<p>This would be trivial to brute force, if you had software to do so.</p>
<p>WIF private keys are encoding a number in base 58. If the last 10 characters are missing, that's <em>58<sup>10</sup> = 430,804,206,899,405,824</em> potential numbers. That's a huge amount, but thankfully, every WIF key also contains a 32-bit checksum. That means that only 1 in <em>2<sup>32</sup></em> of those has to be tried, leaving us with just <em>100,304,421</em> keys left. Even fairly naive implementations could compute the corresponding addresses for those in a few hours of CPU time, to see which one might have been used.</p>

CC BY-SA 4.0
2024-01-02T19:57:54.353Z
Pieter Wuille, Can you please explain how a checksumm can reduce the number of combinations?
Is it relevant in case of unknown characters at the beginning of the
Konstantin luchkovskiy 
CC BY-SA 4.0
2022-04-06T03:50:19.373Z
2022-04-06T03:50:19.373Z
Pieter Wuille
<p>Is it possible to guess missing characters of a wif? Let's say there are 62 characters (lowercase + uppercase + digits) and the last 10 characters are missing. Is it possible to guess the WIF with powerful cloud computing in a reasonable amount of time? Guessing more than 10 characters (when we need to try all combinations) sounds utopic, but how is the WIF generated? Can we infer something about the math behind it to exclude specific combinations?</p>

<wallet-import-format>
Wallet Import Format - Guessing missing characters
goge
# Wallet Import Format - Guessing missing characters

(Asked by: username_0 on 2022-04-05)

Is it possible to guess missing characters of a wif? Let's say there are 62 characters (lowercase + uppercase + digits) and the last 10 characters are missing. Is it possible to guess the WIF with powerful cloud computing in a reasonable amount of time? Guessing more than 10 characters (when we need to try all combinations) sounds utopic, but how is the WIF generated? Can we infer something about the math behind it to exclude specific combinations?

**Tags:** wallet-import-format


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## 1 answers

### Answer by username_1 on 2022-04-06. Score: 2

This would be trivial to brute force, if you had software to do so.

WIF private keys are encoding a number in base 58. If the last 10 characters are missing, that's *5810 = 430,804,206,899,405,824* potential numbers. That's a huge amount, but thankfully, every WIF key also contains a 32-bit checksum. That means that only 1 in *232* of those has to be tried, leaving us with just *100,304,421* keys left. Even fairly naive implementations could compute the corresponding addresses for those in a few hours of CPU time, to see which one might have been used.

> **Comments:**
> **username_2**: username_1, Can you please explain how a checksumm can reduce the number of combinations? Is it relevant in case of unknown characters at the beginning of the  

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