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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 231869 times)
mabdlmonem
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December 12, 2023, 01:39:26 AM
 #4141

hello again , how much time and resource to find a public key in range 40000000000000000000000000000...7ffffffffffffffffffffffffffff  115 puzzle with kangaroo
appreciate your replies , thanks
A 115 bit puzzle has already been solved. If you are looking at JLPs Kangaroo program; In that search, the person ran 256+ Tesla V100 GPUs for (I think) 11-13 days, using JLPs Kangaroo program.
I got 1024 public in 115 one of them is the division of 130 .
WanderingPhilospher
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December 12, 2023, 03:25:12 AM
 #4142

hello again , how much time and resource to find a public key in range 40000000000000000000000000000...7ffffffffffffffffffffffffffff  115 puzzle with kangaroo
appreciate your replies , thanks
A 115 bit puzzle has already been solved. If you are looking at JLPs Kangaroo program; In that search, the person ran 256+ Tesla V100 GPUs for (I think) 11-13 days, using JLPs Kangaroo program.
I got 1024 public in 115 one of them is the division of 130 .

Don't we all man, don't we all.

Why stop there, why not bump up pubkeys and reduce the search range even more?
mabdlmonem
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December 12, 2023, 10:16:03 AM
 #4143

hello again , how much time and resource to find a public key in range 40000000000000000000000000000...7ffffffffffffffffffffffffffff  115 puzzle with kangaroo
appreciate your replies , thanks
A 115 bit puzzle has already been solved. If you are looking at JLPs Kangaroo program; In that search, the person ran 256+ Tesla V100 GPUs for (I think) 11-13 days, using JLPs Kangaroo program.
I got 1024 public in 115 one of them is the division of 130 .

Don't we all man, don't we all.

Why stop there, why not bump up pubkeys and reduce the search range even more?
More range, means double the public keys, 1023 will 2048 and so on
yellowstripes
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December 13, 2023, 09:08:36 AM
 #4144


I appreciate the apology and thank you for the link to Kangaroo.
It won't work with 130 bit. Several program functions need to be rewritten.

Thank you for letting me know. I guess its luck if anyone has come across the public keys I posted. Has anyone successfully scanned up to the 116bit key range and if yes...how long did it take and what program did you use?

I really am looking for the private keys of these public keys
03bd303397e428bf036db49510b70b86d0025f73744b6324a579de16dc591e2056
0213bedb35636bea09e628f3f02d99531a4aeda3fc7ace6bb0a2a80a2a1fc03a70

The sum of the above when added together adds up to this private key 88239662793269832304564822427566081 (decimal format) whose public key is 0352c408ee78229a44b34b1c1672ae5bdf568d865e3c3aefffb07639202f10c795.






The difference between the two keys posted above is public key 02d961be3ab20a70447347f0e487c676d47f61af97115236c38c37aae92df795a6.

The sum of 02d961be3ab20a70447347f0e487c676d47f61af97115236c38c37aae92df795a6 + 0226a400fc8918655bf835ae2beceb8e0c309e8db5a70ae796748f886c936660f2 = 03710c674b0de42c2e5adf1faa4054ddec98db7983c82aaeda53e8109e23b06513


03710c674b0de42c2e5adf1faa4054ddec98db7983c82aaeda53e8109e23b06513 = 44119831396634916152282411213783040 (private key in decimal value) which is half value of 0352c408ee78229a44b34b1c1672ae5bdf568d865e3c3aefffb07639202f10c795 plus 1.


tmar777
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December 13, 2023, 11:16:04 AM
 #4145


I appreciate the apology and thank you for the link to Kangaroo.
It won't work with 130 bit. Several program functions need to be rewritten.

Thank you for letting me know. I guess its luck if anyone has come across the public keys I posted. Has anyone successfully scanned up to the 116bit key range and if yes...how long did it take and what program did you use?

