Algorithms for programmers phần 3

tức là những nguyên tố cùng nhau để m. Rõ ràng, tổng số đơn vị được cho bởi φ (m): | Z / MZ * | = Nếu m factorizes là m = 2k0 · pk1 · Nó chỉ ra rằng R thứ tự tối đa của một phần tử có thể bằng hoặc ít hơn | Z / MZ * |, * Z / MZ vòng sau đó được gọi là chu kỳ hoặc noncyclic, tương ứng. | CHAPTER 2. CONVOLUTIONS 44 - 1 -- 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0- 2 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0- 1- 3 3 4 5 6 7 8 9 10 11 12 13 14 15 0- 1- 2- 4 4 5 6 7 8 9 10 11 12 13 14 15 0- 1- 2- 3- 5 5 6 7 8 9 10 11 12 13 14 15 0- 1- 2- 3- 4- 6 6 7 8 9 10 11 12 13 14 15 0- 1- 2- 3- 4- 5- 7 7 8 9 10 11 12 13 14 15 0- 1- 2- 3- 4- 5- 6- 8 8 9 10 11 12 13 14 15 0- 1- 2- 3- 4- 5- 6- 7- 9 9 10 11 12 13 14 15 0- 1- 2- 3- 4- 5- 6- 7- 8- 10 10 11 12 13 14 15 0- 1- 2- 3- 4- 5- 6- 7- 8- 9- 11 11 12 13 14 15 0- 1- 2- 3- 4- 5- 6- 7- 8- 9- 10- 12 12 13 14 15 0- 1- 2- 3- 4- 5- 6- 7- 8- 9- 10- 11- 13 13 14 15 0- 1- 2- 3- 4- 5- 6- 7- 8- 9- 10- 11- 12- 14 14 15 0- 1- 2- 3- 4- 5- 6- 7- 8- 9- 10- 11- 12- 13- 15 15 0- 1- 2- 3- 4- 5- 6- 7- 8- 9- 10- 11- 12- 13- 14- Here the products that enter with negative sign are indicated with a postfix minus at the corresponding entry. With right-angle convolution the minuses have to be replaced by i ỵ ĩ which means the wrap-around . h 1 elements go to the imaginary part. With real input one thereby effectively separates h 0 and h 1 . Note that once one has routines for both cyclic and negacyclic convolution the parts h 0 and h 1 can be computed as sum and difference respectively. Thereby all expressions of the form ah 0 3h 1 can be trivially computed. Half cyclic convolution for half the price The computation of h 0 from formula without computing h 1 is called half cyclic convolution. Apparently one asks for less information than one gets from the acyclic convolution. One might hope to find an algorithm that computes h 0 and uses only half the memory compared to the linear convolution or that needs half the work possibly both. It may be a surprise that no such algorithm seems to be known currently5. Here is a clumsy attempt to find h 0 alone Use the weighted transform with the weight sequence vx Vx where Vn is very small. Then h 1 will in the result be .