在2D上三角，下三角线性结构的映射元素 [英] Mapping elements in 2D upper triangle and lower triangle to linear structure

问题描述

我有一个矩阵M是N×N个尺寸，对其中M（I，J）= M（J，I）

我想重新present这种结构为（N²+ N）/ 2线阵K，以节省空间。我的问题是想出的公式，将映射M（分（I，J），分钟（I，J））转换成范围[0，（N ^ 2）/ 2）

下面是与索引中K线性阵列的3×3矩阵的映射，所述X表示的那些细胞中不存在，取而代之的转置是用于：

  0123
X456
XX78
XXX9
 

下面是索引用于K线阵

7×7矩阵

  0 1 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月25日26
 6月27日
 

目前我有以下

  INT主要（）
{
   const的无符号整数N = 10;
   INT M [N] [N];

   为int * M_ =安培（M [0] [0]）;

   断言（M [I] [j]的= M_〔N *分（I，J）+最大（I，J）]）;

   //为int * K = .....
   //断言（M [I] [J] = K [...]）;

   返回0;
}
 

解决方案

要往相反的方向而这正是我需要的：

 无效printxy（INT指数）
{
    INT Y =（int）的（（ -  1 + SQRT（8 *指数+ 1））/ 2）;
    INT X =指数 -  Y *（Y + 1）/ 2;
}
 

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

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)

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

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

at the moment I have the following

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;
}

解决方案

To go the opposite direction which is what I needed:

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

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