Can You Recover a Wallet with 6 or 8 Known Words?
Recovering a cryptocurrency wallet when only part of the 12-word mnemonic phrase is known is a daunting challenge. The mnemonic, based on the BIP-39 standard, is a cornerstone of wallet security, designed to be nearly impossible to brute-force. Let’s explore why brute-forcing a mnemonic is plausible when 8 words are known but practically impossible when only 6 words are known.
The Basics of BIP-39 Mnemonics
- Wordlist: The BIP-39 standard uses a fixed list of 2048 unique words.
- Phrase Structure: A typical mnemonic phrase has 12 words.
- Checksum: The last word encodes a checksum that validates the entire phrase, slightly reducing the total number of valid combinations.
Each word is critical. Missing words exponentially increase the difficulty of brute-forcing the phrase.
8 Known Words: Feasible with Enough Power
If you know 8 out of 12 words, there are 4 unknown words to brute-force. Each unknown word can be one of 2048 words. The total number of possible combinations is: 20484=244≈1.1 trillion. $$ 2048^4 = 2^{44} \approx 1.1 \, \text{trillion} $$.
This might sound like an enormous number, but it’s within the range of modern computational power. Let’s break it down:
- Assume: A powerful machine can check 1 million combinations per second.
- Calculation: Time (in seconds)=244106=1.1×1012÷106≈1.1×106 seconds.$$ \text{Time (in seconds)} = \frac{2^{44}}{10^6} = 1.1 \times 10^{12} \div 10^6 \approx 1.1 \times 10^6 \, \text{seconds}.$$ This equals roughly 4.88 hours.
With sufficient computing resources, brute-forcing 4 unknown words is feasible.
6 Known Words: Virtually Impossible
If you know 6 out of 12 words, there are 6 unknown words to brute-force. Each of these 6 words has 2048 possibilities, resulting in: $$ 20486=266≈73 quintillion.2048^6 = 2^{66} \approx 73 \, \text{quintillion} $$.
Now, let’s examine the brute-force feasibility:
- Assume: The same powerful machine can check 1 million combinations per second.
- Calculation: Time (in seconds)=266106=73×1018÷106=73×1012 seconds.$$ \text{Time (in seconds)} = \frac{2^{66}}{10^6} = 73 \times 10^{18} \div 10^6 = 73 \times 10^{12} \, \text{seconds}. $$ This equals approximately 2.31 million years.
Even with the most advanced technology, brute-forcing 6 unknown words is entirely unrealistic.
Why the Difference?
The reason for this dramatic difference lies in the exponential nature of the problem. Each additional unknown word multiplies the search space by 2048, making brute-forcing exponentially harder. Here’s a quick comparison:
| Known Words | Unknown Words | Total Combinations | Feasibility at 1M checks/sec |
|---|---|---|---|
| 8 | 4 | $$ 2442^{44} (~1.1 trillion) $$ | ~4.88 hours |
| 6 | 6 | $$ 2662^{66} (~73 quintillion) $$ | ~2.31 million years |
Conclusion
If you know 8 words of your 12-word mnemonic, there’s a slim chance of brute-forcing the remaining 4 words with the right computational resources. However, if you only know 6 words, the search space becomes so vast that brute-forcing is practically impossible—even for the most advanced computers.
The takeaway? Keep your mnemonic phrase safe and secure. Losing even a portion of it could make recovery infeasible.
https://github.com/bitcoin/bips/blob/master/bip-0039/english.txt
