bestie1549
Jr. Member
Offline
Activity: 75
Merit: 5
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 18, 2023, 02:23:09 PM |
|
Just info.
Think in your own way and your own point of view about this WIF range
Do we have GPU crackers for WIF Private Keys? or do we have someone that can build 1? these have more potentials for brute forcing puzzle 66 from the Random Point of view for probably the last 16 - 18 characters ranges scanned sequentially, compared to the hexadecimal private keys because of the checksum involved in the WIF Private keys... this might be it, every hexadecimal key is valid but not all WIFs are which brings us to the point where scanning might be fun to play with What is the point of this if what you see in every row is just representation of padded zeros? That is the point right there "padded zeros" as explained earlier, the hexadecimal is just from 0 to F base16 whereas the WIF is a base58 equivalent of that base 16 hexadecimal representation... so what point does this make? Well, I can have 3.625 of the hexadecimal character represented as just 1 character of the WIF... this in otherwords simply means if you are able to hit the correct first 3 WIF characters, then you probably don't have too much range to scan compared to hitting the correct first 3 characters of the hexadecimal representation of the same key... you've got a long way to go bro... So i See some potentials in this WIF bruteforcing for the puzzle 66
|
|
|
|
james5000
Jr. Member
Offline
Activity: 69
Merit: 2
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 18, 2023, 04:37:31 PM |
|
It really seems interesting to think a bit about keys in the WIF format. Let's start with base58. One behavior I noticed is that when you input the key in hexadecimal, the digits in base58 don't remain the same. For example, ... hexadecimal aa = base58 3wbut hexadecimal aaaa = base58 DzHWe can notice that the representation of aa isn't aaaa. I believe that if we play around with this, we might get somewhere. However, nonetheless, we would still need to test each key to derive the hash160. ![Sad](https://bitcointalk.org/Smileys/default/sad.gif)
|
|
|
|
sssergy2705
Copper Member
Jr. Member
Offline
Activity: 197
Merit: 1
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 18, 2023, 05:24:18 PM |
|
Just info.
Think in your own way and your own point of view about this WIF range
Do we have GPU crackers for WIF Private Keys? or do we have someone that can build 1? these have more potentials for brute forcing puzzle 66 from the Random Point of view for probably the last 16 - 18 characters ranges scanned sequentially, compared to the hexadecimal private keys because of the checksum involved in the WIF Private keys... this might be it, every hexadecimal key is valid but not all WIFs are which brings us to the point where scanning might be fun to play with What is the point of this if what you see in every row is just representation of padded zeros? That is the point right there "padded zeros" as explained earlier, the hexadecimal is just from 0 to F base16 whereas the WIF is a base58 equivalent of that base 16 hexadecimal representation... so what point does this make? Well, I can have 3.625 of the hexadecimal character represented as just 1 character of the WIF... this in otherwords simply means if you are able to hit the correct first 3 WIF characters, then you probably don't have too much range to scan compared to hitting the correct first 3 characters of the hexadecimal representation of the same key... you've got a long way to go bro... So i See some potentials in this WIF bruteforcing for the puzzle 66 Key search range 66 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q ZVfMsBQggk69993Lj3p KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q aCtgAeZDbST4e5pMroG We can't do anything with this information. In fact, it will be the same brute force of keys. There is no difference in what format to sort through the keys, even in binary, even in hex, even in base58.
|
|
|
|
digaran
Copper Member
Hero Member
![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif)
Offline
Activity: 1330
Merit: 899
🖤😏
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 18, 2023, 09:48:06 PM |
|
If by any chance anyone here has the WIF checksum for #66, post it here so we could find the key in a few hours. The only advantage if WIF is when you know the checksum, the checksum changes in different bit ranges, I mean the base58 representation of keys with the same last 8 characters is different, so we could use wifsolver to search for the key if we had the checksum.
The world of base58 and WIF checksum is an entirely different world but it is amazing once you get to know it's secrets.😉
Speaking of spamming this thread with useless content, please, you should consider the size of the post you are quoting!
