publicclassComplexNumberMultiplicationSolution {publicstaticString complexNumberMultiply(String a, String b) {intp1 = Integer.parseInt(a.split("\\+")[0]);intn1 = Integer.parseInt(a.split("\\+")[1].split("i")[0]);intp2 = Integer.parseInt(b.split("\\+")[0]);intn2 = Integer.parseIn...
I didn't manage to do the recursion but working through the referenced Wikipedia article was able to reduce the large DFT matrix multiplication into smaller matrix products. Even so, 4096 = 64x64 elements took just 0.1s compared to over 10s using the built-in Fourier analysis tool...
I transformed the part-processed result as each stage so that the blocks of the right-hand side vector were stacked horizontally and then the multiplication by a
Among the above, the size of the input image, total number of epochs, and batch size have a major impact on the performance of the network. Size of the input image must be a multiple of 32 to fully transform the feature map; size of the input image determines the resolution of each d...
The improved method first calculates the inverse matrix of the covariance matrix S, S−1, and then computes the product of the difference vector d and the inverse covariance matrix S−1 through matrix multiplication, that is, dTS−1d. Secondly, the original method utilized pre-established ...
Multiplication(a + bi) · (c + di) = (ac–bd) + (ad+bc)i Note using the fact that a real numberccan be expressed asc+ 0i, we see that the multiplication of a complex number by a real number can be expressed as c· (a + bi) =ac+ (bc)i ...
The inverse property of multiplication states that for any number, a, where a is not equal to zero, there exists a number {eq}\frac{1}{a} {/eq} such that {eq}a \times \frac{1}{a} = \frac{1}{a} \times a = 1 {/eq}. Thus, the multiplicative inverse property states that ev...
So, even though we are multiplying both the top and bottom of the fraction by the complex conjugate, we really aren't changing the original number itself; we are just manipulating it so we can simplify it. Okay, so let's get to work: This first step shows multiplication by the complex...
recursion would be pure guesswork on my part. At the more detailed level, I transformed the part-processed result as each stage so that the blocks of the right-hand side vector were stacked horizontally and then the multiplication by a block diagonal matrix reduced to a single broadc...
The convolution seems to be pretty flexible in terms of having both a number of different applications and allowing experimentation to look at the effect of selecting different kernels for the smoothing. In my latest attempt I chose to use the binomial coefficients which I then looked ...