On Nov. 1st 2024 at 0:00 AM UTC I will post here a BTC P2PKH address (compressed) associated with an 80-bits secure private key.
The address owns 0.005 BTC (~340 $)
Starting from that time, at some point during the next 24 hours an outgoing transaction, containing unspent outputs from that address, will be broadcasted on the BTC network (public mempool, not Mara).
Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
Together with the target address I will also provide the range of the search interval.
This will be in the form (just an example):
The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.
If your first instinct is to yell "this is not a 80-bits secure private key", please note this competition is NOT for you, do not expect a response from me.
Get your tools ready boys.
let someone be lucky) if a transaction takes place from that address and the public key is known...The case is 5 minutes. Even with my capabilities, it will be very fast to find him. I won't participate, just watch.(OK, not 5 minutes, but 10, sorry...)))The address owns 0.005 BTC (~340 $)
Starting from that time, at some point during the next 24 hours an outgoing transaction, containing unspent outputs from that address, will be broadcasted on the BTC network (public mempool, not Mara).
Let's see who breaks it. I only ask one thing if that happens: what method was used to break the key.
Together with the target address I will also provide the range of the search interval.
This will be in the form (just an example):
Code:
minKey = 0xf2e542b46066c4e6f91abc80000000000000000000185e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133
The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.
If your first instinct is to yell "this is not a 80-bits secure private key", please note this competition is NOT for you, do not expect a response from me.
Get your tools ready boys.
Total range will be 256 - 80. Too big, you not think so ?
Code:
def hex_to_bin(hex_value):
# Convert hex to binary string, without the "0b" prefix
return bin(int(hex_value, 16))[2:]
def hamming_length(hex1, hex2):
# Convert both hex values to binary
bin1 = hex_to_bin(hex1)
bin2 = hex_to_bin(hex2)
# Calculate the Hamming distance
distance = sum(b1 != b2 for b1, b2 in zip(bin1, bin2))
return distance
# Example usage
hex_value_1 = "0xf2e542b46066c4e6f91abc80000000000000000000185e689447431d74c5b133"
hex_value_2 = "0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133"
hamming_dist = hamming_length(hex_value_1, hex_value_2)
print(f"Hamming length: {hamming_dist} bits")
# Convert hex to binary string, without the "0b" prefix
return bin(int(hex_value, 16))[2:]
def hamming_length(hex1, hex2):
# Convert both hex values to binary
bin1 = hex_to_bin(hex1)
bin2 = hex_to_bin(hex2)
# Calculate the Hamming distance
distance = sum(b1 != b2 for b1, b2 in zip(bin1, bin2))
return distance
# Example usage
hex_value_1 = "0xf2e542b46066c4e6f91abc80000000000000000000185e689447431d74c5b133"
hex_value_2 = "0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133"
hamming_dist = hamming_length(hex_value_1, hex_value_2)
print(f"Hamming length: {hamming_dist} bits")
80BITS!
not info there is 80 bit, in start or at and of 256.
0.........2e542b46066c4e6f91abcb0fffffffffffffffffff
fffffffffffffffffff2e542b46066c4e6f91abcb
1111001011100101010000101011010001100000011001101100010011100110111110010001101 010111100100000000000000000000000000000000000000000000000000000000000000000000000000000000 00110000101111001101000100101000100011101000011000111010111010011000101101100010 0110011
1111001011100101010000101011010001100000011001101100010011100110111110010001101 010111100101111111111111111111111111111111111111111111111111111111111111111111111111111111 10110000101111001101000100101000100011101000011000111010111010011000101101100010 0110011
EDIT- They said
Together with the target address I will also provide the range of the search interval.
if they didn't provide the range along with the address then it would be 256, but since it is a puzzle they are providing us with the range. Please correct me if I am wrong kTimesG 
