The cost of breaking 135 is closer to 1 million $, for a 99.999% chance of success. That is only the price for paying the electricity bill. You also need a bullet-proof distributed management system to sync data.
Relying on luck without a risk strategy will just end up as losing time and money.
Tens of thousands will get you nowhere, unless your risk ratio is around 1 to 1000. Very low chances you'd get to the solution with that amount.
[/quote]

I have never used GPU for puzzle searching.
Only keyhunt, I like it and I enjoy playing with the "spaces".

Question... with a GPU system is it possible to search for multiple public addresses together or just one?
do software that use GPU accept a .txt file like the one generated by keysubtracter?

no technique, no artifact, no math.
Only the IMPROBABLE luck will give the chance to discover the puzzles 135-140-145-150-155-160.

Unless you invest tens of thousands of $ in GPU and electricity + time.

"here" do you mean in this thread or is there a link?
thanks

How do bots anticipate the transfer transaction if a puzzle wallet is found ?

KEYHUNT in BSGS mode work well

for keyhunt
both, cpu and ram

Stop looking for logic, all the addresses of the puzzle are random.
And using python is 100% a waste of energy.
Don't dream idly.
Search in a small space (for 66 or 130) and hope for luck.

99.99999999999999999% JUST FOR FUN

I see that I'm not the only one who made this table :-)

best way to search with keyhunt is to go for sectors of 5% each randomly.
Maybe starting from the end between 100%-95% and 95%-90% of the range.....

All wallets in the puzzle were simply created randomly within their exponent.
There is no point in engaging in any type of hypothesis/study.

This is my opinion.

and 2^130 which requires a search between 2^129 and 2^130 equals 680,564,733,841,876,926,926,749,214,863,536 km.
therefore 71,986,683,526 light years (almost 72 billion).
I agree with the great luck with random and many subkeys....but many.
[/quote]

If you tried brute forcing 130, then yes, however, since we know the public key, you would use Kangaroo or BSGS. With Kangaroo, you’d be looking at roughly the same amount of ops as 66.
[/quote]

yeah, i mean with BSGS and a lot of subkeys

and 2^130 which requires a search between 2^129 and 2^130 equals 680,564,733,841,876,926,926,749,214,863,536 km.
therefore 71,986,683,526 light years (almost 72 billion).
I agree with the great luck with random and many subkeys....but many.

after 6 months on first space of puzzle 130
200000000000000000000000000000000:280000000000000000000000000000000

new space (from 25 to 75%)

./keyhunt -m bsgs -f 130.txt -r 280000000000000000000000000000000:380000000000000000000000000000000 -q -S -B random -k 256 -t 8

• Version 0.2.230519 Satoshi Quest, developed by AlbertoBSD
• Quiet thread output
• K factor 256
• Threads : 8
• Mode BSGS random
• Opening file 130.txt
• Added 20000001 points from file
• Range
• -- from : 0x280000000000000000000000000000000
• -- to     : 0x380000000000000000000000000000000
• N = 0x100000000000

130.txt 20M keys:
./keysubtracter -p 03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852 -n 20000000 -r 0:10000000000000000 >>130.txt

can open your mind to new ideas such as keymath and keydivision..

similitudes with prime numbers...

https://www.youtube.com/watch?v=pnYkK6PqUro&ab_channel=voltagedrop

EC is a spiral curve like spring

try this 3d graph generator:



import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np

# parameters
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
a = 0
b = 7
Gx = 55066263022277343669578718895168534326250603453777594175500187360389116729240
Gy = 32670510020758816978083085130507043184471273380659243275938904335757337482424

# start + end decimal key
start = int(input("input start key: "))
end = int(input("input end key: "))

# generation
x = []
y = []
z = []
for i in range(start, end + 1):
    # point calculation index i
    Px = i * Gx
    Py = i * Gy
    x.append(Px % p)
    y.append(Py % p)
    z.append(i)

# 3D chart
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(x, y, z, c='r', linewidth=2)

# add points on chart
for i, (xi, yi, zi) in enumerate(zip(x, y, z)):
    ax.text(xi, yi, zi, str(i), color='blue')

ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Index')
ax.set_title('point on elliptic curve secp256k1 (3D graph)')

plt.show()

see the Chart of BTC address where did the 27 BTC come from.
first transaction 13 december 2023.

https://www.blockchain.com/explorer/addresses/btc/bc1qlccksaaehjkdv4tgf032pvx8n76uhazqt4rgy70y4drmqwh5espqwx89f9