Yeah sell now when 99% of the population doesn't even own a satoshi. You will cry in 2-3 months when BTC is back to $900. I cried 1 month after December when I sodl and saw BTC back up in one month.
it those situations HODL is needed.
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seems like
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I would not recommend MtGox So maybe just stop using it ;P
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Today wall seems bad Its like go back in time do 13 dec 2013.
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Who cares about SilkRoad2.0... I just realy hope there would be bo 3.0
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The price is just a number, it is low? Then it is a great moment to buy.
+1 right but it seems to be a long long day .... and another day with panic and speculations according to price drops
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Now I'm a little worried, but fairly content. I hit my fear mark right at the bottom of the crash, when the bitcoin people responded to mt gox saying it was their problem not bitcoin. I knew I should buy, but was afraid to. So I bought as much as I could afford down at 600, and I'm holding that now.
hold it hold it. Tis is not a good time to sell.
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Very nice Credit card size would be cool! But plate like this seems cool as weel to put on a bar or something ;p Barmans could use that to collect tips
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mark. is that you?
who is mark? Karpeles - Of mtgox. Thanks well i think that mtgox is like queen said in his song "another one bites the dust" now we only have to wait and see what exchange all that 1 million people will go to and the price will be up again MtGox was slow anyway. We wont cry.
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Im going to return it now guys whats the typical fee that people put in whilst sending? Is there a minimum as such?
Thanks
0.0001 BTC https://en.bitcoin.it/wiki/Transaction_fees"mintxfee 0.0001 (BTC)" you can put no fee at all... that is why it was funny for me when I posted above "will you send it back"
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444 usd at gox!
602.00000000 @ Vircurex
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mark. is that you?
Lol HAhahahahahaha. Can't stop laughin'
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Trolls help us to support bitcoin stronger!
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Time will show.
Nothing is that simple as you think.
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He told that he will give it back Let see if he will resend it back xD OP: "(...) then i of course will return it back obviously. "
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Just as a hint: Python is realllllly slow, so if you really like to start bruteforcing I would suggest to use a C++ implementation. I could provide one tomorrow, but it is not very promising. Better would be some mathematical approach No we dont think we will bruteforce it. I think - that this is not possible at all. Maybe possible IF you show price so we can truely see what we are fighting for.
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Is their anyway to make the python script simply loop and keep adding + 1 to k? Or choose a random number for k?
Yes, this would be the following code. This will however (at least I think so) not be very promising: #! /usr/bin/env python
import random import array import cPickle import struct
class CurveFp( object ): def __init__( self, p, a, b ): self.__p = p self.__a = a self.__b = b
def p( self ): return self.__p
def a( self ): return self.__a
def b( self ): return self.__b
def contains_point( self, x, y ): return ( y * y - ( x * x * x + self.__a * x + self.__b ) ) % self.__p == 0
class Point( object ): def __init__( self, curve, x, y, order = None ): self.__curve = curve self.__x = x self.__y = y self.__order = order if self.__curve: assert self.__curve.contains_point( x, y ) if order: assert self * order == INFINITY def __add__( self, other ): if other == INFINITY: return self if self == INFINITY: return other assert self.__curve == other.__curve if self.__x == other.__x: if ( self.__y + other.__y ) % self.__curve.p() == 0: return INFINITY else: return self.double()
p = self.__curve.p() 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 return Point( self.__curve, x3, y3 )
