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<h1 class="post-title" itemprop="name headline">中国有嘻哈决赛冠军平票的概率问题(代码实现)</h1>
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<p>上一篇博客讨论了“中国有嘻哈”决赛平票的概率问题，分简单情况和复杂情况给出了计算方法和结果，但是留了一个小尾巴，就是复杂情况中的$f(x)$到底怎么算的。由于我只是为了快速算出结果，实现$f(x)$的代码非常丑陋，因此在上一篇博客中没有好意思把代码贴出来。感谢“Tianheng”同学的回复，给了我分享代码的勇气，如果我的代码有问题，欢迎讨论。下面就简述下我的代码实现，代码基于Python。</p>
<p>回顾上一篇博客的计算公式：$$\frac{\sum_{r=0}^{100} C_{100}^{r} f(125-r)}{2^{100}\times 51^{3}}$$<br>我们需要组合函数$C_n^k$和整数规划函数$f(x)$。</p>
<h2 id="组合函数"><a href="#组合函数" class="headerlink" title="组合函数"></a>组合函数</h2><p>组合函数我直接从网上找了一段代码：</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><div class="line">1</div><div class="line">2</div><div class="line">3</div><div class="line">4</div><div class="line">5</div><div class="line">6</div><div class="line">7</div><div class="line">8</div></pre></td><td class="code"><pre><div class="line"><span class="comment"># 组合函数</span></div><div class="line"><span class="keyword">import</span> operator</div><div class="line"><span class="keyword">import</span> math</div><div class="line"><span class="function"><span class="keyword">def</span> <span class="title">c</span><span class="params">(n,k)</span>:</span></div><div class="line"> <span class="keyword">if</span> k == <span class="number">0</span> <span class="keyword">or</span> k == n:</div><div class="line"> <span class="keyword">return</span> <span class="number">1</span></div><div class="line"> <span class="keyword">else</span>:</div><div class="line"> <span class="keyword">return</span> reduce(operator.mul, range(n - k + <span class="number">1</span>, n + <span class="number">1</span>)) / reduce(operator.mul, range(<span class="number">1</span>, k +<span class="number">1</span>))</div></pre></td></tr></table></figure>
<h2 id="整数规划函数"><a href="#整数规划函数" class="headerlink" title="整数规划函数"></a>整数规划函数</h2><p>接下来就是求解 $$x+y+z = v, x_{min} \leq x \leq x_{max}, y_{min} \leq y \leq y_{max}, z_{min} \leq z \leq z_{max}$$ 这个整数规划问题，实际上我们只需要统计可行解的数目即可。原谅我一下子没想出来3个变量的这个问题怎么求解，(感觉直接穷举x,y,z有点太暴力了)，我就先从简单点的问题开始考虑，如何求解包含2个变量的这种问题，也就是$$x+y = v, x_{min} \leq x \leq x_{max}, y_{min} \leq y \leq y_{max}$$ 这个问题。</p>
<h3 id="两变量问题："><a href="#两变量问题：" class="headerlink" title="两变量问题："></a>两变量问题：</h3><p>两个变量的问题比较容易，在给定$v$和$y$的取值范围后，$x$的实际可取值范围是可以计算出来的，$x$不能小于$x_{min}$，也不能小于$v-y_{max}$；$x$不能大于$x_{max}$，也不能大于$v-y_{min}$。在确定$x$后，因为$y=v-x$，实际上同时也确定了$y$，因此，只需要得出$x$的实际可取值范围，就可以知道有多少个可行解。下面是代码实现（请忽略我丑陋的函数和变量命名）：</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><div class="line">1</div><div class="line">2</div><div class="line">3</div><div class="line">4</div><div class="line">5</div><div class="line">6</div><div class="line">7</div><div class="line">8</div><div class="line">9</div><div class="line">10</div><div class="line">11</div><div class="line">12</div><div class="line">13</div><div class="line">14</div></pre></td><td class="code"><pre><div class="line"><span class="function"><span class="keyword">def</span> <span class="title">solution2</span><span class="params">(inputs, value)</span>:</span></div><div class="line"> <span class="string">"""</span></div><div class="line"><span class="string"> find solutions of x + y = value</span></div><div class="line"><span class="string"> inputs is a list of tuples,</span></div><div class="line"><span class="string"> [(x_min, x_max), (y_min, y_max)]</span></div><div class="line"><span class="string"> each tuple gives the domain of each variable</span></div><div class="line"><span class="string"> return is a tuple, indicates the valid range of the first variable</span></div><div class="line"><span class="string"> """</span></div><div class="line"> x_min = max(inputs[<span class="number">0</span>][<span class="number">0</span>], value - inputs[<span class="number">1</span>][<span class="number">1</span>])</div><div class="line"> x_max = min(inputs[<span class="number">0</span>][<span class="number">1</span>], value - inputs[<span class="number">1</span>][<span class="number">0</span>])</div><div class="line"> <span class="keyword">if</span> x_min > x_max:</div><div class="line"> <span class="keyword">return</span> <span class="keyword">False</span></div><div class="line"> <span class="keyword">else</span>:</div><div class="line"> <span class="keyword">return</span> (x_min, x_max)</div></pre></td></tr></table></figure>
<p>这里程序的返回结果不是可行解的个数，而是$x$的可取值范围，可行解的个数就是$x_{max} - x_{min} + 1$，这么写是为了方便写三变量问题代码，</p>
