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A straightforward algorithm to multiply numbers in Montgomery form is therefore to multiply aR mod N, bR mod N, and R′ as integers and reduce modulo N. For example, to multiply 7 and 15 modulo 17 in Montgomery form, again with R = 100, compute the product of 3 and 4 to get 12 as above.
Graphs of functions commonly used in the analysis of algorithms, showing the number of operations versus input size for each function. The following tables list the computational complexity of various algorithms for common mathematical operations.
The definition of matrix multiplication is that if C = AB for an n × m matrix A and an m × p matrix B, then C is an n × p matrix with entries = =. From this, a simple algorithm can be constructed which loops over the indices i from 1 through n and j from 1 through p, computing the above using a nested loop:
The outer product is equivalent to a matrix multiplication, provided that is represented as a column vector and as a column vector (which makes a row vector). [ 2 ] [ 3 ] For instance, if m = 4 {\displaystyle m=4} and n = 3 , {\displaystyle n=3,} then [ 4 ]
The optimal number of field operations needed to multiply two square n × n matrices up to constant factors is still unknown. This is a major open question in theoretical computer science. As of January 2024, the best bound on the asymptotic complexity of a matrix multiplication algorithm is O(n 2.371339). [2]
Decomposition: = where C is an m-by-r full column rank matrix and F is an r-by-n full row rank matrix Comment: The rank factorization can be used to compute the Moore–Penrose pseudoinverse of A , [ 2 ] which one can apply to obtain all solutions of the linear system A x = b {\displaystyle A\mathbf {x} =\mathbf {b} } .
Naïve matrix multiplication requires one multiplication for each "1" of the left column. Each of the other columns (M1-M7) represents a single one of the 7 multiplications in the Strassen algorithm. The sum of the columns M1-M7 gives the same result as the full matrix multiplication on the left.
Example: 100P can be written as 2(2[P + 2(2[2(P + 2P)])]) and thus requires six point double operations and two point addition operations. 100P would be equal to f(P, 100). This algorithm requires log 2 (d) iterations of point doubling and addition to compute the full point multiplication. There are many variations of this algorithm such as ...