stevewig
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February 11, 2019, 01:01:50 PM |
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Good work.
Some of those phrases aren’t CamelCase though. Not criticising, just an observation
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AB de Royse777
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Campaign Manager. My Telegram @Royse777
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February 11, 2019, 01:08:57 PM |
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-----BEGIN BITCOIN SIGNED MESSAGE----- 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! -----BEGIN SIGNATURE----- 179sxfh6rw6bHSo5wVUhLP96k46QaEzVP G/cbms/K/DNzcRin5v2B03iXdbpdVoZbTebt7KG95j3FUqnJvcP9rDYcGpSV27RLspR7SlPjqma4h0tDAMwovIo= -----END BITCOIN SIGNED MESSAGE----- The coin is still in the address which means no one could not solve the puzzle hehe I am getting an inspiration to learn about the coding side of Bitcoin I think. I can not guarantee anything coz I easily give up LOL Have fun guys. Cheers :-)
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..Stake.com.. | | | ▄████████████████████████████████████▄ ██ ▄▄▄▄▄▄▄▄▄▄ ▄▄▄▄▄▄▄▄▄▄ ██ ▄████▄ ██ ▀▀▀▀▀▀▀▀▀▀ ██████████ ▀▀▀▀▀▀▀▀▀▀ ██ ██████ ██ ██████████ ██ ██ ██████████ ██ ▀██▀ ██ ██ ██ ██████ ██ ██ ██ ██ ██ ██ ██████ ██ █████ ███ ██████ ██ ████▄ ██ ██ █████ ███ ████ ████ █████ ███ ████████ ██ ████ ████ ██████████ ████ ████ ████▀ ██ ██████████ ▄▄▄▄▄▄▄▄▄▄ ██████████ ██ ██ ▀▀▀▀▀▀▀▀▀▀ ██ ▀█████████▀ ▄████████████▄ ▀█████████▀ ▄▄▄▄▄▄▄▄▄▄▄▄███ ██ ██ ███▄▄▄▄▄▄▄▄▄▄▄▄ ██████████████████████████████████████████ | | | | | | ▄▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▄ █ ▄▀▄ █▀▀█▀▄▄ █ █▀█ █ ▐ ▐▌ █ ▄██▄ █ ▌ █ █ ▄██████▄ █ ▌ ▐▌ █ ██████████ █ ▐ █ █ ▐██████████▌ █ ▐ ▐▌ █ ▀▀██████▀▀ █ ▌ █ █ ▄▄▄██▄▄▄ █ ▌▐▌ █ █▐ █ █ █▐▐▌ █ █▐█ ▀▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄▀█ | | | | | | ▄▄█████████▄▄ ▄██▀▀▀▀█████▀▀▀▀██▄ ▄█▀ ▐█▌ ▀█▄ ██ ▐█▌ ██ ████▄ ▄█████▄ ▄████ ████████▄███████████▄████████ ███▀ █████████████ ▀███ ██ ███████████ ██ ▀█▄ █████████ ▄█▀ ▀█▄ ▄██▀▀▀▀▀▀▀██▄ ▄▄▄█▀ ▀███████ ███████▀ ▀█████▄ ▄█████▀ ▀▀▀███▄▄▄███▀▀▀ | | | ..PLAY NOW.. |
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qbvbsite
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February 11, 2019, 02:50:41 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 https://bitcointalk.org/index.php?topic=5096267.msg49599982#msg49599982there 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.
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qbvbsite
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February 11, 2019, 03:00:38 PM |
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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.
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qbvbsite
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February 11, 2019, 03:19:59 PM |
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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 Also, a few billion is nothing and quite easy to test for brain wallets.
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iq_armando
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Tips: 3DhgXE1BedBJY6uxjxai3Nsaj8sXGU4ite
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February 11, 2019, 04:41:15 PM |
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I wrote Natasha Otomoski on youtube and this single video came out. I do not know if it helps solve the puzzle or not. https://www.youtube.com/watch?v=AABkJ55Zz3A&t=11s
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BitThinker42
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February 11, 2019, 05:27:46 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 wow! it is a brilliant way to reduce number of combinations
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Trinitas
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February 12, 2019, 07:33:59 AM Last edit: February 12, 2019, 08:24:03 AM by Trinitas |
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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
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PlusOne88
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“Tackling Climate Change Using Blockchain”
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February 12, 2019, 02:01:18 PM |
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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!
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seoincorporation
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February 12, 2019, 03:02:47 PM |
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...
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 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.
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marciks
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February 14, 2019, 10:17:30 PM |
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Coincidence? lol.. I dont think so...
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fecell
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February 15, 2019, 05:18:07 AM |
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I wrote Natasha Otomoski on youtube and... many videos was found.
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iq_armando
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February 15, 2019, 01:25:01 PM |
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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.
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greg007r
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February 16, 2019, 01:53:23 PM Last edit: February 16, 2019, 02:12:27 PM by greg007r |
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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
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runlola
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February 16, 2019, 06:57:13 PM |
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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.
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greg007r
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February 16, 2019, 07:21:41 PM Last edit: February 16, 2019, 08:00:23 PM by greg007r |
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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,
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muhtesem
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February 16, 2019, 09:17:18 PM |
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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.
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🇩🇪 🇩🇪 🇩🇪 𝗔𝗟𝗖𝗘𝗗𝗢 🇩🇪 🇩🇪 🇩🇪 Making Cryptocurrencies Accessible For Everyone! http://alcedoplatform.com
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Meysa_richa
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February 16, 2019, 11:29:26 PM |
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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
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greg007r
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February 16, 2019, 11:48:33 PM |
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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 : 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(); }
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greg007r
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Activity: 6
Merit: 1
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February 17, 2019, 12:25:08 AM Last edit: February 17, 2019, 01:19:34 AM by greg007r Merited by vapourminer (1) |
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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 : 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.pngAnd 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 !
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