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AVprozaik [17]
2 years ago
12

Why can't you use Friedman's attack on block ciphers?

Computers and Technology
1 answer:
n200080 [17]2 years ago
4 0

Answer:

   The Friedman's attack basically uses the same matrices in between the input of the blocks.

The similarities between the two given blocks is basically used to find the similar entries and hence, the deciphering is analyze the given pattern of the block in the system.

The input of the actual block pattern is encrypted into the different types of pattern in text of cipher. The basic parameters of the cipher block cannot be use for performing different types of attacks in the cipher without knowing the encrypted blocks.

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When excel follows the order of operations the formula 8 * 3 + 2 equals?
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Carl knows that water moves through different kinds of soil at different rates. How easily water moves through a soil is known a
Serggg [28]

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To get the same same results from all pots "amount of water should be same" for all pots.

Explanation:

As Carl want to measure and compare the amount of water that flows in pot in one minute from all all pots. He should keep the amount of water constant for all pots to get the desired results.

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Given the macro definition and global declarations shown in the image below, provide answers to the following questions:
galben [10]

Answer:

A. 243

B. True

C. 0

Explanation:

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4 0
2 years ago
PLEASE HELP PROGRAMMING WILL GIVE BRAINLIEST
JulijaS [17]
The fourth choice is correct.
5 0
2 years ago
Using Amdahl’s Law, calculate the speedup gain of an application that has a 60 percent parallel component for (a) two processing
Nina [5.8K]

Answer:

a) Speedup gain is 1.428 times.

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Explanation:

in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:

t=\frac{1 }{(S + (1- S)/N) }

Where S is the portion of the application that must be performed serially, and N is the number of processing cores.

(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4

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(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4

Thus, the speedup is:

t=\frac{1}{(0.4 + (1-0.4)/4)} = 1.8181

Speedup gain is 1.81 times.

8 0
2 years ago
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