123456789组成的3×3的矩阵的行列式最大的值是多少？

文章目录[隐藏]

回复内容：

123456789怎样运算等于1？ – abccsss 的回答
假定每个数字只能出现一次。

回复内容：

Mathematica代码

较简洁

<code class="language-text">Det/@N@Range@9~Permutations~{9}~ArrayReshape~{9!,3,3}//Max</code>

以上用Matlab暴力破解（枚举种情形），暂未输出行列式相同的其他情形，貌似基本秒出。

max_det%20=%200;init_perm%20=%20reshape(1:9,%20[3,%203]);all_perms%20=%20perms(1:9);for%20i%20=%201:size(all_perms,%201)%20%20%20%20matrix%20=%20all_perms(i,%20:);%20%20%20%20matrix%20=%20reshape(matrix,%20[3,%203]);%20%20%20%20det_value%20=%20det(matrix);%20%20%20%20%20%20%20%20if%20det_value%20>%20max_det%20%20%20%20%20%20%20%20max_det%20=%20det_value;%20%20%20%20%20%20%20%20init_perm%20=%20matrix;%20%20%20%20end%20%20%20%20end

list = Permutations[Range[9], {9}];list = Permutations[Range[9], {9}];
matrix = Partition[#, 3] & /@ list;
answer = Det /@ matrix;
m = Max[answer];
pos = Flatten[Position[answer, m]];
matrix[[#]] & /@ pos贴个毫无技术含量暴力程度max的python版。。。

<code class="language-text">import itertoolsimport timedef max_matrix():	begin = time.time()	elements = [1, 2, 3, 4, 5, 6, 7, 8, 9]	maxdet = 0	maxmat = []	for i in itertools.permutations(elements, 9):		det = i[0] * i[4] * i[8] + i[1] * i[5] * i[6] + i[2] * i[3] * i[7] - i[2] * i[4] * i[6] - i[1] * i[3] * i[8] - i[0] * i[5] * i[7]		if(det > maxdet):			maxdet = det			maxmat = []			for j in range(0, 9):				maxmat.append(i[j])	print "|" + str(maxmat[0]) + " " + str(maxmat[1]) + " " + str(maxmat[2]) + "|"	print "|" + str(maxmat[3]) + " " + str(maxmat[4]) + " " + str(maxmat[5]) + "| = " + str(maxdet)	print "|" + str(maxmat[6]) + " " + str(maxmat[7]) + " " + str(maxmat[8]) + "|"	end = time.time()	print str(end - begin) + 's used.'if __name__ == '__main__':	max_matrix()</code>

题目应该改成1 2 3 …n^2组成n阶行列式的最大值。并求最优解的时间复杂度才有意思。C++：

<code class="language-cpp"><span class="cp">#include </span><span class="cp">#include </span><span class="k">using</span> <span class="k">namespace</span> <span class="n">std</span><span class="p">;</span><span class="kt">int</span> <span class="n">ans</span><span class="p">,</span> <span class="n">a</span><span class="p">[]</span> <span class="o">=</span> <span class="p">{</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">8</span><span class="p">,</span> <span class="mi">9</span><span class="p">};</span><span class="kt">int</span> <span class="nf">main</span><span class="p">()</span> <span class="p">{</span>  <span class="k">do</span>    <span class="n">ans</span> <span class="o">=</span> <span class="n">max</span><span class="p">(</span><span class="n">ans</span><span class="p">,</span> <span class="n">a</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">a</span><span class="p">[</span><span class="mi">4</span><span class="p">]</span> <span class="o">*</span> <span class="n">a</span><span class="p">[</span><span class="mi">8</span><span class="p">]</span> <span class="o">-</span> <span class="n">a</span><span class="p">[</span><span class="mi">5</span><span class="p">]</span> <span class="o">*</span> <span class="n">a</span><span class="p">[</span><span class="mi">7</span><span class="p">])</span> <span class="o">+</span>                   <span class="n">a</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">a</span><span class="p">[</span><span class="mi">5</span><span class="p">]</span> <span class="o">*</span> <span class="n">a</span><span class="p">[</span><span class="mi">6</span><span class="p">]</span> <span class="o">-</span> <span class="n">a</span><span class="p">[</span><span class="mi">3</span><span class="p">]</span> <span class="o">*</span> <span class="n">a</span><span class="p">[</span><span class="mi">8</span><span class="p">])</span> <span class="o">+</span>                   <span class="n">a</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">a</span><span class="p">[</span><span class="mi">3</span><span class="p">]</span> <span class="o">*</span> <span class="n">a</span><span class="p">[</span><span class="mi">7</span><span class="p">]</span> <span class="o">-</span> <span class="n">a</span><span class="p">[</span><span class="mi">4</span><span class="p">]</span> <span class="o">*</span> <span class="n">a</span><span class="p">[</span><span class="mi">6</span><span class="p">]));</span>  <span class="k">while</span> <span class="p">(</span><span class="n">next_permutation</span><span class="p">(</span><span class="n">a</span><span class="p">,</span> <span class="n">a</span> <span class="o">+</span> <span class="mi">9</span><span class="p">));</span>  <span class="n">printf</span><span class="p">(</span><span class="s">"%d</span><span class="se">\n</span><span class="s">"</span><span class="p">,</span> <span class="n">ans</span><span class="p">);</span><span class="p">}</span></code>

