digaran
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June 03, 2023, 05:15:05 AM |
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Do people really think Satoshi is reading all the posts here every day? I don't know if anyone here is a giver, unfortunately we are here to take the loots.
Welcome to the jungle, and happy hunting.😉
Wow guys! I just discovered something, everytime I double a key I can get double the distance with my original base key!
I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit? A good thing about mod function in EC, you could keep halving for eternity without reaching destination, as we are moving towards bitcoin halving, we should start key halving to solve some puzzles before the halving event! 😉
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elvis13
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June 03, 2023, 07:02:15 AM |
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Digaran @ You can't offer anything useful. An endless stream of insanity.!
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vjudeu
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June 03, 2023, 08:02:33 AM |
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I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit? You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation. SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7 mask125=0000000000000000000000000000000020000000000000000000000000000000 (hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7 key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814 key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6 (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773 See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.
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batareyka
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June 03, 2023, 10:00:26 AM |
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digaran If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve. There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve. There are special numbers on the curve that can be used to get exactly 1 digit less. A third of the number you are looking for. A quarter of the sought number. You can get the mirror part of the number you are looking for, which can also be compared with the original. And this is not even a 10th part of what is there. You can get a lot.
But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.
I wish you success.
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digaran
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June 03, 2023, 12:55:01 PM |
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Digaran @ You can't offer anything useful. An endless stream of insanity.!
Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.
I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit? You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation. SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7 mask125=0000000000000000000000000000000020000000000000000000000000000000 (hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7 key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814 key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6 (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773 See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits. I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉
digaran If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve. There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve. There are special numbers on the curve that can be used to get exactly 1 digit less. A third of the number you are looking for. A quarter of the sought number. You can get the mirror part of the number you are looking for, which can also be compared with the original. And this is not even a 10th part of what is there. You can get a lot.
But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.
I wish you success.
Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle. I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods. Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣
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lordfrs
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June 03, 2023, 11:36:16 PM |
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Digaran @ You can't offer anything useful. An endless stream of insanity.!
Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.
I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit? You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation. SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7 mask125=0000000000000000000000000000000020000000000000000000000000000000 (hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7 key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814 key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6 (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773 See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits. I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉
digaran If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve. There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve. There are special numbers on the curve that can be used to get exactly 1 digit less. A third of the number you are looking for. A quarter of the sought number. You can get the mirror part of the number you are looking for, which can also be compared with the original. And this is not even a 10th part of what is there. You can get a lot.
But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.
I wish you success.
Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle. I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods. Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣 My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm ![Smiley](https://bitcointalk.org/Smileys/default/smiley.gif) . I don't like working, my brain is almost numb from the blood pressure pills i can't think I'm sorry my English is bad, I'm writing from translate, but I hope you understand what I mean.
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If you want to buy me a coffee
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digaran
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June 04, 2023, 01:35:32 PM |
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My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm ![Smiley](https://bitcointalk.org/Smileys/default/smiley.gif) . If your algorithm is working for one key, it should work for all other keys, unless it's not working, if you can solve a known key and not any other key, you are doing things in a wrong way. Manage to find a key which you don't know it's private key, then you can claim success. Hope you feel better.
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Gianluca95
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June 04, 2023, 04:40:44 PM |
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I think that there isn't any way to find a solution for this puzzle; the only thing to solve this is the strength of community. It requires many and many power of hash/sec to be solved and with maths you can't do many things.
It would be cool to use something with a speed of an ASIC, in this way you'll be able to discover keys from 66 to 76 in very easy way !
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Denis_Hitov
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June 04, 2023, 06:31:09 PM |
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Digaran @ You can't offer anything useful. An endless stream of insanity.!
Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.
I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit? You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation. SHA-256("125-bit")=c383a1ae19ff4401f72fdfbb4ffeb6fc7a38c6692b07b188edcbdc31b0160ee7 mask125=0000000000000000000000000000000020000000000000000000000000000000 (hash%mask125)=key=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee7 key/2=7fffffffffffffffffffffffffffffff6a73d1a7ed2828e256cf1d5f40262814 key-1=000000000000000000000000000000001a38c6692b07b188edcbdc31b0160ee6 (key-1)/2=000000000000000000000000000000000d1c63349583d8c476e5ee18d80b0773 See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits. I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉
digaran If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve. There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve. There are special numbers on the curve that can be used to get exactly 1 digit less. A third of the number you are looking for. A quarter of the sought number. You can get the mirror part of the number you are looking for, which can also be compared with the original. And this is not even a 10th part of what is there. You can get a lot.
