12*12方阵填数问题

wwdwwd117

将1-144这144个自然数填入12*12的方阵，要求相邻的自然数在相邻的方格中。（上下左右为相邻）且144与1相邻。
求主对角线之和的最大值。

wwdwwd117

wwdwwd117

想了两天。

kuing

“主对角线之和”我还以为两条对角线都要考虑

wwdwwd117

realnumber

回复 4# kuing


    应该加上“一条”，我也想成2条了

Infinity

回复 4# kuing
矩阵里面的“主对角线”指的就是转置轴，即左上到右下的那条对角线，“副对角线”则是指右上到左下那条对角线。因此题目表述得很清楚。

kuing

回复 7# Infinity

好吧，没怎么学过矩阵的就是吃亏[笑哭][捂脸]

Infinity

回复 8# kuing
也就是那些学术同行一个约定俗成的叫法，比如还有次对角线（主对角线往下移一行），超对角线（主对角线上移一行）。可以熟悉一下，方便交流。

青青子衿

回复 9# Infinity
其实，应该称“主对角线元素之和”更合理一些

青青子衿

回复 10# 青青子衿
Oeis上有一个数列(A265914)记录着
Number of Hamiltonian paths on an n X n grid reduced for symmetry,
i.e., where rotations and reflections are not counted as distinct.
在对称变换下，\(\,n\times\,\!n\,\)网格中哈密顿路径的数目。
\begin{align*}
a_1&=1  &&a_2=1  \\
a_3&=3  &&a_4=38\\
a_5&=549  &&a_6=28728\\
a_7&=1692417  &&a_8=377919174\\
a_9&=93177169027  &&a_{10}=91255604983167\\
a_{11}&=98333935794279062&&a_{12}=431583106977641773651
\end{align*}
在对称变换下，\(\,3\times3\,\)网格中哈密顿路径（一笔可画完路径）的数目只有三种（可以记作G型、LS型、S型）。
\begin{align*}
\begin{array}{|c|c|c|c|}   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\downarrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\uparrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\downarrow}}\!\!\!{\substack{\phantom\leftarrow}}\\   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\downarrow\\\downarrow}}\!\!\!{\substack{\phantom\rightarrow}}&
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\uparrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\uparrow}}\!\!\!{\substack{\phantom\rightarrow}}\\   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\downarrow\\\phantom\downarrow}}\!\!\!{\substack{\rightarrow}}&
{\substack{\rightarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\uparrow}}\!\!\!{\substack{\rightarrow}}&
{\substack{\rightarrow}}\!\!\!{\substack{\uparrow\\\phantom\uparrow}}\!\!\!{\substack{\phantom\rightarrow}}\\
\hline   
\end{array}
\end{align*}
\begin{align*}
\begin{array}{|c|c|c|c|}   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\uparrow}}\!\!\!{\substack{\phantom\leftarrow}}&
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\downarrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\downarrow}}\!\!\!{\substack{\phantom\leftarrow}}\\   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\uparrow\\\uparrow}}\!\!\!{\substack{\phantom\rightarrow}}&
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\downarrow\\\phantom\uparrow}}\!\!\!{\substack{\rightarrow}}&
{\substack{\rightarrow}}\!\!\!{\substack{\phantom\uparrow\\\downarrow}}\!\!\!{\substack{\phantom\leftarrow}}\\   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\uparrow\\\phantom\downarrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\uparrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\downarrow\\\phantom\uparrow}}\!\!\!{\substack{\phantom\rightarrow}}\\
\hline   
\end{array}
\end{align*}
\begin{align*}
\begin{array}{|c|c|c|c|}   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\downarrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\downarrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\downarrow}}\!\!\!{\substack{\phantom\leftarrow}}\\   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\downarrow\\\phantom\downarrow}}\!\!\!{\substack{\rightarrow}}&
{\substack{\rightarrow}}\!\!\!{\substack{\phantom\downarrow\\\phantom\uparrow}}\!\!\!{\substack{\rightarrow}}&
{\substack{\rightarrow}}\!\!\!{\substack{\phantom\uparrow\\\downarrow}}\!\!\!{\substack{\phantom\leftarrow}}\\   
\hline   
{\substack{\phantom\leftarrow}}\!\!\!{\substack{\phantom\downarrow\\\phantom\downarrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\phantom\uparrow\\\phantom\uparrow}}\!\!\!{\substack{\leftarrow}}&
{\substack{\leftarrow}}\!\!\!{\substack{\downarrow\\\phantom\uparrow}}\!\!\!{\substack{\phantom\rightarrow}}\\
\hline   
\end{array}
\end{align*}

Infinity

回复 10# 青青子衿
是的，这样表述更清楚。