|
🖤😏
|
|
|
Tepan
Jr. Member
Offline
Activity: 49
Merit: 1
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 02:53:58 AM |
|
Just info. KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Yjgd9aFJuCJDo5F6Jm7 # 40 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Yjgd9gCD9CBomewdcUD # 41 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Yjgd9zzYEemjCVJ3vo9 # 42 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjgdB23bP5LsYN8Krv7 # 43 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjgdCpdGyxm7PWoNQdr # 44 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjgdDrgx63tbte4scLA # 45 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjgdLwb42sQhwTBJDnG # 46 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjgdcLTenJYVVVTRzxN # 47 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjgdtUGWxHzvDi28MfR # 48 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjgejcjwprJ4MAYLw8e # 49 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjgfXBMYdNVA4EjUMzg # 50 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Yjgm9fa43QnU2CpULaK # 51 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjguYJTZsntce5oQQrh # 52 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Yjh5SmqrUoK2mhwSYFV # 53 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjhHvuTMSDchRp5hktc # 54 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjidyYKE5NcJQZYvknA # 55 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Yjjb65nDHeqyjiBaJXv # 56 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YjqtiAvYTJzYEmqup7b # 57 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Yjv2kTamEYQT9BNC7o1 # 58 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YkHpsTBP19HvTFqiU6i # 59 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YkzijLsc5qE43yZ5eLV # 60 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YmLDHsih379uP9zbHSD # 61 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q YpCemuaUp7NigjvtJug # 62 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Yv5Z9J7hv7VYYN3XL3Y # 63 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q Z6FxoaD5r1kYegmtbaT # 64 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q ZM21gaY8WN2CdwnTG57 # 65 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 66 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 67 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 68 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 69 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q t5RQNhFCn1r58in2E78 # 70 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 71 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 72 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 73 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 74 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3y HuxB7NfERRuDTNy2kbr # 75 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 76 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 77 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 78 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q # 79 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi APRB6KvN7FnKsw69PP7vW # 80 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi # 81 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi # 82 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi # 83 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi # 84 KwDiBf89QgGbjEhKnhXJuH7LrciVrZ kCnRjpBspr8M5K8YnUtDNa # 85 KwDiBf89QgGbjEhKnhXJuH7LrciVrZ # 86 KwDiBf89QgGbjEhKnhXJuH7LrciVrZ # 87 KwDiBf89QgGbjEhKnhXJuH7LrciVrZ # 88 KwDiBf89QgGbjEhKnhXJuH7LrciVrZ # 89 KwDiBf89QgGbjEhKnhXJuH7LrciVr bEiEhLWVU4Yn1RCfnX51KrT # 90 KwDiBf89QgGbjEhKnhXJuH7LrciVr # 91 KwDiBf89QgGbjEhKnhXJuH7LrciVr # 92 KwDiBf89QgGbjEhKnhXJuH7LrciVr # 93 KwDiBf89QgGbjEhKnhXJuH7LrciVr # 94 KwDiBf89QgGbjEhKnhXJuH7LrciV sLfMYB8ydZ3QXMqEra1sapfT # 95 KwDiBf89QgGbjEhKnhXJuH7LrciV # 96 KwDiBf89QgGbjEhKnhXJuH7LrciV # 97 KwDiBf89QgGbjEhKnhXJuH7LrciV # 98 KwDiBf89QgGbjEhKnhXJuH7LrciV # 99 KwDiBf89QgGbjEhKnhXJuH7Lrci WJvjTPYdp4sSLwstc63avmfRA # 100 KwDiBf89QgGbjEhKnhXJuH7Lrci # 101 KwDiBf89QgGbjEhKnhXJuH7Lrci # 102 KwDiBf89QgGbjEhKnhXJuH7Lrci # 103 KwDiBf89QgGbjEhKnhXJuH7Lrci # 104 KwDiBf89QgGbjEhKnhXJuH7Lrci m5eBMkFQwQtRbW6wxT1ajoNqE # 105 KwDiBf89QgGbjEhKnhXJuH7Lrci # 106 KwDiBf89QgGbjEhKnhXJuH7Lrci # 107 KwDiBf89QgGbjEhKnhXJuH7Lrci # 108 KwDiBf89QgGbjEhKnhXJuH7Lrci # 109 KwDiBf89QgGbjEhKnhXJuH7Lrc tLcZtumiizzSmSckgpxQu6ieu6 # 110 KwDiBf89QgGbjEhKnhXJuH7Lrc # 111 KwDiBf89QgGbjEhKnhXJuH7Lrc # 112 KwDiBf89QgGbjEhKnhXJuH7Lrc # 113 KwDiBf89QgGbjEhKnhXJuH7Lrc # 114 KwDiBf89QgGbjEhKnhXJuH7Lr hcVkatbczm7Y9ZaB56U1jLNU2Xh # 115 KwDiBf89QgGbjEhKnhXJuH7Lr # 116 KwDiBf89QgGbjEhKnhXJuH7Lr # 117 KwDiBf89QgGbjEhKnhXJuH7Lr # 118 KwDiBf89QgGbjEhKnhXJuH7Lr # 119 Think in your own way and your own point of view about this WIF rangeWhat is this, @Tepan? it's just bunch the WIF puzzle already cracked, seems odd and we just need to search the range from WIF (possible ?).
|
|
|
|
Tepan
Jr. Member
Offline
Activity: 49
Merit: 1
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 03:01:02 AM |
|
If by any chance anyone here has the WIF checksum for #66, post it here so we could find the key in a few hours. The only advantage if WIF is when you know the checksum, the checksum changes in different bit ranges, I mean the base58 representation of keys with the same last 8 characters is different, so we could use wifsolver to search for the key if we had the checksum.
