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最大公�U�数

ytl — Thu, 21 Mar 2013 01:39:00 GMT

摘要: �Ƨ几里�d��法阅读全文

ytl 2013-03-21 09:39 发表评论

jvm的内存模型之eden�? 转蝲

ytl — Thu, 01 Mar 2012 10:12:00 GMT

��谈java内存模型

不同的��^収ͼ�内存模型是不一��L��Q�但是jvm的内存模型规范是�l�一的。其实java的多�U�程�q�发问题最�l�都会反映在java的内存模型上�Q�所谓线�E�安全无非是要控制多个线�E�对某个资源的有序访问或修改。�ȝ��java的内存模型，要解决两个主要的问题�Q�可见性和有序性。我们都知道计算机有高速缓存的存在�Q�处理器�q�不是每�ơ处理数据都是取内存的。JVM定义了自��q��内存模型�Q�屏蔽了底层�q�_��内存��理�l�节�Q�对于java开发�h员，要清楚在jvm内存模型的基�� 上，如果解决多线�E�的可见性和有序性�?br /> 那么�Q�何�?strong>可见�?/strong>�Q?多个�U�程之间是不能互�怼�递数据通信的，它们之间的沟通只能通过�׃�n变量来进行。Java内存模型�Q�JMM�Q�规定了jvm有主内存�Q�主内存是多个线�E�共�?的。当new一个对象的时候，也是被分配在��d��存中�Q�每个线�E�都有自��q��工作内存�Q�工作内存存储了��d��的某些对象的副本�Q�当然线�E�的工作内存大小是有限制的。当�U�程操作某个对象�Ӟ��执行��序如下�Q?br /> (1) 从主存复制变量到当前工作内存 (read and load)
(2) 执行代码�Q�改变共享变量�?(use and assign)
(3) 用工作内存数据刷��C��存相兛_��?(store and write)

JVM规范定义了线�E�对��d��的操作指令：read�Q�load�Q�use�Q�assign�Q�store�Q�write。当一个共享变量在多个�U�程的工作内存中都有副本�Ӟ��如果一个线�E�修改了�q�个�׃�n 变量�Q�那么其他线�E�应该能够看到这个被修改后的��|��q�就是多�U�程的可见性问题�?br />        那么�Q�什么是有序�?/strong>�?�Q�线�E�在引用变量时不能直接从��d��存中引用,如果�U�程工作内存中没有该变量,则会从主内存中拷贝一个副本到工作内存�?�q�个�q�程为read-load,�?成后�U�程会引用该副本。当同一�U�程再度引用该字�D�|��,有可能重��C��d��中获取变量副�?read-load-use),也有可能直接引用原来的副�?(use),也就是说 read,load,use��序可以由JVM实现�pȝ��军_��?br />        �U�程不能直接��Z��存中中字�D�赋��|��它会��值指定给工作内存中的变量副本(assign),完成后这个变量副本会同步��C��存储�?store- write)�Q�至于何时同步过去，�Ҏ��JVM实现�pȝ��军_��.有该字段,则会从主内存中将该字�D�赋值到工作内存�?�q�个�q�程为read-load,完成后线 �E�会引用该变量副本，当同一�U�程多次重复对字�D�赋值时,比如�Q?br />Java代码
for(int i=0;i<10;i++)
a++;
�U�程有可能只对工作内存中的副本进行赋�?只到最后一�ơ赋值后才同步到��d��储区�Q�所以assign,store,weite��序可以由JVM实现�pȝ��?定。假设有一个共享变量x�Q�线�E�a执行x=x+1。从上面的描�q�C��可以知道x=x+1�q�不是一个原子操作，它的执行�q�程如下�Q?br />1 从主存中��d��变量x副本到工作内�?br />2 �l�x�?
3 ��x�?后的值写回主 �?br />如果另外一个线�E�b执行x=x-1�Q�执行过�E�如下：
1 从主存中��d��变量x副本到工作内�?br />2 �l�x�?
3 ��x�?后的值写回主�?nbsp;
那么昄��Q�最�l�的x的值是不可靠的。假设x现在�?0�Q�线�E�a�?�Q�线�E�b�?�Q�从表面上看�Q�似乎最�l�x�q�是�?0�Q�但是多�U�程情况下会有这�U�情况发生：
1�Q�线�E�a从主存读取x副本到工作内存，工作内存中x��gؓ10
2�Q�线�E�b从主存读取x副本到工作内存，工作内存中x��gؓ10
3�Q�线�E�a��工作内存中x�?�Q�工作内存中x��gؓ11
4�Q�线�E�a��x提交��d��中，��d��中x�?1
5�Q�线�E�b��工作内存中x值减1�Q�工作内存中x��gؓ9
6�Q�线�E�b��x提交��C��d��中，��d��中x�?/p>

