NotATether
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October 22, 2021, 06:08:50 AM |
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Most likely it is the last line (privkey0 = random.randint(L, U)). L and U are the range of the kangaroos of course. t and w are just the tame and wild points respectively (actually they are represented by numbers here).
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Andre_25
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October 26, 2021, 07:33:50 PM |
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Hello. Friends, I have a question. I can put many public keys (to study, myself) in the kangaroo input file, for example about 1000. Will this make the job more difficult? because, for example, if you find a possible collision of one of them, the others will no longer be searched and you are left with only one? (the possible found). Or it carries a set of possible collisions, independently for each of the public keys put in the input.
Thank you for the help that you may be able to provide
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a.a
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October 26, 2021, 07:38:08 PM |
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They are not searched in parallel. So it first searches the first one, if not found searches the next one... Etc.
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fxsniper
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October 30, 2021, 05:34:52 AM |
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How long JeanLucPons use time for solve puzzle #105 ,#110 and #115 each ? info on GitHub tell only #120
็How to can calculate range for my GPU can search? example [139.25 MK/s][GPU 117.25 MK/s][Count 2^39.32][Dead 0][01:34:25 (Avg 17.4802y)][2.0/4.0MB]
24 hour = 24*60=1,440*60=86,400 sec
117.25 MK = 117000000 117000000*86400=77,760,000,000,000 log2(77760000,0000000) 49.466021547366765 49bit range
just try spilt search D5F00C73EA6B3394A6BB77FFFFFFFF D5FE01DEDF5D348E540611C71C71C7 02CEB6CBBCDBDF5EF7150682150F4CE2C6F4807B349827DCDBDD1F2EFA885A2630
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John_Ahmet
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November 05, 2021, 03:33:27 PM Last edit: November 05, 2021, 03:49:39 PM by John_Ahmet |
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Hello, thank you to everyone who contributed to this project. I want to try something and I need ModPow function for Int type. All Mod functions use P. Can a faster version be made for b=2? I want to calculate 2 ^ e % m For int type like this: int ModPow(int b, int e, int m) { int result = 1; if (1 & e) result = b; while (1) { if (!e) break; e >>= 1; b = (b * b) % m;//ModMul if (e & 1) result = (result * b) % m;//ModMul } return result; }
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NotATether
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November 05, 2021, 04:35:40 PM |
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Hello, thank you to everyone who contributed to this project. I want to try something and I need ModPow function for Int type. All Mod functions use P. Can a faster version be made for b=2? I want to calculate 2 ^ e % m For int type like this: int ModPow(int b, int e, int m) { int result = 1; if (1 & e) result = b; while (1) { if (!e) break; e >>= 1; b = (b * b) % m;//ModMul if (e & 1) result = (result * b) % m;//ModMul } return result; } There is not one that I know of. You could've been able to replace the (b*b) with b << (b >> 1) which uses only bitwise operators, and other basic bitwise optimizations in other statements, if it wasn't for the presence of the possibly odd modulus n.
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John_Ahmet
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November 09, 2021, 12:23:49 AM Last edit: November 09, 2021, 12:41:38 PM by mprep |
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I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method? GenerateRSAKeys() p => random big prime, below 64bits q => random big prime, below 64bits e => random small prime, below 16bits k => random small integer, below 16 bits N = p * q d = (k * ((p - 1) * (q - 1)) + 1) / e
(e, N) => Public key (e, d, p, q) => Private key
RSAEncode(m, e, N) // m => message, e and N are public key h ≡ m ^ e mod N
RSADecode(h, d, N) // h => encrypted message, d is part of private key m ≡ h ^ d mod N
RSACrack_GetPrivateKeyFromPublicKey(e, N, minPrimeBitsLength) bitLength = GetBitsLength(N) k = 2 // iterator uses sieve algorithm primes = all64BitsPrimes.Select(p => GetBitsLength(p) >= minPrimeBitsLength && GetBitsLength(p) <= (bitLength - minPrimeBitsLength)) // Example Primes buffer uses 4GB memory, all below 64Bits primes use 1TB memory // use only odd numbers bit array // max 256 cycles index = -1 for primesBuffer1 of primes index++ for primesBuffer2 of primes.NextTo(index) // next buffers from index, including index. buffer. if you use ModMul, remove this line for p of primesBuffer1 for q of primesBuffer2 // if you use ModMul, remove this line bitLengthA = GetBitsLength(p) bitLengthB = GetBitsLength(q) bitSum = bitLengthA + bitLengthB if (max(bitLengthA, bitLengthB) >= bitLength && bitSum <= bitLength) if (p * q == N) // or (N % p == 0) or ModMul(N+1, 10, p) == 10 d = (k * ((p - 1) * (q - 1)) + 1) / e return (e, d, p, q)
For RSA this code results much faster. I expect you to write basic pseudocode for Kangaroo. Thanks. RSAHacking_GetPrivateKeyFromPublicKeyUsingModPowNoPrimeTable(e, N) k = 2 SqrtN = Sqrt(N) if((SqrtN & 1) == 0) SqrtN++ for(p = SqrtN; p > 2; p-=2) if (true/*You can use more filters for p*/ && ModPow(2, p-1, p) == 1 && ModMul(N+1, 10, p) == 10) q = N / p d = (k * ((p - 1) * (q - 1)) + 1) / e return (e, d, p, q)
[moderator's note: consecutive posts merged]
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NotATether
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November 10, 2021, 07:47:44 AM |
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I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method?
You cannot solve RSA problems with the Kangaroo method because the ECDLP problem is not applicable for RSA, as there are no logarithms to solve for.
