24 word seed question : is splitting it in half dangerous?

NotATether

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Quote from: elliptic joe on April 09, 2024, 04:05:25 AM

I don't even understand why you talk of CS collision... CS collision is sure: given a known CS, there are about 2^248 combo (2^256 / 2^8) that give the same CS.

The reason I brought it up is that obtaining the second half of the mnemonic with a hypothetically larger seed would reduce its usefulness (while making the first half more useful), but then I realized that the CS is not an independent variable which can be modified without changing the entropy.

elliptic joe

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Quote from: NotATether on April 09, 2024, 03:40:41 AM

Quote from: elliptic joe on April 08, 2024, 09:25:34 AM

Quote from: pooya87 on March 10, 2021, 01:54:59 AM

Quote from: HCP on March 09, 2021, 05:12:42 PM

In both instances, you still need to bruteforce 12 words to get the actual seed phrase... So, you're working with a 2028^12 search space... which is 5444517870735015415413993718908291383296.

I don't think we can call it the same. If the total words are 12 and you are missing all 12 then the search space is 2¹²⁸. But if the total words are 24 and you have 12 then you still have half of the entropy and depending on which half it could be a lot simpler. For example if you have the second half of the words then the bulk of the search space is suddenly reduced by roughly 94% because of the checksum.

I think this is incorrect.

When someone knows the last 12 words of the 24 phrase, they know 132 bits. But since 8 bits are for the checksum, they gain only 124 bits of information. So, there are still 256 - 124 = 132 bits left to attack. Now, the brute attacker knows the checksum, so we need to understand how many combinations among the 2^132 generate the same checksum: those that don't can be immediately discarded. The combinations that yield the same CS are about 2^124 (2^132 / 2^8).
Instead for an unknown 12-word phrase, the search space is 2^128. So, the former is slightly less secure, but negligibly so.

That's actually assuming the checksum is 8 bits long. It doesn't necessarily have to be, ~~as the checksum length can be set to an arbitrary value that satisfies the formula in~~ https://github.com/bitcoin/bips/blob/master/bip-0039.mediawiki#user-content-Generating_the_mnemonic (where MS=24 is the length of the resulting mnemonic phrase). But as far as I know, all of the wallets I know of use 8 bit checksums, and I haven't heard of a situation where a larger checksum length might be required to prevent checksum collision.

*OK, I realized that even the checksum and thereby the mnemonic phrase length is constrained by the entropy, so this doesn't actually hold, but it does raise the question on whether 8 bits of entropy will be enough to prevent a checksum collision from happening, particularly in a mass-adoption scenario where trillions of wallets are generated per day by businesses, software, apps, etc.

Then there is always the possibility of writing gibberish words in front of the mnemonic to make it appear like it's longer & non-standard.

I don't understand this.
In the documentation it is clearly written: 256 bit of entropy + 8 bit of CS.

I don't even understand why you talk of CS collision... CS collision is sure: given a known CS, there are about 2^248 combo (2^256 / 2^8) that give the same CS.

NotATether

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Quote from: elliptic joe on April 08, 2024, 09:25:34 AM

Quote from: pooya87 on March 10, 2021, 01:54:59 AM

Quote from: HCP on March 09, 2021, 05:12:42 PM

In both instances, you still need to bruteforce 12 words to get the actual seed phrase... So, you're working with a 2028^12 search space... which is 5444517870735015415413993718908291383296.

I don't think we can call it the same. If the total words are 12 and you are missing all 12 then the search space is 2¹²⁸. But if the total words are 24 and you have 12 then you still have half of the entropy and depending on which half it could be a lot simpler. For example if you have the second half of the words then the bulk of the search space is suddenly reduced by roughly 94% because of the checksum.

I think this is incorrect.

When someone knows the last 12 words of the 24 phrase, they know 132 bits. But since 8 bits are for the checksum, they gain only 124 bits of information. So, there are still 256 - 124 = 132 bits left to attack. Now, the brute attacker knows the checksum, so we need to understand how many combinations among the 2^132 generate the same checksum: those that don't can be immediately discarded. The combinations that yield the same CS are about 2^124 (2^132 / 2^8).
Instead for an unknown 12-word phrase, the search space is 2^128. So, the former is slightly less secure, but negligibly so.

