Ok, here's a 1BTC puzzle. - page 12.

BrewMaster

legendary

Activity: 2128

Merit: 1293

There is trouble abrewing

Quote from: fecell on February 17, 2019, 08:25:21 PM

Quote from: greg007r on February 16, 2019, 06:48:33 PM

You can always validate your output on brain wallet : https://brainwalletx.github.io/

think it's wrong way.
wrote: 8 camel case 32 long plain-text (private key).
not a hash, not a WIF, not a BrainWallet, not a string for generating private key.
said exactly 'private key'.
so this string is a key, and must be used to generate 'public key' and WIF directly, without hashing it before.

is not it so?

it is impossible to say because the "hint" in OP is vague at best so there is a lot of ways that you can get a private key from the supposed set of words that are the answer to this question.

i think people should first focus on solving the "puzzle" and after that when they at least had part of the answer they should work on brute forcing it. in which case if you have a big chunk of the answer you can simply try all the possible ways like using it directly, using its hash, using it in a brainwallet,...

fecell

jr. member

Activity: 137

Merit: 2

Quote from: greg007r on February 16, 2019, 06:48:33 PM

You can always validate your output on brain wallet : https://brainwalletx.github.io/

think it's wrong way.
wrote: 8 camel case 32 long plain-text (private key).
not a hash, not a WIF, not a BrainWallet, not a string for generating private key.
said exactly 'private key'.
so this string is a key, and must be used to generate 'public key' and WIF directly, without hashing it before.

is not it so?

greg007r

newbie

Activity: 6

Merit: 1

Quote from: ?? on ??

Quote from: greg007r on February 16, 2019, 06:48:33 PM

Quote from: ?? on ??

Quote from: greg007r on February 16, 2019, 02:21:41 PM

Quote from: runlola on February 16, 2019, 01:57:13 PM

Quote from: greg007r on February 16, 2019, 08:53:23 AM

My script has already processed 1/10 of the permutations based on BIP39 word list, nothing
Another script has already processed 1/5 of the permutations based on the transcript of the https://www.youtube.com/watch?v=AABkJ55Zz3A page, nothing

considering there are 2048^8 permutations of the bip39 list, i doubt that. The number of valid possibilities using that list is ~2000x smaller (161,735,049,399,563,309,103,078 - see this post for how it was derived). my program can check 100K phrases/s and it would take ~5B years to check 10% of those.

Yes, you are right but you can downsize drastically the amount of possible permutations by :

- Keeping 1/10 of the BIP39 list (from 2048 to 230), removing all not related words (eg: Drift,Fish and Fiscal) and keeping interrsting ones (eg : satoshi, million,coin,supply, tooth, one ...)
- Preventing permutations where two words are duplicated (eg: ThisThisOneToothThisThisThisThis symptom)
- Allowing only phrase of 8 words having 32 chars
- Preventing permutations where no more than 4 instances of a word having 4 letters
- For those who believe in the sequence W T C O N O H T would be even faster to check
- many more

Sample extract of last 10 lines of my logs :

Sat Feb 16 19:51:11 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 FixUseWorthWillMillionCoinAllFit : 12YVLipqmKfPwNwoy1Wpemja89Jf2q1vxh
Sat Feb 16 19:51:11 CET 2019 : 346/4830|3 3 6 4 5 3 5 3 FitAllAnswerCoinWhaleUseMatchOne : 1CRgZoDGdTRfF679hdKXyZgbGuEMPaZXyF
Sat Feb 16 19:51:59 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 OneCanFinalOnlySegmentThisAllFit : 1Fug6cBx487thwS1Zp5nSCJAaAJxWfXrqL
Sat Feb 16 19:52:47 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 OneFixFinalCoinControlOnlyCanFit : 1PKHseF7iqNWMv5At1YdCDVCvUW9C9ados
Sat Feb 16 19:53:34 CET 2019 : 346/4830|3 3 6 4 5 3 5 3 CanFitExcessGoldWhaleAllThereOne : 12MdRtKqKsT8Tw2AyR6JZhmXL8ZJcsXGVw
Sat Feb 16 19:54:22 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 WantSoonHaveOneUseThisFinalExact : 17bcNocTMuHksCbRVRcMiLMMLvaoTMGdSm
Sat Feb 16 19:54:22 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 ThisFixCanWorthAllWhaleGoldThere : 1GwNvXvrbxgmmQY27YL4gwcjjJEyX42LKw
Sat Feb 16 19:55:10 CET 2019 : 347/4830|3 3 5 4 7 4 3 3 AllCanBuildWantControlHaveOneFit : 17azNQ3Z5GqpvrEgo4YemT9Rc7DCBnRi6n
Sat Feb 16 19:55:58 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 WantUseOneTokenCanMimicLikeThere : 17oWja4HkduPeEsQSWC1bgdSLubdnT5pwk
Sat Feb 16 19:55:58 CET 2019 : 347/4830|3 4 4 3 4 3 3 8 UseLikeWaitCanWantFixAllProperty : 17uRQ63APaDNncu2cmKXiQsjff3kNKnXWB
Sat Feb 16 19:56:46 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 LikeHaveWaitAllUseWillLimitExact : 1YuVK4QA2K44sycmF3hDFFy8DCi6ne7sF
Sat Feb 16 19:57:33 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 SoonUseFitBuildCanWhaleOnlyThere : 16PMF61M6G8SQrXVAmgcRCZP2CcmsksnxC
Sat Feb 16 19:58:21 CET 2019 : 347/4830|3 4 4 3 4 3 3 8 CanWillCoinFixWaitUseAllRemember : 1JqCymN2wdVKM1zJUaiTgUTZPwkNXb1mVD
Sat Feb 16 19:59:09 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 SoonThatOnlyAllOneLikeMimicExact : 15rrST9vDQhBwqE6XTcsUfixmZFDMVXpri
Sat Feb 16 19:59:57 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 OnlyCanOneWorthFixExactThatThere : 1DiHHkHMTaQkSsbtV6To7YWF4LWuT329j
Sat Feb 16 20:00:45 CET 2019 : 348/4830|4 4 4 3 3 4 5 5 ThatGoldViewOneAllHaveProofExact : 18DWm63gZ9gYUS5ZnRP7Hp3J3tJGQps3yY
Sat Feb 16 20:01:29 CET 2019 : 348/4830|3 3 5 4 7 4 3 3 CanAllFinalThisSatisfyHaveOneFix : 1MaAgoj6AJ1T4NHoZLM3g5EohWwJ2n5MR2

