中英对照组合问题.htm

软考英语精典短文每日一练汇总 软考英语精典短文每日一练汇总

      examples can be used to solve the general problem．When we use this 
      strategy，we hope to gain insight and understanding by investigating small 
      versions of the same type of problem．<BR><BR>One very common method of 
      counting the elements of a set is that of determining a sequence of 
      well-defined steps for constructing all members of the set but no other 
      objects．The number of objects constructed can then be calculated from the 
      numbers of outcomes of the various steps in the construction process．The 
      following example provides an illustration of counting by 
      construction．<BR><BR>[ Example 1 ] How many three-digit decimal numbers 
      have three different digits，one of which is 6 ?<BR><BR>We can construct 
      any such number by filling each of three adjacent positions with three 
      different digits，where one digit is 6 and the left-hand digit is not 0．If 
      6 is placed in the left-hand position，then any of nine different digits 
      can fill the middle position，and any of eight digits can fill the 
      right-hand position．Thus，（9）（8）= 72 such numbers begin with 6．If 6 is 
      placed in the middle position，eight nonzero digits are available for the 
      left-hand position and eight remaining digits are available for the 
      right-hand position．Hence，（8）（8）= 64 numbers can have 6 as the middle 
      digit．A moment’s reflection [4] should convince us that there also are 64 
      numbers having 6 as the right-hand digit．Thus，the total count of numbers 
      of the prescribed type is<BR><BR>72十64＋64 = 200<BR><BR>Another general 
      approach to counting uses analogy and matching by recognizing that the set 
      to be counted has the same number of elements as some other set that is 
      easier to count．<BR><BR>[ Example 2 ] How many matches need to be 
      scheduled to determine the champion of a tennis tournament in which there 
      are 83 entrants？<BR><BR>Our first inclination might be to draw a large 
      chart showing a possible pairing of players，where the winner of each match 
      survives to play in the next round [5] ．But if we recall that each match 
      determines one loser and that there is only one champion at the end of the 
      tournament（and hence exactly 82 losers），we can deduce that exactly 82 
      matches are needed．<BR><BR>The previous example also illustrates the 
      technique for counting a complement of a set．The set of nonenampions in a 
      tennis tournament is the complement of the set of champions within the set 
      of entrants．<BR><BR>Induction and the related concept of recursion also 
      provide means of counting．The Tower of Hanoi problem can be solved by 
      induction：One move is needed for the 1-ring puzzle，three moves for the 
      2-ring puzzle，seven moves for the 3-ring puzzle，and we guess that 2n —1 
      moves will suffice for the n-ring puzzle．Indeed，we can solve the n-ring 
      puzzle by initially solving the（n—1）-ring puzzle in 2n —1 moves，using the 
      top n—1rings；we then use one move to transfer the largest ring to the 
      empty post，and we repeat the sequence of 2n-1 —1 moves needed to place the 
      other n—1 rings on the largest ring．The total number of moves 
      is<BR><BR>[2n—1—1] + 1 + [2n—1—1] = 2 [2n—1] + 1—2 = 2n—1 <BR><BR>Another 
