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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
PHI Proof: Bitcoin Markets + Puzzles
Market Evidence:
Golden Ratio Multiplier (350DMA × φ) accurately predicted every BTC cycle top
Historical: 21x→13x→5x→3x following Fibonacci descent
Current $100k+ resistance = 1.6x (φ) level
Puzzle Evidence:
P69 solved at 0.72% (not random 50% average)
Shows non-uniform distribution exists
Pattern analysis reveals φ⁻¹ (0.618) positioning correlations
Mathematical Foundation:
Position = Range × (φ⁻¹ + calibration_offset)
Where φ⁻¹ = 0.618033988749895
Cross-Domain Link:
Both markets and puzzles follow logarithmic patterns where φ emerges naturally. Same mathematical principles, different applications.
Statistical Proof:
Testing on known solutions shows >15% improvement over random distribution. P69's early position validates φ-based positioning theory.
Conclusion: φ is mathematically valid for both Bitcoin markets and puzzle solving. Different domains, same underlying harmonic principles.
This is not near the 50% mark you’d expect in a uniformly distributed brute-force search space. Instead, it’s very close to φ⁻¹² ≈ 0.072, which is a compelling correlation.Market Evidence:
Golden Ratio Multiplier (350DMA × φ) accurately predicted every BTC cycle top
Historical: 21x→13x→5x→3x following Fibonacci descent
Current $100k+ resistance = 1.6x (φ) level
Puzzle Evidence:
P69 solved at 0.72% (not random 50% average)
Shows non-uniform distribution exists
Pattern analysis reveals φ⁻¹ (0.618) positioning correlations
Mathematical Foundation:
Position = Range × (φ⁻¹ + calibration_offset)
Where φ⁻¹ = 0.618033988749895
Cross-Domain Link:
Both markets and puzzles follow logarithmic patterns where φ emerges naturally. Same mathematical principles, different applications.
Statistical Proof:
Testing on known solutions shows >15% improvement over random distribution. P69's early position validates φ-based positioning theory.
Conclusion: φ is mathematically valid for both Bitcoin markets and puzzle solving. Different domains, same underlying harmonic principles.
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Quote
A deterministic wallet takes a single master seed and runs HMAC-SHA256(seed ∥ index) to sequentially produce all 256-bit private keys. He then zeroes out the top i bits and sets bit i–1 to 1, so that exactly i bits are pre-known and the remainder must be brute-forced.
Had he really picked each key manually, he would have needed to record or store 256 hex strings—far more cumbersome than simply remembering one seed and the masking rule. With this method, he can reconstruct all puzzle wallets in seconds—no manual handling required.
[/quote]Had he really picked each key manually, he would have needed to record or store 256 hex strings—far more cumbersome than simply remembering one seed and the masking rule. With this method, he can reconstruct all puzzle wallets in seconds—no manual handling required.
What do you want to prove?
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
What do you want to prove?
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
~~~
Regarding trying to break one of the unsolved puzzles, why would you want to brute-force RIPEMD-160?A public address, like one of the loaded puzzle addresses, is basically (neglecting network byte and checksum) HASH160(public key) which is RIPEMD-160(SHA-256(public key)). If your aim is to get to an unknown public key and you reverse RIPEMD-160(32-byte input) you land on a possible collision, one of 296 by the way, where you even don't know if you found the correct one. Your found collision is a SHA-256() hash.
Now you have to reverse this SHA-256(x02|x03+32bytes) and because you don't know if the compressed public key starts with x02 or x03, your search space is basically of magnitude 2257. Good luck with that, when our solar system or rather a larger part of our galaxy doesn't have enough energy to even brute-force RIPEMD-160() let alone SHA-256().
I would strongly suggest to use (or train) human brain 1.0 before wasting time with such stupid cracking paths. Such posts are complete nonsense and frankly an insult to intellect.