The world of base58 and WIF checksum is an entirely different world but it is amazing once you get to know it's secrets.😉
Speaking of spamming this thread with useless content, please, you should consider the size of the post you are quoting!
Nah u just need the public key of 66, not the checksum. How you know the checksum if you cannot reverse the math. just using the publick key + the lost password padded.
|
|
|
|
Tepan
Jr. Member
Offline
Activity: 49
Merit: 1
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 03:24:46 AM Last edit: August 19, 2023, 04:20:10 AM by Tepan |
|
Just info.
Think in your own way and your own point of view about this WIF range
Do we have GPU crackers for WIF Private Keys? or do we have someone that can build 1? these have more potentials for brute forcing puzzle 66 from the Random Point of view for probably the last 16 - 18 characters ranges scanned sequentially, compared to the hexadecimal private keys because of the checksum involved in the WIF Private keys... this might be it, every hexadecimal key is valid but not all WIFs are which brings us to the point where scanning might be fun to play with What is the point of this if what you see in every row is just representation of padded zeros? That is the point right there "padded zeros" as explained earlier, the hexadecimal is just from 0 to F base16 whereas the WIF is a base58 equivalent of that base 16 hexadecimal representation... so what point does this make? Well, I can have 3.625 of the hexadecimal character represented as just 1 character of the WIF... this in otherwords simply means if you are able to hit the correct first 3 WIF characters, then you probably don't have too much range to scan compared to hitting the correct first 3 characters of the hexadecimal representation of the same key... you've got a long way to go bro... So i See some potentials in this WIF bruteforcing for the puzzle 66 Key search range 66 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q ZVfMsBQggk69993Lj3p KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q aCtgAeZDbST4e5pMroG We can't do anything with this information. In fact, it will be the same brute force of keys. There is no difference in what format to sort through the keys, even in binary, even in hex, even in base58. How you can explain this ? Puzzle 118 address is 1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6this is the 0000000000000000000000000000000000af0f4d11574f5deee49961d9609ac6 address 1PEuXxTXfc3qJM4H7EKRbSb8TSPFtds9A5Public key 02bc5f15678ceb70dad97a6b695b9e0df7c405142586931801c2df664563042fe7 I just random search WIF range on that, and i have lock the range area by Hex padded. yeah it's suck it's like 0.0000001% chance hit the 4 same HEX value / WIF. [EDIT 10:41 Indonesia Time] Target 1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6 # 118 this, i found 3 word same address range, from range padded hex. 031e841aa39d5d92ae850400a2d8be24a245d53e9ba29dc86b86696e65290c32d7 = 1PJgNwNbpr1KRxdmPxPXMCWqdBT1bwgDEe
|
|
|
|
digaran
Copper Member
Hero Member
![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif)
Offline
Activity: 1330
Merit: 899
🖤😏
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 04:27:15 AM |
|
Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys. Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it. First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations. Target pub: 03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8 Target priv: Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa. Scalar : aka n/2+1 half n +1 or 1 and half of n. 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2 If you multiply our target pub with scalar above, you will get this : Pub2 : 02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96 Priv : Now if we subtract target from pub2, we will get : Pub3, half of target : 0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da Priv : We didn't even use division, *chop chop and good luck diving.😅 * = hurry! Get to work.
