将二维上三角和下三角中的元素映射到线性结构 [英] Mapping elements in 2D upper triangle and lower triangle to linear structure

问题描述

我有一个矩阵 M，它有 NxN 维，其中 M(i,j) = M(j,i)

I have a matrix M which is of NxN dimensions, where M(i,j) = M(j,i)

我想将此结构表示为 (N²+N)/2 线性阵列 K，以节省空间.我的问题是提出将 M(min(i,j),min(i,j)) 映射到范围 [0,(N^2)/2)

I would like to represent this structure as a (N²+N)/2 linear array K, to save space. My problem is coming up with the formula that will map a M(min(i,j),min(i,j)) into a range [0,(N^2)/2)

下面是带有 K 线性阵列索引的 3x3 矩阵的映射，X 表示这些单元不存在，而是使用它们的转置:

Below is a mapping of a 3x3 matrix with indexes for K linear array, the X means those cells don't exist and instead their transpose is to be used:

0123
X456
XX78
XXX9

这是一个带有 K 线性数组索引的 7x7 矩阵

Here is a 7x7 matrix with indexes for the K linear array

     0  1  2  3  4  5  6
 0  00 01 02 03 04 05 06
 1     07 08 09 10 11 12
 2        13 14 15 16 17
 3           18 19 20 21
 4              22 23 24
 5                 25 26
 6                    27

目前我有以下内容

int main()
{
   const unsigned int N = 10;
   int M[N][N];

   int* M_ = &(M[0][0]);

   assert(M[i][j] = M_[N * min(i,j) + max(i,j)]);

   //int* K = .....
   //assert(M[i][j] = K[.....]);

   return 0;
}

推荐答案

我需要的相反方向:

void printxy(int index)  
{  
    int y = (int)((-1+sqrt(8*index+1))/2);  
    int x = index - y*(y+1)/2;  
}

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