jvm的内存模型之eden�?/strong>
所谓线�E�的“工作内存”到底是个什么东西？有的��为是�U�程的栈�Q�其实这�U�理解是不正��的。看看JLS�Q�java语言规范�Q�对�U�程工作内存的描�q�ͼ��U�程的working memory只是cpu�?strong>寄存器和高速缓存的抽象描述�?/p>
可能很多人都觉得莫名其妙�Q�说JVM的内存模型，怎么会扯到cpu上去呢？在此�Q�我认�ؓ很有必要阐述下，�?得很多�h看得不明不白的。先抛开java虚拟��Z��谈，我们都知道，现在的计��机�Q�cpu在计��的时候，�q�不��L��从内存读取数据，它的数据��d��序优先�U?是：寄存器－高速缓存－内存。线�E�耗费的是CPU�Q�线�E�计��的时候，原始的数据来自内存，在计��过�E�中�Q�有些数据可能被频繁��d��Q�这些数据被存储在寄存器和高速缓存中�Q�当�U�程计算完后�Q�这些缓存的数据在适当的时候应该写回内存。当个多个线�E�同时读写某个内存数据时�Q�就会��生多�U�程�q�发问题�Q�涉及到三个�?性：原子性，有序性，可见性。在《线�E�安全�ȝ��》这��文章中�Q��ؓ了理解方便，我把原子性和有序性统一叫做“多线�E�执行有序�?#8221;。支持多�U�程的��^台都会面�?�q�种问题�Q�运行在多线�E��^��C��支持多线�E�的语言应该提供解决该问题的�Ҏ��?/p>
synchronized, volatile,锁机�Ӟ��如同步块�Q�就�l�队列，��d��队列�Q?/strong>�{�等。这些方案只是语法层面的�Q�但我们要从本质上去理解它，不能仅仅知道一�?synchronized 可以保证同步��完了�?nbsp; 在这里我说的是jvm的内存模型，是动态的�Q�面向多�U�程�q�发的，沿袭JSL�?#8220;working memory”的说法，只是不想牉|��到太多底层细节，因�ؓ 《线�E�安全�ȝ��》这��文章意在说明怎样从语法层面去理解java的线�E�同步，知道各个关键字的使用�?景�?/p>
说说JVM的eden区吧。JVM的内存，被划分了很多的区域：
1.�E�序计数�?br />每一个Java�U�程都有一个程序计数器来用于保存程序执行到当前�Ҏ��的哪一个指令�?br />2.�U�程�?br />�U�程的每个方法被执行的时候，都会同时创徏一个��Q�Frame�Q�用于存储本地变量表、操作栈、动态链接、方法出入口�{�信息。每一个方法的调用臛_��成，��意味着一个��在VM栈中的入栈至出栈的过�E�。如果线�E�请求的栈深度大于虚拟机所允许的深度，��抛出StackOverflowError异常�Q�如果VM栈可以动态扩展（VM Spec中允许固定长度的VM栈）�Q�当扩展时无法申请到��_��内存则抛出OutOfMemoryError异常�?br />3.本地�Ҏ��?br />4.�?br />每个�U�程的栈都是该线�E�私有的�Q�堆则是所有线�E�共享的。当我们new一个对象时�Q�该对象��p��分配��C��堆中。但是堆�Q��ƈ不是一个简单的概念�Q�堆区又划分了很多区域，��Z��么堆划分成这么多区域�Q�这是�ؓ了JVM的内存垃圾收集，��g��扯��远了，扯到垃圾攉��了，现在的jvm的gc都是按代攉��Q�堆区大致被分�ؓ三大块：新生代，旧生代，持久代（虚拟的）�Q�新生代又分为eden区，s0区，s1区。新��Z��个对象时�Q�基本小的对象，生命周期短的对象都会攑֜�新生代的eden��Z��Q�eden区满�Ӟ��有一个小范围的gc�Q�minor gc�Q�，整个新生代满�Ӟ��会有一个大范围的gc�Q�major gc�Q�，��新生代里的部分对象转到旧生代里�?br />5.�Ҏ��?nbsp;
其实��是�怹�代（Permanent Generation�Q�，�Ҏ��Z��存放了每个Class的结构信息，包括帔R��池、字�D�|��q�、方法描�q�等�{�。VM Space描述中对�q�个区域的限刉��常宽松，除了和Java堆一样不需要连�l�的内存�Q�也可以选择固定大小或者可扩展外，甚至可以选择不实现垃圾收集。相�Ҏ��_��垃圾攉��行�ؓ在这个区域是相对比较��发生的�Q�但�q�不是某些描�q�那��h��久代不会发生GC�Q�至 ��对当前��L��的商业JVM实现来说是如此）�Q�这里的GC主要是对帔R��池的回收和对�cȝ��卸蝲�Q�虽然回收的“成�W”一般也比较差强人意�Q�尤其是�c�d��载，条�g相当苛刻�?br />6.帔R��?br /> Class文�g中除了有�cȝ��版本、字�D�c��方法、接口等描述�{�信息外�Q�还有一��信息是帔R��?constant_pool table)�Q�用于存攄��译期已可知的帔R��Q�这部分内容��在�c�d��载后�q�入�Ҏ��区（�怹�代）存放。但是Java语言�q�不要求帔R��一定只有编译期预置入Class的常量表的内�Ҏ��能进入方法区帔R��池，�q�行期间也可��新内容攑օ�帔R��池（最典型的String.intern()�Ҏ��Q��?/p>

