You don't understand. It's better for you when you have the first chunk of characters of your WIF than otherwise, because this would limit the range of private keys to search in a much more favourable way (if I don't get it wrong, I haven't had to do such a recovery myself so far, only experimented a bit to gain some knowledge). Um, not sure if my logic holds water here, but I had probably a Gin Tonic too much to think clearly.
What LoyceV mentions is when you have a limited bit-range that needs to be explored and you have a public key exposed and known of that private key, you can use methods like Kangaroo to find the correct private key much faster than a conventional brute-force search in such a range.
Off the top of my head I can't estimate what 17-18 missing trailing characters of the WIF translate to with respect of missing bit-range. IIRC, five of the trailing characters are the embedded checksum, so you're actually missing 12-13 characters of your WIF (WIF private keys are 51 or 52 characters long, Base58check).
Roughly you're missing around a quarter of your private key, if my napkin math is right, let's assume around 65-75 bits, maybe 80 at worst. That's a bit-range size which seems very doable with a modified RCKangaroo solver or similar at very first glance. (I might be wrong as I compare it to the Bitcoin puzzle challenge where limited bit-range private keys have been "easily" broken for ranges below 100 bits with an exposed public key. The last largest bit-range cracked with exposed public key is currently puzzle #130 (129 bits range) but this one took months with thousands of GPUs, IIRC.)
Be careful: users here might promise you they can help and ask for your 34 characters you have. You shouldn't give them away lightly or at all!
hello,
so since i have the first 34 chars and the public key, i could use rckangaroo if i understand right. because i can short the bit range by a lot . but how ?
i put a 1 in the missing parts and then
i did exactly what this web said :
https://en.bitcoin.it/wiki/Wallet_import_formatand also make one for z (example L14Cf...zzz). is this right ?
now i have two long hex chars , i checked in what bit range those two hexare like where the value changes and its 80 bits.
so could i just input the small right part of the first hex and use it as start range no?