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Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 109. (Read 215705 times)

jr. member
Activity: 76
Merit: 4
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()

https://ibb.co/zXfjbwB">https://i.ibb.co/BysdjRQ/quantum.jpg" alt="quantum" border="0">
jr. member
Activity: 82
Merit: 2
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.
member
Activity: 93
Merit: 16
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.
jr. member
Activity: 32
Merit: 1
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
vhh
newbie
Activity: 14
Merit: 2
Who else cracked down out of laughing because digaran is talking and replying to himself with his accounts?  Grin Grin

Agree  Grin Grin
newbie
Activity: 14
Merit: 0
Who else cracked down out of laughing because digaran is talking and replying to himself with his accounts?  Grin Grin


I am not Digaran.
member
Activity: 177
Merit: 14
Who else cracked down out of laughing because digaran is talking and replying to himself with his accounts?  Grin Grin
newbie
Activity: 14
Merit: 0
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.
newbie
Activity: 14
Merit: 0

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.
jr. member
Activity: 50
Merit: 1

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.
newbie
Activity: 7
Merit: 0

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?
newbie
Activity: 14
Merit: 0

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.


jr. member
Activity: 32
Merit: 1
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
full member
Activity: 1162
Merit: 237
Shooters Shoot...
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?
jr. member
Activity: 32
Merit: 1
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 .
full member
Activity: 1162
Merit: 237
Shooters Shoot...
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.
full member
Activity: 431
Merit: 105
The easiest way currently known to brute force is using bit with my method applied to the different algorithms for solving the discrete logarithm.
I only ask that you mention me in your repositories, credit to whom it is deserved! Surely, those who understand the subject are secretly cooking this up, so I recommend not waiting for them (there is no need to judge them either, we are all after the same thing) . learn C, it is most likely that those repositories will come to light once it is unlikely to find a key using bits.
There are even more efficient methods, but honestly I don't want to throw all the cards at once, because I am also after some puzzle of this, I'm sorry, you can't be so generous in a world where nobody gives you anything, it's obvious, But since strength alone is of no use to me because my resources are limited, I share it so that whoever can take advantage of it.

Hi there mcdouglasx,,

how are you? could you maybe explain what parameters or lines to change for letting it search and create db with different range then this
30 bits example could you explain that would be great, gonna test after i get your explanation again, thanks man.
jr. member
Activity: 32
Merit: 1
hello again , how much time and resource to find a public key in range 40000000000000000000000000000...7ffffffffffffffffffffffffffff  115 puzzle with kangaroo
appreciate your replies , thanks
member
Activity: 239
Merit: 53
New ideas will be criticized and then admired.
The easiest way currently known to brute force is using bit with my method applied to the different algorithms for solving the discrete logarithm.
I only ask that you mention me in your repositories, credit to whom it is deserved! Surely, those who understand the subject are secretly cooking this up, so I recommend not waiting for them (there is no need to judge them either, we are all after the same thing) . learn C, it is most likely that those repositories will come to light once it is unlikely to find a key using bits.
There are even more efficient methods, but honestly I don't want to throw all the cards at once, because I am also after some puzzle of this, I'm sorry, you can't be so generous in a world where nobody gives you anything, it's obvious, But since strength alone is of no use to me because my resources are limited, I share it so that whoever can take advantage of it.
member
Activity: 462
Merit: 24
they never have anything constructive to say anyways

How many accounts do you have here? 5? Do you ask yourself questions from one account and then answer from another?
You probably have a lot of fun thanking yourself for the advice.  Grin

then do your homework and don't beg others to do it for you. This is not a job market portal

This is an infestation of those seeking  for The sum of two public keys. It's like they cloned themselves.  Roll Eyes
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