Address without public key and private key?

[mjojo]

Maybe somebody can explain, I run script use some chars (not public key) as input sha256 then hashing to RMD160 then to base58 address and the address is active and have balance. Then I run same script use real public key as input with same above sequence and give active btc address too. my question is possible create address without private key and public key?

[Ambatman]

No. A valid Bitcoin address are derived from the Pubkey through SHA 256 Which in turn is gotten from a private Key.
You can not create an address that's useable without a private key.
You can indeed create an arbitrary Bitcoin address
But it's useless without having a private and pubkey
So Yes to your Question and a No if it's a valid address.


An address without a Pubkey or private key
Is like a Gun without a bullet or a user.

[mjojo]

No. Bitcoin address are derived from the Pubkey through SHA 256 Which in turn is gotten from a private Key.
You can not create an address that's useable without a private key.
You can indeed create an arbitrary Bitcoin address
But it's useless without having a private and pubkey.


An address without a Pubkey or private key
Is like a Gun without a bullet or a user.

But as I said above I can use some chars as input sha256, can you explain this? this address 1HT7xU2Ngenf7D4yocz2SAcnNLW7rK8d4E, 12zqc8j8Btu8fvDbTTky9PXEMkd9ub5K5s never sent out the coin but I can find the hash sha256 from this address.

[Ambatman]

But as I said above I can use some chars as input sha256, can you explain this? this address 1HT7xU2Ngenf7D4yocz2SAcnNLW7rK8d4E, 12zqc8j8Btu8fvDbTTky9PXEMkd9ub5K5s never sent out the coin but I can find the hash sha256 from this address.

For clarification my definition of valid address is if you in possession of the private key.

You indeed created a Bitcoin address which is possible with the method you took.
The address could still have balance because of several reasons
Maybe it was a burnt address or was mistakenly sent there
Someone did the same thing you did and sent some coins there as an experiment.




P.S can you create an address without a pubkey or private key?
Yes, you can generate an address from an hashable data or even random characters
But you can't spend from it without a private key

https://bitcointalk.org/index.php?topic=1225430.0 ( for these address 1JcJZUPiFkitmWACZ3ihQtFVwpeNx9D7Kz)

https://bitcointalk.org/index.php?topic=5457320.0 ( 1JcJZUPiFkitmWACZ3ihQtFVwpeNx9D7Kz)


Check these topics out and they will help you understand the problem.
I knew it should be an existing address since it had some balance.

And it turns out none of them are new.

[hosemary]

my question is possible create address without private key and public key?

Yes.
To generate a valid bitcoin address, you don't have to start with a private key. If your string has a correct format and it has a correct checksum, you have a valid address and you can send bitcoin to that.
You probably know that if you send fund to such bitcoin address, you won't be able to spend that and the fund will be lost.

[khaled0111]

What hosrmary said is correct.
Take these addresses as an example:
1BitcoinEaterAddressDontSend8MUo1T
1CounterpartyXXXXXXXXXXXXXXXUWLpVr
...
Those addresses weren’t derived from private keys but they are valid and you can see that they already received multiple transactions.
There are many tools that can help you create such addresses but there is no way to verify that they didn’t derive them from private keys they own, So better use your own tool.

[BitMaxz]

But as I said above I can use some chars as input sha256, can you explain this? this address 1HT7xU2Ngenf7D4yocz2SAcnNLW7rK8d4E, 12zqc8j8Btu8fvDbTTky9PXEMkd9ub5K5s never sent out the coin but I can find the hash sha256 from this address.

Actually, you can create BTC addresses as many as you like, but what is your main purpose? Since you don't have full control of these addresses. Sending any BTC to these addresses is considered lost because you can't sign a transaction without the private keys.

Why not start first to share how exactly you created them or give us the open source code of the script you are using, and maybe some of the professionals here or developers can verify the script and how it generated a valid BTC address?

[mjojo]

But as I said above I can use some chars as input sha256, can you explain this? this address 1HT7xU2Ngenf7D4yocz2SAcnNLW7rK8d4E, 12zqc8j8Btu8fvDbTTky9PXEMkd9ub5K5s never sent out the coin but I can find the hash sha256 from this address.

Actually, you can create BTC addresses as many as you like, but what is your main purpose? Since you don't have full control of these addresses. Sending any BTC to these addresses is considered lost because you can't sign a transaction without the private keys.