That's actually assuming the checksum is 8 bits long. It doesn't necessarily have to be, ~~as the checksum length can be set to an arbitrary value that satisfies the formula in~~ https://github.com/bitcoin/bips/blob/master/bip-0039.mediawiki#user-content-Generating_the_mnemonic (where MS=24 is the length of the resulting mnemonic phrase). But as far as I know, all of the wallets I know of use 8 bit checksums, and I haven't heard of a situation where a larger checksum length might be required to prevent checksum collision.

*OK, I realized that even the checksum and thereby the mnemonic phrase length is constrained by the entropy, so this doesn't actually hold, but it does raise the question on whether 8 bits of entropy will be enough to prevent a checksum collision from happening, particularly in a mass-adoption scenario where trillions of wallets are generated per day by businesses, software, apps, etc.

Then there is always the possibility of writing gibberish words in front of the mnemonic to make it appear like it's longer & non-standard.

elliptic joe

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Quote from: pooya87 on March 10, 2021, 01:54:59 AM

Quote from: HCP on March 09, 2021, 05:12:42 PM

In both instances, you still need to bruteforce 12 words to get the actual seed phrase... So, you're working with a 2028^12 search space... which is 5444517870735015415413993718908291383296.

I don't think we can call it the same. If the total words are 12 and you are missing all 12 then the search space is 2¹²⁸. But if the total words are 24 and you have 12 then you still have half of the entropy and depending on which half it could be a lot simpler. For example if you have the second half of the words then the bulk of the search space is suddenly reduced by roughly 94% because of the checksum.

I think this is incorrect.

When someone knows the last 12 words of the 24 phrase, they know 132 bits. But since 8 bits are for the checksum, they gain only 124 bits of information. So, there are still 256 - 124 = 132 bits left to attack. Now, the brute attacker knows the checksum, so we need to understand how many combinations among the 2^132 generate the same checksum: those that don't can be immediately discarded. The combinations that yield the same CS are about 2^124 (2^132 / 2^8).
Instead for an unknown 12-word phrase, the search space is 2^128. So, the former is slightly less secure, but negligibly so.

elliptic joe

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Quote from: o_e_l_e_o on March 07, 2021, 02:13:26 PM

Quote from: Oshosondy on March 07, 2021, 01:54:54 PM

The chances that hackers can brute force the whole seed phrase from just twelve seed phrase is possible with powerful computional algorithmic tools, very possible tools like btcrecover can be able to do it with high computational power.

No, it isn't.

A 24 word BIP39 phrase has 256 bits of entropy, with 8 bits of checksum. Depending on which 12 words the attacker knows, then, the remaining 12 words have either 132 bits or 124 bits of entropy. Both are still far outside the realms of possibilities, with the time taken to brute force measured in billions of years even with huge amounts of cloud computing dedicated to the task.

Yes, but if 2^132 is the initial searching space when someone knows the last 12 words of a 24 phrase, you have to consider the fact that only 2^124 combinations generate the same known checksum.
In other words the brute attacker can immediately discard 2^8 combinations without derive the addresses to check if they contain some tokens.

pooya87

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Quote from: HCP on March 09, 2021, 05:12:42 PM

In both instances, you still need to bruteforce 12 words to get the actual seed phrase... So, you're working with a 2028^12 search space... which is 5444517870735015415413993718908291383296.

I don't think we can call it the same. If the total words are 12 and you are missing all 12 then the search space is 2¹²⁸. But if the total words are 24 and you have 12 then you still have half of the entropy and depending on which half it could be a lot simpler. For example if you have the second half of the words then the bulk of the search space is suddenly reduced by roughly 94% because of the checksum.

HCP

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Quote from: jerry0 on March 07, 2021, 05:07:36 PM

Bit confused with this. Electrum only has a 12 word seed. The nano ledger s has a 24 word seed. So wouldn't just having half of the nano ledger seed which is 12 word seed essentially the same thing as having no word in an electrum seed?

That is essentially correct.

In both instances, you still need to bruteforce 12 words to get the actual seed phrase... So, you're working with a 2028^12 search space... which is 5444517870735015415413993718908291383296.

o_e_l_e_o

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Quote from: pooya87 on March 08, 2021, 12:01:04 AM

-snip-

In terms of an attacker trying to brute force a 12 word seed or a 24 word seed with 12 words known, jerry0 is correct though.

