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zalisa [80]
2 years ago
14

Modify the closest pair of points algorithm so that the separating line L now separates the first n/4 points (sorted according t

o their r coordinates) from the remaining 3n/4 points. Write the recurrence relation that gives the running time of the modified algorithm. Is the running time of your algorithm still O(nlog n)? Specify the best asymptotic running time you can get for your algorithm and briefly justify.
Now let the line L separate the first √ n points (according to their x- coordinates) from the remaining n - √ n points. Write the recurrence relation that gives the running time of this modification of the algorithm. Is the running time of your algorithm still O(n log n)? If your answer is yes, provide a brief justification; if your answer is no, provide a (asymptotic) lower bound on the running time of the modified algorithm that should be enough to justify your answer.

Computers and Technology
1 answer:
Nutka1998 [239]2 years ago
3 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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A contact list is a place where you can store a specific contact with other associated information such as a phone number, email
NikAS [45]

Answer:

C++.

Explanation:

<em>Code snippet.</em>

#include <map>

#include <iterator>

cin<<N;

cout<<endl;

/////////////////////////////////////////////////

map<string, string> contacts;

string name, number;

for (int i = 0; i < N; i++) {

   cin<<name;

   cin<<number;

   cout<<endl;

   contacts.insert(pair<string, string> (name, number));

}

/////////////////////////////////////////////////////////////////////

map<string, string>::iterator it = contacts.begin();

while (it != contacts.end())  {

   name= it->first;

   number = it->second;

   cout<<word<<" : "<< count<<endl;

   it++;

}

/////////////////////////////////////////////////////////////////////////////////////////////////////////

I have used a C++ data structure or collection called Maps for the solution to the question.

Maps is part of STL in C++. It stores key value pairs as an element. And is perfect for the task at hand.

8 0
2 years ago
In this lab, you use the pseudocode in figure below to add code to a partially created Python program. When completed, college a
Elza [17]

Answer:

Python code is given below with appropriate comments

Explanation:

#Prompt the user to enter the test score and class rank.

testScore = input()

classRank = input()

#Convert test score and class rank to the integer values.

testScore = int(testScore)

classRank = int(classRank)

#If the test score is greater than or equal to 90.

if(testScore >= 90):

   #If the class rank is greater than or equal to 25,

   #then print accept message.

   if(classRank >= 25):

       print("Accept")

   

   #Otherwise, display reject message.

   else:

       print("Reject")

#Otherwise,

else:

   #If the test score is greater than or equal to 80.

   if(testScore >= 80):

       #If class rank is greater than or equal to 50,

       #then display accept message.

       if(classRank >= 50):

           print("Accept")

       

       #Otherwise, display reject message.

       else:

           print("Reject")

   

   #Otherwise,

   else:

       #If the test score is greater than or equal to

       #70.

       if(testScore >= 70):

           #If the class rank is greater than or equal

           #to 75, then display accept message.

           if(classRank >= 75):

               print("Accept")

           

           #Otherwise, display reject message.

           else:

               print("Reject")

       

       #Otherwise, display reject message.

       else:

           print("Reject")

4 0
2 years ago
There are some processes that need to be executed. Amount of a load that process causes on a server that runs it, is being repre
kykrilka [37]

Answer:ummmm

Explanation:

3 0
1 year ago
Marissa works at a company that makes perfume. She noticed many samples of the perfume were not passing inspection. She conducte
garik1379 [7]
Writing a business letter as if she gets her point across to the head of department then he could change the way they made the perfume so they would pass inspection and standard.
6 0
2 years ago
Read 2 more answers
Generating a signature with RSA alone on a long message would be too slow (presumably using cipher block chaining). Suppose we c
boyakko [2]

Answer:

Following are the algorithm to this question:

Explanation:

In the RSA algorithm can be defined as follows:  

In this algorithm, we select two separate prime numbers that are the "P and Q", To protection purposes, both p and q combines are supposed to become dynamically chosen but must be similar in scale but 'unique in length' so render it easier to influence. Its value can be found by the main analysis effectively.  

Computing N = PQ.  

In this, N can be used for key pair, that is public and private together as the unit and the Length was its key length, normally is spoken bits. Measure,

\lambda (N) = \ lcm( \lambda  (P), \lambda (Q)) = \ lcm(P- 1, Q - 1)  where \lambda is the total function of Carmichaels. It is a privately held value. Selecting the integer E to be relatively prime from 1and gcd(E,  \lambda (N) ) = 1; that is E \ \ and  \ \ \lambda (N).  D was its complex number equivalent to E (modulo \lambda (N) ); that is d was its design multiplicative equivalent of E-1.  

It's more evident as a fix for d provided of DE ≡ 1 (modulo \lambda (N) ).E with an automatic warning latitude or little mass of bigging contribute most frequently to 216 + 1 = 65,537 more qualified encrypted data.

In some situations it's was shown that far lower E values (such as 3) are less stable.  

E is eligible as a supporter of the public key.  

D is retained as the personal supporter of its key.  

Its digital signature was its module N and the assistance for the community (or authentication). Its secret key includes that modulus N and coded (or decoding) sponsor D, that must be kept private. P, Q, and \lambda (N) will also be confined as they can be used in measuring D. The Euler totient operates \varphi (N) = (P-1)(Q - 1) however, could even, as mentioned throughout the initial RSA paper, have been used to compute the private exponent D rather than λ(N).

It applies because \varphi (N), which can always be split into  \lambda (N), and thus any D satisfying DE ≡ 1, it can also satisfy (mod  \lambda (N)). It works because \varphi (N), will always be divided by \varphi (N),. That d issue, in this case, measurement provides a result which is larger than necessary (i.e. D >   \lambda (N) ) for time - to - time). Many RSA frameworks assume notation are generated either by methodology, however, some concepts like fips, 186-4, may demand that D<   \lambda (N). if they use a private follower D, rather than by streamlined decoding method mostly based on a china rest theorem. Every sensitive "over-sized" exponential which does not cooperate may always be reduced to a shorter corresponding exponential by modulo  \lambda (N).

As there are common threads (P− 1) and (Q – 1) which are present throughout the N-1 = PQ-1 = (P -1)(Q - 1)+ (P-1) + (Q- 1)), it's also possible, if there are any, for all the common factors (P -1) \ \ \ and \ \ (Q - 1)to become very small, if necessary.  

Indication: Its original writers of RSA articles conduct their main age range by choosing E as a modular D-reverse (module \varphi (N)) multiplying. Because a low value (e.g. 65,537) is beneficial for E to improve the testing purpose, existing RSA implementation, such as PKCS#1, rather use E and compute D.

8 0
1 year ago
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