|
🖤😏
|
|
|
Tepan
Jr. Member
Offline
Activity: 49
Merit: 1
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 05:17:19 AM Last edit: August 19, 2023, 10:56:11 AM by hilariousandco |
|
Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys. Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it. First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations. Target pub: 03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8 Target priv: Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa. Scalar : aka n/2+1 half n +1 or 1 and half of n. 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2 If you multiply our target pub with scalar above, you will get this : Pub2 : 02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96 Priv : Now if we subtract target from pub2, we will get : Pub3, half of target : 0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da Priv : We didn't even use division, *chop chop and good luck diving.😅 * = hurry! Get to work. I rather be use the target public key is then (using ecdsa arithmetic) reduced subseqentially until hitting one of the rendezvous point. Given G is the generator point, and R=x*G the rendezvous point (which private key is known). and BSGS. u just like a teacher on class math in 4th grade lmao. ---------- There's i got brute the WIF and it's manually not any application or program i use. I just use called MATH. i just make key range more better for fast scan at the lower and upper range. Puzzle 130 addresses 1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua 2baf7cf4d51574f5deee49961d9609ac6 # 1FHMVBF6Z5ebm5LJKRpkZs8RjNJXUMyUVV # KwDiBf89QgGbjEhKnhXJuH85q VZKQivFjnRMPNnqDyFoSqrSaKfd 2baf78fec51be4f59eaf6966bd9709ac6 # 1FXUvZnrcqRnJV1Cvre2ribDz6yKqQugjR # KwDiBf89QgGbjEhKnhXJuH85q VMiSmxd4F8urBwimS72wrJD2XYS 2baf78fac5b3e4f6feaf6962bd9709ac6 # 1FqZVS2u9iqcycye9BmtVEcCGM6wCYXHM2 # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvjexRvRbBdvgeEZd 2baf78fac5b3e4f6feaf6942bd9709ac6 # 1Fe7X3dKymfuae7xLDZ9Hk5dn9W5pf4SE8 # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvjAC4AFCc3v8Pooi 2baf78fac5b3e4f6feaf69c2bf1709ac6 # 1FnUH4z8cgLuWBLFFWvXBDdQ1VfFc4SM1k # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvm9FrvdFhuoEb9k1 2baf78fac5b3e4f6feaf69f2bf9a8a46c # 1FfUFwHRS4cEESP3JDBMQMaZxLCZTN8tY # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvmtQXdpe3Zfy6UtD 2baf78fac5b3e4d018100000000000000 # 1F2nhqcHQRKwjjokgkKkcSpLH4uucuQZSV # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPuxaPiBGv4PL2aFAnuFR8 3baf70fac037e51011500040000000000 # 1FpLqZyycLiaLWQ6Lbe5zcXuypbX6F7q4C # KwDiBf89QgGbjEhKnhXJuH8Ma WVpmffaLXxzgafdFBXKhAAvtpYn 3baf7efac999e3c95c4ea7660b9946c94 # 1Fcmac1xrLNMMr286BsFnWN35pFNgZeWWb # KwDiBf89QgGbjEhKnhXJuH8Ma XCqFfLS8nKhgd7RpHR2XZB5AKhx 3baf7efac999e3cddc6ba764fb9b66cb4 # 1FqfgJmp8AKvDkAKbUcncgMA7kmDn9dJ2v # KwDiBf89QgGbjEhKnhXJuH8Ma XCqFfLS9Eo6LP4p7NDXN5sx82Js 3bbff6cba4bd242d6ceb2777fb9866ca4 # 1FfPWje1WcZ47fAKuGBijKsqhvF8bHzf3f # KwDiBf89QgGbjEhKnhXJuH8Me C7XhwLTtDcX5XKNNEHm7o8DUj98 3bbff6cb95bca56cdee22756fb970bcaa # 1F2UYUqkShEp8aa9CCE6mKN3yLHbKpEPud # KwDiBf89QgGbjEhKnhXJuH8Me C7XfgWcWBrqsVyALfRFgYf7PCe9 i just hunt the pub and address by privK variation with 16 bit per each line of HEX. just reminder, if you BSGS the full range this 130 Puzzle, it takes 13.8 billion years lmao.
|
|
|
|
sssergy2705
Copper Member
Jr. Member
Offline
Activity: 197
Merit: 1
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 08:00:02 AM |
|
Just info.
Think in your own way and your own point of view about this WIF range
Do we have GPU crackers for WIF Private Keys? or do we have someone that can build 1? these have more potentials for brute forcing puzzle 66 from the Random Point of view for probably the last 16 - 18 characters ranges scanned sequentially, compared to the hexadecimal private keys because of the checksum involved in the WIF Private keys... this might be it, every hexadecimal key is valid but not all WIFs are which brings us to the point where scanning might be fun to play with What is the point of this if what you see in every row is just representation of padded zeros? That is the point right there "padded zeros" as explained earlier, the hexadecimal is just from 0 to F base16 whereas the WIF is a base58 equivalent of that base 16 hexadecimal representation... so what point does this make? Well, I can have 3.625 of the hexadecimal character represented as just 1 character of the WIF... this in otherwords simply means if you are able to hit the correct first 3 WIF characters, then you probably don't have too much range to scan compared to hitting the correct first 3 characters of the hexadecimal representation of the same key... you've got a long way to go bro... So i See some potentials in this WIF bruteforcing for the puzzle 66 Key search range 66 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q ZVfMsBQggk69993Lj3p KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q aCtgAeZDbST4e5pMroG We can't do anything with this information. In fact, it will be the same brute force of keys. There is no difference in what format to sort through the keys, even in binary, even in hex, even in base58. How you can explain this ? Puzzle 118 address is 1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6this is the 0000000000000000000000000000000000af0f4d11574f5deee49961d9609ac6 address 1PEuXxTXfc3qJM4H7EKRbSb8TSPFtds9A5Public key 02bc5f15678ceb70dad97a6b695b9e0df7c405142586931801c2df664563042fe7 I just random search WIF range on that, and i have lock the range area by Hex padded. yeah it's suck it's like 0.0000001% chance hit the 4 same HEX value / WIF. [EDIT 10:41 Indonesia Time] Target 1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6 # 118 this, i found 3 word same address range, from range padded hex. 031e841aa39d5d92ae850400a2d8be24a245d53e9ba29dc86b86696e65290c32d7 = 1PJgNwNbpr1KRxdmPxPXMCWqdBT1bwgDEe The public address has no binding either to the public key or to the private key. You need to carefully study the process of obtaining bitcoin addresses.
