Pollard Kangaroo approach is for the know private key range and for the best, within 125 bit range and of course since you have the Multibit wallet handy all you have to do is "Once you located the files, if you have access to an old multibit app on you computer you need to import the wallet and export the private keys to a new wallet. If you do not have a copy of Multibit, you can use OpenSSL together with your password to decrypt the files and recover the keys."
You need the password to export them. I still have the wallet open every single day since then. I just keep looking at what value i've missed out on  I'm confused, do you have the wallet or you don't? How come the wallet is lost and you have it open? Maybe I'm missing something here? |
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hi members!
I have a multibit classic wallet that currently holds 17BTC
Somewhere in 2014/2015 i lost my wallet password.
Many attempts have been tried to bruteforce the wallet password, without any success.
KeychainX, dave bitcoin, everyone already tried to guess or break my password.
Would you asume there is a possibility to use software to try and recover my wallet using the Pollard approach?
Can someone point me in the right direction on what sofware and settings i should be using?
I still have access to the Multibit wallet, only the password is missing.
I have my wallet address and it's public key.
Maybe this could come in handy to extract certain transaction data?
Thank you
I have experience working with multibit wallets. Maybe I can help. |
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I'm sorry, but if pk = 3; then pk - 10 =/= 7, pk - 10 = -7, very different.
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BSGS? sorry but not for 125 and more bit.
the memory will be huge.
No, for those we have Pollard Kangaroo, and they are math too  |
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....
You are right, I didn't even think about it. On the other hand, I understand his point. I can make another pub key and see if he's willing to test his method with me or not.
Just sign a message with the privatekey to proof that you own the privatekey for the pubkey without exposing it. Post the signature with the used message. Without knowledge of the message the signature can be forged. Excellent idea!, thank you; now I just need to know if mcdouglasx is willing to play along. |
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Sorry, If I manage to solve any puzzle I will give all the information at the moment I do not have a good economic position for it.
Anyone could add x quantity to the puzzle key, generating a different public key and camouflage the puzzle with it, and by giving them information I would be solving the puzzle without knowing it.
then simply subtract that same amount from the camouflaged public key and go back to the puzzle and solve it.
Well, if that is really the problem, here is the private key for that public key: priv key HEX: 0x155d735095df3f8c24ee33e4dfe955da priv key DEC: 28399010381067550399127235982780880346 pub key: 0343b7d69e8372746596980d678d6cdecbffb2927916b3dbd3bc3d27bf5366b166 As you can see, I won't solve any puzzle with the info you provide; I only want to test if your script works or not. But if you decide not to test it with me, that's OK. Cheers. With a known private key, each person can reduce the key to any size. Here is an example: 115 bit pub key: 02903740144e9f301f62362813dec638747153d4f18467f32286d09b9b4412768b priv key: 0x5735095df3f8c24ee33e4dfe955da 85 bit pub key: 02f50518c0b808dff5bd75f10e146d33677dccbfd18370e64be0c4cc36abc9675f priv key: 0x1f3f8c24ee33e4dfe955da 40 bit pub key: 03a0f8f7d72f6199bb09762c8500ed22eba5253f9661f752db338bbf5eeefb1852 priv key: 0xe4dfe955da By revealing the private key, you will not know whether the method works or not. You are right, I didn't even think about it. On the other hand, I understand his point. I can make another pub key and see if he's willing to test his method with me or not. |
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Sorry, If I manage to solve any puzzle I will give all the information at the moment I do not have a good economic position for it.
Anyone could add x quantity to the puzzle key, generating a different public key and camouflage the puzzle with it, and by giving them information I would be solving the puzzle without knowing it.
then simply subtract that same amount from the camouflaged public key and go back to the puzzle and solve it.
