Topic: Private Key missing 4 characters (Read 1100 times)

Indeed... anecdotally, I have seen so many stories on this forum over the last 4-5 years of "small children" that destroy backups (ripping paper, scribbling in notebooks etc)... and, from memory, one or two involving animals. Also, a few with "bad handwriting" where they can't decipher certain characters... and spilling of liquid onto papers is another.

Any software that is not 100% open source and available to public to be reviewed is not trustworthy.

It can be a lot of different things, you already though of the printer issue and typos. There can also be damages to the storage medium (eg. paper) where some characters have become unreadable now.
4 characters is nothing, the paper could have been torn and half of it could be missing.

what i dont understand is how the private key ended up 4 characters short. that's really confusing. did they print it out and the printer messed up or ...i guess they wrote it down wrong on paper but that's kind of hard to believe. one mistake is undestandable but 4?

Nothing like that, see https://en.bitcoin.it/wiki/Wallet_import_format for the theory.

7 chars allnum 25terabyte of data, they probably have big IO throughput at hand and did the same as the script, import the entire batch to a wallet or scan it with a block explorer or match the list against one checksum - now idea what else would do, where else could that be derived?

The last 7 characters of the WIF private key are a checksum. I have successfully used this method to import a private key without the last 7 characters into blockchain.info, and when I export it again, it's complete.
If blockchain.info can do that, it would be very nice to have a local program that does the same.

Checksum for what? Same as restoring the key, no?
Dont exactly get what you mean, bit tired

It's less efficient but easier to understand than the Private Key Restorer in the second post. I'd still be interested in a way to just restore the checksum, that should be quite easy and very fast, but I haven't been able to find anything for it.

here a very simple bash to guess last 5 chars allnum of a string and add the string as prefix, then import prefixed string to bitcoin-core wallet.

I don't understand where you come up with 46 * 47 * 48 * 49 and (58 choose 4)
we're not choosing 4 out of 58, but we pick 1 of 58, do 4 times (positions) = 58*58*58*58 = 58^4
 
why 58 choose 4? the 4 chars are not guaranteed distinct
what I do is C(n,k) = n choose k = 49 choose 4 , but not for chars
I'm doing this ( 49! / 4! (49-4)! ) for possible position combinations of the missing 4 chars
then for each possible position combination, there are 58^4 possible char combinations

49 positions, 45 known chars (order sequence assumed) and trying to fill 4 chars in any 4 positions
so total = (49 choose 4) * 58^4 = 211,876 * 58^4 = 2,397,693,906,496 = 2.39 * 10¹²

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small example:
4 positions, 2 known chars (order sequence assumed) and trying to fill 2 chars in any 2 positions
4 choose 2 = ( 4! / 2! (4-2)! ) = ( 1*2*3*4 / 1*2*1*2 ) = 6 possible position combinations
so total = 6*58*58 = (4 choose 2) * 58^2

Yes.

Unfortunately not as common sense as it should be 
But you are right. The expression itself is a bit odd.

I just wanted to say that you don't actually have to seach the whole space (because thats what most people believe when they hear they are that much combination).
On average, it is enough to seach 50% of the space.




It is not about a 50% chance to find one correct key, but to find '2 same keys'.
~3,458,770 would be the amount of guesses it would take to have a 50% chance for having generated the same key twice.




Actually this should be a reason for vanitygen to not use randomly.
My guess would be that it is just way faster to randomly generate them than to create them sequentially + memory storage and I/O.
And since vanitygen is used to create addresses with a relatively small prefix (compared to the overall length), it probably doesn't make a huge difference. 

This can't be right. The birthday paradox is about finding a collision, in other words testing the same private key twice.

I can believe it takes about 3 million guesses for a 50% chance (out of 8 trillion possibilities) to guess the same key twice. But that's not going to help you, as you'll still need to search 4 trillion keys for 50% chance of finding the right one.
Using a random search only means you test the same key several times, while you may never test some of the other keys.

Just say, that when i read this information, i starting to go in shock - it is your own security!
What is your opinion in it? Why you didn't put this information on a safe place???