But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.
I wish you success.
Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle. I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods. Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣 My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm ![Smiley](https://bitcointalk.org/Smileys/default/smiley.gif) . I don't like working, my brain is almost numb from the blood pressure pills i can't think I'm sorry my English is bad, I'm writing from translate, but I hope you understand what I mean. Hello. Could you share your algorithm? I wonder how you managed this: "Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds". Sincerely, be healthy.
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GR Sasa
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June 04, 2023, 08:02:54 PM |
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fake
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albert0bsd
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June 05, 2023, 12:38:20 AM |
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Hello. Could you share your algorithm? I wonder how you managed this: "Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds".
Sincerely, be healthy.
Don't believe in that, it is fake. fake
Agree. Why from 115 to 28 bits? in that case why not up to 1 bit? Guys please don't believe in that, in any case we only need to know one bit on the right side to break ECDSA
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lordfrs
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June 05, 2023, 09:30:23 PM |
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Hello. Could you share your algorithm? I wonder how you managed this: "Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds".
Sincerely, be healthy.
Don't believe in that, it is fake. fake
Agree. Why from 115 to 28 bits? in that case why not up to 1 bit? Guys please don't believe in that, in any case we only need to know one bit on the right side to break ECDSA Private Key Public key 1090246098153987172547740458951748 # puzzle 110 0309976ba5570966bf889196b7fdf5a0f9a1e9ab340556ec29f8bb60599616167d 545123049076993586273870229475874 272561524538496793136935114737937 136280762269248396568467557368968 68140381134624198284233778684484 34070190567312099142116889342242 17035095283656049571058444671121 8517547641828024785529222335560 4258773820914012392764611167780 2129386910457006196382305583890 1064693455228503098191152791945 532346727614251549095576395972 266173363807125774547788197986 133086681903562887273894098993 66543340951781443636947049496 33271670475890721818473524748 16635835237945360909236762374 8317917618972680454618381187 4158958809486340227309190593 2079479404743170113654595296 1039739702371585056827297648 519869851185792528413648824 259934925592896264206824412 129967462796448132103412206 64983731398224066051706103 32491865699112033025853051 16245932849556016512926525 8122966424778008256463262 4061483212389004128231631 2030741606194502064115815 1015370803097251032057907 507685401548625516028953 253842700774312758014476 126921350387156379007238 63460675193578189503619 31730337596789094751809 15865168798394547375904 7932584399197273687952 3966292199598636843976 1983146099799318421988 991573049899659210994 495786524949829605497 247893262474914802748 123946631237457401374 61973315618728700687 30986657809364350343 15493328904682175171 7746664452341087585 3873332226170543792 1936666113085271896 968333056542635948 484166528271317974 242083264135658987 121041632067829493 60520816033914746 30260408016957373 15130204008478686 7565102004239343 3782551002119671 1891275501059835 945637750529917 472818875264958 236409437632479 118204718816239 59102359408119 29551179704059 14775589852029 7387794926014 3693897463007 1846948731503 923474365751 461737182875 230868591437 115434295718 57717147859 28858573929 14429286964 7214643482 3607321741 1803660870 901830435 450915217 225457608 112728804 56364402 28182201 14091100 7045550 3522775 1761387 880693 440346 220173 110086 55043 27521 13760 6880 3440 1720 860 430 215 107 53 26 13 6 3 1 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
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If you want to buy me a coffee
Btc = 3246y1G9YjnQQNRUrVMnaeCFrymZRgJAP7
Doge = DGNd8UTi8jVTVZ2twhKydyqicynbsERMjs
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albert0bsd
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June 05, 2023, 09:51:54 PM Last edit: June 05, 2023, 10:57:06 PM by albert0bsd |
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Private Key Public key
1090246098153987172547740458951748 # puzzle 110 0309976ba5570966bf889196b7fdf5a0f9a1e9ab340556ec29f8bb60599616167d ... 1 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
That is easy if you have the privatekey (X = X % 2 ? X/2 : (X-1)/2), but we can't do that if we don't have the privatekey, because there is no way that know if the publickey belongs to an even or odd privatekey. But proof your self and show us.. here is a Publickey in the 110 bit range: 0301040c4e33b6ab297ba2a9c2858d15be5f6844febfc003ef7646796608f7f819
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GR Sasa
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June 05, 2023, 10:22:09 PM |
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1090246098153987172547740458951748 # puzzle 110 0309976ba5570966bf889196b7fdf5a0f9a1e9ab340556ec29f8bb60599616167d 1 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
What is this mess? This doesn't prove anything XD
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albert0bsd
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June 05, 2023, 10:29:47 PM |
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What is this mess? This doesn't prove anything XD
Maybe he is trying to reduce the privatekey based only on the current information of the privatekey without do any calculation with the publickey.