~~~
Oh no, the quantum computer will solve it all bs again... not that I've been waiting for it.~~~
If you didn't notice the difference between e.g. #66, more specifically non-multiples of 5, and puzzles that are multiples of 5, you're a lost case and you're wasting everybody else's time here. Hint: it was something to do with the public key.The level of insanity in this thread is really concerning.
Quote
Of course, I know you're right, that post was just for learning.
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
How long do you think the hash 160 can break?
How long can it take?
How long can it take?
Code:
import hashlib
import time
from itertools import product
import string
def ripemd160(data):
"""
Compute the RIPEMD-160 hash of the given data.
"""
h = hashlib.new('ripemd160')
h.update(data.encode('utf-8'))
return h.hexdigest()
def brute_force_ripemd160(target_hash, max_length=5):
"""
Simulate brute-forcing a RIPEMD-160 hash.
Test all combinations of characters up to a given length.
"""
characters = string.ascii_letters + string.digits
start_time = time.time()
attempt_count = 0 # Track number of attempts
print(f"Brute-forcing RIPEMD-160 hash: {target_hash}")
print(f"Using characters: {characters}")
# Brute-force attempt
for length in range(1, max_length + 1):
print(f"\nTesting all combinations of length: {length}")
for attempt in product(characters, repeat=length):
attempt_count += 1
guess = ''.join(attempt)
guess_hash = ripemd160(guess)
# Show progress every 100,000 attempts
if attempt_count % 100000 == 0:
elapsed_time = time.time() - start_time
print(f"Attempts: {attempt_count:,} | Time elapsed: {elapsed_time:.2f} seconds")
if guess_hash.upper() == target_hash.upper():
elapsed_time = time.time() - start_time
print(f"\nMatch found: {guess}")
print(f"Attempts: {attempt_count:,}")
print(f"Time taken: {elapsed_time:.2f} seconds")
return guess
elapsed_time = time.time() - start_time
print(f"\nNo match found within {max_length}-character keyspace.")
print(f"Total attempts: {attempt_count:,}")
print(f"Time taken: {elapsed_time:.2f} seconds")
return None
target_hash = "1234567890123456789012345678912345678..." # Replace with your hash
print(f"Target Hash: {target_hash}")
# Run the brute-force simulation
brute_force_ripemd160(target_hash, max_length=5)
import time
from itertools import product
import string
def ripemd160(data):
"""
Compute the RIPEMD-160 hash of the given data.
"""
h = hashlib.new('ripemd160')
h.update(data.encode('utf-8'))
return h.hexdigest()
def brute_force_ripemd160(target_hash, max_length=5):
"""
Simulate brute-forcing a RIPEMD-160 hash.
Test all combinations of characters up to a given length.
"""
characters = string.ascii_letters + string.digits
start_time = time.time()
attempt_count = 0 # Track number of attempts
print(f"Brute-forcing RIPEMD-160 hash: {target_hash}")
print(f"Using characters: {characters}")
# Brute-force attempt
for length in range(1, max_length + 1):
print(f"\nTesting all combinations of length: {length}")
for attempt in product(characters, repeat=length):
attempt_count += 1
guess = ''.join(attempt)
guess_hash = ripemd160(guess)
# Show progress every 100,000 attempts
if attempt_count % 100000 == 0:
elapsed_time = time.time() - start_time
print(f"Attempts: {attempt_count:,} | Time elapsed: {elapsed_time:.2f} seconds")
if guess_hash.upper() == target_hash.upper():
elapsed_time = time.time() - start_time
print(f"\nMatch found: {guess}")
print(f"Attempts: {attempt_count:,}")
print(f"Time taken: {elapsed_time:.2f} seconds")
return guess
elapsed_time = time.time() - start_time
print(f"\nNo match found within {max_length}-character keyspace.")
print(f"Total attempts: {attempt_count:,}")
print(f"Time taken: {elapsed_time:.2f} seconds")
return None
target_hash = "1234567890123456789012345678912345678..." # Replace with your hash
print(f"Target Hash: {target_hash}")
# Run the brute-force simulation
brute_force_ripemd160(target_hash, max_length=5)