Hope this help,

greg007r, I wonder how You Generate this addresses.
Should one not just hex output the text and use this as the ECDSA private key?

If I take for example your last example; CanAllFinalThisSatisfyHaveOneFix : 1MaAgoj6AJ1T4NHoZLM3g5EohWwJ2n5MR2
I get 1NSPwaCCTcA8JS99Em6nhyW1vHiX2F2pUg as the Address?

How do you get that 1MaAgoj6AJ1T4NHoZLM3g5EohWwJ2n5MR2 address?

Thanks in Advance

You can always validate your output on brain wallet : https://brainwalletx.github.io/

I use part of the api of bitcoinj :

Code:

import java.security.MessageDigest;
import org.bitcoinj.core.ECKey;
import org.bitcoinj.core.LegacyAddress;
import org.bitcoinj.params.MainNetParams;
byte[] hash = null;
try {
   MessageDigest md = MessageDigest.getInstance("SHA-256");
   md.update("CanAllFinalThisSatisfyHaveOneFix".getBytes("UTF-8"));
   hash = md.digest();
   ECKey key = new ECKey(hash, (byte[]) null);
   LegacyAddress address = LegacyAddress.fromKey(MainNetParams.get(), key);
   //if ("179sxfh6rw6bHSo5wVUhLP96k46QaEzVP".equals(address.toString())) {
   System.out.println("--------------------------------------------");
System.out.println("privkey hex :" + key.getPrivateKeyAsHex());
System.out.println("privkey wif :" + key.getPrivateKeyAsWiF(MainNetParams.get()));
   System.out.println("privkey int :" + key.getPrivKey());
   System.out.println("base58 :" + address.toBase58());
   System.out.println("string :" + address.toString());
   System.out.println("hex hash :" + Hex.encodeHexString(address.getHash()));
   //}
} catch (NoSuchAlgorithmException e) {
e.printStackTrace();
} catch (UnsupportedEncodingException e) {
   e.printStackTrace();
}

Ok, Thanks

I tried what you suggested and indeed the address fits the one you showed before.

But in an earlier post, someone suggested to take the 32 Char String, get the hex output for it (by for example paste it in https://www.online-toolz.com/tools/text-hex-convertor.php) and then use that HEX output as the ECDSA Private Key (for example paste it in https://gobittest.appspot.com/Address) and then it gives the answer I showed you before.

So, now are my questions:

- Which is the way to follow?
- Which of both VALID addresses is the one we search for?

Thanks In Advance

As far as I understand, a private key is a 256bit unsigned integer that can be represented as well as HEX, BASE64, WIF or PLAIN.
In the concept of BrainWallet, we want to use a phrase instead of 64 digits long number.
So you need to SHA256 hash the 32 char 'phrase' to get your 256bit unsigned integer or private key
From there, it will be possible to derive the public key and then the uncompressed Base58 Address.
Producing the uncompressed base58 address from the private key involves multiple sha256 and RIPEMD160 hashing in a specific sequence.

https://en.bitcoin.it/w/images/en/9/9b/PubKeyToAddr.png

And for https://www.online-toolz.com/tools/text-hex-convertor.php converts your text into the corresponding HEX representation of code point of the character. (eg : a=61,b=62 )
If you provide, a 32 letters long word to the converter, it will return a 64 long digit (32x2) exactly the same length as the return of sha256(phrase) but completely different !

greg007r

newbie

Activity: 6

Merit: 1

Quote from: ?? on ??