Why not start first to share how exactly you created them or give us the open source code of the script you are using, and maybe some of the professionals here or developers can verify the script and how it generated a valid BTC address?

Code:

import hashlib

def sha256(data):
    """Hashes the input data using SHA-256"""
    return hashlib.sha256(data).digest()

def ripemd160(data):
    """Hashes the input data using RIPEMD-160"""
    h = hashlib.new('ripemd160')
    h.update(data)
    return h.digest()

input_data = bytes.fromhex("0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798")

# Step 1: SHA-256 Hash
sha256_result = sha256(input_data)
print("SHA-256 Hash:", sha256_result.hex())

# Step 2: RIPEMD-160 Hash
ripemd160_result = ripemd160(sha256_result)
print("RIPEMD-160 Hash:", ripemd160_result.hex())

Thank for all explaining.
script is just normal script as usual. and this hash sha256 for both address f299791cddd3d6664f6670842812ef6053eb6501bd6282a476bbbf3ee91e750c and e3b0c44298fc1c149afbf4c8996fb92427ae41e4649b934ca495991b7852b855

[nc50lc]

I saw that you also posted this on the "Puzzle transaction thread", if you can clarify the main goal for asking this, people can give you more tailored answers.
If it's simply creating an address without a pubKey or prvKey, then as others' have said: Yes.
If there's more to it, then please expand the question.

In fact, you can just use RNG to generate a pseudo-random 160-bit number, prepend with 0x00 and append with 0x01 then "base58check" encode that, and you got a compressed legacy address.
You can skip all those steps in your script (no hash required) and you'll still be able to generate an address that has no public/private key.

[mjojo]

I saw that you also posted this on the "Puzzle transaction thread", if you can clarify the main goal for asking this, people can give you more tailored answers.
If it's simply creating an address without a pubKey or prvKey, then as others' have said: Yes.
If there's more to it, then please expand the question.

In fact, you can just use RNG to generate a pseudo-random 160-bit number, prepend with 0x00 and append with 0x01 then "base58check" encode that, and you got a compressed legacy address.
You can skip all those steps in your script (no hash required) and you'll still be able to generate an address that has no public/private key.

I just wonder is there possible collision in hash sha256 or in hash RMD160? because with different input method (bytes, string or integer) will give different output too. or public key is possible to substitute with some integer that represent public key it self (compressed, uncomp or in X,Y coordinate). Maybe all above response is correct if  that both address is have no pvkey & pubkey and just create by hash only (in other words unspendable address). Or we are just waiting quantum computer to execute shor algo for brute all address in 256 bits space to prove that address have pvkey and pubkey or not.

**PS
sorry for my english, just translated

[nc50lc]

If there's more to it, then please expand the question.

I just wonder is there possible collision in hash sha256 or in hash RMD160?

Oh I see, then the answer is there's little to no chance of collision.
There's a possibility, yes because 2^160 (the "pubKeyhash") is a finite number but still too large to have a reasonable chance of collision happening in our lifetime. (at least with the current hardware)
Not to mention, SHA256 hash which is 2^256.

Take note those hashes are not exclusive to Bitcoin or Cryptocurrencies so if ever there's a legitimate collision happened in the past,
It would've been in the news by now.

[mjojo]

If there's more to it, then please expand the question.

I just wonder is there possible collision in hash sha256 or in hash RMD160?

Oh I see, then the answer is there's little to no chance of collision.
There's a possibility, yes because 2^160 (the "pubKeyhash") is a finite number but still too large to have a reasonable chance of collision happening in our lifetime. (at least with the current hardware)
Not to mention, SHA256 hash which is 2^256.

Take note those hashes are not exclusive to Bitcoin or Cryptocurrencies so if ever there's a legitimate collision happened in the past,
It would've been in the news by now.

