Is it possible to recover private key or mnemonic phrase data?

[Geka041]

Hello. Everyone knows that it is impossible to calculate the private key from the address. But this question interests. For example, I clicked the generate button on this site (iancoleman.io/bip39/) to generate a mnemonic phrase automatically.
As a result, the site issues a mnemonic phrase (catalog situate syrup awkward feature squirrel baby venture carry upon vintage hamster moment divert gadget) and all keys, addresses for receiving bitcoin
Bitcoin address list BIP44:
m/44'/0'/0'/0/0   19oKvHWLCW9Y1uM9mUaTM3QECGa2i5iGR6      
m/44'/0'/0'/0/1   153zq6hKyaYPGFW5Gi1FLCWhUsZ34jBWR4      
m/44'/0'/0'/0/2   15mapW5Kq5MpuyYsWvRniJk7hq4juu7m4w      
m/44'/0'/0'/0/3   1CuVTWnLGMwF2SUYzFD2bUiUcaGSBB8bZx      
m/44'/0'/0'/0/4   1LN6ayuALC2dxmQfjRegrD7Wdmp9fKYTgM      
m/44'/0'/0'/0/5   1Q1w8osa63mMwpjoy8mZpm4Z926XW3xkFC      
m/44'/0'/0'/0/6   1Bj7eVG25YCJwugnxeNzG57h8bDrteLbpy      
m/44'/0'/0'/0/7   16Xyk52PoWL5XZNctK3C8KnuiJmpxrsstN      
m/44'/0'/0'/0/8   1JLXvCMm4yq3o8GtS7YwfUukCxSG8t6VZf      
m/44'/0'/0'/0/9   1EidLMD4UBAjaUC98q6HXni1pttnHz2CP7      
m/44'/0'/0'/0/10   18iRA6yRTzgAFQKibTZWWuMPYQ72Psg1y2      
m/44'/0'/0'/0/11   14rn4BYf4SaDu9qsXePwiG6Z4oQEJiVYxF      
m/44'/0'/0'/0/12   13AUsYABAh5NCnoyBV9BPUYBTwjK1xqVT7      
m/44'/0'/0'/0/13   1Pi2on2aQu1ngByh2ixP5NbEJKdJbbagQC      
m/44'/0'/0'/0/14   19ytiiPPxJCwLaqdbQCyBWKhSSHM3wcF5P      
m/44'/0'/0'/0/15   14EKFJPuhwnNg5pzWF7fVkAK3S214yTxte      
m/44'/0'/0'/0/16   1Kgp4gZwXVDvidRr3oz6w8iy3jtH8ZCMdm      
m/44'/0'/0'/0/17   1BubsVnCpU2ddG9aP1FK1omM5PEiHCoETS      
m/44'/0'/0'/0/18   1GzmsAUr5mbGWCHUcqruvjvaXUo8cTDYUj      
m/44'/0'/0'/0/19   1JekiSDJLjkts6XmMDYiGvrFMcnx4tWRMg      
Now the actual question. Knowing the exact sequence of addresses (there are infinitely many addresses), it is possible to recover a mnemonic phrase or any other information?

[BlackHatCoiner]

Now the actual question. Knowing the exact sequence of addresses (there are infinitely many addresses), it is possible to recover a mnemonic phrase or any other information?

No, there isn't. Every derived address is distinct with each other, because they all get hashed with a different index each time. That being said, whether you have the RIPEMD-160 hash of 1000 addresses or even their revealed public keys, you can't go further to recover any information that would expose they're all derived from the same seed or anything similar.

You obviously can't recover a mnemonic phrase from its hashed derivations.

[ranochigo]

Not possible. You need at least a master public key and one of the private key of the addresses.**

The most you can figure out from the addresses is the public key, which requires breaking two one way functions.

** This allows you to derive the master private key of a non hardened derivation.

[Geka041]

Not possible. You need at least a master public key and one of the private key of the addresses.

The most you can figure out from the addresses is the public key, which requires breaking two one way functions.

Thanks a lot for the explanation.

[Geka041]

Now the actual question. Knowing the exact sequence of addresses (there are infinitely many addresses), it is possible to recover a mnemonic phrase or any other information?

No, there isn't. Every derived address is distinct with each other, because they all get hashed with a different index each time. That being said, whether you have the RIPEMD-160 hash of 1000 addresses or even their revealed public keys, you can't go further to recover any information that would expose they're all derived from the same seed or anything similar.

You obviously can't recover a mnemonic phrase from its hashed derivations.

Thanks a lot for the explanation.

[Charles-Tim]

Now the actual question. Knowing the exact sequence of addresses (there are infinitely many addresses), it is possible to recover a mnemonic phrase or any other information?

This is not possible, because to even derive private key from public key is impossible with the current technology.

The size of bitcoin’s private key space, (22^56) is an unfathomably large number. It is approximately 10^77 in decimal. For comparison, the visible universe is estimated to contain 10^80 atoms.

So, brute forcing private key is impossible with the current day technology, also the the derivation of resulting public key from the private key makes use of asymmetric cryptography, which means it is in one way direction.

K = k * G
where k is the private key
G is the generator point
K is the public key which is a point on the curve

Dividing the K with G will not result to the corresponding private key.

Not possible. You need at least a master public key and one of the private key of the addresses.**

The master public key and private key of a non hardened address can result to derivation of the master private key. If an address derivation path is hardened, the master private key can not be derived.

[BlackHatCoiner]

Dividing the K with G will not result to the corresponding private key.

Actually, there isn't such thing as the division (neither subtraction) of two points in elliptic curves as far as I'm concerned, due to the fact that both addition and multiplication aren't like the regular ones. In other words, you don't have to solve a discrete algorithm in addition and multiplication, they're ordinary maths.

I just don't know if “dividing” can be used properly in the sentence. Surely, if you divide K with G in regular division, you won't get the corresponding private key.

[csgopolygoncode]

 Not possible. sorry

[BrewMaster]

Actually, there isn't such thing as the division (neither subtraction) of two points in elliptic curves as far as I'm concerned,

the division of two elliptic curve points is not defined but subtraction is. P - Q is the same as P + -Q where you negate the y coordinate of Q as the defined by negative point then add the two points.

[2chase]

If this were possible, then a long time ago this task would have been solved by someone, but since Bitcoin has been functioning safely for more than 12 years, the answer to this question is more than obvious.