Quote from: greg007r on February 16, 2019, 02:21:41 PM

Quote from: runlola on February 16, 2019, 01:57:13 PM

Quote from: greg007r on February 16, 2019, 08:53:23 AM

My script has already processed 1/10 of the permutations based on BIP39 word list, nothing
Another script has already processed 1/5 of the permutations based on the transcript of the https://www.youtube.com/watch?v=AABkJ55Zz3A page, nothing

considering there are 2048^8 permutations of the bip39 list, i doubt that. The number of valid possibilities using that list is ~2000x smaller (161,735,049,399,563,309,103,078 - see this post for how it was derived). my program can check 100K phrases/s and it would take ~5B years to check 10% of those.

Yes, you are right but you can downsize drastically the amount of possible permutations by :

- Keeping 1/10 of the BIP39 list (from 2048 to 230), removing all not related words (eg: Drift,Fish and Fiscal) and keeping interrsting ones (eg : satoshi, million,coin,supply, tooth, one ...)
- Preventing permutations where two words are duplicated (eg: ThisThisOneToothThisThisThisThis symptom)
- Allowing only phrase of 8 words having 32 chars
- Preventing permutations where no more than 4 instances of a word having 4 letters
- For those who believe in the sequence W T C O N O H T would be even faster to check
- many more

Sample extract of last 10 lines of my logs :

Sat Feb 16 19:51:11 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 FixUseWorthWillMillionCoinAllFit : 12YVLipqmKfPwNwoy1Wpemja89Jf2q1vxh
Sat Feb 16 19:51:11 CET 2019 : 346/4830|3 3 6 4 5 3 5 3 FitAllAnswerCoinWhaleUseMatchOne : 1CRgZoDGdTRfF679hdKXyZgbGuEMPaZXyF
Sat Feb 16 19:51:59 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 OneCanFinalOnlySegmentThisAllFit : 1Fug6cBx487thwS1Zp5nSCJAaAJxWfXrqL
Sat Feb 16 19:52:47 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 OneFixFinalCoinControlOnlyCanFit : 1PKHseF7iqNWMv5At1YdCDVCvUW9C9ados
Sat Feb 16 19:53:34 CET 2019 : 346/4830|3 3 6 4 5 3 5 3 CanFitExcessGoldWhaleAllThereOne : 12MdRtKqKsT8Tw2AyR6JZhmXL8ZJcsXGVw
Sat Feb 16 19:54:22 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 WantSoonHaveOneUseThisFinalExact : 17bcNocTMuHksCbRVRcMiLMMLvaoTMGdSm
Sat Feb 16 19:54:22 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 ThisFixCanWorthAllWhaleGoldThere : 1GwNvXvrbxgmmQY27YL4gwcjjJEyX42LKw
Sat Feb 16 19:55:10 CET 2019 : 347/4830|3 3 5 4 7 4 3 3 AllCanBuildWantControlHaveOneFit : 17azNQ3Z5GqpvrEgo4YemT9Rc7DCBnRi6n
Sat Feb 16 19:55:58 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 WantUseOneTokenCanMimicLikeThere : 17oWja4HkduPeEsQSWC1bgdSLubdnT5pwk
Sat Feb 16 19:55:58 CET 2019 : 347/4830|3 4 4 3 4 3 3 8 UseLikeWaitCanWantFixAllProperty : 17uRQ63APaDNncu2cmKXiQsjff3kNKnXWB
Sat Feb 16 19:56:46 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 LikeHaveWaitAllUseWillLimitExact : 1YuVK4QA2K44sycmF3hDFFy8DCi6ne7sF
Sat Feb 16 19:57:33 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 SoonUseFitBuildCanWhaleOnlyThere : 16PMF61M6G8SQrXVAmgcRCZP2CcmsksnxC
Sat Feb 16 19:58:21 CET 2019 : 347/4830|3 4 4 3 4 3 3 8 CanWillCoinFixWaitUseAllRemember : 1JqCymN2wdVKM1zJUaiTgUTZPwkNXb1mVD
Sat Feb 16 19:59:09 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 SoonThatOnlyAllOneLikeMimicExact : 15rrST9vDQhBwqE6XTcsUfixmZFDMVXpri
Sat Feb 16 19:59:57 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 OnlyCanOneWorthFixExactThatThere : 1DiHHkHMTaQkSsbtV6To7YWF4LWuT329j
Sat Feb 16 20:00:45 CET 2019 : 348/4830|4 4 4 3 3 4 5 5 ThatGoldViewOneAllHaveProofExact : 18DWm63gZ9gYUS5ZnRP7Hp3J3tJGQps3yY
Sat Feb 16 20:01:29 CET 2019 : 348/4830|3 3 5 4 7 4 3 3 CanAllFinalThisSatisfyHaveOneFix : 1MaAgoj6AJ1T4NHoZLM3g5EohWwJ2n5MR2

Hope this help,

greg007r, I wonder how You Generate this addresses.
Should one not just hex output the text and use this as the ECDSA private key?

If I take for example your last example; CanAllFinalThisSatisfyHaveOneFix : 1MaAgoj6AJ1T4NHoZLM3g5EohWwJ2n5MR2
I get 1NSPwaCCTcA8JS99Em6nhyW1vHiX2F2pUg as the Address?

How do you get that 1MaAgoj6AJ1T4NHoZLM3g5EohWwJ2n5MR2 address?