Thank for clarification, so I can conclusion hash RMD160 is possible to collision the reason;
1. The input have 2**256 combination (SHA256)
2. The output only have 2**160 combination (RMD160)

Lets break with this key;
- pvkey
00000000000000000000000000000000000000000000000300000e4692d2217b
00000000000000000000000000000000000000000000000300005a15da2a15b1

- pubkey (comp)
0358ec223b4c813a10033d5519da9ca50f0c8754b914fbd5500c6cbe9ea290cff5
0355a3cc6b5399da908a19fac59abf1382c7dd599198cb323a5fcbde7fad39e8cf

- SHA256
9ec24e06cacf3033c8afaf1ae266c9aec885523144992c82dc1e736b8219dda2
53e94b4ccadab2d3a3ad3323b7a07bc9752e7fcdcb25ccc1384b332b558630ef

- RMD160
20d45a6a7627 d06f3ad6e0f34707d19a2f9f12f1  (12 prefixes collision)
20d45a6a7627 38faae370bae7cc4517b0ccf01b2 (12 prefixes collision)

but this just paper calculation and until full collision found this is only hypothesis, maybe ratio 1:2 between hash sha256 with hash RMD160 is like coordinate in curve secp256k1 which one X coordinate have two Y coordinate

**
Thank for all response

[nc50lc]

Thank for clarification, so I can conclusion hash RMD160 is possible to collision the reason;
-snip-
but this just paper calculation and until full collision found this is only hypothesis, maybe ratio 1:2 between hash sha256 with hash RMD160 is like coordinate in curve secp256k1 which one X coordinate have two Y coordinate

Just don't get your hopes up on this if you're going to use this for something like the puzzle transaction
since you wont likely get a full collision of the HASH160 result even if you procure a high number of GPUs since that possibility of that is too low.

You only demonstrated a partial 48-bit collision which is expectedly common since it's relatively small.

[DannyHamilton]

maybe ratio 1:2 between hash sha256 with hash RMD160

There are 2²⁵⁶ possible SHA256 results when hashing the public key

There are 2¹⁶⁰ possible RIPEMD160 results when hashing the SHA256 hash result.

Since there are more SHA256 results than there are RIPEMD160 results, there must be some SHA256 results which, when hashed with RIPEMD160 will provide identical results (this is known as the pigeonhole principal).

Dividing the number of possible SHA256 results by the number of possible RIPEMD160 results, we find that, on average, each RIPEMD160 result can be computed from 2⁹⁶ different SHA256 results. That implies that there is an average of 2⁹⁶ valid private keys for each address computed with RIPEMD160 (that's 1 : 2⁹⁶ ratio, and not the 1:2 ratio that you suggested).

The odds of finding any one of those 2⁹⁶ private keys is vanishingly small. It just isn't going to happen. With nearly 2²⁵⁶ valid private keys,  and only 2⁹⁶ of them being valid for a given address, you have a 0.0000000000000000000000000000000000000000000000684% chance of a collision on each attempt.

[Catenaccio]

The odds of finding any one of those 2⁹⁶ private keys is vanishingly small. It just isn't going to happen. With nearly 2²⁵⁶ valid private keys,  and only 2⁹⁶ of them being valid for a given address, you have a 0.0000000000000000000000000000000000000000000000684% chance of a collision on each attempt.

I don't understand it technically deep but I see some illustrative explanation with similar information on impossibility of finding Bitcoin private keys like finding an atom in a universe.

Bitcoin private keys.

The size of Bitcoin’s private key space (2²⁵⁶) is an unfathomably large number. It is approximately 10⁷⁷ in decimal. For comparison, the visible universe is estimated to contain 10⁸⁰ atoms.

Private keys.

The range of 256-bit numbers (and therefore the number of possible private keys) is unfathomably large. Just as it's impossible for the human mind to visualize the true scale of the universe, it's impossible for the human mind to comprehend the sheer size of 256-bit numbers.

This may seem hard to believe, but to give you some perspective, there are roughly 2²⁵⁶ or 10⁷⁷ possible private keys, and there are roughly 10⁷⁸ atoms in the universe[1]. So it's like asking two different people to randomly select an atom in the universe and for both to choose the exact same one.

To put it another way, it's like two people choosing the same grain of sand from anywhere on the earth (10¹⁸)[2]. Except each grain of sand contains another earth-worth of sand, and each grain of that sand contains another earth-worth of sand, and each grain of that sand also contains another earth worth of sand.

And even this amount of sand is still far, far less than the number of possible private keys.

[nc50lc]

I don't understand it technically deep but I see some illustrative explanation with similar information on impossibility of finding Bitcoin private keys like finding an atom in a universe.

Those articles are close to that reply's explanation but those are about "private key collision".
Which explained the whole private key range's search space and the possibility of randomly generating the same private key.

On the other hand, the reply is about "pubKeyHash collision" (a.k.a address) which is about (in summary):
The possibility of multiple different private keys to generate the same pubKeyHash.