stevewig
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February 05, 2019, 09:07:01 PM |
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To save writing your own code, just run brainwallet offline. That’s what I do. Easy & saves re-inventing the wheel.
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iq_armando
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Tips: 3DhgXE1BedBJY6uxjxai3Nsaj8sXGU4ite
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February 05, 2019, 09:55:59 PM |
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I meant to solve the puzzle
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stevewig
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February 05, 2019, 10:01:19 PM |
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Basically, you supply brainflayer with a word list, and the public key of the wallet.
If any of those words in the list, match any of the supplied public keys, it tells you which word (or phrase) is a match.
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bulleteyedk
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February 05, 2019, 10:02:26 PM |
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Another 230 million passwords tried using the brainflayer tool, using the tool really don't take up too much time, compiling the wordlists is another matter, haha...
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BitThinker42
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February 06, 2019, 07:24:18 AM |
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I think you are in the wrong way. the answer of this puzzle in not a brainwallet. let me explain: each 64 byte hexadecimal number can be a bitcoin private key. for generating a bitcoin private key from a 32 character plain text we have 2 options: option 1: a brainwallet: we calculate sha256 hash of text. the result is 64 byte hexadecimal number. in fact result of calculation of hsa256 of any text with any length always is a 64 byte hex. then in this option the length of text is not necessary be 32 character. option 2: we convert each letter to its ascii code. the result of converting each English letter to its ascii code is a 2 byte hexadecimal number. then converting a 32 character text to its ascii code is a 64 byte hexadecimal number. according to the OP post I think the option 2 is the right way.
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bulleteyedk
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February 06, 2019, 07:43:28 AM |
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Another 230 million passwords tried using the brainflayer tool, using the tool really don't take up too much time, compiling the wordlists is another matter, haha... I think you are in the wrong way. the answer of this puzzle in not a brainwallet. let me explain: each 64 byte hexadecimal number can be a bitcoin private key. for generating a bitcoin private key from a 32 character plain text we have 2 options: option 1: a brainwallet: we calculate sha256 hash of text. the result is 64 byte hexadecimal number. in fact result of calculation of hsa256 of any text with any length always is a 64 byte hex. then in this option the length of text is not necessary be 32 character. option 2: we convert each letter to its ascii code. the result of converting each English letter to its ascii code is a 2 byte hexadecimal number. then converting a 32 character text to its ascii code is a 64 byte hexadecimal number. according to the OP post I think the option 2 is the right way. This is actually a pretty good explanation, i can see what you mean - and i think you are right, that being said, it's until now been pretty interesting and learning for me to dig into this puzzle. Is there a way to automate the process of testing larger "option 2" possibilities?
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stevewig
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February 06, 2019, 09:38:56 AM |
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Useful sites https://www.browserling.com/tools/text-to-asciithen this formula in excel (each number in its own cell). =DEC2HEX(A1) Don't have time to look into the automation at the moment unfortunately
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bitcoin_collector
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February 06, 2019, 11:29:57 AM |
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I think you are in the wrong way. the answer of this puzzle in not a brainwallet. let me explain: each 64 byte hexadecimal number can be a bitcoin private key. for generating a bitcoin private key from a 32 character plain text we have 2 options: option 1: a brainwallet: we calculate sha256 hash of text. the result is 64 byte hexadecimal number. in fact result of calculation of hsa256 of any text with any length always is a 64 byte hex. then in this option the length of text is not necessary be 32 character. option 2: we convert each letter to its ascii code. the result of converting each English letter to its ascii code is a 2 byte hexadecimal number. then converting a 32 character text to its ascii code is a 64 byte hexadecimal number. according to the OP post I think the option 2 is the right way. I believe it is option 2 as well. (Except I believe you meant 32 bytes, not 64. A private key is 256 bits which is 32 bytes, or 64 characters in the range 0-9 or A-F in hexadecimal.) It could also be sha256 of a .txt which could be formated in unicode/bigendian or utf8 (edit, but then the puzzle is not correct by saying the 32 characters string is the private key).