ytl 2012-03-01 18:12 发表评论

Java 原码代码学习

ytl — Sat, 24 Sep 2011 07:30:00 GMT
ArrayList
关于Java中的transient�Q�volatile和strictfp关键�?nbsp;http://www.iteye.com/topic/52957
(1), ArrayList底层使用Object数据实现�Q?nbsp;private transient Object[] elementData;且在使用不带参数的方式实例化�Ӟ��生成数组默认的长度是10�?br /> (2), add�Ҏ��实现
public boolean add(E e) {
//ensureCapacityInternal判断��d��新元素是否需要重新扩大数�l�的长度�Q�需要则扩否则不
ensureCapacityInternal(size + 1); // 此�ؓJDK7调用的方�?JDK5里面使用的ensureCapacity�Ҏ��
elementData[size++] = e; //把对象插入数�l�，同时把数�l�存储的数据长度size�?
return true;
}
JDK 7�?nbsp;ensureCapacityInternal实现
private void ensureCapacityInternal(int minCapacity) {
modCount++;修改�ơ数
// overflow-conscious code
if (minCapacity - elementData.length > 0)
grow(minCapacity);//如果需要扩大数�l�长�?/div>
}
/**
* The maximum size of array to allocate. --甌��新数�l�最大长�?/div>
* Some VMs reserve some header words in an array.
* Attempts to allocate larger arrays may result in
* OutOfMemoryError: Requested array size exceeds VM limit --如果甌��的数�l�占用的内心大于JVM的限制抛出异�?/div>
*/
private static final int MAX_ARRAY_SIZE = Integer.MAX_VALUE - 8;//��Z��么减�?看注释第2�?/div>
/**
* Increases the capacity to ensure that it can hold at least the
* number of elements specified by the minimum capacity argument.
*
* @param minCapacity the desired minimum capacity
*/
private void grow(int minCapacity) {
// overflow-conscious code
int oldCapacity = elementData.length;
int newCapacity = oldCapacity + (oldCapacity >> 1); //新申��L��长度为old�?/2倍同时��用位�U�运��更高效�Q�JDK5中： (oldCapacity *3)/2+1
if (newCapacity - minCapacity < 0)
newCapacity = minCapacity;
if (newCapacity - MAX_ARRAY_SIZE > 0) //你懂�?/div>
newCapacity = hugeCapacity(minCapacity);
// minCapacity is usually close to size, so this is a win:
elementData = Arrays.copyOf(elementData, newCapacity);
}
//可以甌��的最大长�?/div>
private static int hugeCapacity(int minCapacity) {
if (minCapacity < 0) // overflow
throw new OutOfMemoryError();
return (minCapacity > MAX_ARRAY_SIZE) ?
Integer.MAX_VALUE :
MAX_ARRAY_SIZE;
}