I really am looking for the private keys of these public keys
03bd303397e428bf036db49510b70b86d0025f73744b6324a579de16dc591e2056
0213bedb35636bea09e628f3f02d99531a4aeda3fc7ace6bb0a2a80a2a1fc03a70

The sum of the above when added together adds up to this private key 88239662793269832304564822427566081 (decimal format) whose public key is 0352c408ee78229a44b34b1c1672ae5bdf568d865e3c3aefffb07639202f10c795.





The difference between the two keys posted above is public key 02d961be3ab20a70447347f0e487c676d47f61af97115236c38c37aae92df795a6.

The sum of 02d961be3ab20a70447347f0e487c676d47f61af97115236c38c37aae92df795a6 + 0226a400fc8918655bf835ae2beceb8e0c309e8db5a70ae796748f886c936660f2 = 03710c674b0de42c2e5adf1faa4054ddec98db7983c82aaeda53e8109e23b06513


03710c674b0de42c2e5adf1faa4054ddec98db7983c82aaeda53e8109e23b06513 = 44119831396634916152282411213783040 (private key in decimal value) which is half value of 0352c408ee78229a44b34b1c1672ae5bdf568d865e3c3aefffb07639202f10c795 plus 1.




Man, I cannot understand what you are trying to achieve... Even though I see some arithmetic logic in your calculations, this doesn't mean anything (at least to me who I don't know how you ended up with these numbers). Also, where 0226a400fc8918655bf835ae2beceb8e0c309e8db5a70ae796748f886c936660f2 is coming from?
kalos15btc
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December 13, 2023, 11:49:22 AM
 #4146


So what is it you want? You explained in detail about 2+7=9, which is half of 18, but what are you getting at? Because you are looking for X, which is unknown, do you have any value related to X?

Never mind that, let me give you a hint:
Let your X be 35, your end range be 50, now you can sub 35 -50 = 15, sub 15 from half of 50, so 15-25 = 10, then sub 10 from half of 25, so 10-12.5 = 2.5, now you should multiply by 4 to get 10, by 5 to get 12.5, by 6 to get 15 and so on.

Someone here was talking about being good in math, this should be easy for mathematicians to figure out, after all we have the start and end range.

bro we dont need your advices, you are the professor of maths and crypto here, please dont wright comments that dont have any meanings, for 1 year and you are typing nothing in this thread. except big talk and there is nothing, this topic become useless i swear, , if any mod here please close this thread.
newbie now cant see all +200 pages on this thread thats why we will have lot of newbie asking for pk of their substracted public keys without knowloadge they must see all replys and this topic become useless.
yellowstripes
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December 13, 2023, 12:23:48 PM
 #4147


So what is it you want? You explained in detail about 2+7=9, which is half of 18, but what are you getting at? Because you are looking for X, which is unknown, do you have any value related to X?

Never mind that, let me give you a hint:
Let your X be 35, your end range be 50, now you can sub 35 -50 = 15, sub 15 from half of 50, so 15-25 = 10, then sub 10 from half of 25, so 10-12.5 = 2.5, now you should multiply by 4 to get 10, by 5 to get 12.5, by 6 to get 15 and so on.

Someone here was talking about being good in math, this should be easy for mathematicians to figure out, after all we have the start and end range.

bro we dont need your advices, you are the professor of maths and crypto here, please dont wright comments that dont have any meanings, for 1 year and you are typing nothing in this thread. except big talk and there is nothing, this topic become useless i swear, , if any mod here please close this thread.
newbie now cant see all +200 pages on this thread thats why we will have lot of newbie asking for pk of their substracted public keys without knowloadge they must see all replys and this topic become useless.