Thanks in Advance

You can always validate your output on brain wallet : https://brainwalletx.github.io/

I use part of the api of bitcoinj :

Code:

import java.security.MessageDigest;
import org.bitcoinj.core.ECKey;
import org.bitcoinj.core.LegacyAddress;
import org.bitcoinj.params.MainNetParams;
byte[] hash = null;
try {
   MessageDigest md = MessageDigest.getInstance("SHA-256");
   md.update("CanAllFinalThisSatisfyHaveOneFix".getBytes("UTF-8"));
   hash = md.digest();
   ECKey key = new ECKey(hash, (byte[]) null);
   LegacyAddress address = LegacyAddress.fromKey(MainNetParams.get(), key);
   //if ("179sxfh6rw6bHSo5wVUhLP96k46QaEzVP".equals(address.toString())) {
   System.out.println("--------------------------------------------");
System.out.println("privkey hex :" + key.getPrivateKeyAsHex());
System.out.println("privkey wif :" + key.getPrivateKeyAsWiF(MainNetParams.get()));
   System.out.println("privkey int :" + key.getPrivKey());
   System.out.println("base58 :" + address.toBase58());
   System.out.println("string :" + address.toString());
   System.out.println("hex hash :" + Hex.encodeHexString(address.getHash()));
   //}
} catch (NoSuchAlgorithmException e) {
e.printStackTrace();
} catch (UnsupportedEncodingException e) {
   e.printStackTrace();
}

Meysa_richa

member

Activity: 458

Merit: 10

I think this is a very confusing way, but this also needs to be studied and I am also very happy with the code you provided

muhtesem

jr. member

Activity: 73

Merit: 1

Quote from: blockladder on January 12, 2019, 05:33:13 AM

Ok, here's a 1BTC puzzle.

The question is the following:

WhyTheCombOfNatashaOtomoskiHas21Teeth?.txt

The solution is a 32 characters long plain-text (the private key).

Hint: 8 camel case english words, no special symbols

Happy puzzling!

G/cbms/K/DNzcRin5v2B03iXdbpdVoZbTebt7KG95j3FUqnJvcP9rDYcGpSV27RLspR7SlPjqma4h0tDAMwovIo=

txid 39ae730abf9190f1985a3600e35b6451efc51bfb885bc9d1c3b91d502de3907d

I don't understand anything from this puzzle. Is it really able to get this bitcoin that finds the answer? if so I'll find 10 people and make them think for 1 month: D Are you going to give me some more clues, or is it just that? It's hard.

greg007r

newbie

Activity: 6

Merit: 1

Quote from: runlola on February 16, 2019, 01:57:13 PM

Quote from: greg007r on February 16, 2019, 08:53:23 AM

My script has already processed 1/10 of the permutations based on BIP39 word list, nothing
Another script has already processed 1/5 of the permutations based on the transcript of the https://www.youtube.com/watch?v=AABkJ55Zz3A page, nothing

considering there are 2048^8 permutations of the bip39 list, i doubt that. The number of valid possibilities using that list is ~2000x smaller (161,735,049,399,563,309,103,078 - see this post for how it was derived). my program can check 100K phrases/s and it would take ~5B years to check 10% of those.

Yes, you are right but you can downsize drastically the amount of possible permutations by :

- Keeping 1/10 of the BIP39 list (from 2048 to 230), removing all not related words (eg: Drift,Fish and Fiscal) and keeping interrsting ones (eg : satoshi, million,coin,supply, tooth, one ...)
- Preventing permutations where two words are duplicated (eg: ThisThisOneToothThisThisThisThis symptom)
- Allowing only phrase of 8 words having 32 chars
- Preventing permutations where no more than 4 instances of a word having 4 letters
- For those who believe in the sequence W T C O N O H T would be even faster to check
- many more

Sample extract of last 10 lines of my logs :