def negative (self): negative_self = Point( self.__curve, self.__x, -self.__y, self.__order ) return negative_self
def __mul__( self, other ): def leftmost_bit( x ): assert x > 0 result = 1L while result <= x: result = 2 * result return result / 2
e = other if self.__order: e = e % self.__order if e == 0: return INFINITY if self == INFINITY: return INFINITY assert e > 0 e3 = 3 * e negative_self = Point( self.__curve, self.__x, -self.__y, self.__order ) i = leftmost_bit( e3 ) / 2 result = self while i > 1: result = result.double() if ( e3 & i ) != 0 and ( e & i ) == 0: result = result + self if ( e3 & i ) == 0 and ( e & i ) != 0: result = result + negative_self i = i / 2 return result
def __rmul__( self, other ): return self * other
def __str__( self ): if self == INFINITY: return "infinity" return "(%d,%d)" % ( self.__x, self.__y )
def double( self ): if self == INFINITY: return INFINITY
p = self.__curve.p() a = self.__curve.a() l = ( ( 3 * self.__x * self.__x + a ) * \ inverse_mod( 2 * self.__y, p ) ) % p x3 = ( l * l - 2 * self.__x ) % p y3 = ( l * ( self.__x - x3 ) - self.__y ) % p return Point( self.__curve, x3, y3 )
def halve( self ): if self == INFINITY: return INFINITY
p = self.__curve.p() a = self.__curve.a() # next three lines must be reverted somehow, then I am a multi millionaire :-) # as a=0 in this case, I have eliminated it! l = ( ( 3 * self.__x * self.__x ) * inverse_mod( 2 * self.__y, p ) ) % p x3 = ( l * l - 2 * self.__x ) % p y3 = ( l * ( self.__x - x3 ) - self.__y ) % p
return Point( self.__curve, x3, y3 )
def x( self ): return self.__x
def y( self ): return self.__y
def curve( self ): return self.__curve def order( self ): return self.__order INFINITY = Point( None, None, None )
def inverse_mod( a, m ): if a < 0 or m <= a: a = a % m c, d = a, m uc, vc, ud, vd = 1, 0, 0, 1 while c != 0: q, c, d = divmod( d, c ) + ( c, ) uc, vc, ud, vd = ud - q*uc, vd - q*vc, uc, vc assert d == 1 if ud > 0: return ud else: return ud + m
_p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2FL _r = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141L _b = 0x0000000000000000000000000000000000000000000000000000000000000007L _a = 0x0000000000000000000000000000000000000000000000000000000000000000L _Gx = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798L _Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8L
class Public_key( object ): def __init__( self, generator, point ): self.curve = generator.curve() self.generator = generator self.point = point n = generator.order() if not n: raise RuntimeError, "Generator point must have order." if not n * point == INFINITY: raise RuntimeError, "Generator point order is bad." if point.x() < 0 or n <= point.x() or point.y() < 0 or n <= point.y(): raise RuntimeError, "Generator point has x or y out of range."
sex = CurveFp( _p, _a, _b ) ass = Point( sex, _Gx, _Gy, _r ) g = ass
if __name__ == "__main__": print '=======================================================================' ### generate privkey challenge = Public_key(g, Point( sex, 0x4641b45737ee8e11ae39899060160507d61a30928b0d3e37b6aede29b4ed807bL, 0xb61b706b81dbb5512c556dfd16815cced84e2fa12b5c8b6440057355f0df2a12L)) ppp=challenge.point
# find the correct k k=random.randrange(1,2**255) # !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ppp=ppp + k*g
while True: ppp=ppp+g k=k+1 if ppp.x() == g.x(): print "found!!!!!!! k=" + hex(k) else: print hex(ppp.x()) + " not matching " + hex(g.x())
Are you sure that will actully stop when it finds something? Just because from looking quickly it seems like even if i find it it will keep going. Also i assume K= a real number meaning 1 or 5959493 and not 58484.4835 I post code for this a while ago And it will stop but anyway... (...)
# find the correct k k=1200000000000000000000000000000000000000000000000000000000000000000000000000000 aNumber=1
while aNumber==1:
ppp=ppp + k*g if ppp.x() == g.x():
print "HORRAY! You are the WINNER !! :) You just won 500BTC but you must wait till guys will buy them and transfer to your wallet, because they did not have them yet"
print "K=", k aNumber=2 else: k=k-1 print k
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Alright! Good job! I hope it works!
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My friend always say that you can do everything with a box of matches. Maybe you can give a try too (better just cut...) It will hold a rig anyway.
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