基本上,双轨搜索算法仅将其工作空间(按数量(顺序排列)的体积分为零。 因此,这无疑是部署“divide战略”,因此时间复杂性降低到O(lg n)。 因此,这涵盖了其中的“divide”部分。

As can be noticed, the final solution is obtained from the last comparison made, that is, when we are left with only one element for comparison. Binary search does not merge or combine solution.

简言之,双轨搜索将problem(必须工作)的大小分为二分之四,但在借方和碎块中却未找到<>solution<>>,因此无需将解决办法合并!

我知道它太长了,但希望它有助于:

也可从以下网站获得一些想法:。 https://www.khanacademy.org/computing/computer-science/algorithms/binary-search/a/running-time-of-binary-search

Apparently some people consider binary search a divide-and-conquer algorithm, and some are not. I quickly googled three references (all seem related to academia) that call it a D&C algorithm: http://www.cs.berkeley.edu/~vazirani/algorithms/chap2.pdf http://homepages.ius.edu/rwisman/C455/html/notes/Chapter2/DivConq.htm http://www.csc.liv.ac.uk/~ped/teachadmin/algor/d_and_c.html

我认为,大家一致认为,D&C算法至少应分这三步的前两个阶段:

• divide, i.e. decide how the whole problem is separated into sub-problems;
• conquer, i.e. solve each of the sub-problems independently;
• [optionally] combine, i.e. merge the results of independent computations together.

第二阶段——连续阶段——应重新采用同样的方法,通过分成小小小小小小小小小小子问题等,解决小问题。 然而,在实践中,往往使用一些门槛来限制休养的做法,因为对于小面积问题,不同的做法可能更快。 例如,在阵列面积小时,快速执行往往使用泡沫。

第三阶段可能不可行,我认为它不会将算法定为D&C。 一个常见的实例是,将“> 的<代码>重新定位,即与纯粹与独立数据项目(即不减少任何形式)合作的所有迭代。 它可能无用地看着眼光,但事实上,它非常有力,例如平行地实施 lo,并且利用Cilk和Intels TBB等框架。

回到原来的问题:请考虑实施算法的一些法典(我使用C++;如果这不是为了你所喜欢的语言,则担心:

int search( int value, int* a, int begin, int end ) {
  // end is one past the last element, i.e. [begin, end) is a half-open interval.
  if (begin < end)
  {
    int m = (begin+end)/2;
    if (value==a[m])
      return m;
    else if (value<a[m])
      return search(value, a, begin, m);
    else
      return search(value, a, m+1, end);
  }
  else // begin>=end, i.e. no valid array to search
    return -1;
}

这里的分部分为<代码>int m = (begin+end)/2;,其余部分为序部分。 计算法以回馈式D&C形式明确写,尽管只采用了一个分行。 然而,也可以用假形式写:

int search( int value, int* a, int size ) {
  int begin=0, end=size;
  while( begin<end ) {
    int m = (begin+end)/2;
    if (value==a[m])
      return m;
    else if (value<a[m])
      end = m;
    else
      begin = m+1;
  }
  return -1;
}

我认为,采用双轨搜索法是一种非常常见的做法;我故意使用与休养一样的变称,以便更容易看到共同点。 因此,我们可以说,计算中间点再次是分裂的一部分,而 lo体的其余部分是 con。

但是,当然,如果你的审查者认为不同,他们可能难以说服他们打D&C。

Update: just had a thought that if I were to develop a generic skeleton implementation of a D&C algorithm, I would certainly use binary search as one of API suitability tests to check whether the API is sufficiently powerful while also concise. Of course it does not prove anything :)

<代码>Merge Sort和> Quick Sort 算法使用divide and conquer技术(由于存在2个子问题)和Binary search在下出现减少和 conquer(因此有1个子问题)。

因此,本金搜索实际上使用了减少和混合技术,而不是鸿沟和混合技术。

资料来源:。 https://www.geeksforgeeks.org/减少-and-conquer/。

进行双轨搜索是为了描述分裂和趋同,因为征服步骤并不明确。 计算法的结果是沙斯特的针索引和纯粹的D&C执行将退回最小海斯特的针头指数(单项清单中的<>0),然后在分界线的较大沙滩中重新添加冲抵。

Pseudocode to interpret:

function binary_search has arguments needle and haystack and returns index
    if haystack has size 1
       return 0
    else 
        divide haystack into upper and lower half
        if needle is smaller than smallest element of upper half
            return 0 + binary_search needle, lower half
        else
            return size of lower half + binary_search needle, upper half

The addition (0 + or size of lower half) is the conquer part. Most people skip it by providing indices into a larger list as arguments, and thus it is often not readily available.

当然,这一鸿沟将这块土地分成一半。

组成部分是确定在所处理部分是否有搜查要素,以及该部分的位置。

Dichotomic in computer science refers to choosing between two antithetical choices, between two distinct alternatives. A dichotomy is any splitting of a whole into exactly two non-overlapping parts, meaning it is a procedure in which a whole is divided into two parts. It is a partition of a whole (or a set) into two parts (subsets) that are: 1. Jointly Exhaustive: everything must belong to one part or the other, and 2. Mutually Exclusive: nothing can belong simultaneously to both parts.

多样化和 con化工作,重新将问题分解成两个或两个以上相同的次要问题,直到这些问题变得简单,可以直接解决。

因此,双轨搜索将每个回收点检查的物项数量减半,并确定如果能够确定关键点不在,该半部是否有机会找到“关键”项目,或将它移至另一半。 由于算法是分辨率,因此双轨搜索将认为,“钥匙”必须部分地存在,直到它返回了钥匙缺失的撤离状态。

多样化和公平算法基于以下三个步骤:

1. Divide
2. Conquer
3. Combine

Binary Search problem can be defined as finding x in the sorted array A[n]. According to this information:

1. Divide: compare x with middle
2. Conquer: Recurse in one sub array. (Finding x in this array)
3. Combine: it is not necessary.