Well, if that is really the problem, here is the private key for that public key: priv key HEX: 0x155d735095df3f8c24ee33e4dfe955da priv key DEC: 28399010381067550399127235982780880346 pub key: 0343b7d69e8372746596980d678d6cdecbffb2927916b3dbd3bc3d27bf5366b166 As you can see, I won't solve any puzzle with the info you provide; I only want to test if your script works or not. But if you decide not to test it with me, that's OK. Cheers. |
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if I have a pub key that represents an unknown pk1
pk1=957273957373958372947
my script reduces the bits for example to
pk2=958372947
that is, remove the first 12 numbers for the example.
Among approximately 1500 keys there is one of them true that corresponds, I look for the key by brute force and once I get pk2 it is easy to deduce the private key of pk1
@mcdouglasx: Hey man, please don't be offended by my request, but can you please give me the pubkeys that your script generate out of this pubkey? 0343b7d69e8372746596980d678d6cdecbffb2927916b3dbd3bc3d27bf5366b166 You can send me a DM with the results, I'm trying to make a test. Thanks in advance, cheers!. Have you thought about it? You can post them here if you prefer. |
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@batareyka
If you really have a way to find out the range in which the private key is from the public key, then contact me. I will show you how using this method of yours you can get a private key from any public.
I'm pretty sure that someone who can manage to obtain the range of the private key out of the public key can find a method to get such private key, don't you think so? Agree. But suddenly he really found such a way but does not understand how to apply it? Maybe he just think he found it. As he is not willing to explain what he really found, who knows? |
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@batareyka
If you really have a way to find out the range in which the private key is from the public key, then contact me. I will show you how using this method of yours you can get a private key from any public.
I'm pretty sure that someone who can manage to obtain the range of the private key out of the public key can find a method to get such private key, don't you think so? |
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if I have a pub key that represents an unknown pk1
pk1=957273957373958372947
my script reduces the bits for example to
pk2=958372947
that is, remove the first 12 numbers for the example.
Among approximately 1500 keys there is one of them true that corresponds, I look for the key by brute force and once I get pk2 it is easy to deduce the private key of pk1
@mcdouglasx: Hey man, please don't be offended by my request, but can you please give me the pubkeys that your script generate out of this pubkey? 0343b7d69e8372746596980d678d6cdecbffb2927916b3dbd3bc3d27bf5366b166 You can send me a DM with the results, I'm trying to make a test. Thanks in advance, cheers!. |
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Then he must be the one who solved #120 and #125; otherwise, he is missing on the "prizes" |
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There is even a way to find out the range in which the key is located, ...
Really? So tell us what the range is where #66 is lying I apologize for not providing clarification. You can find out the range from the public key. Then it will not be difficult for you to determine in what range this public key is located. 0359c777fb4a0bf9c16ba308259eb4f5ef2abfd2928a8f499cbdc47300f9dcb8d5 What strange people. You don't take my word for it, you always want confirmation. And then this and this. I regret that I again fell into the trap I fell into a year ago on the same forum. Everyone laughed at me. No more comments. Good luck to everyone. 1000000000000000000000000000000000000000 F000000000000000000000000000000000000000 8000000000000000000000000000000000000023 I don't understand how, but you're absolutely right. But could you share a way to determine the range? At least a hint. If you can determine the range of the private key from it's public key, ECC is broken. So, I don't believe that, yet. |
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I see this thing is still going, has anyone solved any of the puzzles yet? Or has anyone made any type of progress? Can I get a TLDR of the events over the years from the start of the puzzle until now?