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BitThinker42
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February 06, 2019, 01:11:04 PM |
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I think you are in the wrong way. the answer of this puzzle in not a brainwallet. let me explain: each 64 byte hexadecimal number can be a bitcoin private key. for generating a bitcoin private key from a 32 character plain text we have 2 options: option 1: a brainwallet: we calculate sha256 hash of text. the result is 64 byte hexadecimal number. in fact result of calculation of hsa256 of any text with any length always is a 64 byte hex. then in this option the length of text is not necessary be 32 character. option 2: we convert each letter to its ascii code. the result of converting each English letter to its ascii code is a 2 byte hexadecimal number. then converting a 32 character text to its ascii code is a 64 byte hexadecimal number. according to the OP post I think the option 2 is the right way. I believe it is option 2 as well. (Except I believe you meant 32 bytes, not 64. A private key is 256 bits which is 32 bytes, or 64 characters in the range 0-9 or A-F in hexadecimal.) It could also be sha256 of a .txt which could be formated in unicode/bigendian or utf8 (edit, but then the puzzle is not correct by saying the 32 characters string is the private key). yes, you are right. a private key is 32 byte and 64 hexadecimal digit.
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BitThinker42
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February 06, 2019, 01:41:49 PM |
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I think you are in the wrong way. the answer of this puzzle in not a brainwallet. let me explain: each 64 byte hexadecimal number can be a bitcoin private key. for generating a bitcoin private key from a 32 character plain text we have 2 options: option 1: a brainwallet: we calculate sha256 hash of text. the result is 64 byte hexadecimal number. in fact result of calculation of hsa256 of any text with any length always is a 64 byte hex. then in this option the length of text is not necessary be 32 character. option 2: we convert each letter to its ascii code. the result of converting each English letter to its ascii code is a 2 byte hexadecimal number. then converting a 32 character text to its ascii code is a 64 byte hexadecimal number. according to the OP post I think the option 2 is the right way. This is actually a pretty good explanation, i can see what you mean - and i think you are right, that being said, it's until now been pretty interesting and learning for me to dig into this puzzle. Is there a way to automate the process of testing larger "option 2" possibilities? to doing that I've written a python script for myself. you can download it from this link. for running it you need Python 3.6 or newer and "blocksmith" module for Python. after installing Python open Command Pormpt and write: in next step you need a word list file. the structure of your file should be like this: each word in a new line. this script needs 8 copy of your word list file. copy your word list file eight times and past them into the scripts folder and rename them to "list1.txt", "list2.txt","list3.txt",..."list8.txt". then run the script. if you are a Windows user you can open it with IDLE (a simple python editor that installs with python) and press F5 button for run. it generates all combinations of 8 words and if the length of a combination be 32 character it generates its public key and if that public key be equal to this puzzle public key it shows a message that found right words.
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BitThinker42
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February 06, 2019, 07:19:06 PM |
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to doing that I've written a python script for myself. you can download it from this link. for running it you need Python 3.6 or newer and "blocksmith" module for Python. after installing Python open Command Pormpt and write: in next step you need a word list file. the structure of your file should be like this: each word in a new line. this script needs 8 copy of your word list file. copy your word list file eight times and past them into the scripts folder and rename them to "list1.txt", "list2.txt","list3.txt",..."list8.txt". then run the script. if you are a Windows user you can open it with IDLE (a simple python editor that installs with python) and press F5 button for run. it generates all combinations of 8 words and if the length of a combination be 32 character it generates its public key and if that public key be equal to this puzzle public key it shows a message that found right words. Generating public addresses to check this way seems too slow. Like 5 per second (if you feed it only 32 char-rs long strings instead of going though all of them) It generates public key just for 32 character strings, all other strings are omitted. I tested it on win 7 and ubuntu, on ubuntu it has better performance .Its slowness is for slowness of python language.
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pooya87
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February 07, 2019, 02:14:30 AM |
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option 1: a brainwallet: we calculate sha256 hash of text. the result is 64 byte hexadecimal number. in fact result of calculation of hsa256 of any text with any length always is a 64 byte hex. then in this option the length of text is not necessary be 32 character.
that is not the only way! there are literary hundreds of methods to use for brainwallet because there is no standard for it. for example you can perform multiple rounds of SHA256 on it. or this "password" you use can be the password of a PBKD2 with some iteration count and then derive a 32 byte result from it.
