My new public key search system is almost ready. I had to reinvent my binary database system because, although the database was lightweight
https://bitcointalk.org/index.php?topic=5475626, I had efficiency issues with binary search. This is now a thing of the past. I have designed a system that stores 100 million public keys in an
80 KB file, yes, what you read
80KB!(in the future it will be smaller) that meets maximum efficiency. We would only be limited by the current speed of Secp256k1 when generating the 100 million or more public keys while creating the database. I am finishing designing the search script after months of being stuck due to personal issues, I am finally back on track.
fo get key from 2^27 need a
2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2×2 = 134217728 pubkeys. this downgrade pubkey 2^27 to pubkey with privkey 1
same situation with downgrade 2^130 to 2^103 etc
if substract 1 bit, from 2^30 need initial 2^27 pubkeys to take one of them in result with privkey 1. Because all step generate 50/50 probability what will go to - areal or you steal in + area
I think what way of kangaroo or bsgs has no resources for find DLP of lage privkey, and qantum computers is more fare problem for btc securety then neuronetworks. So maybe this DB can help make network with enough mutations for solve problem.
enothe interesting I think what bits in privkeys has a paterns, I thin youse paters can speed up finding privkeys, for ex lets see to privkey 2^100:
https://privatekeys.pw/key/000000000000000000000000000000000000000af55fc59c335c8ec67ed24826ib bits:
1010111101010101111111000101100111000011001101011100100011101100011001111110110
100100100100000100110
Sequence 1010 appears 6 times
Sequence 0101 appears 6 times
Sequence 1011 appears 6 times
Sequence 0111 appears 6 times
Sequence 1111 appears 8 times
Sequence 1110 appears 6 times
Sequence 1101 appears 5 times
Sequence 1100 appears 7 times
Sequence 1000 appears 5 times
Sequence 0001 appears 5 times
Sequence 0010 appears 6 times
Sequence 0110 appears 7 times
Sequence 1001 appears 8 times
Sequence 0011 appears 7 times
Sequence 0000 appears 3 times
Sequence 0100 appears 6 times
so using sequences is more fast then generate all 2^100
scrypt for get sequences:
def count_sequences(number):
sequences = {}
for i in range(len(number) - 3):
sequence = number[i:i+4] #repace 4 to get sequnces with another lenght
if sequence in sequences:
sequences[sequence] += 1
else:
sequences[sequence] = 1
return sequences
number = "1010111101010101111111000101100111000011001101011100100011101100011001111110110
100100100100000100110"
sequences = count_sequences(number)
for sequence, count in sequences.items():
print(f"Sequence {sequence} appears {count} times")
posible to check puzzle privkey and find what pizzle haz a 38 combinations from 64 total posible combinations with each vombinations lenght 6. Total combinations is a same to brute only then try all cases with 38 combinations, but probkey vill be foumded not after check all cases, but example after half of them ...
another way:
imput privkey
0x2ca447fa844948c661d1e35ada56713d7
for get pubkey of 0x17c93cb4fa4bfbb from input key
need from this sequence
1234567890123456789012345678901234567890123456789012345678901234567890123456789
012345678901234567890123456789012345678901234
remove jast this numbers(put "," between all numbers for batter understanding
91231161189747822734455
you get
1234567804567890234567890234578902345678902345670123456890123569013456789013456
89012567890123678901234
after sunstract all numbers in sequnce from pubkey and divide to each step to 9, you get
0x17c93cb4fa4bfbb pubkey in 91231161189747822734455 operations
OДИH 91231161189747822734455 23
ДBA 1234567804567890234567890234578902345678902345670123456890123569013456789013456
89012567890123678901234 102
TPИ 1234567890123456789012345678901234567890123456789012345678901234567890123456789
0123456789012345678901234567890123456789012345 125
your DB is real powerful DB !!! Thank you for DB.