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	<title>Key4ever&#039;s 窝 &#187; 推理</title>
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		<title>一道小智力题（二）枪声</title>
		<link>http://www.key-4ever.com/2009/04/%e4%b8%80%e9%81%93%e5%b0%8f%e6%99%ba%e5%8a%9b%e9%a2%98%ef%bc%88%e4%ba%8c%ef%bc%89%e6%9e%aa%e5%a3%b0/</link>
		<comments>http://www.key-4ever.com/2009/04/%e4%b8%80%e9%81%93%e5%b0%8f%e6%99%ba%e5%8a%9b%e9%a2%98%ef%bc%88%e4%ba%8c%ef%bc%89%e6%9e%aa%e5%a3%b0/#comments</comments>
		<pubDate>Tue, 07 Apr 2009 17:29:01 +0000</pubDate>
		<dc:creator>Key4ever</dc:creator>
				<category><![CDATA[Intelligence]]></category>
		<category><![CDATA[博弈论]]></category>
		<category><![CDATA[思考]]></category>
		<category><![CDATA[推理]]></category>
		<category><![CDATA[智力]]></category>

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		<description><![CDATA[看了上一题，想起高中时看过的另一道类似的小题,大概是这样的。 一个村子里住着N个猎人，每一个猎人都有一条猎狗。后来发现他们的猎狗中至少有一条得了怪病，他们要尽快找出病狗并枪毙它。他们不能判断自己的猎狗有没 有患病，只能判断别人的猎狗，当他们发现病狗后不能告诉病狗的主人，也不能枪毙别人的狗，他们只能推算自己的猎狗是不是患病的狗，只有枪毙自己猎狗的权 利，第一天早上他们巡视各家判断有多少条患病的猎狗，若推算出自己的猎狗患病晚上得立刻枪毙。若第M天晚上响起了一阵枪声，那有多少条患病的狗呢？ 用上一道题的方法，先把问题的规模缩小，建立一个小模型进行分析。 ①设N=2，只有A,B两户猎人，a，b分别是他们的猎狗，若a正常，b患病，那么A看到患病的狗有一只，b看到0只，B就可以推算出自己的狗患病，第 一天晚上就会把自己的狗枪决。    &#62;&#62;&#62;&#62;&#62;&#62;&#62;&#62; ②若a，b都患病。A,B各看到一条患病的猎狗，但不知自己的猎狗是否患病，从B的角度看，他要听第一天晚上有没有枪声，若有，即上一种情况①发生，只 有a患病，即自己的猎狗正常，若没听到枪声，则自家猎狗也患病，第二天晚上要枪决。A的推算方式也一样。（当然，若果两条猎狗都正常，也会发生同样情 况，但要记得题目有一个条件，至少有一条得了怪病） ③扩大样本空间，在②中加入C，c正常，结果和②的一样，就是说若新加入的样本都正常，都是②的结果。 ④若在②中加入C，c患病，那么A,B,C都各自看到两条患病的猎狗，从C的角度出发（设想你是C），确定a，b都患病，若自己的猎狗正常，将会发生③ 的情况，第二天晚上听到枪声。若第二天晚上没听到枪声，即自己的猎狗患病，第三天晚上要把自家的猎狗解决，B,C也同样做法。 如此归类类推，即第M天听到枪响即有M条患病的猎狗。 相关日志2009/04/06 -- 一道小智力题：五个聪明的海盗（附英文版） (0)<table class="wumii-related-items" cellspacing="0" cellpadding="3" border="0"  style="clear: both;">
    
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			<content:encoded><![CDATA[<p>看了上一题，想起高中时看过的另一道类似的小题,大概是这样的。</p>
<p>一个村子里住着N个猎人，每一个猎人都有一条猎狗。后来发现他们的猎狗中至少有一条得了怪病，他们要尽快找出病狗并枪毙它。他们不能判断自己的猎狗有没<br />
有患病，只能判断别人的猎狗，当他们发现病狗后不能告诉病狗的主人，也不能枪毙别人的狗，他们只能推算自己的猎狗是不是患病的狗，只有枪毙自己猎狗的权<br />
利，第一天早上他们巡视各家判断有多少条患病的猎狗，若推算出自己的猎狗患病晚上得立刻枪毙。若第M天晚上响起了一阵枪声，那有多少条患病的狗呢？</p>
<p>用上一道题的方法，先把问题的规模缩小，建立一个小模型进行分析。<br />
①设N=2，只有A,B两户猎人，a，b分别是他们的猎狗，若a正常，b患病，那么A看到患病的狗有一只，b看到0只，B就可以推算出自己的狗患病，第<br />
一天晚上就会把自己的狗枪决。    <span style="color: #ff0000;">&gt;&gt;&gt;&gt;&gt;&gt;&gt;&gt;</span><span id="more-71"></span></p>
<p>②若a，b都患病。A,B各看到一条患病的猎狗，但不知自己的猎狗是否患病，从B的角度看，他要听第一天晚上有没有枪声，若有，即上一种情况①发生，只<br />
有a患病，即自己的猎狗正常，若没听到枪声，则自家猎狗也患病，第二天晚上要枪决。A的推算方式也一样。（当然，若果两条猎狗都正常，也会发生同样情<br />
况，但要记得题目有一个条件，至少有一条得了怪病）</p>
<p>③扩大样本空间，在②中加入C，c正常，结果和②的一样，就是说若新加入的样本都正常，都是②的结果。</p>
<p>④若在②中加入C，c患病，那么A,B,C都各自看到两条患病的猎狗，从C的角度出发（设想你是C），确定a，b都患病，若自己的猎狗正常，将会发生③<br />
的情况，第二天晚上听到枪声。若第二天晚上没听到枪声，即自己的猎狗患病，第三天晚上要把自家的猎狗解决，B,C也同样做法。</p>
<p>如此归类类推，即第M天听到枪响即有M条患病的猎狗。</p>
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