Sat Feb 16 19:51:11 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 FixUseWorthWillMillionCoinAllFit : 12YVLipqmKfPwNwoy1Wpemja89Jf2q1vxh
Sat Feb 16 19:51:11 CET 2019 : 346/4830|3 3 6 4 5 3 5 3 FitAllAnswerCoinWhaleUseMatchOne : 1CRgZoDGdTRfF679hdKXyZgbGuEMPaZXyF
Sat Feb 16 19:51:59 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 OneCanFinalOnlySegmentThisAllFit : 1Fug6cBx487thwS1Zp5nSCJAaAJxWfXrqL
Sat Feb 16 19:52:47 CET 2019 : 346/4830|3 3 5 4 7 4 3 3 OneFixFinalCoinControlOnlyCanFit : 1PKHseF7iqNWMv5At1YdCDVCvUW9C9ados
Sat Feb 16 19:53:34 CET 2019 : 346/4830|3 3 6 4 5 3 5 3 CanFitExcessGoldWhaleAllThereOne : 12MdRtKqKsT8Tw2AyR6JZhmXL8ZJcsXGVw
Sat Feb 16 19:54:22 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 WantSoonHaveOneUseThisFinalExact : 17bcNocTMuHksCbRVRcMiLMMLvaoTMGdSm
Sat Feb 16 19:54:22 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 ThisFixCanWorthAllWhaleGoldThere : 1GwNvXvrbxgmmQY27YL4gwcjjJEyX42LKw
Sat Feb 16 19:55:10 CET 2019 : 347/4830|3 3 5 4 7 4 3 3 AllCanBuildWantControlHaveOneFit : 17azNQ3Z5GqpvrEgo4YemT9Rc7DCBnRi6n
Sat Feb 16 19:55:58 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 WantUseOneTokenCanMimicLikeThere : 17oWja4HkduPeEsQSWC1bgdSLubdnT5pwk
Sat Feb 16 19:55:58 CET 2019 : 347/4830|3 4 4 3 4 3 3 8 UseLikeWaitCanWantFixAllProperty : 17uRQ63APaDNncu2cmKXiQsjff3kNKnXWB
Sat Feb 16 19:56:46 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 LikeHaveWaitAllUseWillLimitExact : 1YuVK4QA2K44sycmF3hDFFy8DCi6ne7sF
Sat Feb 16 19:57:33 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 SoonUseFitBuildCanWhaleOnlyThere : 16PMF61M6G8SQrXVAmgcRCZP2CcmsksnxC
Sat Feb 16 19:58:21 CET 2019 : 347/4830|3 4 4 3 4 3 3 8 CanWillCoinFixWaitUseAllRemember : 1JqCymN2wdVKM1zJUaiTgUTZPwkNXb1mVD
Sat Feb 16 19:59:09 CET 2019 : 347/4830|4 4 4 3 3 4 5 5 SoonThatOnlyAllOneLikeMimicExact : 15rrST9vDQhBwqE6XTcsUfixmZFDMVXpri
Sat Feb 16 19:59:57 CET 2019 : 347/4830|4 3 3 5 3 5 4 5 OnlyCanOneWorthFixExactThatThere : 1DiHHkHMTaQkSsbtV6To7YWF4LWuT329j
Sat Feb 16 20:00:45 CET 2019 : 348/4830|4 4 4 3 3 4 5 5 ThatGoldViewOneAllHaveProofExact : 18DWm63gZ9gYUS5ZnRP7Hp3J3tJGQps3yY
Sat Feb 16 20:01:29 CET 2019 : 348/4830|3 3 5 4 7 4 3 3 CanAllFinalThisSatisfyHaveOneFix : 1MaAgoj6AJ1T4NHoZLM3g5EohWwJ2n5MR2

Hope this help,

runlola

newbie

Activity: 24

Merit: 2

Quote from: greg007r on February 16, 2019, 08:53:23 AM

My script has already processed 1/10 of the permutations based on BIP39 word list, nothing
Another script has already processed 1/5 of the permutations based on the transcript of the https://www.youtube.com/watch?v=AABkJ55Zz3A page, nothing

considering there are 2048^8 permutations of the bip39 list, i doubt that. The number of valid possibilities using that list is ~2000x smaller (161,735,049,399,563,309,103,078 - see this post for how it was derived). my program can check 100K phrases/s and it would take ~5B years to check 10% of those.

greg007r

newbie

Activity: 6

Merit: 1

This is my best guess so far with no luck :

OneToothFitForEveryMillionOfCoin
21MillionCoinsIsTheAsymptoticCap
ItWillNeverReach21MillionOfCoins
1ToothAMillionCoinFitSatoshiView

My script has already processed 1/10 of the permutations based on BIP39 word list, nothing
Another script has already processed 1/5 of the permutations based on the transcript of the https://www.youtube.com/watch?v=AABkJ55Zz3A page, nothing

I think I will forgive shortly

Good luck to all

17Zgna852tADxchdWuAsqXezpbW5ByKReP

iq_armando

full member

Activity: 379

Merit: 112

Tips: 3DhgXE1BedBJY6uxjxai3Nsaj8sXGU4ite

Quote from: fecell on February 15, 2019, 12:18:07 AM

Quote from: iq_armando on February 11, 2019, 11:41:15 AM

I wrote Natasha Otomoski on youtube and this single video came out. I do not know if it helps solve the puzzle or not. Grin

https://www.youtube.com/watch?v=AABkJ55Zz3A&t=11s

I wrote Natasha Otomoski on youtube and... many videos was found.

I wrote Natasha Otomoski again on YouTube and this time I also saw many videos, but most of them, if not, they are all about Bitcoin and about 21 million.

fecell

jr. member

Activity: 137

Merit: 2

Quote from: iq_armando on February 11, 2019, 11:41:15 AM

I wrote Natasha Otomoski on youtube and this single video came out. I do not know if it helps solve the puzzle or not. Grin

https://www.youtube.com/watch?v=AABkJ55Zz3A&t=11s

I wrote Natasha Otomoski on youtube and... many videos was found.

marciks

jr. member

Activity: 108

Merit: 1

Quote from: iq_armando on February 11, 2019, 11:41:15 AM

I wrote Natasha Otomoski on youtube and this single video came out. I do not know if it helps solve the puzzle or not. Grin

https://www.youtube.com/watch?v=AABkJ55Zz3A&t=11s

Coincidence? lol.. I dont think so...

seoincorporation

legendary

Activity: 3346

Merit: 3130

Quote from: BitThinker42 on February 11, 2019, 12:27:46 PM

Quote from: seoincorporation on February 11, 2019, 07:35:25 AM

...

wow! it is a brilliant way to reduce number of combinations

Thanks but wasn't enough to get the right answer. At least i had developed a system to make brainwallet brute force to text files, maybe the code is useful for someone to solve this puzzle...