      principle of counting seems almost too obvious to mention，but its 
      appilcations are sometimes quite subtle [6]．<BR><BR>The Pigeonhole 
      Principle．If n pigeons roost in k pigeonholes（n＞k），then at least one 
      pigeonhole must contain more than one pigeon．<BR><BR>This principle，for 
      example，allows us to deduce that two or more residents of Detroit have the 
      same number of hairs on their heads，because the number of 
      residents（pigeons）is much larger than the maximum number of 
      hairs（pigeonholes）that can grow on any human head．<BR><BR>Now suppose that 
      n pigeons roost in k pigeonholes and that n&gt;2k；then at least one 
      pigeonhole must contain more than two pigeons．To prove that assertion，we 
      use an indirect proof．Suppose that n of the pigeons are roosting，but that 
      none of the k pigeonholes contains more than two pigeons．Then the number 
      of roosting pigeons does not exceed 2k，and 2k<N，CONTRADICTING 
      principle．<br pigeonhole the of statement general more following prove can 
      argument，we same essentially roosting．Using are pigeons all that 
      assumption><BR>The Generalized Pigeonhole Principle．If n pigeons roost in 
      k pigeonholes，and if n = rnk＋r，where m and r are positive integers with 
      r<K，THEN pigeonhole more <br pigeons． m than contains one least 
      at><BR>有限集的成员可以按多种方式计数。如果集合较小，可以简单地列出其成员并用正整数编排列出的成员数，从1开始，以自然数的次序往下直至结束。但对于一个较大的集合，这一方法就显得慢而且易于出错。因此人们制定出许多更复杂的策略和方法来对较大且定义复杂的集合进行计数。成功的计数大多依赖于选择一种适用的解决该问题的方法时所用的技巧和洞察力。人们可以通过完全熟悉各种各样的计数方法以及使用这些计数方法广泛地解决一些特定问题从而获得经验，进而完善计数的技巧。<BR><BR>首先要强调推广和试验的重要性。例如需要确定从美国参议院的100个议员中选出由11人组成的委员会的数目，就应该清楚这一问题是下面更一般问题的一个特例：<BR><BR>从一个有n个可区别的成员的集合中可以选出多少个不同的具有k个成员的子集？<BR><BR>可以通过使用某些较小的k值和n值来解决这一问题，其中k≤n。通常在较小的例子中所用的推理可以用于解决一般的问题。当我们使用这一策略时，希望通过观察同类型问题的较小模式来获取对问题的认识与理解。<BR><BR>对集合元素进行计数的非常通用的方法是确定一个严格定义的步骤顺序来构造该集合的所有成员而不含其他对象，然后由构造过程的不同步骤所产生的结果数目计算出所构造的对象数。下面给出一个通过构造进行计数的例子。<BR><BR>［例1］有多少个3位十进制数，它们有3个不同的数字，其中一位是6 
      ?<BR><BR>我们可以通过用3个不同的数字去填充3个相邻的位置来构造任何一个这样的数，该数中一位是6而其左边位不为0。如果将6放在最左的一位上，那么余下的9个数中的任何一个可以填在中间一位，余下的8个数中的任何一个可填在最右边的一位上。于是，共有（9） 
      （8）=72个以6开始的数。如果6放在中间一位上，则8个非零数字可以放在左侧一位，而余下的8个数字可以放在右侧位置。由此共有（8）（8）=64，中间一位是6的数。很快我们知道同样也有64个右边一位是6的数。于是，这种类型的数的总数是<BR><BR>72十64 
      ＋ 64 = 
      200<BR><BR>另一个用于计数的普遍方法是类比和匹配，它将要计数的集合看成是另一个具有相同数目但易于计算的集合。<BR><BR>［例2］在有83位选手参加的网球锦标赛中，需要安排多少场比赛才能产生出冠军？<BR><BR>人们首先会想到画一个大图显示选手间可能的配对，其中每场比赛的胜者进入下一轮角逐。但是如果注意到每场比赛定出一个败者而在赛末只有一个冠军产生（因此正好有82个败者），那么我们能够推出恰恰需要82场比赛。<BR><BR>上一例子还表明了对一个集合的补集进行计数的方法，一场网球锦标赛中，对于参赛选手的集合，非冠军的集合是冠军集合的补集。<BR><BR>归纳法及与之有关的递归概念也可以提供计数的方法。河内塔问题可以通过归纳法解决：一个圆环需要一次搬动；两个圆环的难题需要3次搬动；3个圆环难题共需搬动7次；进而可以推测对于n个圆环的难题共需搬动2n一1次。实际上，对于n个圆环的难题，可以先通过2n一1次搬动解决其顶上（n-1）个环的搬动问题；然后用一次搬动将最大的圆环送到空柱上，然后再重复2n-1一1次搬动的过程将其他（n-1）个环放在最大环上面。搬动的总数为：<BR><BR>（2n-1一1）＋1＋（2n-1一1）＝2（2n-1）＋1一2＝2n一1<BR><BR>另一个用于计数的原理明显似乎不必提及，但它的应用有时却相当微妙。<BR><BR>鸽巢原理：如果有n只鸽子栖息在k个鸽巢中（n&gt;k），那么至少有一个鸽巢中有一只以上的鸽子。<BR><BR>例如，这一原理使我们能推断底特律的居民中有两个或更多个居民的头发数目相同，这是因为居民的数目（鸽子）远远大于任何一个人头上能生长的头发的数目（鸽巢）。<BR><BR>现在假定n只鸽子栖息在k个鸽巢中并且n&gt;2k；那么至少有一个鸽巢中有多于两只鸽子。为了证明这一点，我们用间接证法。假定所有n只鸽子都栖息于鸽巢中，而k个鸽巢中没有一个多于两只鸽子，那么栖息的鸽子的数目不会超过2k，而（此时）2k<N，于是与所有鸽子都栖息于鸽巢中的假定相矛盾。用实质上相同的论证，可以证明下面关于鸽巢原理更普遍的结论。<BR><BR>广义鸽巢原理：如果n只鸽子栖息在k个鸽巢中，且n 
      = mk＋r，此处m和r均为正整数，r＜k，那么至少有一个鸽巢有多于m只的鸽子。<BR><BR></DIV></TD></TR></TBODY></TABLE>
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