Over how many years exactly? 😂, #65 and #120 been solved. Prize has been increased 10 fold, now there is around 1000 bitcoins for us to loot ( yeah right). You forgot one important detaill: puzzle #65 was solved before #64. Idk why!!!  That was because public key of #65 was revealed, so Kangaroo was used to find its private key. |
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i found this address in substraction of 125 13zb1hQ664XDDhM2LkWLLa23cdbhHwya7c # 19.................... hash 160 : 20d45a6a51bc082aeb4d344be937b1bd4ea1f238 am i close if i search in the range 66 cuz the address looks like its in that range it starts with same 7 digits with 66 puzzle and same 8 first digits of hash160... 02c0643bd28d11d650b24dae3143b1d3578e6a3597a9395f43c25ead8ed2a9298f who can scan in kangaroo 66 bits in 5 min or less ? if you find this address we will split the 12 btc  , of course if it is in that range im not sure im asking you guys ? Sorry to tell you this, but there are tons of 13zb1hQ addresses in all the ranges, so no, that doesn't mean it's in 66 range. |
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The thing is: not all people need money incentives (specially real scientists or researchers): Grigori «Grisha» Yákovlevich Perelmán.
Cheers.
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For example, puzzle #65: Private key: 00000000000000000000000000000000000000000000001a838b13505b26867 = 000000000000000000000000000000000000000000000000000000000000000 + 1a838b13505b26867Public key: 0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b Private key: fffffffffffffffffffffffffffffebaaedce6af48a03a1799ad57ca83d8da = fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 - 1a838b13505b26867Public key: 0330210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b As you can see, the key is the same, only in one case it is added, in the other it is subtracted. Now do you understand? Sorry man, but you are getting confuse: Private key #65 = 00000000000000000000000000000000000000000000001a838b13505b26867 Public key #65 = 0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b On the second point you are talking about, what you are doing is subtracting PK #65 to N, and the result is: Private key for second point = fffffffffffffffffffffffffffffebaaedce6af48a03a1799ad57ca83d8da Public key for second point = 0330210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b PK #65 =/= PK second point Your confusion strive on the fact that both points share the same x-coordinate, because one is the "reflection" of the other, but the PKs are not the same; and as you can see, none of them are "negatives". There are no negative numbers on ECC. Good luck to you too, man. |
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@digaran:
Man, this is the last I'm going to say about the subject: You are doing the calculations wrong, like alberto said and WP repeated to you, you can't do operations on public keys and privet keys at once, you can't mix them, they are not in the same context. One makes reference to the other, but they are not in the same category (excuse my writings, English is not my native language).
To prove my point, you should use any script that manages public keys/privet keys and you will realize that you are doing it wrong.
It's not criticizing, I just want you to advise that you are wasting your time doing that kind of operations.
Good luck, man.
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but I'm asking for the closest private key to work with on the BSGS. I'm still learning how to convert these public keys to private keys
You can't convert public keys to privet keys (at least not yet). What you can do is what BSGS do: try a privet key, convert that one to public key, and see if it match the public key you are looking for; if it don't match, try the next one, and so on. |
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The thing is that the "x" coordinate of #125 is: 33709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e; so I don't see how do you add "F" to 03 x coordinate of #125. Not only that, but if you switch "02" to "03" on a public key, you are talking about a very different privet key, so 0333709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e is no longer #125.
And BSGS don't use dp, it actually works in a sequential manner, deterministic, and not probabilistic like Kangaroo.
Thanks anyway man, but I think you should re-learn ECC, I can see you are confused on how it works.
Cheers.
Hello. You apparently did not carefully read what @digaran wrote to you. If you change #125 02 to 03, you will get the same private key as #125 but with a minus sign "-"! The public key 0333709eb11e0d4439a729f21c2c443dedb727528229713f0065721ba8fa46f00e is also on the EC, but on the other hand. For example, puzzle #65: Private key: 00000000000000000000000000000000000000000000001a838b13505b26867Public key: 0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1bPrivate key: fffffffffffffffffffffffffffffebaaedce6af48a03a1799ad57ca83d8daPublic key: 0330210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1bNow do you understand? Well, it's amazing that even you wrote it yourself, you don't realize what your saying. As you can clearly see on your own example "0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b" and "0330210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b" have different privet keys, which is what I wrote, Now do you understand? BTW, there are no privet keys 'with a minus sign "-"!' |
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