Code:

#Code in Unix BashScript
rm -rf bfResult.txt bfClean.txt;
cat bf.txt | tr -dc '[:alnum:]\n\r''\n ' | tr -d [0-9] >> bfClean.txt
for y in {a..z};
do
sed -i "s/ $y/ $(echo "$y" | tr '[:lower:]' '[:upper:]')/g" "bfClean.txt"
done
z1=$(cat bfClean.txt | wc -w)
z=$(echo "$z1-8" | bc)
for a in $(seq 1 $z)
do
c=$(echo "$a+7" | bc)
b=$(cat bfClean.txt | cut -d " " -f$a-$c)
if [ "$(echo "$b" | wc -m)" = "40" ]; #40 because is the 32+spaces
then
echo $b;
echo $b >> bfResult.txt;
fi
done

By the way, i think Op should drop another hint. Grin

PlusOne88

member

Activity: 420

Merit: 10

“Tackling Climate Change Using Blockchain”

Post like these is somehow interesting enough to hook up somebody into thinking. Its good to keep the brain working sometimes. Its just enjoying, hope everyone else feels the same way too. By the way its fun to see how one forum member searched a photo with 21 teeth but failed because 20 teeth was only seen. Very nice and funny effort. OP maybe you can give some of your bitcoin to him? Just joking though!

Trinitas

newbie

Activity: 6

Merit: 0

If it is a puzzle
Then we only need these characters "WhyTheCombOfNatashaOtomoskiHas21Teethtxt"

If it is a riddle
it can be any character and I don't think that's possible to solve

If it's both (what I think) then we need
WhyTheCombOfNatashaOtomoskiHas21Teethtxt
To be 32 chars example:
hyheombfatashatomoskias21eethtxt
Then make words out of that

BitThinker42

newbie

Activity: 19

Merit: 2

Quote from: seoincorporation on February 11, 2019, 07:35:25 AM

Quote from: BitThinker42 on February 11, 2019, 12:11:43 AM

Quote from: seoincorporation on February 10, 2019, 02:56:27 PM

Isn't an easy puzzle...

Quote

WhyTheCombOfNatashaOtomoskiHas21Teeth?.txt

NatashaOtomoski > Satoshi Nakamoto...
21 > Total bitcoin supply in millions
.txt > Satoshi White paper?
Teeth > Huh

I think the answer of this puzzle is in the bitcoin white paper... i will try to brute force that.

brute forcing is not good way for solving this puzzle. assume the white paper has 100 distinct words, total number of combinations is: 100! / (100-8)! = 7.50306389818e+15 . assume you have a good computer and it can compute 1000 combinations per second, total time of computing will be: 7.50306389818e+12 seconds= 237920 years.

I had in mind another kind of brute force. What i did was:

Code:

a1=(words count in wp.txt)
a=(a1 - 8)
for b in (sequence 1 to $a)
do
take words from $b to $b+8, if the sum is 32 char then save it on a file
done

And what i get was this:

Code:

