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Math problems

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 shgh
(@shgh)
Active Member
Joined: 15 years ago
Posts: 15
Topic starter  

Hi
With the help of a PC in a brute force attack on an encrypted file, how many months it will take and find the right key if the computer can analyze 10 million keys per second, if the file is encrypted using an algorithm with 40 bit key length, and we know nothing about the format, language or content?

I want to know how to solve the problem in detail.

thanks
SHGh


   
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(@d1g14n6)
Active Member
Joined: 14 years ago
Posts: 12
 

40 bit keys have 2^40 possibilities
This equals out to 1099511627776
So 1099511627776 possibilities/ 10000000 per second = 109951.1627776 seconds
109951.1627776 seconds / 60 = 1832.51937962666667 minutes
1832.51937962666667 minutes / 60 = 30.5419896604444445 hours
I'm sure you can do the math from there..about 1 1/4 days


   
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 shgh
(@shgh)
Active Member
Joined: 15 years ago
Posts: 15
Topic starter  

40 bit keys have 2^40 possibilities
This equals out to 1099511627776
So 1099511627776 possibilities/ 10000000 per second = 109951.1627776 seconds
109951.1627776 seconds / 60 = 1832.51937962666667 minutes
1832.51937962666667 minutes / 60 = 30.5419896604444445 hours
I'm sure you can do the math from there..about 1 1/4 days

Thanks d1g14n6


   
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(@c-r-s)
Estimable Member
Joined: 14 years ago
Posts: 170
 

With the help of a PC in a brute force attack on an encrypted file, how many months it will take and find the right key if the computer can analyze 10 million keys per second, if the file is encrypted using an algorithm with 40 bit key length, and we know nothing about the format, language or content?

If this is an exercise in theoretical cryptography, not in CF, the answer is
Without knowing anything about the content, the right key can't be found.


   
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(@xennith)
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Joined: 15 years ago
Posts: 177
 

CRS is spot on, how can you evaluate any keys at all if you dont know the algorithm used? The question is meaningless.


   
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 shgh
(@shgh)
Active Member
Joined: 15 years ago
Posts: 15
Topic starter  

CRS is spot on, how can you evaluate any keys at all if you dont know the algorithm used? The question is meaningless.

Hi Xennith,
The question is an exercise in theoretical cryptography, not in CF.


   
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(@audio)
Estimable Member
Joined: 19 years ago
Posts: 149
 

40 bit keys have 2^40 possibilities
This equals out to 1099511627776
So 1099511627776 possibilities/ 10000000 per second = 109951.1627776 seconds
109951.1627776 seconds / 60 = 1832.51937962666667 minutes
1832.51937962666667 minutes / 60 = 30.5419896604444445 hours
I'm sure you can do the math from there..about 1 1/4 days

Correct me if I'm wrong, but 1 1/4 days is assuming the last key tried is the correct key. Couldn't you equally assume the correct key would be the first key tried? The average time will be around 15 hours, but you can factor in other things like the entropy of the key. A lot of people choose poor passwords so you probably wouldn't have to go through the entire key space.


   
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PaulSanderson
(@paulsanderson)
Honorable Member
Joined: 19 years ago
Posts: 651
 

There is a difference between the complexity of the passphrase and the entropy of the key. The key would typically be generated from the passphrase by a hashing algorithm Two passwords a and aa would have vastly different keys.


   
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jaclaz
(@jaclaz)
Illustrious Member
Joined: 18 years ago
Posts: 5133
 

To be picky 😯 , the original question was

… how many months it will take and find the right key if ….

To which the actual answer is

  • Less than 1.

with a degree of accuracy proportional to the question, and independent from entropy of the key or complexity of the passphrase. wink

D

jaclaz


   
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(@audio)
Estimable Member
Joined: 19 years ago
Posts: 149
 

There is a difference between the complexity of the passphrase and the entropy of the key. The key would typically be generated from the passphrase by a hashing algorithm Two passwords a and aa would have vastly different keys.

I knew the password goes through an algorithm but was under the impression you could use "key" or "password" interchangeably. Guess not… thanks. )


   
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