Actually am on a path to solve puzzle 130. The numbers am posting here are connected to the private key of puzzle 130. I however, am the only one who can trace back how. Once I solve it, I will let you in on my nonsensical calculations.
yellowstripes
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December 13, 2023, 01:00:23 PM
 #4148

Actually am on a path to solve puzzle 130. The numbers am posting here are connected to the private key of puzzle 130. I however, am the only one who can trace back how. Once I solve it, I will let you in on my nonsensical calculations.
I told you to ignore anyone here complaining, if you read their posts you'll realize they have nothing substantial to add to the conversation. They also think you and other newbies are my alt accounts, so whatever insult they throw is directed to me, don't take offense.

Let me give you a bit of advice, 6 months ago I was exactly where you are now, using keysubtracter has it's own tricks, but it's also very confusing. you'll understand later. Regardless I wish you get what you want if you really deserve it. Good luck.

I have seen that. I don't know why someone would think you are talking to yourself!!!!  I have been through the loops you are talking about and that's why past here am not using a key subtracter anymore. I can easily find myself with negative values. At least I am in a range that can be scanned.
GR Sasa
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December 13, 2023, 01:22:08 PM
Merited by citb0in (1), alek76 (1)
 #4149

Who else cracked down out of laughing because digaran is talking and replying to himself with his accounts?  Grin Grin
yellowstripes
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December 13, 2023, 02:21:19 PM
 #4150

Who else cracked down out of laughing because digaran is talking and replying to himself with his accounts?  Grin Grin


I am not Digaran.
vhh
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December 13, 2023, 03:30:11 PM
 #4151

Who else cracked down out of laughing because digaran is talking and replying to himself with his accounts?  Grin Grin

Agree  Grin Grin
mabdlmonem
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December 13, 2023, 05:14:06 PM
 #4152

Actually am on a path to solve puzzle 130. The numbers am posting here are connected to the private key of puzzle 130. I however, am the only one who can trace back how. Once I solve it, I will let you in on my nonsensical calculations.
I told you to ignore anyone here complaining, if you read their posts you'll realize they have nothing substantial to add to the conversation. They also think you and other newbies are my alt accounts, so whatever insult they throw is directed to me, don't take offense.

Let me give you a bit of advice, 6 months ago I was exactly where you are now, using keysubtracter has it's own tricks, but it's also very confusing. you'll understand later. Regardless I wish you get what you want if you really deserve it. Good luck.

I have seen that. I don't know why someone would think you are talking to yourself!!!!  I have been through the loops you are talking about and that's why past here am not using a key subtracter anymore. I can easily find myself with negative values. At least I am in a range that can be scanned.
I downgrade the public 130 to 104 , and got a 15 public each one substracted from constant value and each public has his constant values. For puzzle 66 that doesn't have public key yet , I have 33 million value needs to be scanned in range 2**40
alek76
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December 13, 2023, 07:19:15 PM
 #4153

Who else cracked down out of laughing because digaran is talking and replying to himself with his accounts?  Grin Grin
I can't laugh Smiley I can tell you where he screwed up. He said: - I am close to the decision 130. This is Digaran, I have no doubt.
Woz2000
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December 13, 2023, 08:30:07 PM
 #4154

Three accounts so far. Sadly its not very funny. Its driving people away and we lose the purpose of this forum.


Who else cracked down out of laughing because digaran is talking and replying to himself with his accounts?  Grin Grin
I can't laugh Smiley I can tell you where he screwed up. He said: - I am close to the decision 130. This is Digaran, I have no doubt.
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December 13, 2023, 11:47:10 PM
 #4155

I wrote a program to connect to IBM quantum to test looking for puzzle 66 and am currently connected to IBM_Kyoto Japan super computer with the free 10 minutes outcome unknown, is 10 minutes enough?. System puts you in and out of queue with 2 to 3 second snipets so the 10 minutes could take hours before any results if any. anyone who wants to try it go to my github page to download  script/instructions https://github.com/unclevito2017 If like me I have already tried everything

import qiskit
from qiskit import execute, Aer, IBMQ
from qiskit.circuit.library import PhaseOracle
from qiskit_ibm_provider import IBMProvider
from qiskit import QuantumCircuit

def sha256_compression_function(qc, message_bits, expression):
    # Ensure the length of message_bits is 256
    assert len(message_bits) == 256, "Message must be 256 bits long"

