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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 255584 times)
yellowstripes
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December 16, 2023, 04:14:42 AM
 #4161

Here, I give you 2 which are really close to each other, lets go.😉


Code:
03b4f75b9205bdcfcf269e604e27066ea422f11e7c1cf50dee6d6b005510e4deaf
02f92b4119e2ecdc2924ff6983f65c3b5af9ed036ce5a6469f3462965840e3daef

The difference between the two public keys is 6.

If I had more time I would find the sum. Once you know the difference and sum then you can solve for the private keys individually. Simple algebra really.
WanderingPhilospher
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December 16, 2023, 06:09:59 AM
 #4162

Here, I give you 2 which are really close to each other, lets go.😉


Code:
03b4f75b9205bdcfcf269e604e27066ea422f11e7c1cf50dee6d6b005510e4deaf
02f92b4119e2ecdc2924ff6983f65c3b5af9ed036ce5a6469f3462965840e3daef
03b4f75b9205bdcfcf269e604e27066ea422f11e7c1cf50dee6d6b005510e4deaf
Actually, if this were possible, bitcoin was underground.
I would say it is above
1000000000000000000000000000000000000000
but my calculations have errors, so if I'm right I'll take it as a coincidence., but I will continue investigating.
Although I am more than 99% sure that this is not possible, we always have to explore all angles in science..
Both of these are definitely above 1000000000000000000000000000000000000000.

I developed a script about a year ago that you can change the Generator and it will get you within 43 bits of a pubkeys location.

These are roughly 21,474,836,480 to 49,123,688,448 above 1000000000000000000000000000000000000000.
nomachine
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December 16, 2023, 06:35:41 AM
Last edit: December 16, 2023, 06:58:13 AM by nomachine
 #4163

Hi mate,

how it is going so far? How much is the real rate of keys/s?

Quantum computers, in theory, could provide a speedup for certain types of search problems, but finding hash collisions for cryptographic hash functions is not one of them. Cryptographic primitives like hash functions and ECDSA are designed to be resistant to preimage attacks, and quantum computers do not break this resistance.
Quantum computation do not pose an immediate threat to the security of Bitcoin or other widely used cryptographic systems.
This approach is more of a proof-of-concept or educational exercise rather than a practical implementation for breaking Bitcoin's security.
Quantum algorithm/hardware for solving such problems would need to be carefully designed to provide actual advantages over classical approaches.
Example:
https://arxiv.org/pdf/2302.06639.pdf
IBM Quantum Backend don't have the hardware (arithmetic circuits) available for "126 133 Cat Qubits"  
If someone succeeds in doing this, they will know exactly who is capable of it in the scientific community and they have their addresses where they live. Grin
All other talk is just good advertising for these quantum backend providers.

bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
AlanJohnson
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December 16, 2023, 07:55:54 AM
 #4164

Hi mate,

how it is going so far? How much is the real rate of keys/s?

Quantum computers, in theory, could provide a speedup for certain types of search problems, but finding hash collisions for cryptographic hash functions is not one of them. Cryptographic primitives like hash functions and ECDSA are designed to be resistant to preimage attacks, and quantum computers do not break this resistance.
Quantum computation do not pose an immediate threat to the security of Bitcoin or other widely used cryptographic systems.
This approach is more of a proof-of-concept or educational exercise rather than a practical implementation for breaking Bitcoin's security.
Quantum algorithm/hardware for solving such problems would need to be carefully designed to provide actual advantages over classical approaches.
Example:
https://arxiv.org/pdf/2302.06639.pdf
IBM Quantum Backend don't have the hardware (arithmetic circuits) available for "126 133 Cat Qubits"  
If someone succeeds in doing this, they will know exactly who is capable of it in the scientific community and they have their addresses where they live. Grin
All other talk is just good advertising for these quantum backend providers.

Are you sure ?

From what i know ECDSA is vulnerable to quantum computers attack. Maybe not that one IBM allows to use for free for 10 minutes. But generally : IT'S VULNERABLE.

https://security.stackexchange.com/questions/34940/is-ecdsa-breakable-by-quantum-computers
tmar777
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December 16, 2023, 09:20:16 AM
Last edit: May 01, 2024, 08:36:30 PM by Mr. Big
 #4165

Here, I give you 2 which are really close to each other, lets go.😉


Code:
03b4f75b9205bdcfcf269e604e27066ea422f11e7c1cf50dee6d6b005510e4deaf
02f92b4119e2ecdc2924ff6983f65c3b5af9ed036ce5a6469f3462965840e3daef

The difference between the two public keys is 6.