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VinIVaderr
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June 07, 2023, 03:17:12 AM |
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I was trying to figure out a good way to break the key search into ranges. I decided to use columns, ie. 17 columns. And group them into 4.
group 1 2000 - 2fff group 2 0000 - ffff group 3 0000 - ffff group 4 0000 - ffff group 5 0123456789abcdef
output: group 1 has 4096 lines group 2 has 65536 lines group 3 has 65536 lines group 4 has 65536 lines group 5 is 16 lines. My idea is to create 2000 0000 0000 0000 0 But I'm still not sure how to bring them together in a string.
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Evillo
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Two things you should never abandon: Family & BTC
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June 08, 2023, 12:00:17 AM |
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I was trying to figure out a good way to break the key search into ranges. I decided to use columns, ie. 17 columns. And group them into 4.
group 1 2000 - 2fff group 2 0000 - ffff group 3 0000 - ffff group 4 0000 - ffff group 5 0123456789abcdef
output: group 1 has 4096 lines group 2 has 65536 lines group 3 has 65536 lines group 4 has 65536 lines group 5 is 16 lines. My idea is to create 2000 0000 0000 0000 0 But I'm still not sure how to bring them together in a string.
On linux or WSL type: paste file1.txt file2.txt > fullrange.txt Obviously change file names in your command according to your actual file names
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Cool Story Bro. BTC: 1EviLLo1Y5VeNn2Lajv9tdZTkUuVgePVYN
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digaran
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June 08, 2023, 10:19:44 AM |
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Guys I need some help over here! So here is what I have so far.
I have subtracted this from #125. 0000000000000000000000000000000010****************************** The result is : 02c473281c8531d524c1a2b92b9192ffe3e8124448e47ec306f146f2f4f35269ee
Now if I add that result to this one : 0387a3951644f3bf9ee08135bcb9a00a31ed7b09e26ff7850b83713eb6f0582920
I get this key : 000000000000000000000000000000000c******************************
Ok, now I have the distance between the green key above and #125, which is : 000000000000000000000000000000000 4******************************
So why can't I figure the correct key while I have my double triangle, 3 known keys plus 3 unknown keys?
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NotATether
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June 08, 2023, 11:06:59 AM |
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Guys I need some help over here! So here is what I have so far.
I have subtracted this from #125. 0000000000000000000000000000000010****************************** The result is : 02c473281c8531d524c1a2b92b9192ffe3e8124448e47ec306f146f2f4f35269ee
Now if I add that result to this one : 0387a3951644f3bf9ee08135bcb9a00a31ed7b09e26ff7850b83713eb6f0582920
I get this key : 000000000000000000000000000000000c******************************
Ok, now I have the distance between the green key above and #125, which is : 000000000000000000000000000000000 4******************************
So why can't I figure the correct key while I have my double triangle, 3 known keys plus 3 unknown keys?
Are the result and green keys public keys or private keys? Because without context, they could be either one.
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kalos15btc
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June 08, 2023, 11:42:37 AM |
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Guys I need some help over here! So here is what I have so far.
I have subtracted this from #125. 0000000000000000000000000000000010****************************** The result is : 02c473281c8531d524c1a2b92b9192ffe3e8124448e47ec306f146f2f4f35269ee
Now if I add that result to this one : 0387a3951644f3bf9ee08135bcb9a00a31ed7b09e26ff7850b83713eb6f0582920
I get this key : 000000000000000000000000000000000c******************************
Ok, now I have the distance between the green key above and #125, which is : 000000000000000000000000000000000 4******************************
So why can't I figure the correct key while I have my double triangle, 3 known keys plus 3 unknown keys?
i told you multiple times, to test that on any other puzzle 100 or 115 you will figure out how it works,, get the private keys of substracted puzzle 110 and you will understand bro, then test it on 125,, when you substract told you to do this range 0:1fffffffffffffffffffffffffffffff you will have 125 substracted maybe to 100 or even 80 bit, but you need to know where exactly you can do that -f address ![Smiley](https://bitcointalk.org/Smileys/default/smiley.gif)
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