|
|
|
|
Tepan
Jr. Member
Offline
Activity: 49
Merit: 1
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 09:05:45 AM |
|
Just info.
Think in your own way and your own point of view about this WIF range
Do we have GPU crackers for WIF Private Keys? or do we have someone that can build 1? these have more potentials for brute forcing puzzle 66 from the Random Point of view for probably the last 16 - 18 characters ranges scanned sequentially, compared to the hexadecimal private keys because of the checksum involved in the WIF Private keys... this might be it, every hexadecimal key is valid but not all WIFs are which brings us to the point where scanning might be fun to play with What is the point of this if what you see in every row is just representation of padded zeros? That is the point right there "padded zeros" as explained earlier, the hexadecimal is just from 0 to F base16 whereas the WIF is a base58 equivalent of that base 16 hexadecimal representation... so what point does this make? Well, I can have 3.625 of the hexadecimal character represented as just 1 character of the WIF... this in otherwords simply means if you are able to hit the correct first 3 WIF characters, then you probably don't have too much range to scan compared to hitting the correct first 3 characters of the hexadecimal representation of the same key... you've got a long way to go bro... So i See some potentials in this WIF bruteforcing for the puzzle 66 Key search range 66 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q ZVfMsBQggk69993Lj3p KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q aCtgAeZDbST4e5pMroG We can't do anything with this information. In fact, it will be the same brute force of keys. There is no difference in what format to sort through the keys, even in binary, even in hex, even in base58. How you can explain this ? Puzzle 118 address is 1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6this is the 0000000000000000000000000000000000af0f4d11574f5deee49961d9609ac6 address 1PEuXxTXfc3qJM4H7EKRbSb8TSPFtds9A5Public key 02bc5f15678ceb70dad97a6b695b9e0df7c405142586931801c2df664563042fe7 I just random search WIF range on that, and i have lock the range area by Hex padded. yeah it's suck it's like 0.0000001% chance hit the 4 same HEX value / WIF. [EDIT 10:41 Indonesia Time] Target 1PJZPzvGX19a7twf5HyD2VvNiPdHLzm9F6 # 118 this, i found 3 word same address range, from range padded hex. 031e841aa39d5d92ae850400a2d8be24a245d53e9ba29dc86b86696e65290c32d7 = 1PJgNwNbpr1KRxdmPxPXMCWqdBT1bwgDEe The public address has no binding either to the public key or to the private key. You need to carefully study the process of obtaining bitcoin addresses. Might be you need double the reading of my progress, it's not binding to pub or priv, just calculated the random and padded HEX on the same range, it's like HEX range -> check 1 by 1 hex to WIF -> Segwit Base 58 (If the address have the same targeted addresses, keep the range, if not skip the hex range *not saved for the next batch*). This is the demo for 130 Puzzle Puzzle 130 addresses 1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua 2baf7cf4d51574f5deee49961d9609ac6 # 1FHMVBF6Z5ebm5LJKRpkZs8RjNJXUMyUVV # KwDiBf89QgGbjEhKnhXJuH85q VZKQivFjnRMPNnqDyFoSqrSaKfd 2baf78fec51be4f59eaf6966bd9709ac6 # 1FXUvZnrcqRnJV1Cvre2ribDz6yKqQugjR # KwDiBf89QgGbjEhKnhXJuH85q VMiSmxd4F8urBwimS72wrJD2XYS 2baf78fac5b3e4f6feaf6962bd9709ac6 # 1FqZVS2u9iqcycye9BmtVEcCGM6wCYXHM2 # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvjexRvRbBdvgeEZd 2baf78fac5b3e4f6feaf6942bd9709ac6 # 1Fe7X3dKymfuae7xLDZ9Hk5dn9W5pf4SE8 # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvjAC4AFCc3v8Pooi 2baf78fac5b3e4f6feaf69c2bf1709ac6 # 1FnUH4z8cgLuWBLFFWvXBDdQ1VfFc4SM1k # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvm9FrvdFhuoEb9k1 2baf78fac5b3e4f6feaf69f2bf9a8a46c # 1FfUFwHRS4cEESP3JDBMQMaZxLCZTN8tY # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvmtQXdpe3Zfy6UtD 2baf78fac5b3e4d018100000000000000 # 1F2nhqcHQRKwjjokgkKkcSpLH4uucuQZSV # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPuxaPiBGv4PL2aFAnuFR8 3baf70fac037e51011500040000000000 # 1FpLqZyycLiaLWQ6Lbe5zcXuypbX6F7q4C # KwDiBf89QgGbjEhKnhXJuH8Ma WVpmffaLXxzgafdFBXKhAAvtpYn 3baf7efac999e3c95c4ea7660b9946c94 # 1Fcmac1xrLNMMr286BsFnWN35pFNgZeWWb # KwDiBf89QgGbjEhKnhXJuH8Ma XCqFfLS8nKhgd7RpHR2XZB5AKhx 3baf7efac999e3cddc6ba764fb9b66cb4 # 1FqfgJmp8AKvDkAKbUcncgMA7kmDn9dJ2v # KwDiBf89QgGbjEhKnhXJuH8Ma XCqFfLS9Eo6LP4p7NDXN5sx82Js 3bbff6cba4bd242d6ceb2777fb9866ca4 # 1FfPWje1WcZ47fAKuGBijKsqhvF8bHzf3f # KwDiBf89QgGbjEhKnhXJuH8Me C7XhwLTtDcX5XKNNEHm7o8DUj98 3bbff6cb95bca56cdee22756fb970bcaa # 1F2UYUqkShEp8aa9CCE6mKN3yLHbKpEPud # KwDiBf89QgGbjEhKnhXJuH8Me C7XfgWcWBrqsVyALfRFgYf7PCe9 it's BSGS and manually count by my magic math. ![Tongue](https://bitcointalk.org/Smileys/default/tongue.gif)
|
|
|
|
Denis_Hitov
Newbie
Offline
Activity: 49
Merit: 0
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 09:49:26 AM |
|
Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys. Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it. First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations. Target pub: 03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8 Target priv: Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa. Scalar : aka n/2+1 half n +1 or 1 and half of n. 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2 If you multiply our target pub with scalar above, you will get this : Pub2 : 02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96 Priv : Now if we subtract target from pub2, we will get : Pub3, half of target : 0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da Priv : We didn't even use division, *chop chop and good luck diving.😅 * = hurry! Get to work. Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.
|
|
|
|
digaran
Copper Member
Hero Member
![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif)
Offline
Activity: 1330
Merit: 899
🖤😏
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 10:43:33 AM Last edit: August 19, 2023, 10:53:39 AM by digaran |
|
To install the `sympy` module on Windows, type CMD in the search bar and open Command Prompt and type pip install mpmath, first and then pip install sympy. put ice dll and secp256k1.py in the same folder where you have the script.
Thanks, I did what you said and learned a few things about python commands, after a few errors now it says module secp256k1 has no attribute 'pub2upub' even though I correctly pasted the upub in it's place. I went to secp256k1.py and there was no mention of pub2upub. Lol this thing burned more energy than a week of dealing with EC math.
Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.
You can't figure out if a key is odd or even, that would make it easy to find every other digits to see if they are also odd or even. But there is a way to make sure that the correct results are in your files and your tool had all the incorrect and correct keys. Here is how, you know some keys can be divided by 2 a few times in a row, so when you divide a key you need to divide it by 2 at least 4 times, subtract 1 from all and do the same with all the keys. This should occupy less than 1 GB of storage/ RAM, and it should solve the low bit range keys in minutes, but I don't know why nobody is trying it.🤔
|
🖤😏
|
|
|
bestie1549
Jr. Member
Offline
Activity: 75
Merit: 5
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 10:49:17 AM |
|
Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys. Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it. First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations. Target pub: 03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8 Target priv: Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa. Scalar : aka n/2+1 half n +1 or 1 and half of n. 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2 If you multiply our target pub with scalar above, you will get this : Pub2 : 02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96 Priv : Now if we subtract target from pub2, we will get : Pub3, half of target : 0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da Priv : We didn't even use division, *chop chop and good luck diving.😅 * = hurry! Get to work. Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours. The tools already developed achieve these operations better... all we need is something to work better than these tools or programs... if we can write all these ideas out in a code and run the scripts, we would be able to work faster on the CPU 100 times faster than the codes already built to run on GPUs. BSGS works great, Kangaroo is awesomely perfect... these programs are great but we can programmatically edit their functions to our desires or the codes will completely break down which eventually brings us to the point where all we need is functions to calculate these operations programmatically to our own tastes and commands. the 1s and 0s idea is a great idea that worked but still, we needed it to go faster... the BitCrack, Kangaroo programs works super fast but what do we get when the bit ranges keeps increasing? these programs, though very fast now feels super slow because the way the keys starts to get even bigger and bigger and the ranges wider and wider which now brings us to the moment of truth... There are only two possible explanations: either no one told us, or no one knows. Because as we are undoubtedly gathering, the anomaly is not systemic, and it's not creating any fluctuations in even the most simplistic equations. The problem is choice. Our function is now to return to the source code, allowing a temporary dissemination of the code, reinserting the prime program. Failure to comply with this process will result in a cataclysmic system crash in the brain because we won't be able to break the curve without getting to it's source. Which brings us at last to the moment of truth, wherein the fundamental flaw is ultimately expressed, and the anomaly revealed as both beginning, and end... and the curve at one point is a reflection of another point... wherein we can call it a mirror. There are two doors. One door is bruteforcing or as we all call it, BSGS, Kangaroo... whatever you like you can call it... and the other door is owning, buying, mining. this door only costs you some resources like money or vis-a-vis finance but the first door only gives us Hope, it is the quintessential human delusion, simultaneously the source of our greatest strength, and our greatest weakness.