ytl 2011-09-24 15:30 发表评论

计量炏V��计量分�cȝ��

ytl — Sun, 07 Aug 2011 03:02:00 GMT
     摘要:   阅读全文

ytl 2011-08-07 11:02 发表评论

Alogrithms to quicksort

ytl — Sun, 08 May 2011 06:13:00 GMT

Quicksort

Quicksort is a fast sorting algorithm, which is used not only for educational purposes, but widely applied in practice. On the average, it has O(n log n) complexity, making quicksort suitable for sorting big data volumes. The idea of the algorithm is quite simple and once you realize it, you can write quicksort as fast as bubble sort.

Algorithm
The divide-and-conquer strategy is used in quicksort. Below the recursion step is described:

Choose a pivot value. We take the value of the middle element as pivot value, but it can be any value, which is in range of sorted values, even if it doesn't present in the array.

Partition. Rearrange elements in such a way, that all elements which are lesser than the pivot go to the left part of the array and all elements greater than the pivot, go to the right part of the array. Values equal to the pivot can stay in any part of the array. Notice, that array may be divided in non-equal parts.

Sort both parts. Apply quicksort algorithm recursively to the left and the right parts.

Partition algorithm in detail

There are two indices i and j and at the very beginning of the partition algorithm i points to the first element in the array andj points to the last one. Then algorithm moves i forward, until an element with value greater or equal to the pivot is found. Index j is moved backward, until an element with value lesser or equal to the pivot is found. If i ≤ j then they are swapped and i steps to the next position (i + 1), j steps to the previous one (j - 1). Algorithm stops, when i becomes greater than j.

After partition, all values before i-th element are less or equal than the pivot and all values after j-th element are greater or equal to the pivot.
Example. Sort {1, 12, 5, 26, 7, 14, 3, 7, 2} using quicksort.

Notice, that we show here only the first recursion step, in order not to make example too long. But, in fact, {1, 2, 5, 7, 3} and {14, 7, 26, 12} are sorted then recursively.

Why does it work?
On the partition step algorithm divides the array into two parts and every element a from the left part is less or equal than every element b from the right part. Also a and b satisfy a ≤ pivot ≤ b inequality. After completion of the recursion calls both of the parts become sorted and, taking into account arguments stated above, the whole array is sorted.

Complexity analysis

On the average quicksort has O(n log n) complexity, but strong proof of this fact is not trivial and not presented here. Still, you can find the proof in [1]. In worst case, quicksort runs O(n²) time, but on the most "practical" data it works just fine and outperforms other O(n log n) sorting algorithms.

Code snippets

Java

int partition(int arr[], int left, int right)

{

int i = left;

int j = right;

int temp;

int pivot = arr[(left+right)>>1];

while(i<=j){

while(arr[i]>=pivot){

i++;

}

while(arr[j]<=pivot){

j--;

}

if(i<=j){

temp = arr[i];

arr[i] = arr[j];

arr[j] = temp;

i++;

j--;

}

}

return i

}

void quickSort(int arr[], int left, int right) {

      int index = partition(arr, left, right);

if(left

quickSort(arr,left,index-1);

}

if(index

quickSort(arr,index,right);

}

}

python

def quickSort(L,left,right) {

      i = left

j = right

if right-left <=1:

return L

      pivot = L[(left + right) >>1];

      /* partition */

      while (i <= j) {

            while (L[i] < pivot)

                  i++;

            while (L[j] > pivot)

                  j--;

            if (i <= j) {

                  L[i],L[j] = L[j],L[i]

                  i++;

                  j--;

            }

      };

      /* recursion */

      if (left < j)

            quickSort(L, left, j);

      if (i < right)

            quickSort(L, i, right);

}

ytl 2011-05-08 14:13 发表评论

Algorithms to Insertion Sort

ytl — Sun, 08 May 2011 04:24:00 GMT

Insertion Sort

Insertion sort belongs to the O(n²) sorting algorithms. Unlike many sorting algorithms with quadratic complexity, it is actually applied in practice for sorting small arrays of data. For instance, it is used to improve quicksort routine. Some sources notice, that people use same algorithm ordering items, for example, hand of cards.

