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<title>2(2N+1) 魔方阵</title>
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<h3><a href="http://caterpillar.onlyfun.net/GossipCN/index.html">From
Gossip@caterpillar</a></h3>
<h1><a href="AlgorithmGossip.htm">Algorithm Gossip: 2(2N+1) 魔方阵</a></h1>
<h2>说明</h2>
方阵的维度整体来看是偶数，但是其实是一个奇数乘以一个偶数，例如6X6，其中6=2X3，我们也称这种方阵与单偶数方阵。<br>
<h2>解法</h2>
如果您会解奇数魔术方阵，要解这种方阵也就不难理解，首先我们令n=2(2m+1)，并将整个方阵看作是数个奇数方阵的组合，如下所示： <br>
<div style="text-align: center;"><img style="width: 393px; height: 236px;" alt="2(2N+1)魔方阵" title="2(2N+1)魔方阵" src="images/twoNOneArray-1.jpg"><br>
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首先依序将A、B、C、D四个位置，依奇数方阵的规则填入数字，填完之后，方阵中各行的和就相同了，但列与对角线则否，此时必须在A-D与C- B之间，作一些对应的调换，规则如下：<br>
<ol>
<li>将A中每一列(中间列除外)的头m个元素，与D中对应位置的元素调换。</li>
<li>将A的中央列、中央那一格向左取m格，并与D中对应位置对调</li>
<li>将C中每一列的倒数m-1个元素，与B中对应的元素对调</li>
</ol>
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举个实例来说，如何填6X6方阵，我们首先将之分解为奇数方阵，并填入数字，如下所示： <br>
<div style="text-align: center;"><img style="width: 202px; height: 258px;" alt="2(2N+1)魔方阵" title="2(2N+1)魔方阵" src="images/twoNOneArray-2.jpg"></div>
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接下来进行互换的动作，互换的元素以不同颜色标示，如下：<br>
<div style="text-align: center;"><img style="width: 194px; height: 192px;" alt="2(2N+1)魔方阵" title="2(2N+1)魔方阵" src="images/twoNOneArray-3.jpg"></div>
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由于m-1的数为0，所以在这个例子中，C-B部份并不用进行对调。<br>
<h2> 实作</h2>
<ul>
<li> C
</li>
</ul>
<pre>#include <stdio.h> <br>#include <stdlib.h> <br><br>#define N 6 <br>#define SWAP(x,y) {int t; t = x; x = y; y = t;} <br><br>void magic_o(int [][N], int); <br>void exchange(int [][N], int); <br><br>int main(void) { <br> int square[N][N] = {0}; <br> int i, j; <br><br> magic_o(square, N/2); <br> exchange(square, N); <br><br> for(i = 0; i < N; i++) { <br> for(j = 0; j < N; j++) <br> printf("%2d ", square[i][j]); <br> printf("\n"); <br> } <br><br> return 0; <br>} <br><br>void magic_o(int square[][N], int n) { <br> int count, row, column; <br><br> row = 0; <br> column = n / 2; <br><br> for(count = 1; count <= n*n; count++) { <br> square[row][column] = count; // 填A <br> square[row+n][column+n] = count + n*n; // 填B <br> square[row][column+n] = count + 2*n*n; // 填C <br> square[row+n][column] = count + 3*n*n; // 填D <br> if(count % n == 0) <br> row++; <br> else { <br> row = (row == 0) ? n - 1 : row - 1 ; <br> column = (column == n-1) ? 0 : column + 1; <br> } <br> } <br>} <br><br>void exchange(int x[][N], int n) { <br> int i, j; <br> int m = n / 4; <br> int m1 = m - 1; <br><br> for(i = 0; i < n/2; i++) { <br> if(i != m) { <br> for(j = 0; j < m; j++) // 处理规则 1 <br> SWAP(x[i][j], x[n/2+i][j]); <br> for(j = 0; j < m1; j++) // 处理规则 2 <br> SWAP(x[i][n-1-j], x[n/2+i][n-1-j]); <br> } <br> else { // 处理规则 3 <br> for(j = 1; j <= m; j++) <br> SWAP(x[m][j], x[n/2+m][j]); <br> for(j = 0; j < m1; j++) <br> SWAP(x[m][n-1-j], x[n/2+m][n-1-j]); <br> } <br> } <br>} <br></pre>
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<ul>
<li> Java
</li>
</ul>
<pre>public class Matrix {<br> public static int[][] magic22mp1(int n) {<br> int[][] square = new int[n][n]; <br><br> magic_o(square, n/2); <br> exchange(square, n); <br> <br> return square;<br> }<br> <br> private static void magic_o(int[][] square, int n) {<br> int row = 0; <br> int column = n / 2; <br><br> for(int count = 1; count <= n*n; count++) { <br> square[row][column] = count; // 填A <br> square[row+n][column+n] = count + n*n; // 填B <br> square[row][column+n] = count + 2*n*n; // 填C <br> square[row+n][column] = count + 3*n*n; // 填D <br> if(count % n == 0) <br> row++; <br> else { <br> row = (row == 0) ? n - 1 : row - 1 ; <br> column = (column == n-1) ? 0 : column + 1; <br> } <br> }<br> }<br> <br> private static void exchange(int[][] x, int n) {<br> int i, j; <br> int m = n / 4; <br> int m1 = m - 1; <br><br> for(i = 0; i < n/2; i++) { <br> if(i != m) { <br> for(j = 0; j < m; j++) // 处理规则 1 <br> swap(x, i, j, n/2+i, j); <br> for(j = 0; j < m1; j++) // 处理规则 2 <br> swap(x, i, n-1-j, n/2+i, n-1-j); <br> } <br> else { // 处理规则 3 <br> for(j = 1; j <= m; j++) <br> swap(x, m, j, n/2+m, j); <br> for(j = 0; j < m1; j++) <br> swap(x, m, n-1-j, n/2+m, n-1-j); <br> } <br> } <br> }<br> <br> private static void swap(int[][] number,<br> int i, int j, int k, int l) {<br> int t; <br> t = number[i][j]; <br> number[i][j] = number[k][l]; <br> number[k][l] = t;<br> }<br> <br> public static void main(String[] args) {<br> int[][] magic = Matrix.magic22mp1(6);<br> <br> for(int k = 0; k < magic.length; k++) {<br> for(int l = 0; l < magic[0].length; l++) {<br> System.out.print(magic[k][l] + " ");<br> }<br> System.out.println();<br> }<br> }<br>} </pre>
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