2 billion computers worldwide1
3 million addresses checked per day on a computer2
2160 / ( 2e9 * 3e6 * 365 ) = 667e27 years
...
Simple crude calculation:
2160 bitcoin addresses
2 billion computers worldwide1
3 million addresses checked per day on a computer2
2160 / ( 2e9 * 3e6 * 365 ) = 667e27 years
...
2 billion that is.
Of course it is arguable on how many computers to add each year, and about how fast they are going to get.
But even if you let is grow in a few orders of size, the point of my explanation stays the same. It is just too hard to do.
689,434,752,464?
That is very nice and perhaps this scale is not bad at all. I would just change 1 thing
You just changed it back to where it was before, which was the whole point of the edit and perhaps this scale is not bad at all. I would just change 1 thing
(scroll up)
Ah, but one crucial fact you neg nellies are missing is that it depends how many computers you throw at it. If it would take a 1000 years on one PC, then it would take 1 year on a 1000. Especailly if using the earlier distributed idea where ranges are assigned. Yes, it may not be economically feasible to pay for that electricity versus just buying bitcoins, but it would make solving the problem quicker.
Simple crude calculation:
Timeline:
(timeline destroyed by quote font)
Entropy is a bitch.
The crudeness in this is really the timeline. While we have a pretty good idea of roughly (in billions of years or so) when the big bang was, the sun accumulated enough mass, when it'll burn out and eventually the heat death of the universe, these are relatively fixed points in time and they aren't going to move around on that timeline very much.Timeline:
(timeline destroyed by quote font)
Entropy is a bitch.
For finding a key collision,
it could just as easily be: , you finding k
big bang sun was born now. / sun dies existing stars burn out all matter evaporated
|-------------|---------------||-------------|---------------|-----------------------------------------|
0 9.2e9 13.8e9 19e9 1e15 1e34
( not to scale
)It's just that the odds of that being the case are very, very, very (rinse and repeat a lot) low. Of course the odds are exactly the same for your point along the timeline, or for any other point along the timeline within the bounds as proposed, to the point where statistically speaking it doesn't make sense to even try.
That is very nice
and perhaps this scale is not bad at all. I would just change 1 thing,you finding k
big bang sun was born now. sun dies existing stars burn out / all matter evaporated
|-------------|---------------|-------------|---------------|----------------------------|-------------|
0 9.2e9 13.8e9 19e9 1e15 1e34
Off topic (because the topic is about very big and very small numbers, and I think it's mind boggling):
1 + 2 + 3 + 4 + 5 + ... = -1/121
Only if you use a very specific treatment of infinity that also happens to be useful in physics (edit: or was it quantum mechanics.. hmmm, already forgot) because it actually ends up matching what they're seeing 1 + 2 + 3 + 4 + 5 + ... = -1/121
( saw that vid a while back - there's all sorts of mathematical oddities in those channels, well worth subscribing. )
In one of the videos it was mentioned that it appears in the beginning of quantum mechanics books. It is one of the reasons why the number of dimensions is what it is in string theory, because of that outcome.
I want to make a remark that pure mathematically it can be deduced from the Riemann zeta function. So maybe they found it because of physics, but it also exists in the abstract world of mathematics.
Off topic (because the topic is about very big and very small numbers, and I think it's mind boggling):
1 + 2 + 3 + 4 + 5 + ... = -1/121
Only if you use a very specific treatment of infinity that also happens to be useful in physics (edit: or was it quantum mechanics.. hmmm, already forgot) because it actually ends up matching what they're seeing 1 + 2 + 3 + 4 + 5 + ... = -1/121
( saw that vid a while back - there's all sorts of mathematical oddities in those channels, well worth subscribing. )
...
It's just that the odds of that being the case are very, very, very (rinse and repeat a lot) low. Of course the odds are exactly the same for your point along the timeline, or for any other point along the timeline within the bounds as proposed, to the point where statistically speaking it doesn't make sense to even try.
It's just that the odds of that being the case are very, very, very (rinse and repeat a lot) low. Of course the odds are exactly the same for your point along the timeline, or for any other point along the timeline within the bounds as proposed, to the point where statistically speaking it doesn't make sense to even try.
Yep, you're right of course. After 667e27 years the chance of finding the right k is 100%. In statistics the expected value or the mean would be halve of that, 333.5e27 years. But the odds or so low, it could be zero in a human life time for that matter.
Off topic (because the topic is about very big and very small numbers, and I think it's mind boggling):
1 + 2 + 3 + 4 + 5 + ... = -1/121
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