He/they solved puzzle #120 27.02.2023
Most likely after that they immediately moved on to solving puzzle #125 that was solved 09.7.2023
In total it took them about 130 days = 11232000s = 2^23.42s
Expected op for #125 2^63.4 so solving speed around 2^40 op/s
If for ex. GPU speed 2^32(~4Gkey/s) than need 256 GPUs to solve #125 puzzle in 130 days.
I don't see anything out of the ordinary here.
As far as I understand, these people have access to a large number of GPUs
In addition, they already have additional funds to rent additional GPUs to speed up the process of finding #130 puzzles.
In order to defeat this monopoly, it is necessary to unite.Creating a pool does not make sense for many reasons.
I propose an alternative solution that will also require the participation of each hunter.
We can use fraction-kangaroo or just a python script that will divide #130 puzzles into 2^20 pieces.
Each hunter will look at least for 1 key in the range of 109 bits.
Search position in the range from 0 to 2^20-1 everyone can determine himself randomly.
If you have multiple rigs you can use multiple random positions for each rig.
For example, a rig of 6 3070 will be able to find or say with a probability of 75% that there is no key in this range for 90 days
Thus, if we can cover 2^20 positions at the same time, then one of us will find the key in 90 days.
Naturally, the prize will remain with the winner. But each participant will at least have a chance and will not require too many resources from each.
3Emiwzxme7Mrj4d89uqohXNncnRM15YESs will need at least 1500GPUs to solve #130 puzzle in this time.
Otherwise, if everyone tries to look for a puzzle #130 with their few rigs in the entire range 2^129, then I can definitely say that these guys 3Emiwzxme7Mrj4d89uqohXNncnRM15YESs will find it earlier.Here is python scrypt that devide 2^129 bit range by 2^20 with random position:
import random
import math
def inverse(x, p):
"""
Calculate the modular inverse of x ( mod p )
"""
inv1 = 1
inv2 = 0
n=1
while p != 1 and p!=0:
quotient = x // p
inv1, inv2 = inv2, inv1 - inv2 * quotient
x, p = p, x % p
n = n+1
return inv2
def dblpt(pt, p):
"""
Calculate pt+pt = 2*pt
"""
if pt is None:
return None
(x,y)= pt
if y==0:
return None
slope= 3*pow(x,2,p)*pow(2*y,p-2,p)
xsum= pow(slope,2,p)-2*x
ysum= slope*(x-xsum)-y
return (xsum%p, ysum%p)
def addpt(p1,p2, p):
"""
Calculate p1+p2
"""
if p1 is None or p2 is None:
return None
(x1,y1)= p1
(x2,y2)= p2
if x1==x2:
return dblpt(p1, p)
# calculate (y1-y2)/(x1-x2) modulus p
slope=(y1-y2)*pow(x1-x2,p-2,p)
xsum= pow(slope,2,p)-(x1+x2)
ysum= slope*(x1-xsum)-y1
return (xsum%p, ysum%p)
def ptmul(pt,a, p):
"""
Calculate pt*a
"""
scale= pt
acc=None
while a:
if a&1:
if acc is None:
acc= scale
else:
acc= addpt(acc,scale, p)
scale= dblpt(scale, p)
a >>= 1
return acc
def ptdiv(pt,a,p,n):
"""
Calculate pt/a
"""
divpt=inverse(a, n)%n
return ptmul(pt, divpt, p)
def getuncompressedpub(compressed_key):
"""
returns uncompressed public key
"""
y_parity = int(compressed_key[:2]) - 2
x = int(compressed_key[2:], 16)
a = (pow(x, 3, p) + 7) % p
y = pow(a, (p+1)//4, p)
if y % 2 != y_parity:
y = -y % p
return (x,y)
def compresspub(uncompressed_key):
"""
returns compressed public key
"""
(x,y)=uncompressed_key
y_parity = y&1
head='02'
if y_parity ==1:
head='03'
compressed_key = head+'{:064x}'.format(x)
return compressed_key
#secp256k1 constants
Gx=0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798
Gy=0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8
n=0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
p = 2**256 - 2**32 - 977
g= (Gx,Gy)
rangePower=129
rb=2**rangePower
re=2**(rangePower+1)-1
print ("Bit range 2^",math.log2(re-rb))
print ("Begin > 0x%x"%rb,">DEC ",rb)
print ("End > 0x%x"%re,">DEC ",re)
#MANUAL
compressed_key='03633CBE3EC02B9401C5EFFA144C5B4D22F87940259634858FC7E59B1C09937852'
point=getuncompressedpub(compressed_key)
divisor = pow(2,20)
print("")
rbdiv = rb//divisor
rediv = re//divisor
print ("Div Begin > 0x%x"%rbdiv,">DEC ",rbdiv)
print ("Div End > 0x%x"%rediv,">DEC ",rediv)
print ("Div Bit range 2^",math.log2(rediv-rbdiv))
newpub=ptdiv(point,divisor,p,n)
(partGx,partGy)=ptdiv(g,divisor,p,n)
idx=random.randrange(0,divisor-1)
#Randomly generate position in range 0..divisor-1
print("pos > 0x%x"%idx)
if idx if idx==0:
(searchpubx,searchpuby)=newpub
(fracX, fracY)=(partGx,partGy)
else:
(fracX, fracY)=ptmul((partGx,partGy),idx,p)
(searchpubx,searchpuby) = addpt(newpub,(fracX,p-fracY), p)
(fracX, fracY) = addpt((fracX, fracY),(partGx,partGy), p)
print("searchpub > ",compresspub((searchpubx,searchpuby)))
else:
print("idx>=divisor")