If I had more time I would find the sum. Once you know the difference and sum then you can solve for the private keys individually. Simple algebra really.

Can you please explain this part how it works?
"Once you know the difference and sum then you can solve for the private keys individually. Simple algebra really."

Also, can somebody explain to me about which mode of searching for private keys is more efficient?
I mean is it more efficient to search sequentially each possible binary (of private key), the address (sequentially), the rmd160, the Xpoint or what else?
Is there any research paper about this?



Hi mate,

how it is going so far? How much is the real rate of keys/s?

Quantum computers, in theory, could provide a speedup for certain types of search problems, but finding hash collisions for cryptographic hash functions is not one of them. Cryptographic primitives like hash functions and ECDSA are designed to be resistant to preimage attacks, and quantum computers do not break this resistance.
Quantum computation do not pose an immediate threat to the security of Bitcoin or other widely used cryptographic systems.
This approach is more of a proof-of-concept or educational exercise rather than a practical implementation for breaking Bitcoin's security.
Quantum algorithm/hardware for solving such problems would need to be carefully designed to provide actual advantages over classical approaches.
Example:
https://arxiv.org/pdf/2302.06639.pdf
IBM Quantum Backend don't have the hardware (arithmetic circuits) available for "126 133 Cat Qubits"  
If someone succeeds in doing this, they will know exactly who is capable of it in the scientific community and they have their addresses where they live. Grin
All other talk is just good advertising for these quantum backend providers.

Thanks for your reply mate :-)
citb0in
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December 16, 2023, 04:09:53 PM
 #4166

Here, I give you 2 which are really close to each other, lets go.😉


Code:
03b4f75b9205bdcfcf269e604e27066ea422f11e7c1cf50dee6d6b005510e4deaf
02f92b4119e2ecdc2924ff6983f65c3b5af9ed036ce5a6469f3462965840e3daef

The difference between the two public keys is 6.

If I had more time I would find the sum. Once you know the difference and sum then you can solve for the private keys individually. Simple algebra really.

it is 4

  _      _   _       __  _          _  _   __
 |_) |  / \|/   (_  / \ | \  / |_ |_) (_ 
 |_) |_ \_/ \_ |\   __) \_/ |_ \/  |_ | \ __)
--> citb0in Solo-Mining Group <--- low stake of only 0.001 BTC. We regularly rent about 5 PH/s hash power and direct it to SoloCK pool. Wanna know more? Read through the link and JOIN NOW
nomachine
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December 17, 2023, 06:11:19 AM
Last edit: December 17, 2023, 07:19:03 AM by nomachine
 #4167

Hi mate,

how it is going so far? How much is the real rate of keys/s?

Quantum computers, in theory, could provide a speedup for certain types of search problems, but finding hash collisions for cryptographic hash functions is not one of them. Cryptographic primitives like hash functions and ECDSA are designed to be resistant to preimage attacks, and quantum computers do not break this resistance.
Quantum computation do not pose an immediate threat to the security of Bitcoin or other widely used cryptographic systems.
This approach is more of a proof-of-concept or educational exercise rather than a practical implementation for breaking Bitcoin's security.
Quantum algorithm/hardware for solving such problems would need to be carefully designed to provide actual advantages over classical approaches.
Example:
https://arxiv.org/pdf/2302.06639.pdf
IBM Quantum Backend don't have the hardware (arithmetic circuits) available for "126 133 Cat Qubits"  
If someone succeeds in doing this, they will know exactly who is capable of it in the scientific community and they have their addresses where they live. Grin
All other talk is just good advertising for these quantum backend providers.

Are you sure ?

From what i know ECDSA is vulnerable to quantum computers attack. Maybe not that one IBM allows to use for free for 10 minutes. But generally : IT'S VULNERABLE.

https://security.stackexchange.com/questions/34940/is-ecdsa-breakable-by-quantum-computers

Did you read the link above?  Theoretically it is. Practically not. Equipment required for something like this goes beyond home garages conditions. Even ordinary quantum computers. A special laboratory is needed for such a computer.
Not to mention that laser generated (radiated) random numbers are needed combined with a quantum computer. And such a quantum computer must have a special kind of qubit for Shor's algorithm, special efficiency power supplies, sub-zero cooling and a fully controlled environment.
Development and deployment of practical, large-scale quantum computers capable of breaking these algorithms are still in the early stages. Many technical challenges need to be addressed before quantum computers become a practical threat to current cryptographic systems.
Laser-generated random numbers, on the other hand, are not directly related to the security of ECDSA or other cryptographic algorithms. Random number generation is a separate aspect of cryptography that is crucial for key generation and other cryptographic processes.

bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
nomachine
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December 17, 2023, 07:35:46 AM
Merited by digaran (1)
 #4168

One does not become a university professor after posting one liner or few liner generic sh*tposts.

