Algorithm Compute (A[0,...n-1, 0,....n-1])//Input: An matrix A[0,...n-1, 0,.....n-1]) of real numbersfor i 0 to n-2 dofor j i+1 to n-1 doif A[i,j] A[j,i]return falsereturn truea. What does the algorithm compute?b. what is the basic operation?c. How many time is the basic operation executed?d. What is the efficiency class of the algorithm?e. Suggest an improvement or a better algorithm altogether and indicate its efficiency class.
Question
Algorithm Compute (A[0,...n-1, 0,....n-1])//Input: An matrix A[0,...n-1, 0,.....n-1]) of real numbersfor i 0 to n-2 dofor j i+1 to n-1 doif A[i,j] A[j,i]return falsereturn truea. What does the algorithm compute?b. what is the basic operation?c. How many time is the basic operation executed?d. What is the efficiency class of the algorithm?e. Suggest an improvement or a better algorithm altogether and indicate its efficiency class.
Solution
a. The algorithm checks if a given matrix A is symmetric. A matrix is symmetric if it is equal to its transpose. In other words, for all i and j, A[i,j] should be equal to A[j,i].
b. The basic operation in this algorithm is the comparison operation A[i,j] ≠ A[j,i].
c. The basic operation is executed approximately n(n-1)/2 times. This is because the algorithm checks each pair of elements A[i,j] and A[j,i] only once for i ≠ j.
d. The efficiency class of the algorithm is O(n^2). This is because in the worst-case scenario, the algorithm performs a constant amount of work for each pair of elements in the matrix, and there are approximately n^2/2 such pairs.
e. The algorithm is already quite efficient, but one possible improvement could be to stop the process as soon as a pair A[i,j] ≠ A[j,i] is found, rather than continuing to check the remaining pairs. This would not change the worst-case efficiency class, but it could potentially make the algorithm faster in practice if the matrix is not symmetric. The efficiency class would still be O(n^2).
Similar Questions
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