    # Apply controlled-X gates based on the message bits
    for i, bit in enumerate(message_bits[:32]):
        if bit == '1' and i < 16:  # Ensure i is within the valid range
            qc.cx(i, 31)  # Apply CX gate to qubit 16 with control qubit i

    # Manually construct the boolean conditions from the expression
    for i, char in enumerate(expression):
        if char == '1' and i < 16:  # Ensure i is within the valid range
            qc.x(i)  # Apply X gate to qubit i

    # Apply the oracle to each qubit individually
    for i in range(16):
        qc.cx(i, 16)  # Controlled-X gate with control qubit i and target qubit 16
        qc.x(i)  # Reset the control qubit qubit 16

def main():
    # Load IBM Quantum account
    IBMQ.load_account()
    provider = IBMProvider()  # No hub, group, or project parameters

    # Target Bitcoin address
    target_address_hex = "20d45a6a762535700ce9e0b216e31994335db8a5"
    target_address_decimal = int(target_address_hex, 16)

    # Define the range for iteration in hexadecimal
    start_range = 0x2000000000000000
    end_range = 0x3fffffffffffffff

    # Iterate over the range
    for decimal_value in range(int(start_range), int(end_range) + 1):
        # Convert decimal value to bytes and binary string
        hex_value = hex(decimal_value)[2:].zfill(32)  # Ensure a fixed length of 32 characters
        message_bytes = bytes.fromhex(hex_value)
        binary_message = ''.join(format(byte, '08b') for byte in message_bytes)
        binary_message = binary_message.zfill(256)  # Pad to 256 bits

        # Initialize quantum circuit with the initial state based on the binary message
        qc = QuantumCircuit(32, 32, name="qc", global_phase=0)

        # Apply bit operations to encode the initial state and message onto the qubits
        for i, bit in enumerate(binary_message[:16]):
            if bit == '1':
                qc.x(i)

        # Implement the SHA-256 compression function using a quantum oracle search for target address prefix 20d45
        sha256_compression_function(qc, binary_message, expression="message[0] == '1' and message[1] == '0' and message[2] == '1' and message[3] == '1'")

        # Measure the final state of the qubits
        qc.measure_all()

        # Use the IBM Quantum backend
        backend = provider.get_backend('ibm_kyoto')

        # Simulate the circuit on the IBM Quantum backend
        job = execute(qc, backend=backend, shots=1024)

        # Get the results and extract the final state
        counts = job.result().get_counts(qc)
        final_state = int(list(counts.keys())[0].replace(" ", ""), 2)

        # Check if the generated hash matches the target address characters 10
        if hex(final_state)[:10] == hex(target_address_decimal)[:10]:
            print(f"Target address found!")
            print(f"Decimal Value: {decimal_value}")
            print(f"Simulated Bitcoin hash160: {hex(final_state)[2:].zfill(40)}")
            break

    else:
        print("Target address not found within the specified range.")

if __name__ == "__main__":
    main()

<a href="https://ibb.co/zXfjbwB"><img src="https://i.ibb.co/BysdjRQ/quantum.jpg" alt="quantum" border="0"></a>
mabdlmonem
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December 14, 2023, 12:21:48 AM
 #4156

I downgrade the public 130 to 104 , and got a 15 public each one substracted from constant value and each public has his constant values. For puzzle 66 that doesn't have public key yet , I have 33 million value needs to be scanned in range 2**40
I'm sorry but I can't really understand what you are saying, what is a constant value? What is the relation between 2**40 and puzzle 130 with puzzle 66? They are totally separate, you can't possibly solve 66 with public key operations, because you'd need to generate rmd160 on each key to compare, so don't mix up those two. Moreover if you want, you can scan and find any key in 2**40 bit range using kangaroo in a second, I linked to the topic a few posts above.