Algorithm

Insertion sort algorithm somewhat resembles selection sort. Array is imaginary divided into two parts - sorted one andunsorted one. At the beginning, sorted part contains first element of the array and unsorted one contains the rest. At every step, algorithm takes first element in the unsorted part and inserts it to the right place of the sorted one. Whenunsorted part becomes empty, algorithm stops. Sketchy, insertion sort algorithm step looks like this:

becomes

The idea of the sketch was originaly posted here.

Let us see an example of insertion sort routine to make the idea of algorithm clearer.

Example. Sort {7, -5, 2, 16, 4} using insertion sort.

The ideas of insertion

The main operation of the algorithm is insertion. The task is to insert a value into the sorted part of the array. Let us see the variants of how we can do it.

"Sifting down" using swaps

The simplest way to insert next element into the sorted part is to sift it down, until it occupies correct position. Initially the element stays right after the sorted part. At each step algorithm compares the element with one before it and, if they stay in reversed order, swap them. Let us see an illustration.

This approach writes sifted element to temporary position many times. Next implementation eliminates those unnecessary writes.

Shifting instead of swapping

We can modify previous algorithm, so it will write sifted element only to the final correct position. Let us see an illustration.

It is the most commonly used modification of the insertion sort.

Using binary search

It is reasonable to use binary search algorithm to find a proper place for insertion. This variant of the insertion sort is calledbinary insertion sort. After position for insertion is found, algorithm shifts the part of the array and inserts the element. This version has lower number of comparisons, but overall average complexity remains O(n²). From a practical point of view this improvement is not very important, because insertion sort is used on quite small data sets.

Complexity analysis

Insertion sort's overall complexity is O(n²) on average, regardless of the method of insertion. On the almost sorted arrays insertion sort shows better performance, up to O(n) in case of applying insertion sort to a sorted array. Number of writes is O(n²) on average, but number of comparisons may vary depending on the insertion algorithm. It is O(n²) when shifting or swapping methods are used and O(n log n) for binary insertion sort.

From the point of view of practical application, an average complexity of the insertion sort is not so important. As it was mentioned above, insertion sort is applied to quite small data sets (from 8 to 12 elements). Therefore, first of all, a "practical performance" should be considered. In practice insertion sort outperforms most of the quadratic sorting algorithms, like selection sort or bubble sort.

Insertion sort properties

adaptive (performance adapts to the initial order of elements);

stable (insertion sort retains relative order of the same elements);

in-place (requires constant amount of additional space);

online (new elements can be added during the sort).

Code snippets

We show the idea of insertion with shifts in Java implementation and the idea of insertion using python code snippet.

Java implementation

void insertionSort(int[] arr) {

      int i,j,newValue;

for(i=1;i

newValue = arr[i];

j=i;

while(j>0&&arr[j-1]>newValue){

arr[j] = arr[j-1];

j--;

}

arr[j] = newValue;

}

Python implementation

void insertionSort(L) {

      for i in range(l,len(L)):

            j = i

newValue = L[i]

            while j > 0 and  L[j - 1] >L[j]:

L[j] = L[j - 1]

                  j = j-1

            }

L[j] = newValue

      }

}

ytl 2011-05-08 12:24 发表评论

Binary search algorithm

ytl — Fri, 06 May 2011 10:11:00 GMT

Binary search algorithm

Generally, to find a value in unsorted array, we should look through elements of an array one by one, until searched value is found. In case of searched value is absent from array, we go through all elements. In average, complexity of such an algorithm is proportional to the length of the array.

Situation changes significantly, when array is sorted. If we know it, random access capability can be utilized very efficientlyto find searched value quick. Cost of searching algorithm reduces to binary logarithm of the array length. For reference, log₂(1 000 000) �?20. It means, that in worst case, algorithm makes 20 steps to find a value in sorted array of a million elements or to say, that it doesn't present it the array.

Algorithm

Algorithm is quite simple. It can be done either recursively or iteratively:

get the middle element;

if the middle element equals to the searched value, the algorithm stops;

otherwise, two cases are possible:

searched value is less, than the middle element. In this case, go to the step 1 for the part of the array, before middle element.

searched value is greater, than the middle element. In this case, go to the step 1 for the part of the array, after middle element.

Now we should define, when iterations should stop. First case is when searched element is found. Second one is when subarray has no elements. In this case, we can conclude, that searched value doesn't present in the array.
Examples

Example 1. Find 6 in {-1, 5, 6, 18, 19, 25, 46, 78, 102, 114}.

