I thought WIF keys that began with K or L were compressed keys and WIF keys that began with 5 were the uncompressed ones? Also I am a bit confused how I get those big numbers with only the public address and a WIF key that is missing characters. Could you elaborate anymore? I apologize I am fairly new to learning about all this.
Yeah you're right - my memory was a bit rusty. In any case just replace the -u flag with -c.
I simply converted the base58 of the lower characters to decimal (and hex). First I went to this page:
https://www.dcode.fr/base-58-cipherAnd then I pasted the characters after the lost 10 chars inside the page. Before the characters, I pated the 'w' (since you know you have that), followed the 10 characters lowest possible private keys that still base-58 encode into w........JzXaqU2rcFSoaLaehAQHqoQX1cWCo92tAA3ihLJ7 - replace the dots with 10 "1" (the number one) characters. Because 1 is the first digit of base8 number system. The resulting hex gives the starting range.
Then to get the end range, you repeat the process but instead of ten 1 characters, you insert 10 'z' characters (the highest character in base58 is lowercase 'z').
To get the stride, I simply converted the lower part of the base58 you had (the one after the dots).
To determine the start and end ranges and the stride, you only need part of the WIF, not the public address.
These steps will create a range and strie that is suitable to input inside Bitcrack.
It has a difficulty of log2(58**10) = 58.5798 bits, this is doable if you have a few GPUs.
If there is an outgoing transaction, then with the un/compressed public key and kangaroo or pollard, also possible.
Sadly this was an offline wallet, so it only ever had incoming transaction. It has never sent out.
No problem, because Bitcrack is more efficient than Kangaroo for your problem (also, Kangaroo will only work if you have the public key, not the address).
Really? I thought with 10 missing characters that was 58^10th or a large number and it was going to take centuries to solve this. How exactly do I solve this in minutes? What tool should I use, or is there something custom? I am willing to pay a bounty for someone assisting me to set this up on machine.
58**10 is a very large number, however to estimate the difficulty, you need the equivalent power in base 2, so what we do is we take the log2 of the result: log2(58**10). Then if gives us the difficulty in bits: such that 2**bits == 58**10 (here, bits equals 58.a_fractional_part).
Difficulty can also be written as "10 base 8 characters", but its common for programs to estimate it in terms of bits as well.
This will probably
not take minutes unless you have a large GPU farm, but a few weeks is a more accurate estimate. Since Bitcrack can only talk to 1 GPU as far as I know.