<h3 id="三变量问题："><a href="#三变量问题：" class="headerlink" title="三变量问题："></a>三变量问题：</h3><p>有了两变量问题的解决办法，三变量问题就可以比较容易的解决了，我们可以把三变量问题先分解为两变量问题，让$t=y+z$，先求解$x$和$t$的两变量问题，求出$x$的真实取值范围，然后对于每个可能的$x$的取值，求解$y+z=v-x$这个两变量问题。下面是代码实现（同样请忽略我丑陋的函数和变量命名）：<br><figure class="highlight python"><table><tr><td class="gutter"><pre><div class="line">1</div><div class="line">2</div><div class="line">3</div><div class="line">4</div><div class="line">5</div><div class="line">6</div><div class="line">7</div><div class="line">8</div><div class="line">9</div><div class="line">10</div><div class="line">11</div><div class="line">12</div><div class="line">13</div><div class="line">14</div><div class="line">15</div><div class="line">16</div><div class="line">17</div><div class="line">18</div><div class="line">19</div><div class="line">20</div><div class="line">21</div></pre></td><td class="code"><pre><div class="line"><span class="function"><span class="keyword">def</span> <span class="title">solution3</span><span class="params">(inputs, value)</span>:</span></div><div class="line"> <span class="string">"""</span></div><div class="line"><span class="string"> find the count of solutions of x + y + z = value</span></div><div class="line"><span class="string"> """</span></div><div class="line"> x_min = inputs[<span class="number">0</span>][<span class="number">0</span>]</div><div class="line"> x_max = inputs[<span class="number">0</span>][<span class="number">1</span>]</div><div class="line"></div><div class="line"> yz_min = inputs[<span class="number">1</span>][<span class="number">0</span>] + inputs[<span class="number">2</span>][<span class="number">0</span>]</div><div class="line"> yz_max = inputs[<span class="number">1</span>][<span class="number">1</span>] + inputs[<span class="number">2</span>][<span class="number">1</span>]</div><div class="line"></div><div class="line"> sx = solution2([(x_min,x_max), (yz_min,yz_max)], value)</div><div class="line"> <span class="keyword">if</span> <span class="keyword">not</span> sx:</div><div class="line"> <span class="keyword">return</span> <span class="number">0</span></div><div class="line"> nx_min, nx_max = sx </div><div class="line"></div><div class="line"> counts = <span class="number">0</span></div><div class="line"> <span class="keyword">for</span> x <span class="keyword">in</span> range(nx_min, nx_max + <span class="number">1</span>):</div><div class="line"> c = solution2([(inputs[<span class="number">1</span>][<span class="number">0</span>], inputs[<span class="number">1</span>][<span class="number">1</span>]), (inputs[<span class="number">2</span>][<span class="number">0</span>], inputs[<span class="number">2</span>][<span class="number">1</span>])], value-x)</div><div class="line"> <span class="keyword">if</span> c:</div><div class="line"> counts += c[<span class="number">1</span>] - c[<span class="number">0</span>] + <span class="number">1</span></div><div class="line"> <span class="keyword">return</span> counts</div></pre></td></tr></table></figure></p>
<p>函数solution3就是我们需要的$f(x)$。</p>
<h2 id="求解完整问题"><a href="#求解完整问题" class="headerlink" title="求解完整问题"></a>求解完整问题</h2><p>在有了组合函数和整数规划函数，就可以根据公式：$$\frac{\sum_{r=0}^{100} C_{100}^{r} f(125-r)}{2^{100}\times 51^{3}}$$</p>
<p>计算概率了，代码也比较简单：<br><figure class="highlight python"><table><tr><td class="gutter"><pre><div class="line">1</div><div class="line">2</div><div class="line">3</div><div class="line">4</div><div class="line">5</div><div class="line">6</div><div class="line">7</div><div class="line">8</div></pre></td><td class="code"><pre><div class="line">domain = (<span class="number">0</span>, <span class="number">50</span>)</div><div class="line">inputs = [domain,domain,domain]</div><div class="line">counts = <span class="number">0</span></div><div class="line"><span class="keyword">for</span> r <span class="keyword">in</span> range(<span class="number">0</span>, <span class="number">101</span>):</div><div class="line"> counts += c(<span class="number">100</span>,r) * solution3(inputs, <span class="number">125</span> - r)</div><div class="line"></div><div class="line">prob = counts *<span class="number">1.0</span> / (math.pow(<span class="number">2</span>,<span class="number">100</span>) * <span class="number">51</span>**<span class="number">3</span>)</div><div class="line"><span class="keyword">print</span> prob</div></pre></td></tr></table></figure></p>
<p>函数给出的结果是：0.0145193025331</p>
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