|
|
|
|
Tepan
Jr. Member
Offline
Activity: 49
Merit: 1
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 11:14:41 AM Last edit: August 19, 2023, 12:00:24 PM by hilariousandco |
|
Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys. Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it. First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations. Target pub: 03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8 Target priv: Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa. Scalar : aka n/2+1 half n +1 or 1 and half of n. 7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2 If you multiply our target pub with scalar above, you will get this : Pub2 : 02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96 Priv : Now if we subtract target from pub2, we will get : Pub3, half of target : 0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da Priv : We didn't even use division, *chop chop and good luck diving.😅 * = hurry! Get to work. Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours. Yeah absolutely 1 hours 4 minutes, but he's doin nothing with random speculation ahaha, see. just computing use 50% brain and computing power, not 100% brain enough. To install the `sympy` module on Windows, type CMD in the search bar and open Command Prompt and type pip install mpmath, first and then pip install sympy. put ice dll and secp256k1.py in the same folder where you have the script.
Thanks, I did what you said and learned a few things about python commands, after a few errors now it says module secp256k1 has no attribute 'pub2upub' even though I correctly pasted the upub in it's place. I went to secp256k1.py and there was no mention of pub2upub. Lol this thing burned more energy than a week of dealing with EC math.
Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.
You can't figure out if a key is odd or even, that would make it easy to find every other digits to see if they are also odd or even. But there is a way to make sure that the correct results are in your files and your tool had all the incorrect and correct keys. Here is how, you know some keys can be divided by 2 a few times in a row, so when you divide a key you need to divide it by 2 at least 4 times, subtract 1 from all and do the same with all the keys. This should occupy less than 1 GB of storage/ RAM, and it should solve the low bit range keys in minutes, but I don't know why nobody is trying it.🤔 just remember bro, what are you doing in right now it's already done with someone who was not in this forum, believe me or not, u just need develop it no being amateur with code flying inside. just ensure the progress and don't be a jerk about someone idea.
|
|
|
|
digaran
Copper Member
Hero Member
![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif) ![*](https://bitcointalk.org/Themes/custom1/images/star.gif)
Offline
Activity: 1330
Merit: 899
🖤😏
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 12:25:05 PM |
|
just ensure the progress and don't be a jerk about someone idea.
Sorry but I'm not really being a jerk when I say what you are doing is waste of time, I wasted 2 month on WIFs, then I found out there is no shortcut unless we know the checksum which is 6 base58 characters even though the checksum is not part of the hex private key but knowing it could change everything. Apology for calling you a spammer, you could post a one line example instead of 20 lines. Btw, if I wanted to spend time on learning how to code, I couldn't learn the little I know about ECC. Imagine I spent 1 hour to download and install packages and then copy them one by one to the directory of my script just to realize I am not built for such things, every person has his own sets of skills, learning how to code is not mine to master.
|
🖤😏
|
|
|
bestie1549
Jr. Member
Offline
Activity: 75
Merit: 5
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 01:05:22 PM |
|
just ensure the progress and don't be a jerk about someone idea.