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BitThinker42
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February 07, 2019, 01:20:05 PM |
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option 1: a brainwallet: we calculate sha256 hash of text. the result is 64 byte hexadecimal number. in fact result of calculation of hsa256 of any text with any length always is a 64 byte hex. then in this option the length of text is not necessary be 32 character.
that is not the only way! there are literary hundreds of methods to use for brainwallet because there is no standard for it. for example you can perform multiple rounds of SHA256 on it. or this "password" you use can be the password of a PBKD2 with some iteration count and then derive a 32 byte result from it. yes, that isn't the only way, but all brinwallets use hashing to generate private key, some of them use it one time and some of them use it many time. I think the answer of puzzle is the private key straightforwardly, we don't need these indirect ways.
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stevewig
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February 09, 2019, 10:03:55 PM |
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Been bust last few days, I see its still not solved.
Anyone have any more idea?
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bitart
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February 09, 2019, 10:58:14 PM |
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Been bust last few days, I see its still not solved.
Anyone have any more idea?
If you have a promising idea, would you post it here before trying it out first? Just think about the possible solutions, as others, try them and you can still post those methods you have already tried without luck. And of course, keep an eye on the address if it still has the balance
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bulleteyedk
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February 10, 2019, 06:14:06 PM |
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I think we need a hint or two to get this puzzle moving - there is just too many possibilities for brute forcing the password, but maybe the answer is straight up a simple sentence of 8 words, it's a bit hard to figure out right now.
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seoincorporation
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February 10, 2019, 07:56:27 PM |
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Isn't an easy puzzle... WhyTheCombOfNatashaOtomoskiHas21Teeth?.txt NatashaOtomoski > Satoshi Nakamoto... 21 > Total bitcoin supply in millions .txt > Satoshi White paper? Teeth > I think the answer of this puzzle is in the bitcoin white paper... i will try to brute force that.
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BitThinker42
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February 11, 2019, 05:11:43 AM Merited by coinlocket$ (1) |
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Isn't an easy puzzle... WhyTheCombOfNatashaOtomoskiHas21Teeth?.txt NatashaOtomoski > Satoshi Nakamoto... 21 > Total bitcoin supply in millions .txt > Satoshi White paper? Teeth > 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.
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bulleteyedk
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February 11, 2019, 09:37:07 AM |
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Yeah, there is just too many possible combinations, if anyone is going to bruteforce this, we need some hints. Tried these without luck: IJustWonOneBitcoinForSolvingThis DidIJustWinOneBitcoinSolvingThis OnlyNatashaOHaveTheAnswerForThis YouReallyBelieveWeCanTellYouThis NoOneHaveTheAnswerToThisQuestion ThisPuzzleNeedTwoHintsToBeSolved Looked at the question once again, and maybe we should start looking into a "Natasha" in either a book, film og a game? it would make sense if that .txt really meant we should look at at something written. There is a book "Natasha" by David Bezmozgis, in the preview of this book, the surname of Natasha is not mentioned, i've come up emptyhanded looking for a pdf or a epub file of that particular book https://www.amazon.com/Natasha-Other-Stories-David-Bezmozgis/dp/0312423934If we are looking for a person in either a movie or a TV show, there is these 3: https://www.imdb.com/title/tt4563266/https://www.imdb.com/title/tt0453521/https://www.imdb.com/title/tt3866526/the last mentioned movie is based on the book No surname of the persons in these movies is mentioned, so they might be a lead to get this information about that comb? I have not been able to find any games matching a playable figure "Natasha"...
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seoincorporation
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February 11, 2019, 12:35:25 PM |
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Isn't an easy puzzle... WhyTheCombOfNatashaOtomoskiHas21Teeth?.txt NatashaOtomoski > Satoshi Nakamoto... 21 > Total bitcoin supply in millions .txt > Satoshi White paper? Teeth > 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: 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: 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
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