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How many real links are required to store a sparse matrix of 10 rows, 10 columns, and 15 non-zero entries, (Pick up the closest answer)
- a)15
- b)20
- c)50
- d)100
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
How many real links are required to store a sparse matrix of 10 rows, ...
Matrix of 10 rows and 10 columns can have m axim um 1 0 x 1 0 = 100 entries. S ince only 15 non-zero entries are present, so number of real link will be 15 only.
Assuming undirected graph.
Assuming undirected graph.
Most Upvoted Answer
How many real links are required to store a sparse matrix of 10 rows, ...
Explanation:
To store a sparse matrix, we only need to store the non-zero elements of the matrix along with their row and column indices. This approach leads to saving space and time complexity.
Formula:
For a sparse matrix of size m x n with s non-zero elements, the number of real links required can be calculated by the formula:
Number of Real Links = s + 1
The additional 1 is required to store the size of the matrix.
Given:
Size of matrix = 10 x 10
Number of non-zero elements = 15
Calculation:
Using the formula above, we can calculate the number of real links required for the given sparse matrix as:
Number of Real Links = 15 + 1 = 16
Answer:
Therefore, the closest answer to the number of real links required to store the sparse matrix of 10 x 10 size with 15 non-zero elements is option 'A', i.e., 15.
To store a sparse matrix, we only need to store the non-zero elements of the matrix along with their row and column indices. This approach leads to saving space and time complexity.
Formula:
For a sparse matrix of size m x n with s non-zero elements, the number of real links required can be calculated by the formula:
Number of Real Links = s + 1
The additional 1 is required to store the size of the matrix.
Given:
Size of matrix = 10 x 10
Number of non-zero elements = 15
Calculation:
Using the formula above, we can calculate the number of real links required for the given sparse matrix as:
Number of Real Links = 15 + 1 = 16
Answer:
Therefore, the closest answer to the number of real links required to store the sparse matrix of 10 x 10 size with 15 non-zero elements is option 'A', i.e., 15.
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