OfWorkFormingARecordThatCannotBe
ButProofThatItCameFromTheLargest
CameFromTheLargestPoolOfCPUPower
TheLargestPoolOfCPUPowerAslongAs
PowerAslongAsAMajorityOfCPUPower
OfWorkChainAsProofOfWhatHappened
OnTheInternetHasComeToRelyAlmost
DefineAnElectronicCoinAsAChainOf
AndThePublicKeyOfTheNextOwnerand
OwnerandAddingTheseToTheEndOfThe
OfCourseIsThePayeeCantVerifyThat
ReturnedToTheMintToissueANewCoin
SolutionIsThatTheFateOfTheEntire
GoThroughThemJustLikeABankWeNeed
ForThePayeeToKnowThatThePrevious
ToDoubleSpendTheOnlyWayToConfirm
TheMintBasedModelTheMintWasAware
ASystemForParticipantsToAgreeOnA
ServerWorksByTakingAhashOfABlock
ObviouslyInOrderToGetIntoTheHash
ValueThatWhenHashedSuchAsWithSHA
HashedSuchAsWithSHAThehashBegins
InTheblockUntilAValueIsFoundThat
IsFoundThatGivesTheBlocksHashThe
BlocksAreChainedAfterItTheWorkTo
AreChainedAfterItTheWorkToChange
ChainedAfterItTheWorkToChangeThe
OfWorkIsEssentiallyOneCPUOneVote
ProofOfWorkEffortInvestedinItIfA
IntoABlockEachNodeWorksOnFinding
FindsAProofOfWorkItBroadcastsThe
AProofOfWorkItBroadcastsTheBlock
IfAllTransactionsInItAreValidAnd
OfTheBlockByWorkingOnCreatingThe
BlockInThechainUsingTheHashOfThe
UsingTheHashOfTheAcceptedBlockAs
TheLongestChainToBeTheCorrectOne
LongestChainToBeTheCorrectOneAnd
OfWorkIsFoundAndOneBranchBecomes
NodesTheyWillGetIntoABlockBefore
WillGetIntoABlockBeforeLongBlock
ReceiveABlockItWillRequestItWhen
ItWillRequestItWhenItReceivesThe
OfTheBlockThisAddsAnIncentiveFor
AdditionOfAConstantOfAmountOfNew
OurCaseItIsCPUTimeAndElectricity
HelpEncourageNodesToStayHonestIf
ThanAllTheHonestNodesHeWouldHave
UsingItToGenerateNewCoinsHeOught
ItMoreProfitableToPlayByTheRules
ByTheRulesSuchRulesThatFavourHim
InACoinIsBuriedUnderEnoughBlocks
TreeWithOnlyTheRootIncludedInThe
NeedToBeStoredABlockHeaderWithNo
StorageShouldNotBeAProblemEvenIf
RunningAFullNetworkNodeAUserOnly
NetworkNodeAUserOnlyNeedsToKeepa
HeHasTheLongestChainAndObtainThe
BlockItsTimestampedInHeCantCheck
ForhimselfButByLinkingItToAPlace
ByAnAttackerWhileNetworkNodesCan
AsLongAsTheAttackerCanContinueTo
OneReturningTheChangeIfAnyBackTo
ReturningTheChangeIfAnyBackToThe
ProblemHereThereIsNeverTheNeedTo
HereThereIsNeverTheNeedToExtract
InputsWereOwnedByTheSameOwnerThe
WereOwnedByTheSameOwnerTheRiskis
OwnedByTheSameOwnerTheRiskisThat
SameOwnerTheRiskisThatIfTheOwner
ThemAnAttackerCanOnlyTryToChange
ByOneBlockIncreasingItsleadByAnd
AndTheFailureEventIsTheAttackers
OfAnAttackerCatchingUpFromAGiven
UpFromAGivenDeficitIsAnalogousTo
TrialsToTryToReachBreakevenWeCan
HasToCatchUpWithIncreasesWithThe
TheRecipientBelieveHePaidHimForA
OfBlocksAheadOfTimeByWorkingOnit
IsLuckyEnoughToGetFarEnoughAhead
AndBlocksHaveBeenlinkedAfterItHe
BeenlinkedAfterItHeDoesntKnowThe
ToSolveThisWeproposedAPeerToPeer
NetworkUsingProofOfWorkToRecordA
TheyDoNotNeedToBeIdentifiedSince
TheyWereGoneTheyVoteWithTheirCPU

I test those phrases as brainwallets but itdoesn't work Sad

wow! it is a brilliant way to reduce number of combinations

iq_armando

full member

Activity: 379

Merit: 112

Tips: 3DhgXE1BedBJY6uxjxai3Nsaj8sXGU4ite

I wrote Natasha Otomoski on youtube and this single video came out. I do not know if it helps solve the puzzle or not. Grin

https://www.youtube.com/watch?v=AABkJ55Zz3A&t=11s

qbvbsite

newbie

Activity: 6

Merit: 0

Quote from: ?? on ??

Quote from: qbvbsite on February 11, 2019, 10:00:38 AM

There are certain words that have a low probability of being used. using possible hints from the puzzle, among other things.

"low probability" is probability, not help and you think it is "low probability" but OP doesnt think so. Yes after use "possible hints" i see billion billion billion of these sentences. What are "other things"?

Other things are what I'm trying out. Can't give away all the hints Wink

Also, a few billion is nothing and quite easy to test for brain wallets.

qbvbsite

newbie

Activity: 6

Merit: 0

Quote from: ?? on ??

Quote from: qbvbsite on February 11, 2019, 09:50:41 AM

there are many ways of brute forcing that limit the number of results, so don't count it out.

tell me just one way of brute forcing that limit the number of results for this puzzle?

There are certain words that have a low probability of being used. using possible hints from the puzzle, among other things.

qbvbsite

newbie

Activity: 6

Merit: 0

Quote from: ?? on ??

Quote from: seoincorporation on February 11, 2019, 07:35:25 AM

Quote from: BitThinker42 on February 11, 2019, 12:11:43 AM

Quote from: seoincorporation on February 10, 2019, 02:56:27 PM

Isn't an easy puzzle...

Quote

WhyTheCombOfNatashaOtomoskiHas21Teeth?.txt

NatashaOtomoski > Satoshi Nakamoto...
21 > Total bitcoin supply in millions
.txt > Satoshi White paper?
Teeth > Huh

I think the answer of this puzzle is in the bitcoin white paper... i will try to brute force that.

brute forcing is not good way for solving this puzzle. assume the white paper has 100 distinct words, total number of combinations is: 100! / (100-8)! = 7.50306389818e+15 . assume you have a good computer and it can compute 1000 combinations per second, total time of computing will be: 7.50306389818e+12 seconds= 237920 years.

