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Cloud [144]
3 years ago
12

Alice posts a key for encrypting that anyone can access. Bob uses that key to encrypt a message, then sends it to Alice. When Al

ice receives the message, she decrypts it using a private key.What kind of encryption process is this
Computers and Technology
1 answer:
UNO [17]3 years ago
5 0

Answer:

Public key encryption

Explanation:

DescriptionPublic-key cryptography, or asymmetric cryptography, is a cryptographic system that uses pairs of keys: public keys which may be disseminated widely, and private keys which are known only to the owner.

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Award documentation is typically required to be prepared and submitted within how long after the end of a project period of 90 days.

What does Award includes?

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1 year ago
. (a) Prove or disprove carefully and in detail: (i) Θ is transitive and (ii) ω is transitive. (b) Assume n is a positive intege
Sergio [31]

Answer:

The Following are the solution to this question:

Explanation:

In Option a:

In the point (i) \Omega is transitive, which means it converts one action to others object because if \Omega(f(n))=g(n) indicates c.g(n). It's true by definition, that becomes valid. But if \Omega(g(n))=h(n), which implies c.h(n). it's a very essential component. If c.h(n) < = g(n) = f(n) \. They  \Omega(f(n))   will also be h(n).  

In point (ii), The  value of \Theta is convergent since the \Theta(g(n))=f(n). It means they should be dual a and b constant variable, therefore a.g(n) could only be valid for the constant variable, that is  \frac{1}{a}\ \  and\ \ \frac{1}{b}.

In Option b:

In this algorithm, the input size value is equal to 1 object, and the value of  A is a polynomial-time complexity, which is similar to its outcome that is O(n^{2}). It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop(n^{2}). All internal loops operate on a total number of N^{2} generations and therefore the final time complexity is O(n^{2}).

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