Description: sparse matrix A and B were 3 groups, as a storage structure and try to write together the matrix algorithm, results stored in ternary Group C table. Solution : This is a bit complicated algorithm, taking into account the two nonzero sparse matrix element is not one-to-one. the establishment of a new ternary Group C table, in order to enable groups of elements remaining three yuan prioritize OK, So inserted three yuan each group is not necessarily A, in accordance with the matrix elements ranks to find a group of three yuan, and if so, C is added, and if this element is B, plus the B value of the element, Otherwise, the value on the same if not A, then B to find, then insert C, no one will find under the matrix elements.
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