You still haven't answered the question of where your scientific works are. Where can we read online? Grin

bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
citb0in
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December 17, 2023, 10:24:55 AM
 #4169

@nomachine: you quoted him, so I see his messages... damn  Grin it took me a while to find out about the great button IGNORE and I can highly recommend using it. You will find it under his name. Believe me or not - it helps! Wink


  _      _   _       __  _          _  _   __
 |_) |  / \|/   (_  / \ | \  / |_ |_) (_ 
 |_) |_ \_/ \_ |\   __) \_/ |_ \/  |_ | \ __)
--> citb0in Solo-Mining Group <--- low stake of only 0.001 BTC. We regularly rent about 5 PH/s hash power and direct it to SoloCK pool. Wanna know more? Read through the link and JOIN NOW
yellowstripes
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December 17, 2023, 10:36:00 AM
 #4170

Here, I give you 2 which are really close to each other, lets go.😉


Code:
03b4f75b9205bdcfcf269e604e27066ea422f11e7c1cf50dee6d6b005510e4deaf
02f92b4119e2ecdc2924ff6983f65c3b5af9ed036ce5a6469f3462965840e3daef

The difference between the two public keys is 6.

If I had more time I would find the sum. Once you know the difference and sum then you can solve for the private keys individually. Simple algebra really.

it is 4

Yes. My bad. It is 4. I misread the column number on my spreadsheet.
nomachine
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December 17, 2023, 10:36:53 AM
 #4171

@nomachine: you quoted him, so I see his messages... damn  Grin it took me a while to find out about the great button IGNORE and I can highly recommend using it. You will find it under his name. Believe me or not - it helps! Wink



Thanks, Done.  Wink

bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
KiiraStella
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December 17, 2023, 07:49:44 PM
 #4172

I guess it is not big deal. it should be possible to solve in one day or so. But it is true that it is quite boring to copy paste different number 10 hours for example.
SDDD125
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December 17, 2023, 11:52:43 PM
 #4173

Can anyone recover the valid key by reusing r

Priv = 00000000000000000000000000000000000000000000000000180788E47E326C
Pub=040FAAF5F3AFE58300A335874C80681CF66933E2A7AEB28387C0D28BB048BC634965455EF6AFC625E4AEEF1229F052DFD9AB299B02AAA7E659C0E010A795E49E38
r = 0x1fafd110f1b06756ee15a1781daf5fa52d2d548bc772fb61cbbae7effde374db
s1 = 0x1fafd110f1b06756ee15a1781daf5fa52d2d548bc772fb61cbbae7effde374db
z1 = 0x3f5fa221e360ceaddc2b42f03b5ebf4a5a5aa9178ee5f6c39775cfdffbc6e9b6
z2 = 0x13f5fa221e360ceaddc2b42f03b5ebf49150985fe3e2e96ff57482e6ccbfd2af7
s2 = 0xe0502eef0e4f98a911ea5e87e250a0598d81885ae7d5a4d9f417769cd252cc66
V1 Is True!
V2 Is True!
The recovered value of k is correct!
The value of k retrieved 1: 0x180788e47e326e
The recovered k value 2: 0xfffffffffffffffffffffffffffffebaaedce6af48a03bbfba5703ebb80ed3
The R point corresponding to k1: 0x1fafd110f1b06756ee15a1781daf5fa52d2d548bc772fb61cbbae7effde374db
The R point corresponding to k2: 0x1fafd110f1b06756ee15a1781daf5fa52d2d548bc772fb61cbbae7effde374db
r = 0x949bdc126ed9f366d564a3c5e25e425bedf7045ed85b855fccf7904d7b844f90
s1 = 0x4a4dee09376cf9b36ab251e2f12f212df6fb822f6c2dc2afe67bc826bdc227c8
z1 = 0xdee9ca1ba646ed1a4016f5a8d38d6389e4f2868e4489480fb373587439467758
z2 = 0x1dee9ca1ba646ed1a4016f5a8d38d63889fa16374f3d1e84b7345b701097cb899
s2 = 0xb5b211f6c893064c954dae1d0ed0ded0c3b35ab7431add8bd956966612741979
V1 Is True!
V2 Is True!
The recovered value of k is correct!
The value of k retrieved 1: 0x300f11c8fc64db
The value of k retrieved 2: 0xfffffffffffffffffffffffffffffebaaedce6af48a03bbfa24f7b0739dc66
The R point corresponding to k1: 0x949bdc126ed9f366d564a3c5e25e425bedf7045ed85b855fccf7904d7b844f90
The R point corresponding to k2: 0x949bdc126ed9f366d564a3c5e25e425bedf7045ed85b855fccf7904d7b844f90
r = 0x588c57b33f7b758b0d10d1335cee3cec17f1babadfbf4ba96258ffce6c22e34c
s1 = 0x2c462bd99fbdbac586886899ae771e760bf8dd5d6fdfa5d4b12c7fe7361171a6
z1 = 0xb118af667ef6eb161a21a266b9dc79d82fe37575bf7e9752c4b1ff9cd845c698
z2 = 0x1b118af667ef6eb161a21a266b9dc79d6ea92525c6ec7378e84845e29a87c07d9
s2 = 0xd3b9d4266042453a797797665188e188aeb5ff893f68fa670ea5dea59a24cf9b
V1 Is True!
V2 Is True! Huh
Woz2000
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December 18, 2023, 05:00:47 AM
 #4174