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superkatchu
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November 10, 2021, 10:51:45 AM |
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I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method?
You cannot solve RSA problems with the Kangaroo method because the ECDLP problem is not applicable for RSA, as there are no logarithms to solve for. Not ECDLP but DLP so Pollard Rho and Pollard Kangaroo should be applicable no?
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demoinvest1
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November 16, 2021, 09:45:02 AM |
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Can possible Pollard's kangaroo can search multiple pubkey in same time? I mean check multiple pubkey at once
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wedom
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November 16, 2021, 09:48:50 AM |
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Can possible Pollard's kangaroo can search multiple pubkey in same time? I mean check multiple pubkey at once
No, one at a time only
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NotATether
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November 16, 2021, 12:54:01 PM |
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I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method?
You cannot solve RSA problems with the Kangaroo method because the ECDLP problem is not applicable for RSA, as there are no logarithms to solve for. Not ECDLP but DLP so Pollard Rho and Pollard Kangaroo should be applicable no? Rho is specifically designed for Discrete Logarithm (factoring) problem. Also I read the Wikipedia page for Kangaroo which said that it's applicable for any group. In fact the paper for Pollard's Kangaroo originally mentioned it's use for multiplicative [again, RSA/factoring] groups.
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bigvito19
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November 16, 2021, 01:22:43 PM |
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I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method?
You cannot solve RSA problems with the Kangaroo method because the ECDLP problem is not applicable for RSA, as there are no logarithms to solve for. Not ECDLP but DLP so Pollard Rho and Pollard Kangaroo should be applicable no? Rho is specifically designed for Discrete Logarithm (factoring) problem. Also I read the Wikipedia page for Kangaroo which said that it's applicable for any group. In fact the paper for Pollard's Kangaroo originally mentioned it's use for multiplicative [again, RSA/factoring] groups. Are there any Rho programs?
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fxsniper
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November 18, 2021, 12:57:15 AM |
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I not yet deep understand about ECDLP and Pollard's kangaroo Algorithm and about Collision I just know overview that not enough for knowledge can solve puzzle
I am not sure I understand it correctly This Collision method meaning it have Collision every where on range around 2*32 will be found Collision of every key right (but size large 2**256 it is large enough for can not find it)
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NotATether
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November 18, 2021, 04:04:02 AM |
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Are there any Rho programs?
Not that I know of. I not yet deep understand about ECDLP and Pollard's kangaroo Algorithm and about Collision I just know overview that not enough for knowledge can solve puzzle
I am not sure I understand it correctly This Collision method meaning it have Collision every where on range around 2*32 will be found Collision of every key right (but size large 2**256 it is large enough for can not find it)
It is considering as a collision any key that has the same Most Significant Bytes as some other key in the hash table. How many MSBs that have to be identical depends on your DP setting. So if your DP size is 10, all keys with the same 10 beginning bits are considered a collision for hashtable purposes.
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mausuv
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November 19, 2021, 02:09:09 PM |
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guys please tell two point subtract example i need
tell easy understand because, i know tow point add and multipliction but subtract not understand please write example easy math
for example :
c = K(qy-py)/(qx-px) rx1 = K(c^2-px-qx) ry2 = K(c*(px-rx))
your example write this method #sorrymyBADenglish
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jovica888
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November 20, 2021, 12:36:01 AM Last edit: November 20, 2021, 01:22:40 AM by jovica888 |
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@mausuv I am searching for this too... It looks to me that if | public key = private key * G point | then some parts of the whole process is the same for all keys... It does not need to be calculated again... Also what I understood is I have 2 points P and Q and I am searching for x and y of R... So what does generator points have to do with all process... And how do I get values of P and Q if I know the private key and that private key has a value "0x2"... Somebody could explain which variable means what? ![](https://ip.bitcointalk.org/?u=https%3A%2F%2Fi.imgur.com%2FRU7sYqO.png&t=665&c=CAyWAcqZXsbFIw)
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fxsniper
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November 20, 2021, 01:52:09 AM |
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I am not yet understand about subtract method I follow come later May be subtract method it is not works. for now can not solve What is thing make subtract method stuck, power of GPU not help right
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jovica888
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November 20, 2021, 03:02:52 AM |
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And also one more thing. ![](https://ip.bitcointalk.org/?u=https%3A%2F%2Fi.imgur.com%2FRU7sYqO.png&t=665&c=CAyWAcqZXsbFIw) When I look at this picture I see that this one value for x and y of Rhas millions and billions different pairs of x and y of Q and PI mean just change the slope of the line
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mausuv
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November 20, 2021, 04:19:47 AM |
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And also one more thing. https://i.imgur.com/RU7sYqO.pngWhen I look at this picture I see that this one value for x and y of Rhas millions and billions different pairs of x and y of Q and PI mean just change the slope of the line this code write for c in keysubtracter i need python >>>> https://github.com/albertobsd/keysubtracteri know tow point add,multiple in python code, but subtracter not understand https://github.com/albertobsd/keysubtracter/blob/main/keysubtracter.c please modifiy this code c to python convert another c code https://github.com/albertobsd/ecctools subtract please only subtract code get write to python atleast this c code https://github.com/albertobsd/keysubtracter/blob/main/keysubtracter.c decode to mathematical explain for example : c = K(qy-py)/(qx-px) rx1 = K(c^2-px-qx) ry2 = K(c*(px-rx)) #sorrymyBADenglish ----------------------------------------------------------i use this tow point multiple in python P = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1 N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141 Gx = 0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 Gy = 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a Acurve = 0 Cx,Cy = multiple(Gx,Gy) print (hex(Cx)[2:66],hex(Cy)[2:66])
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