A 12 word Electrum seed with no known words has 132 bits of entropy needing brute forced.
A 24 word Ledger Nano seed with 12 words known has either 132 or 124 bits of entropy needing brute forced, depending on whether the checksum word is known or not.

There are other differences in regards to derivation path and so on, but broadly speaking, they are comparably difficult to brute force.

pooya87

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Quote from: jerry0 on March 07, 2021, 05:07:36 PM

Bit confused with this. Electrum only has a 12 word seed. The nano ledger s has a 24 word seed. So wouldn't just having half of the nano ledger seed which is 12 word seed essentially the same thing as having no word in an electrum seed?

No, because the entropy size is a power of 2 number which is growing exponentially. And after a certain entropy size the entropy is considered secure. 12 word mnemonic offers 128 bits of entropy which is secure and more importantly it is offering the same exact level of security as a bitcoin private key (key for a 256-bit elliptic curve which has 128 bits of security).

Cutting 12 words in half is cutting your entropy down to 64 which is not as secure anymore.
This is the difference:
2⁶⁴ = 18446744073709551616 2¹²⁸ =340282366920938463463374607431768211456

jerry0

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Quote from: Oshosondy on March 07, 2021, 01:54:54 PM

Quote from: Wexfgy12 on April 05, 2017, 05:55:04 PM

If I have a 24 word seed, and I split it into 2 - 12 and 12
If an attacker somehow finds either the first 12 words or the 2nd 12 words, is there anyway they can use that to easily derive/find the second 12 words? or am I as safe as If I simply had a 12 word seed (which many wallets use?)

The chances that hackers can brute force the whole seed phrase from just twelve seed phrase is possible with powerful computional algorithmic tools, very possible tools like btcrecover can be able to do it with high computational power. It is not good to split your seed phrase, instead you can use shamir sharing for such in a way the seed phrase can be perfectly divided and splitted into secrets which is best for this.

Bit confused with this. Electrum only has a 12 word seed. The nano ledger s has a 24 word seed. So wouldn't just having half of the nano ledger seed which is 12 word seed essentially the same thing as having no word in an electrum seed?

o_e_l_e_o

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Quote from: Oshosondy on March 07, 2021, 01:54:54 PM

The chances that hackers can brute force the whole seed phrase from just twelve seed phrase is possible with powerful computional algorithmic tools, very possible tools like btcrecover can be able to do it with high computational power.

No, it isn't.

A 24 word BIP39 phrase has 256 bits of entropy, with 8 bits of checksum. Depending on which 12 words the attacker knows, then, the remaining 12 words have either 132 bits or 124 bits of entropy. Both are still far outside the realms of possibilities, with the time taken to brute force measured in billions of years even with huge amounts of cloud computing dedicated to the task.

If we are talking about a 15 or 18 word BIP39 phrase on the other hand, then the remaining entropy in those cases ranges from 28 bits to 66 bits, which is somewhere in the range of "very easy" to "possible in a few weeks/months", depending on the computing power involved.

Yes, brute forcing 12 words is exponentially easier than brute forcing 24, but it is still impossible for the time being.

Oshosondy

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Quote from: Wexfgy12 on April 05, 2017, 05:55:04 PM

If I have a 24 word seed, and I split it into 2 - 12 and 12
If an attacker somehow finds either the first 12 words or the 2nd 12 words, is there anyway they can use that to easily derive/find the second 12 words? or am I as safe as If I simply had a 12 word seed (which many wallets use?)

The chances that hackers can brute force the whole seed phrase from just twelve seed phrase is possible with powerful computional algorithmic tools, very possible tools like btcrecover can be able to do it with high computational power. It is not good to split your seed phrase, instead you can use shamir sharing for such in a way the seed phrase can be perfectly divided and splitted into secrets which is best for this.

o_e_l_e_o

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Quote from: owenohio on February 18, 2021, 12:49:43 PM

Yet in this video, Andreas positively hates the idea: https://www.youtube.com/watch?v=p5nSibpfHYE

It seems his biggest issue here is that without using a system like Shamir's, finding one of the shares reveals two thirds of the secret, and reduces the remaining entropy from 2²⁵⁶ to 2⁸⁰. Whereas finding anything less than m in a m-of-n Shamir's scheme reveals nothing about the secret, every additional share in a m-of-n non-Shamir's scheme makes brute forcing the remaining unknown words progressively easier.