Sorry but I'm not really being a jerk when I say what you are doing is waste of time, I wasted 2 month on WIFs, then I found out there is no shortcut unless we know the checksum which is 6 base58 characters even though the checksum is not part of the hex private key but knowing it could change everything. Apology for calling you a spammer, you could post a one line example instead of 20 lines. Btw, if I wanted to spend time on learning how to code, I couldn't learn the little I know about ECC. Imagine I spent 1 hour to download and install packages and then copy them one by one to the directory of my script just to realize I am not built for such things, every person has his own sets of skills, learning how to code is not mine to master. codes make life easy bro, copying and pasting could be easily done with a few codes... the files your notepads and editors can't open... say 100GB of pubkeys in a single file, you can easily manipulate the data in the file even if you have 1GB RAM... the point is the things you are busy spending hours to calculate, you could just build a program that would run it 1 million times before you are done doing 3. that's why it's called coding or programming... you program it to your taste. you could almost leave your system running all on it's own if you have the programs needed without even opening anything as long as you've written the program to fulfil your choice of operation. double clicking is too slow and I bet if you knew coding, we wouldn't be stuck with puzzle 66 or puzzle 130 by now... You have the ideas needed but you're not flexible enough to write the programs needed to carry out the operations. That's why I said sometimes ideas might be half in the bearer's mind but it takes another mind thinking towards the same direction to make it a full idea... you know all the divisions and G factors to multiply with and but if a code should handle those ideas you'd definitely be solving these puzzles like it's nothing... you don't necessarily need to learn the coding yourself if you don't want to, but just try to get closer to someone that knows how to translate these ideas to programs and you'll see for yourself how you won't be needing to spend hours dividing and multiplying keys manually. just always make sure there's an argparse or a command line args which is defined already in places where you need to constantly make some edits on the code so you don't have to always edit the code itself every time. we all are learning and until we move on to the other side of the world can we ever stop learning.
|
|
|
|
mcdouglasx
Member
![*](https://bitcointalk.org/Themes/custom1/images/star.gif)
Offline
Activity: 247
Merit: 53
New ideas will be criticized and then admired.
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 01:14:18 PM |
|
To install the `sympy` module on Windows, type CMD in the search bar and open Command Prompt and type pip install mpmath, first and then pip install sympy. put ice dll and secp256k1.py in the same folder where you have the script.
Thanks, I did what you said and learned a few things about python commands, after a few errors now it says module secp256k1 has no attribute 'pub2upub' even though I correctly pasted the upub in it's place. I went to secp256k1.py and there was no mention of pub2upub. Lol this thing burned more energy than a week of dealing with EC math.
Now, if it were possible to calculate an even target pub or an odd one, then 130 puzzle could be solved in a few hours.
You can't figure out if a key is odd or even, that would make it easy to find every other digits to see if they are also odd or even. But there is a way to make sure that the correct results are in your files and your tool had all the incorrect and correct keys. Here is how, you know some keys can be divided by 2 a few times in a row, so when you divide a key you need to divide it by 2 at least 4 times, subtract 1 from all and do the same with all the keys. This should occupy less than 1 GB of storage/ RAM, and it should solve the low bit range keys in minutes, but I don't know why nobody is trying it.🤔 target_public_key = "03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852" pub = ice.pub2upub(target_public_key) replace the line with this, the pub should be compressed, don't forget to copy secp256k1.py and ice_secp256k1.dll in the same folder where your script is located.
|
I'm not dead, long story... BTC bc1qxs47ttydl8tmdv8vtygp7dy76lvayz3r6rdahu
|
|
|
unpluggedcoin
Newbie
Offline
Activity: 18
Merit: 0
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 05:10:22 PM |
|
There are two doors. One door is bruteforcing or as we all call it, BSGS, Kangaroo... whatever you like you can call it... and the other door is owning, buying, mining. this door only costs you some resources like money or vis-a-vis finance but the first door only gives us Hope, it is the quintessential human delusion, simultaneously the source of our greatest strength, and our greatest weakness.
About the DOORs, I read on this forum at some other threads about Backdoors, and Oracle Black box,, that may be true or false God Knows, but as far I tried for some one and half year around Elliptic Curve maths, there is no way to somehow calculate or comeup with existing mathematical functions of the curve like Addition, Subtraction, Division and Multiplication that can replace any bruteforce method, NOT A CHANCE!! Few months earlier I started working on the point at infinity! The ultimate master source behind all Elliptic Curves. Though I am still lost in this jungle along with you and didn't reached anywhere but I suspect there might be a third Door behind this point at infinity!!! It is also related with Euclidean key to Universe as Stephen Hawkins said, i guess!!!!!
|
|
|
|
GR Sasa
Member
![*](https://bitcointalk.org/Themes/custom1/images/star.gif)
Offline
Activity: 191
Merit: 14
|
![](https://bitcointalk.org/Themes/custom1/images/post/xx.gif) |
August 19, 2023, 08:33:48 PM |
|
Warning: Please stay On-Topic, otherwise you will be getting banned
|
|
|
|
|