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rjkz [21]
2 years ago
6

which of the following is a key difference between agent-based orchestration and agentless orchestration? (choose two). answer a

gent-based setup is simpler than agentless setup. agentless orchestration does not require a proprietary software agent to be installed on the managed hosts. agent-based requires that proprietary software is installed on each device you wish to monitor. agentless orchestration is more difficult and expensive to deploy.
Computers and Technology
1 answer:
Valentin [98]2 years ago
6 0

The automated configuration, management, and coordination of computer systems, applications, and services is known as orchestration.

<h3>What is orchestration?</h3>
  • The term "orchestration" refers to the automated configuration, management, and coordination of computer systems, applications, and services. IT can more easily manage difficult activities and workflows with the aid of orchestration.
  • While managing numerous servers and applications by hand is necessary, it is not a scalable approach. Managing all the moving parts can be more difficult the more complex an IT system is. In groups of systems or machines, there is a growing need to combine multiple automated tasks and their configurations. Where orchestration can be useful is in this situation.
  • The concepts of orchestration and automation are distinct but related. By reducing or eliminating human interaction with IT systems and replacing it with software to carry out tasks in order to lower costs, complexity, and errors, automation helps your business become more efficient.
  • Automation typically refers to automating a specific task. This is distinct from orchestration, which is a method for automating a process or workflow that comprises numerous steps and multiple dissimilar systems. You may choreograph your processes to run automatically after beginning by incorporating automation into them.
  • Additionally, IT orchestration enables you to streamline and improve routine tasks and workflows, which can support a DevOps strategy and speed up the deployment of apps by your team.
  • IT activities including server provisioning, incident management, cloud orchestration, database administration, application orchestration, and many other jobs and workflows may all be automated with orchestration.

To Learn more About orchestration refer To:

brainly.com/question/11958641

#SPJ1

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A palindrome is a string of characters that is the same when reversed (e.g., ‘affa’). Single characters are palindromes. Suppose
Nina [5.8K]

Answer:

Check the explanation

Explanation:

a) This done can be executed through the use of brute force by the given algorithm: Consider any two indices i and j then we can check if string from s[i....j] is palindrome or not in O(L) where L is length of string.

Algorithm to check if string is palindrome :

Takes String s as argument and the two indices i,j between which the string is to be considered

bool palindromeCheck(s,i,j) :

        y = j

        for x in range(i,j) :

             if(s[x]==s[y]) :

                      y--

                      continue

            else :

                      return false

       return true

Now taking a look at the  part a) we will be calling the function to confirm and execute palindrome for all i<=j that is a total of points of the order of O(n^2) and the string checking will take O(n) in the worst case making it O(n^3).

b) To advance over we became aware of the fact that repeated calculation is being done i.e when we confirmed the string s[i...j] we also tested in between strings like s[i+1,j-1] etc in the same time. Consequently we used memorization to store the outcome that we calculate and reuse them we need them later.

We know that P[i][i] will always be one since a single character string will always be a palindrome.

Using this as our base case to check for P[i][j] we need to check if the ith and the jth character are equal and also if P[i+1][j-1] is true i.e is a palindrome.

We come up with the following recurrence

P[i][j] = 1 if i=j

P[i][j] = P[i+1][j-1] ; if i>=j and s[i]==s[j]

            0; otherwise

There are order of O(n^2) states and each one of the state necessitates only constant and regular time character checking for its evaluation and then assigning the value of 1, P[i+1][j-1],0 as a result of that which is again a constant time operation. Therefore all the P[][] matrix values can be found in O(n^2) time.

Pseudo code :

Let n be the length of the string

for x in range(1,n) :

        P[x][x]=1

for sz in range(2,n) : # Since we have already computed answer for string of size 1

        for x in range (1,n-sz+1):

                   y=x+sz - 1

                   if x == y :

                            if sz==2 :

                                      P[x][y] = 1

                            else :

                                      P[x][y] = P[x+1][y-1]

                  else :

                            P[x][y] = 0

From the implementation also it is clear that the running time of the algorithm is O(n^2).

5 0
3 years ago
An example for Shareware?
Kryger [21]
A software that is free of cost for a certain amount of time such as Adobe programs, or Winzip.
6 0
2 years ago
Read 2 more answers
The alpha channel is a color component that determines the degree of opacity/transparency of a color.
denis-greek [22]
THIS ANSWER IS FALSE......
7 0
4 years ago
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It is a well-researched fact that men in a restroom generally prefer to maximizetheir distance from already occupied stalls, by
Artyom0805 [142]

Answer:

The code below addresses the problem statement with appropriate comments for explanation.

Explanation:

public class Stall

{

public static void main(String[] args)

{

boolean[] arr = new boolean[10];

while(hasFreeSpots(arr))   // as long as there is a free slot

{

arr[position(arr)] = true;  

// position(arr) gets the position of slot a person would most likely enter at(view problem statement)

// setting that position true(position occupied) by arr[position(arr)]

System.out.println(stallView(arr));   // printing current state of the stall using 'X' and '_'

}

}

/**

* Find the empty position that is in the middle of the sub-array of largest group of empty spots

* @param arr an array of boolean values

* @return returns an an integer value. Optimal position to be occupied.

*/

public static int position(boolean[] arr)

{

int first = 0;   // first stores the starting position of "longest sequence of unoccupied places"

int last = 0;   // last stores the ending position of "longest sequence of unoccupied places"

int temp1 = 0;   // temp1 stores the starting position of "current sequence of unoccupied places"

int temp2 = 0; // temp2 stores the ending position of "current sequence of unoccupied places"

 

for(int i = 0; i < arr.length; i++)   // for each slot in the Stall

{    

if(!arr[i]) // if slot i is not occupied

{    

temp1 = i;   // store starting position of "current sequence of unoccupied places" in temp1

while(i < arr.length && !arr[i] )   // if i is a valid index and as long as this slot is not occupied increment i

{

i++;

}

// we break out of the loop either if i is invalid index(reached end of the array)

// or if the slot is occupied by a person already

temp2 = i;   // store ending position of "current sequence of unoccupied places" in temp2

if(Math.abs(first - last) < Math.abs(temp1 - temp2))

  // if current sequence is longer than longest sequence, update it

{

first = temp1;

last = temp2;    

}

}

}

return (first + last) / 2;   // return middle index of first and last

}

/**

* Checks if there are empty spots in an array, basically checks if there is a false boolean value

* thats representing an empty spot in an array

* @param arr the input array, of boolean values

* @return returns a boolean value, true if there are false elements inside the array

*/

public static boolean hasFreeSpots(boolean[] arr)

{

boolean rtn = true;   // initializing rtn to 'true'

for(int i = 0; i < arr.length; i++)   // for every slot in the Stall

{

if(!arr[i]){rtn = false;}   // sets rtn to false if a free spot found

}

return !rtn;   // returns true if a free spot found and returns false if no free spot found

}

/**

* X for true, _ for false. Representing occupied(X) and free stalls(_)

* @param arr a boolean array

* @return returns a string representation if the input

*/

public static String stallView(boolean[] arr)

{

String str = "";       // initializing str to be an empty String

for(int i = 0; i < arr.length; i++)   // for every slot in Stall

{

if(arr[i]){str += " X";}   // Add X to str if slot is occupied

else {str += " _";}           // prints _ to str if slot not occupied

}

return str;       // return str(pictorial representation of Stall with 'X', '_')

}

}

7 0
3 years ago
Read the following sentence from an argumentative essay.
bazaltina [42]
The right choice is D.
5 0
4 years ago
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