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Diano4ka-milaya [45]
2 years ago
8

Consider a system that uses a 32-bit unique salt where users have a 4-digit number as a password (e.g. 6813). Eve wants to crack

the accounts of two users, Alice and Bob. Eve performs an online attack, and is able to guess 1 password per second, though there is no lockout after guessing too many times. In the worst case, in seconds, how long will it take Eve to crack both Alice's and Bob's accounts
Computers and Technology
2 answers:
Alexus [3.1K]2 years ago
8 0

Answer:

5 hr. 33 min. 20 sec.

Explanation:

Let P₁ be the number of possible passwords Alice can choose

Let P₂ be the number of possible passwords Bob can choose

In a 4 digit password, since the passwords are made up of 10 digits from 0 to 9, therefore the user can choose:

P₁ = 10⁴ = 10000

P₂ = 10⁴ = 10000

The total number of possible passwords combinations that both Alice and Bob can choose is therefore P₁ + P₂ = 10000 + 10000 = 20000.

If Eve performs an online attack and is able to guess 1 password per second.

Eve is therefore able to crack both Alice's and Bob's accounts in:

1 × 20000 = 20000 seconds

Converting 20000 seconds to hours, minutes and seconds will give 5 hr. 33 min. 20 sec.

Eve is able to crack both Alice's and Bob's accounts in 5 hr. 33 min. 20 sec.

Alenkasestr [34]2 years ago
6 0

Answer:

18000 seconds or 300 minutes.

Explanation:

In the example given in the question, it is stated that the system uses 32-bit unique salt which is equal to 4 bytes where every digit takes up 1 byte thus forming the 4 digit passwords.

Considering that the passwords are 4 digits, starting from 1000 and up to 9999, there are 9000 possible password combinations.

If Eve has to go through the whole range of possible password combinations and it takes her 1 second to guess 1 password. Then in the worst case scenario, it would take her 18000 seconds or 300 minutes to crack both accounts, assuming that it is possible for them to use the same passwords.

I hope this answer helps.

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An athlete is signed for a season. These days each aspect of an athlete is noted through the effective use of the best technology. However, for detailed study, one or several plays, and certainly not the second half of a game is enough. It's required to collect the details for a complete season. And that is possible, as an athlete is hired for a season. And through such a detailed data set of a complete season, we can now train a machine as well, and it will let the athlete know where she is going wrong. And thus she can improve and remove those faults from her game, and become a better athlete. And even for a coach, one complete season is required, though when he has not seen her playing before that season. It's assumed that this is her first season. All the options mentioned are good, but the best is certainly the one with complete details, and that is a complete season. The rest is good but not the best.

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Def rectangle_area(base,height):
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a password to a certain database consists of digits that cannot be repeated. if the password is known to consist of at least 8 d
BaLLatris [955]

Answer:

\frac{10!}{2}mins

Explanation:

12 seconds to try one combination will be equivalent to  \frac{1}{12}\times 60 = \frac{1}{5} \ mins

Password contain at least 8 digit i.e. password can contain 8, 9, 10 digit.

Password cannot contain more than 10 digit because it will give room for repetition which it is clearly stated that digit cannot be repeated.

Possible digit that can be used: 9,8,7,6,5,4,3,2,1,0.

Total number of passwords combination possible for each position in 8 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3. Total number of passwords combination possible in 8 digit is equivalent to \frac{10!}{2}.

Total number of passwords combination possible for each position in 9 digit.  

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2. Total number of passwords combination possible in 9 digit is equivalent to \frac{10!}{1}.

Total number of passwords combination possible for each position in 10 digit.

1st position = 10, 2nd position = 9, 3rd position = 8, 4th position = 7, 5th position = 6, 6th position = 5, 7th position = 4, 8th position = 3, 9th position = 2, 10th position = 1.  Total number of passwords combination possible in 10 digit is equivalent to 10!.

Adding them up and multiplying by  \frac{1}{5} \ mins  to get the total number of time needed to guarantee access to database =  [\frac{10!}{2}\times\ \frac{10!}{1} \times 10!] \frac{1}{5}\ mins = \frac{10!}{2}

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