I had in mind another kind of brute force. What i did was:

Code:

a1=(words count in wp.txt)
a=(a1 - 8)
for b in (sequence 1 to $a)
do
take words from $b to $b+8, if the sum is 32 char then save it on a file
done

And what i get was this:

Code:

OfWorkFormingARecordThatCannotBe
ButProofThatItCameFromTheLargest
CameFromTheLargestPoolOfCPUPower
TheLargestPoolOfCPUPowerAslongAs
PowerAslongAsAMajorityOfCPUPower
OfWorkChainAsProofOfWhatHappened
OnTheInternetHasComeToRelyAlmost
DefineAnElectronicCoinAsAChainOf
AndThePublicKeyOfTheNextOwnerand
OwnerandAddingTheseToTheEndOfThe
OfCourseIsThePayeeCantVerifyThat
ReturnedToTheMintToissueANewCoin
SolutionIsThatTheFateOfTheEntire
GoThroughThemJustLikeABankWeNeed
ForThePayeeToKnowThatThePrevious
ToDoubleSpendTheOnlyWayToConfirm
TheMintBasedModelTheMintWasAware
ASystemForParticipantsToAgreeOnA
ServerWorksByTakingAhashOfABlock
ObviouslyInOrderToGetIntoTheHash
ValueThatWhenHashedSuchAsWithSHA
HashedSuchAsWithSHAThehashBegins
InTheblockUntilAValueIsFoundThat
IsFoundThatGivesTheBlocksHashThe
BlocksAreChainedAfterItTheWorkTo
AreChainedAfterItTheWorkToChange
ChainedAfterItTheWorkToChangeThe
OfWorkIsEssentiallyOneCPUOneVote
ProofOfWorkEffortInvestedinItIfA
IntoABlockEachNodeWorksOnFinding
FindsAProofOfWorkItBroadcastsThe
AProofOfWorkItBroadcastsTheBlock
IfAllTransactionsInItAreValidAnd
OfTheBlockByWorkingOnCreatingThe
BlockInThechainUsingTheHashOfThe
UsingTheHashOfTheAcceptedBlockAs
TheLongestChainToBeTheCorrectOne
LongestChainToBeTheCorrectOneAnd
OfWorkIsFoundAndOneBranchBecomes
NodesTheyWillGetIntoABlockBefore
WillGetIntoABlockBeforeLongBlock
ReceiveABlockItWillRequestItWhen
ItWillRequestItWhenItReceivesThe
OfTheBlockThisAddsAnIncentiveFor
AdditionOfAConstantOfAmountOfNew
OurCaseItIsCPUTimeAndElectricity
HelpEncourageNodesToStayHonestIf
ThanAllTheHonestNodesHeWouldHave
UsingItToGenerateNewCoinsHeOught
ItMoreProfitableToPlayByTheRules
ByTheRulesSuchRulesThatFavourHim
InACoinIsBuriedUnderEnoughBlocks
TreeWithOnlyTheRootIncludedInThe
NeedToBeStoredABlockHeaderWithNo
StorageShouldNotBeAProblemEvenIf
RunningAFullNetworkNodeAUserOnly
NetworkNodeAUserOnlyNeedsToKeepa
HeHasTheLongestChainAndObtainThe
BlockItsTimestampedInHeCantCheck
ForhimselfButByLinkingItToAPlace
ByAnAttackerWhileNetworkNodesCan
AsLongAsTheAttackerCanContinueTo
OneReturningTheChangeIfAnyBackTo
ReturningTheChangeIfAnyBackToThe
ProblemHereThereIsNeverTheNeedTo
HereThereIsNeverTheNeedToExtract
InputsWereOwnedByTheSameOwnerThe
WereOwnedByTheSameOwnerTheRiskis
OwnedByTheSameOwnerTheRiskisThat
SameOwnerTheRiskisThatIfTheOwner
ThemAnAttackerCanOnlyTryToChange
ByOneBlockIncreasingItsleadByAnd
AndTheFailureEventIsTheAttackers
OfAnAttackerCatchingUpFromAGiven
UpFromAGivenDeficitIsAnalogousTo
TrialsToTryToReachBreakevenWeCan
HasToCatchUpWithIncreasesWithThe
TheRecipientBelieveHePaidHimForA
OfBlocksAheadOfTimeByWorkingOnit
IsLuckyEnoughToGetFarEnoughAhead
AndBlocksHaveBeenlinkedAfterItHe
BeenlinkedAfterItHeDoesntKnowThe
ToSolveThisWeproposedAPeerToPeer
NetworkUsingProofOfWorkToRecordA
TheyDoNotNeedToBeIdentifiedSince
TheyWereGoneTheyVoteWithTheirCPU

I test those phrases as brainwallets but itdoesn't work Sad

https://bitcointalksearch.org/topic/ok-heres-a-1btc-puzzle-5096267

there are billion billion billion of these sentences. This puzzle has no logic, just solve by guessing. So nobody can solve it. OP should solve it or should close this topic to save time of everybody

there are many ways of brute forcing that limit the number of results, so don't count it out.

Topic: Ok, here's a 1BTC puzzle. - page 12. (Read 14685 times)