把yellow的答案重排一下可得
9 4 2
3 8 6
5 1 7

很容易看出思路了。
1.所有数按大小在斜率为-1的对角线上依次排开。（即：987在一条对角线，654在一条，321在一条）很容易看出这是让正向数值最大的方法。
2.对于反向的对角线，排除主对角线之外的任意两个数之和相等，且乘积越大的，相应的主对角线元素越小。（也就是让三个乘积的最大值最小，然后最大的结果再和最小的数相配这样）

但是以上方法仅限于1~9的3×3矩阵，对于其它的矩阵不一定适用。
因为显然这种方法要求正向和负向都只有对角线（或平行于对角线），但是4×4的行列式就开始有拐弯了。。。

然后，我感觉还有三个漏洞，一是贪心法不一定保证正向最大，也不一定保证反向最小，更不一定保证正反向之差最大。（不一定都是漏洞，可能有的是恒成立的）
但是我感觉对3×3的非负矩阵来说，贪心在多数情况下是可以拿到最大值的。

PS：试了很多组数，都是这个解，然后又试了一组[1 2 3 4 5 6 7 8 100]，显然答案发生了变化，因为100的权值比8和7大太多，所以负向的时候直接就把2和1给了100。那么这也就证明了贪心法确实有时候得不到最大值。前面已经有了python，c和MMA的代码了，我来一发matlab的本文来源gaodai$ma#com搞$代*码网2吧

<code class="language-matlab"><span class="n">p</span><span class="p">=</span><span class="nb">perms</span><span class="p">(</span><span class="mi">1</span><span class="p">:</span><span class="mi">9</span><span class="p">);</span><span class="p">[</span><span class="n">n</span><span class="p">,</span><span class="o">~</span><span class="p">]=</span><span class="nb">size</span><span class="p">(</span><span class="n">p</span><span class="p">);</span><span class="n">z</span><span class="p">=</span><span class="nb">zeros</span><span class="p">(</span><span class="n">n</span><span class="p">,</span><span class="mi">1</span><span class="p">);</span><span class="k">for</span> <span class="nb">i</span><span class="p">=</span><span class="mi">1</span><span class="p">:</span><span class="n">n</span>    <span class="n">z</span><span class="p">(</span><span class="nb">i</span><span class="p">)=</span><span class="n">det</span><span class="p">(</span><span class="nb">reshape</span><span class="p">(</span><span class="n">p</span><span class="p">(</span><span class="nb">i</span><span class="p">,:),</span><span class="mi">3</span><span class="p">,</span><span class="mi">3</span><span class="p">));</span><span class="k">end</span><span class="n">max</span><span class="p">(</span><span class="n">z</span><span class="p">)</span><span class="n">id</span><span class="p">=</span><span class="nb">find</span><span class="p">(</span><span class="n">z</span><span class="o">==</span><span class="n">max</span><span class="p">(</span><span class="n">z</span><span class="p">));</span><span class="k">for</span> <span class="nb">i</span><span class="p">=</span><span class="mi">1</span><span class="p">:</span><span class="nb">length</span><span class="p">(</span><span class="n">id</span><span class="p">)</span>    <span class="nb">disp</span><span class="p">(</span><span class="nb">reshape</span><span class="p">(</span><span class="n">p</span><span class="p">(</span><span class="n">id</span><span class="p">(</span><span class="nb">i</span><span class="p">),:),</span><span class="mi">3</span><span class="p">,</span><span class="mi">3</span><span class="p">));</span><span class="k">end</span></code>

对于三阶的穷举，可以不用det函数会比较简单：

<code class="language-matlab"><span class="n">p</span> <span class="p">=</span> <span class="nb">reshape</span><span class="p">(</span><span class="nb">perms</span><span class="p">(</span><span class="mi">1</span><span class="p">:</span><span class="mi">9</span><span class="p">),</span><span class="s">''</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">3</span><span class="p">);</span><span class="n">M</span> <span class="p">=</span> <span class="n">max</span><span class="p">(</span><span class="n">sum</span><span class="p">(</span><span class="n">prod</span><span class="p">(</span><span class="n">p</span><span class="p">,</span><span class="mi">2</span><span class="p">),</span><span class="mi">3</span><span class="p">)</span><span class="o">-</span><span class="n">sum</span><span class="p">(</span><span class="n">prod</span><span class="p">(</span><span class="n">p</span><span class="p">,</span><span class="mi">3</span><span class="p">),</span><span class="mi">2</span><span class="p">));</span></code>

话题的语言还少个Mathematica,就我来吧
直接9!个结果存下来刚正面,0优化

<code class="language-text">Det[Partition[#, 3]] & /@ Permutations[Range[9]] // Max412</code>

搞代码网（gaodaima.com）提供的所有资源部分来自互联网，如果有侵犯您的版权或其他权益，请说明详细缘由并提供版权或权益证明然后发送到邮箱[email protected]‍，我们会在看到邮件的第一时间内为您处理，或直接联系QQ：872152909。本网站采用BY-NC-SA协议进行授权
转载请注明原文链接：123456789组成的3×3的矩阵的行列式最大的值是多少？

回复内容：

Hi，您需要填写昵称和邮箱！