Surely it is just common sense that searching half the space obviously has a 50% chance of finding the right key

The maths of the birthday attack suggests that by randomly generating keys (as opposed to sequentially searching the space), the smallest number of keys we'd need to test to have a 50% chance of finding the right one would be:


Which for your calculated search space is only something like: ~3,458,770 guesses... which isn't a great deal to be honest.

I'm pretty sure that the Birthday Attack is the reason why VanityGen generates keys randomly rather than sequentially

I am pretty sure you made a small mistake there. You do ignore the order of the 4 missing words with your statement: ( 49! / 4! (49-4)! )
You do iterate through each position with your words, but you have to iterate within the words too.

This should be the correct calculation:



There are 58 choose 4 possible combinations to pick the 4 correct chars from the charspace (without considering the order).


Now consider the first two characters as already known. Only looking at the private key without the first two chars here:

To put the first unknown char into the correct place, you can choose between 1) before of each already existing char (45 possibilities) and 2) behind the last one (45+1).
For the second one you choose either before each char in your privkey (46 possibilities) or behind the last one (46+1)...

Overall there are 46 * 47 * 48 * 49 possible combinations to place the 4 chars into a private key with 45 known characters (ignoring the first two) into the right order. This assumes the already known characters are in a correct order already.


The total search space (for the private key without the first 2 chars) therefore is 46 * 47 * 48 * 49 * (58 choose 4) = 2.1574231*10¹².
Multiplied with 4 (combinations the priv key can start with) = total amount of combinations = 8.6296925 * 10¹²


Considering the birthay paradox, there is a 50% chance to find the correct private key after 1/2 of this space.



Feel free to correct me if i have made a mistake!

If someone skips 4 characters while writing down a private key, chances are he also wrote down one or more characters incorrectly.
If you're sure the rest is correct, it could be worth spending some money on Cloud computing, start up enough nodes, and crunch it quickly.

Someone specialized in wallet recoveries may be able to help you with this. I'm thinking about Dave from https://walletrecoveryservices.com/ for instance.

I think your calculation is a bit off. Also we can save computation power by narrowing down search space,
the first 2 chars always start with '5H', '5J', '5K', 51-2 = 49 positions and 49-4 = 45 characters known
if exact position of 4 lost characters are known = 58^4 = 11,316,496 iterations
if exact position is unknown = 58^4 * 4-combination
= (58^4) * ( 49! / 4! (49-4)! ) = (58^4) * (46*47*2*49)
= (58^4) * 211876 = 2,397,693,906,496 iterations = 2.39 * 10¹²
do I get this right?

are you sure we can have that computation power for combinations/second?
if compute power 200,000 combinations/second ~ 137.5 days (4.5 months)
if compute power 1,000,000 combinations/second ~ 27.75 days (1 month)
if by any chance the search space can be narrowed down even more (for e.g. assume the first 15 chars are correct)
it can cut down the number combinations and the needed time drastically

In which case, I think the search space is something like (58chars*47positions)*(58chars*47pos)*(58chars*47pos)*(58chars*47pos)... (58*47)⁴ = 5.5220891*10¹³ possible combinations

It should be possible to modify the code (or write new code) to test all the possible combinations... but it is going to take a LONG time to search all of that. Testing 1,000,000 combinations/second will take nearly 2 years to search all of them. (for reference, my Core i5-3570K, running Ubuntu on the Windows 10 Linux Subsystem seems to average around 200,000 combinations/second.)

Not exactly "impossible" (and for 5 BTC, probably worth it )... but technically it will be difficult to run a task non-stop for that period of time.

I did another test of the algorithm with my private key and got no feedback ...

But I understood the algorithm and I managed to reproduce a certain result, I used a random private key of 51 characters, the algorithm only shows a result with a private key of 51 characters and if the missing positions are filled exactly in the place of origin, in any other position the algorithm does not give a response with a valid private key.

Unfortunately in my case I have no idea of the positions of the lost characters, but I keep trying, because it has been very useful for me to learn new skills and resources.