Don't forget to ignore his imaginary friends as well.

@nomachine: you quoted him, so I see his messages... damn  Grin it took me a while to find out about the great button IGNORE and I can highly recommend using it. You will find it under his name. Believe me or not - it helps! Wink



Thanks, Done.  Wink
yellowstripes
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December 18, 2023, 05:01:41 AM
 #4175

One does not become a university professor after posting one liner or few liner generic sh*tposts.

You still haven't answered the question of where your scientific works are. Where can we read online? Grin

For those solving puzzle 130... I have been able to reduce 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 to a 110 bit search range and have the 1st 4 digits. I have worked on this daily for the past 2 months. and unfortunately do not have the hardware that would allow me to perform an optimal search in the range. I am willing to sell this info for the fair price of 1BTC. PM if interested. If you want a clue if what am offering is real...feel free to ask.
yellowstripes
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December 18, 2023, 12:04:28 PM
 #4176

For those solving puzzle 130... I have been able to reduce 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 to a 110 bit search range and have the 1st 4 digits. I have worked on this daily for the past 2 months. and unfortunately do not have the hardware that would allow me to perform an optimal search in the range. I am willing to sell this info for the fair price of 1BTC. PM if interested. If you want a clue if what am offering is real...feel free to ask.
That public key is useless, why would we need it? Besides, if you have managed to reduce the size, you should continue reducing it further, unless your imaginary method doesn't actually work, but nice try asking 1BTC for nothing.
I will give you a public key and a range, if you can tell me it's first 3 characters, then I will ask you to reveal the 4 characters you claim to know to puzzle creator and receive your coin, deal?


It actually is not useless. 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 is the difference between the public key of puzzle 130 and the end range. Once you know it, you just need to subtract the private key of 026a...from the end range of puzzle 130 and then get the key of puzzle 130.
sssergy2705
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December 18, 2023, 12:41:04 PM
 #4177

For those solving puzzle 130... I have been able to reduce 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 to a 110 bit search range and have the 1st 4 digits. I have worked on this daily for the past 2 months. and unfortunately do not have the hardware that would allow me to perform an optimal search in the range. I am willing to sell this info for the fair price of 1BTC. PM if interested. If you want a clue if what am offering is real...feel free to ask.
That public key is useless, why would we need it? Besides, if you have managed to reduce the size, you should continue reducing it further, unless your imaginary method doesn't actually work, but nice try asking 1BTC for nothing.
I will give you a public key and a range, if you can tell me it's first 3 characters, then I will ask you to reveal the 4 characters you claim to know to puzzle creator and receive your coin, deal?


It actually is not useless. 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 is the difference between the public key of puzzle 130 and the end range. Once you know it, you just need to subtract the private key of 026a...from the end range of puzzle 130 and then get the key of puzzle 130.

Well, decide for yourself, why do you need to share money with someone?
As for the equipment, there are no problems now. You can rent any server, even with huge RAM, or with powerful video cards.
SDDD125
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December 18, 2023, 07:22:43 PM
 #4178

I guess it is not big deal. it should be possible to solve in one day or so. But it is true that it is quite boring to copy paste different number 10 hours for example.
You are absolutely right, it's not that much of a BIG deal, only worth $1 trillion in total. easy peasy.

@SDDD125, please explain your steps, what script you use etc, btw, lattice attack is useless here.

