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[Baitty]

0.9231 is the number 

[1714010622]

1714010622

[1714010622]

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[1714010622]

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[1714010622]

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[1714010622]

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[BawsyBoss]

Interesting... I will be following this to see how everything goes and how it affects BTC.

[roslinpl]

0.9231 is the number 

I am sure it is not

Funny thread, it will never be [closed] I am telling ya.


regards!

[mmitech]

65536

18kCuQtWtsYCJFkB6YEAMP69hAxDhwx1ae     

[Evil-Knievel]

This message was too old and has been purged

[mmitech]

Guys, please try yourself before just wildly posting your "guess" 

opps I though it was a lotto; sorry for spamming   

[TheRealSteve]

If you really did want to brute force it, instead of finding a mathematical/algorithmic solution, you should just take the corresponding address for the given public key and feed its first set of characters into vanitygen (first set of characters because vanitygen doesn't accept full length).  That will do two things: 1. Narrow down the search field considerably.  2. Tell you just how unbelievably long this would take.. on average.  Multiply by several factors more to figure out how long it'd take for the full length

You could even set it up as a partial bounty over at https://vanitypool.appspot.com/ - though the minimum fee for any appreciably useful length (for this) exceeds the bounty.
That vanity pool did spawn oclvanityminer, at least, which is very fast indeed and can automatically receive and send work - so if anybody seriously wanted to attack in a distributed manner, they could set up an alternative vanityminer pool server just for this project (might have to sniff the network traffic first, as I never did see any documentation on vanitypool/oclvanityminer's network communication).

[Raya]

This is impossible without having a pool which stores already guessed things in a database so that each person doesn't reguess and waste a lot of time.

[TheRealSteve]

This is impossible without having a pool which stores already guessed things in a database so that each person doesn't reguess and waste a lot of time.

Well, the above (a vanity miner pool) already reduces the number of wasted guesses if you consider that everybody starts with a random guess in the first place (at least, if I recall correctly that vanitygen doesn't use a fixed seed).

But yes, more ideal still would be to split up the entire work space based on the number of participants, splitting and merging as participants join/leave.  But that would definitely require more work for something that's ultimately pretty futile in the first place

[Raya]

That script will only over heat your computer and potentially fry it instead of finding the value "K" multiple scientific organsations have been trying to solve this for many years without succeeding.

[vnvizow]

Might sound crazy but lets use the latest supercomputer in China(http://en.wikipedia.org/wiki/Tianhe-2)...but wait, what if the Chinese already have the answer and redeem the bounty then what will happen? hmmm........

[FiatKiller]

Darn, ran that addy thru my 876 MB of pairs I have so far and nada...  :-(    

Edit: need to add a "mein laben!" just for EK   lol

[roslinpl]

Might sound crazy but lets use the latest supercomputer in China(http://en.wikipedia.org/wiki/Tianhe-2)...but wait, what if the Chinese already have the answer and redeem the bounty then what will happen? hmmm........

Nothing :-) just world will collapse, red lava will spread around, dinosaurs will reborn and Mark K. Will withdraw.
Nothing will happen :-)

[funbitcoins]

k = 1000 

13eShdtiqfJE6FqLkaMJ5K5jbY896MDmJY

[gogodr]

Might sound crazy but lets use the latest supercomputer in China(http://en.wikipedia.org/wiki/Tianhe-2)...but wait, what if the Chinese already have the answer and redeem the bounty then what will happen? hmmm........

just to make things more graspable

Tianhe-2 can perform 33860000000000000 floating point operations per second
There are 1157920892373161954235709850086900000000000000000000000000000000000000000000000 000000000 combinations for the SHA 256 output and each SHA256 has 64rounds which means that each hash has to be processed 64 times and each of that rounds have a complex set of operations.
So even with the fastest computer on earth, it would take more than just 1000 years (there are 31104000 seconds in a year)

[FiatKiller]

better way to maximize your "dumb luck" chances would be to use a different starting point
or a random starting point...

[DeathProxy]

You just waste your time if you try to bruteforce this.

[FiatKiller]

This is what I have for the end section so far(added a counter & writing the correct answer to a file):

challenge = Public_key(g, Point( sex, 0x4641b45737ee8e11ae39899060160507d61a30928b0d3e37b6aede29b4ed807bL,   0xb61b706b81dbb5512c556dfd16815cced84e2fa12b5c8b6440057355f0df2a12L))
 #testchallenge = Public_key(g, Point( sex, 0x2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01L, 0x5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904L))
  ppp=challenge.point

  # find the correct k
  k=random.randrange(1,2**255)
  count=1
  go=1

  # !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
  #testk=8
  ppp=ppp + k*g  <<<<<<< shouldn't this be removed for looping values of k??

  while go==1:
    <<<<<<<<< don't you need "ppp=challenge.point" here???
    ppp=ppp + g  <<<<<<<<<< shouldn't there be a "k*" in there or no?
    if ppp.x() == g.x():
      print "***1000 BTC are YOURS, Homey!!!***"
      print k,hex(k)
      with open("WINNER.txt", "a+") as myfile:
        myfile.write(k + hex(k) + "\n")
      time.sleep(10)
      go=0
    else:
      print count, k
      #time.sleep(1)
      k=random.randrange(1,2**255)
      count=count+1



Is there a SEED command or something to reduce the faults of the random generator?

[FiatKiller]

For testing purpose you can replace the challenge line in the python script with this:

Code:

challenge = Public_key(g, Point( sex, 0x2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01L, 0x5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904L))

In this case k=8

this test code does not work with the orig script. yes, I did set k to 8.

[pthnmj]

Sh&^! I ve got it

k  = 254634957345987934598347597261842936423946893247.

will I get the price?
please?

nope, because your solution is wrong:

Code:

=======================================================================
0x9772c99f2db406a441000d9288ba9cd7ed05e7a0ae02e09615b7cd39cb8b1cedL not matching 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798L

I got it!

55066263022277343669578718895168534326250603453777594175500187360389116729240