He's not incorrect that brute forcing 2⁸⁰ is exponentially easier than brute forcing an entire seed phrase, and he's also not wrong that brute forcing 2⁸⁰ is potentially possible in the not too distant future, and for those reasons I agree that a Shamir's secret sharing scheme is a better mechanism. However, if you are concerned about your back up being discovered and cannot physically store it in a more secure fashion, than a 2-of-3 scheme is still better than no scheme at all.

And with any secret splitting scheme, you have to take in to account how you would deal with one or more of your shares being lost or destroyed. If using any 2-of-3 scheme, I would make at least 2 copies of each share, so you could lose at least 3 shares and still be able to recover your coins.

owenohio

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I wanted to re-up this topic in case developments since it was last discussed alter people's opinions.

Splitting a 24 word mnemonic into two parts seems like a good way of performing at least one backup.

And the seems to have garnered moderate approval here in this thread.

Yet in this video, Andreas positively hates the idea: https://www.youtube.com/watch?v=p5nSibpfHYE

I'm not sure if that's because he's against having 16/24 in one place and whether his concerns would be assuaged in a 12/24 set-up.

Coin-Keeper

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Quote from: BurtW on April 09, 2017, 03:39:30 PM

I wish I had done this.

When I was arrested they searched my home and office, found my backup with all 24 words, and confiscated all of my Bitcoins. So it can be dangerous to have all 24 words where they can be found by homeland security.

That is why I like my Trezor. Even finding my Trezor SEED words leaves them lacking an enormously long hidden wallet passphrase (in my head only). I keep a short decoy passphrase to "throw them off the scent" if I ever get in that situation. It may cost me 2-4 BTC but that would be it. There is no proof I have a hidden wallet either.

I have always hated reading about your situation.

BurtW

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I wish I had done this.

When I was arrested they searched my home and office, found my backup with all 24 words, and confiscated all of my Bitcoins. So it can be dangerous to have all 24 words where they can be found by homeland security.

Wexfgy12

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OK thanks for replies.

What if someone knows the checksum word - can that help at all, because it means certain combinations are not possible - or are the combinatorics still too vast?

BitcoinNewsMagazine

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Quote from: Wexfgy12 on April 05, 2017, 05:55:04 PM

Hi,

If I have a 24 word seed, and I split it into 2 - 12 and 12
If an attacker somehow finds either the first 12 words or the 2nd 12 words, is there anyway they can use that to easily derive/find the second 12 words? or am I as safe as If I simply had a 12 word seed (which many wallets use?)

Basically do the 2 halves in any way relate / help to derive each other, or do I retain half the entropy with half the seed? and does it matter if its the half including the checksum (can that be used in anyway to derive / guess what words it could be?)

Thanks

*if this is wrong forum / somewhere better to ask please advise.

Having half the seed does not help an attacker. What could harm you is storing the halves separately and losing access. Put the seed someplace safe offline. How safe depends on how much bitcoin you own. If you are talking hardware wallets you can also add a layer of security by using a password to protect the seed. If someone gained access to your Trezor or Ledger seed it would be useless to them, they could not reconstruct your accounts without the password.

pooya87

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to really get the feel of the chances go look at some of the "Electrum Seed recovery tools" out there. you can even download one from Github and try recovering a dummy seed from a new empty wallet or using this:

Quote

constant forest adore false green weave stop guy fur freeze giggle clock

from documents. try with 1 missing word then with 2 and so on and see how long it takes.

Coin-Keeper

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From a math perspective logic would say that of course having the first 12 words of a 24 word puzzle is better than having none of them. However; the math is so complicated that its still beyond any plausible range of calculation that I have ever seen or read about.

As an example (this is my example and not mathematically prove-able by me); I can give you my btc address (view my signature as an example) so you have the entire address numbers. Now I tell you to "sign" that account without me giving you the matching private key. It can't be done.

Topic: 24 word seed question : is splitting it in half dangerous? (Read 2407 times)