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Nataliya [291]
1 year ago
5

true or false? in a known-plaintext attack (kpa), the cryptanalyst hs access only to a segment of encrpted data and has no choic

e as to what that data might be
Computers and Technology
1 answer:
bulgar [2K]1 year ago
6 0

Yes , it’s true. In a known-plaintext attack (kpa), the cryptanalyst can only view a small portion of encrypted data, and he or she has no control over what that data might be.

The attacker also has access to one or more pairs of plaintext/ciphertext in a Known Plaintext Attack (KPA). Specifically, consider the scenario where key and plaintext were used to derive the ciphertext (either of which the attacker is trying to find). The attacker is also aware of what are the locations of the output from key encrypting. That is, the assailant is aware of a pair. They might be familiar with further pairings (obtained with the same key).

A straightforward illustration would be if the unencrypted messages had a set expiration date after which they would become publicly available. such as the location of a planned public event. The coordinates are encrypted and kept secret prior to the event. But when the incident occurs, the attacker has discovered the value of the coordinates /plaintext while the coordinates were decrypted (without knowing the key).

In general, a cipher is easier to break the more plaintext/ciphertext pairs that are known.

To learn more about Plaintext Attack click here:

brainly.com/question/28445346

#SPJ4

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Explanation:

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Read 2 more answers
1. Implement the function dict_intersect, which takes two dictionaries as parameters d1 and d2, and returns a new dictionary whi
stich3 [128]

Answer:

1  

def dict_intersect(d1,d2): #create dictionary

  d3={} #dictionaries

  for key1,value1 in d1.items():       #iterate through the loop  

      if key1 in d2:   #checking condition

          d3[key1]=(d1[key1],d2[key1])   #add the items into the dictionary  

  return d 3

print(dict_intersect({'a': 'apple', 'b': 'banana'}, {'b': 'bee', 'c': 'cat'})) #display

2

def consolidate(*l1):  #create consolidate

  d3={} # create dictionary

  for k in l1:       #iterate through the loop

      for number in k:   #iterate through  the loop                               d3[number]=d3.get(number,0)+1   #increment the value

             return d 3 #return

print(consolidate([1,2,3], [1,1,1], [2,4], [1])) #display

Explanation:

1

Following are  the description of program

  • Create a dictionary i.e"dict_intersect(d1,d2) "   in this dictionary created a dictionary d3 .
  • After that iterated the loop and check the condition .
  • If the condition is true then add the items into the dictionary and return the dictionary d3 .
  • Finally print them that are specified in the given question .

2

Following are  the description of program

  • Create a dictionary  consolidate inside that created a dictionary "d3" .
  • After that iterated the loop outer as well as inner loop and increment the value of items .
  • Return the d3 dictionary and print the dictionary as specified in the given question .

5 0
4 years ago
Bobby is investigating how an authorized database user is gaining access to information outside his normal clearance level. Bobb
Sveta_85 [38]

Answer:

Aggregate Function .

Explanation:

The aggregate function returns the single value of the entire column of the database management system. The Aggregate Function is used in the SELECT statement and the GROUP BY clause.  

The sum(),avg(), etc are some examples of Aggregate Function in the database with the help of aggregate function bobby gains the access to information above the usual clearance point. and summarizes the data.

7 0
4 years ago
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