Quote from: DaCryptoRaccoon on October 29, 2024, 09:10:02 PM

Code:

def calculate_time_to_explore_keyspace(attempts_per_second):
total_keys = 2**256 # Approximately 1.15 x 10^77

seconds_per_year = 60 * 60 * 24 * 365
keys_per_year = attempts_per_second * seconds_per_year
total_years = total_keys / keys_per_year
total_seconds = total_years * seconds_per_year
total_minutes = total_seconds / 60
total_hours = total_minutes / 60
total_days = total_hours / 24

return total_years, total_seconds, total_minutes, total_hours, total_days

def print_detailed_time(total_years, total_seconds, total_minutes, total_hours, total_days):
print(f"Total time to explore the entire keyspace: {total_years:.2e} years")
print(f"Total time in seconds: {total_seconds:.2e} seconds")
print(f"Total time in minutes: {total_minutes:.2e} minutes")
print(f"Total time in hours: {total_hours:.2e} hours")
print(f"Total time in days: {total_days:.2e} days\n")

def present_keyspace_analysis():
print("Keyspace Analysis for Bitcoin Private Keys")
print("=" * 40)

high_attempt_rate = 10**9 # assuming 1 billion attempts per second (very optimistic)

total_years, total_seconds, total_minutes, total_hours, total_days = calculate_time_to_explore_keyspace(high_attempt_rate)

print(f"Assuming {high_attempt_rate:,} keys can be tested per second:")
print(f"Total keys in Bitcoin (2^256): {2**256:.2e} keys")
print_detailed_time(total_years, total_seconds, total_minutes, total_hours, total_days)

hex64_attempts = 10**18 # 1 quintillion HEX64 combinations (also very optimistic)
total_years_hex64, total_seconds_hex64, total_minutes_hex64, total_hours_hex64, total_days_hex64 = calculate_time_to_explore_keyspace(hex64_attempts)

print(f"Assuming {hex64_attempts:,} HEX64 combinations can be tested per second:")
print_detailed_time(total_years_hex64, total_seconds_hex64, total_minutes_hex64, total_hours_hex64, total_days_hex64)
print("Comparison of Bitcoin keyspace sizes:")
print(f"- 256-bit keyspace: 2^256 ≈ {2**256:.2e} keys")
print(f"- 135-bit public key space: 2^135 ≈ {2**135:.2e} keys")
print(f"- 67-bit address space: 2^67 ≈ {2**67:.2e} keys")
print("\nProbability of randomly guessing a Bitcoin private key:")
probability = 1 / (2**256)
print(f"The probability of successfully guessing a valid Bitcoin private key: {probability:.2e}")
print("\nConclusion:")
print("Even with highly optimistic assumptions about the number of keys tested per second,")
print("the time required to brute-force the entire Bitcoin private key space is")
print("astronomically high and completely impractical, making the claims about")
print("recovering lost wallets through this method unfounded.")

if __name__ == "__main__":
present_keyspace_analysis()

Your numbers do not add up for HEX64.

I apologize for writing so late. I will formulate the structure with a different argument without going into detail, because the theory you wrote is just linear search, the method I use is point-to-point, think of it like this, there are trillions of spheres in a circle, there is a dotted communication network between these spheres. I send a different hash generator to the circle furthest away from the collision, like searching at different points.
This search is not something I made up. Think of it as the math equivalent of a mode.
The number of wallet addresses that I think are completely missing from these is 743.
I realize it is incredibly difficult even with these factors, but ASICs are running at TH power (10,000,000,000,000,000) for this block generation. Secp256k1 can be variable. Cuda nvida allows 1/3 of that in Python. If 10,000 users are included in the average network, it still reaches 100 quadrillion throughput.

Even the model approximation can create another formula based on probability and 743 wallet hits given Python Functions. I will spend time to make a few formulas of this, there is an important technical fair. After a few days I will present you the approximate probability with the structure I have.
Also, if there is an ASIC pool, machines like Antminer S9 are running 13.5 TH.
This thread was only opened to investigate the technical possibility of this.
But the P2P model beta test is available.

btcdvl

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