If anyone is wondering about these kids trolling, it's normal ever since the prize were increased after I asked for it, they started their personal attacks, just like 2018 when they realized I was appointed as a merit source after I asked for it. envy burns really deep I suppose.

[insert satoshi and admin's reaction meme here "WTH is he talking about?"] Lol.
I would try that if I have enough resources for kangaroo . And about the relationship between 2**40 and puzzle 66 , you don't have to think about it , I am just playing around
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December 14, 2023, 12:33:29 AM
Last edit: December 14, 2023, 09:58:40 PM by Mr. Big
 #4157

@unclevito, your code looks like a time travel code, and you got a chance at using a quantum computer just to waste it on 66? They are not good at brute forcing, they should be used for public key calculations.

anyone can use it since you are using IBM's resources. any computer that can execute python can get it running. Sooner or later we will all be using super computers so at least get a taste of one



@unclevito, your code looks like a time travel code, and you got a chance at using a quantum computer just to waste it on 66? They are not good at brute forcing, they should be used for public key calculations.
6.6 bitcoins is over $250,000 dollars so 66 is not a waste unless you are an evil billionaire
unclevito
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December 14, 2023, 01:09:38 AM
 #4158

@unclevito, your code looks like a time travel code, and you got a chance at using a quantum computer just to waste it on 66? They are not good at brute forcing, they should be used for public key calculations.
6.6 bitcoins is over $250,000 dollars so 66 is not a waste unless you are an evil billionaire
Public key operations are much faster with QC, anyways you could spend the same amount of energy and resources to solve for 130, which has 13 Bitcoins, unless you are an evil billionaire!😉
I doubt if 10 minutes for 66 even on a super computer is enough and after the 10 minutes its $1.60 per minute so 130 would cost a lot more money than 66. eventually the cost will go down.
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December 14, 2023, 10:31:12 AM
 #4159

I need to make changes to the Kangaroo_CPU.dll file. How can I do this? I cannot see its content. I tried various programs but I could not see the codes. The programs give errors, I think it is encrypted

If you want to buy me a coffee

Btc = 3246y1G9YjnQQNRUrVMnaeCFrymZRgJAP7

Doge = DGNd8UTi8jVTVZ2twhKydyqicynbsERMjs
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December 14, 2023, 11:57:38 AM
Last edit: December 14, 2023, 12:56:09 PM by nomachine
 #4160

I doubt if 10 minutes for 66 even on a super computer is enough and after the 10 minutes its $1.60 per minute so 130 would cost a lot more money than 66. eventually the cost will go down.

Keep in mind the limitations imposed by the IBM Quantum provider, such as the number of allowed jobs, 10 minutes execution time, and queuing times (a shared resources).

Instead of applying Controlled-X gates individually in a loop, you can use Qiskit's QuantumCircuit.mct (Multiple-Control Toffoli) gate to combine multiple Controlled-X gates into a single operation.

qc.mct(list(range(16)), 16)

Code:
def sha256_compression_function(qc, message_bits, expression):
    # Ensure the length of message_bits is 256
    assert len(message_bits) == 256,

    # Apply Controlled-X gates based on the message bits
    qc.mct(list(range(16)), 16)  # Combine Controlled-X gates into a single operation

    # Manually construct the boolean conditions from the expression
    qc.x([i for i, char in enumerate(expression) if char == '1' and i < 16])


Good Luck !

p.s.
I think it takes at least a week or two to hit something with this...... Every day is $1.60x1440.
High class gambling. Grin

I need to make changes to the Kangaroo_CPU.dll file. How can I do this? I cannot see its content. I tried various programs but I could not see the codes. The programs give errors, I think it is encrypted

 DLLs and SOs are binary files that contain compiled code. If you have access to the source code of the library, it is always better to make the necessary changes there and recompile the library. This is the safest and most recommended approach. If not, the question arises as to what is hidden in them?  Huh

bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8
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