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Ostrovityanka [42]
4 years ago
5

Order the following list of functions by the big-Oh notation. Group together those functions that are big θ of one another.6nlog

n 2^100 loglogn log^2n 2log^n2^2^n (√n) n^0.01 1/n 4n^3/23n^0.5 5n 2nlog^2n 2^n nlog4n4^n n^3 n^2logn 4^logn √logn
Computers and Technology
1 answer:
KatRina [158]4 years ago
5 0

Answer:

The increasing order of growth

1. 1/n

2. 2^100

3. Log log n

4. Sqrt(log n)

5. Log^2 n

6. n^0.01

7. ceil(sqrt(n)), 3n^0.5

8. 2^log n , 5n

9. n log4n, 6 n log n

10. floor (2 n log^2 n)

11. 4 n^3/2

12. 4^log n

13. n^2 log n

14. n^3

15. 2^n

16. 4^n

17. 2˄2n

Explanation:

When in doubt about two functions f(n) and g(n), consider log f(n) and log g(n).

Also note that Exponential growth is faster than polynomial.

i.e. the higher the degree of polynomial, the faster the growth.

I hope this helps, cheers.

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Answer:

1) Method calcTotal:

  1. public static long calcTotal(long [][] arr2D){
  2.        long total = 0;
  3.        
  4.        for(int i = 0; i < arr2D.length; i++ )
  5.        {
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  9.            }
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Line 1: Define a public method <em>calcTotal</em> and this method accept a two-dimensional array

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Line 8: Within the inner for-loop, use current row and column index, i and j, to repeatedly extract the value of each element in the array and add it to the variable total.

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2) Method calcAverage:

  1. public static double calcAverage(long [][] arr2D){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            for(int j = 0; j < arr2D[i].length; j++)
  8.            {
  9.                total = total + arr2D[i][j];
  10.                count++;
  11.            }
  12.            
  13.        }
  14.        
  15.        double average = total / count;
  16.        
  17.        return average;
  18.    }

Explanation:

The code in method <em>calcAverage</em> is quite similar to method <em>calcTotal</em>. We just need to add a counter and use that counter as a divisor of total values to obtain an average.

Line 4: Declare a variable, count, as an counter and initialize it to zero.

Line 11: Whenever an element of the 2D array is added to the total, the count is incremented by one. By doing so, we can get the total number of elements that exist in the array.

Line 16: Use the count as a divisor to the total to get average

Line 18: Return the average of all the values in the array as an output.

Answer:

3) calcRowAverage:

  1. public static double calcRowAverage(long [][] arr2D, int row){
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  4.        
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  6.        {
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  13.                }
  14.            }
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Line 1: Add one more parameter, row,

Line 8-15: Check if current row index, i, is equal to the target row number, proceed to sum up the array element in that particular row and increment the counter.

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float t1;

float t2;

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float t4;

float t5;

double grss_Pay;

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   cin>>hrs;

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         grss_Pay=grss_Pay-grss_Pay;

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        cout<<"Final Salary"<<in_hnd;

        return 0;

}

Explanation:

Please check the answer section.

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