I would like to know, because it is not possible to recover the private key d in this type of signature, they are valid, they reuse the same point r but using the formula print (hex(((z1*s2 - z2*s1) * modinv((r*( s1-s2)),p)) % p)) does not give the correct value
yellowstripes
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December 19, 2023, 06:13:22 AM
 #4179

It actually is not useless. 026a0747b3229f32ce2f0f7bd77a7bd306f6c95d27e7c5bee22a417938d9988605 is the difference between the public key of puzzle 130 and the end range. Once you know it, you just need to subtract the private key of 026a...from the end range of puzzle 130 and then get the key of puzzle 130.
Congratulations, you just managed to subtract #130 from it's end range, I wonder is that all you learned in the past 2 month? So what the hell am I even doing here all this time if not teaching at least 1 person something useful? sigh!

Here, think of this as #130 :
Code:
000000000000000000000000000000034551231950b75fc4402da1732fc9bebf  - 00000000000000000000000000000003ffffffffffffffffffffffffffffffff  =  00000000000000000000000000000000baaedce6af48a03bbfd25e8cd0364140

Now you want to sell the third key above for 1BTC? Do you have it? then solve the puzzle already, note the difference between the first key and the third key, if you drop the first char aka "3", you'd realize the rest of the characters of the first key have turned into a mirror version in the third key. E.g, "455123" is mirrored into "baaedc" now what is 0xb? in decimal is 11, so 4 turned into 11, 5 turned into 0xa(10), 1 turned into 0xe(14), 2 turned into 0xd(13) etc.  Work on a solution based on that knowledge, figure out their difference in percent, it is a fixed value, no matter what, any number has a fixed difference value with it's mirror self.
0,1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, f.
f, e, d, c, b, a, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0.

Work on that instead of trying to sell your calculations, then you can solve it yourself.



What 3rd key? Digaran ...please if you don't understand what I posted then ask...and nicely too for that matter. Just because our ideas and methods do not align does not mean yours is superior. And certainly put me in touch with the creator of the puzzle...I'll certainly tell him the first 4 digits of the private key of puzzle 130 for that 1 bitcoin and even gave a 110 bit range for it.
mabdlmonem
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December 19, 2023, 09:57:25 PM
 #4180

Code:
using System;
using System.Numerics;
using System.Security.Cryptography;
using System.Text;

public class Program
{
    public static void Main()
    {
        BigInteger x = BigInteger.Parse("55066263022277343669578718895168534326250603453777594175500187360389116729240");
        BigInteger y = BigInteger.Parse("32670510020758816978083085130507043184471273380659243275938904335757337482424");

        string compressed = CompressPoint(x, y);
        Console.WriteLine("Compressed Form: " + compressed);

        string hash160 = GetHash160(compressed);
        Console.WriteLine("Hash160: " + hash160);
    }

    public static string CompressPoint(BigInteger x, BigInteger y)
    {
        string prefix;
        if (y % 2 == 0)
        {
            prefix = "02";
        }
        else
        {
            prefix = "03";
        }

        string xHex = x.ToString("x");
        string compressed = prefix + xHex;

        return compressed;
    }

    public static string GetHash160(string data)
    {
        byte[] bytes = StringToByteArray(data);
        byte[] hashBytes = SHA256.Create().ComputeHash(bytes);

        using (RIPEMD160 ripemd160 = RIPEMD160.Create())
        {
            byte[] hash160Bytes = ripemd160.ComputeHash(hashBytes);
            return ByteArrayToString(hash160Bytes);
        }
    }

    public static byte[] StringToByteArray(string hex)
    {
        return Enumerable.Range(0, hex.Length)
                         .Where(x => x % 2 == 0)
                         .Select(x => Convert.ToByte(hex.Substring(x, 2), 16))
                         .ToArray();
    }

    public static string ByteArrayToString(byte[] bytes)
    {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < bytes.Length; i++)
        {
            sb.Append(bytes[i].ToString("x2"));
        }
        return sb.ToString();
    }
}

i would be apperciate if anyone has knowldge in c# help me to fix this code
Code:
public static string GetHash160(string data)
    {
        byte[] bytes = StringToByteArray(data);
        byte[] hashBytes = SHA256.Create().ComputeHash(bytes);

        using (RIPEMD160 ripemd160 = RIPEMD160.Create())
        {
            byte[] hash160Bytes = ripemd160.ComputeHash(hashBytes);
            return ByteArrayToString(hash160Bytes);
        }
    }

The type or namespace name 'RIPEMD160' could not be found (are you missing a using directive or an assembly reference?)
The name 'RIPEMD160' does not exist in the current context

this the problem
the expected result is
0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
751e76e8199196d454941c45d1b3a323f1433bd6
thanks again for who would help
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