Step 1 (middle element is 19 > 6):     -1 5 6 18  19 25 46 78 102 114

Step 2 (middle element is 5 < 6):      -1  5 6 18  19 25 46 78 102 114

Step 3 (middle element is 6 == 6):     -1 5  6 18  19 25 46 78 102 114

Example 2. Find 103 in {-1, 5, 6, 18, 19, 25, 46, 78, 102, 114}.

Step 1 (middle element is 19 < 103):   -1 5 6 18  19 25 46 78 102 114

Step 2 (middle element is 78 < 103):   -1 5 6 18 19  25 46  78 102 114

Step 3 (middle element is 102 < 103):  -1 5 6 18 19 25 46 78  102 114

Step 4 (middle element is 114 > 103):  -1 5 6 18 19 25 46 78  102  114

Step 5 (searched value is absent):     -1 5 6 18 19 25 46 78 102 114

Complexity analysis

Huge advantage of this algorithm is that it's complexity depends on the array size logarithmically in worst case. In practice it means, that algorithm will do at most log₂(n) iterations, which is a very small number even for big arrays. It can be proved very easily. Indeed, on every step the size of the searched part is reduced by half. Algorithm stops, when there are no elements to search in. Therefore, solving following inequality in whole numbers:

n / 2^iterations > 0

resulting in

iterations <= log₂(n).

It means, that binary search algorithm time complexity is O(log₂(n)).

Code snippets.
You can see recursive solution for Java and iterative for python below.
Java

int binarySearch(int[] array, int value, int left, int right) {

      if (left > right)

            return -1;

      int middle = left + (right-left) / 2;

      if (array[middle] == value)

            return middle;

      if (array[middle] > value)

            return binarySearch(array, value, left, middle - 1);

      else

            return binarySearch(array, value, middle + 1, right);

}

Python

def biSearch(L,e,first,last):

if last - first < 2: return L[first] == e or L[last] == e

mid = first + (last-first)/2

if L[mid] ==e: return True

if L[mid]> e :

return biSearch(L,e,first,mid-1)

return biSearch(L,e,mid+1,last)



ytl 2011-05-06 18:11 发表评论

Algorithm to merge sorte

ytl — Fri, 06 May 2011 09:05:00 GMT
Merge sort is an O(n log n) comparison-based sorting algorithm. Most implementations produce a stable sort, meaning that the implementation preserves the input order of equal elements in the sorted output. It is a divide and conquer algorithm. Merge sort was invented by John von Neumann in 1945. A detailed description and analysis of bottom-up mergesort appeared in a report byGoldstine and Neumann as early as 1948
divide and conquer algorithm: 1, split the problem into several subproblem of the same type. 2,solove independetly. 3 combine those solutions

Python Implement

def mergeSort(L):

if len(L) < 2 :
return L
middle = len(L)/2
left = mergeSort(L[:mddle])
right = mergeSort(L[middle:])
together = merge(left,right)
return together

ytl 2011-05-06 17:05 发表评论

Algorithm to merge sorted arrays

ytl — Fri, 06 May 2011 08:55:00 GMT

Algorithm to merge sorted arrays

In the article we present an algorithm for merging two sorted arrays. One can learn how to operate with several arrays and master read/write indices. Also, the algorithm has certain applications in practice, for instance in merge sort.

Merge algorithm

Assume, that both arrays are sorted in ascending order and we want resulting array to maintain the same order. Algorithm to merge two arrays A[0..m-1] and B[0..n-1] into an array C[0..m+n-1] is as following:

Introduce read-indices i, j to traverse arrays A and B, accordingly. Introduce write-index k to store position of the first free cell in the resulting array. By default i = j = k = 0.

At each step: if both indices are in range (i < m and j < n), choose minimum of (A[i], B[j]) and write it toC[k]. Otherwise go to step 4.

Increase k and index of the array, algorithm located minimal value at, by one. Repeat step 2.

Copy the rest values from the array, which index is still in range, to the resulting array.

Enhancements

Algorithm could be enhanced in many ways. For instance, it is reasonable to check, if A[m - 1] < B[0] orB[n - 1] < A[0]. In any of those cases, there is no need to do more comparisons. Algorithm could just copy source arrays in the resulting one in the right order. More complicated enhancements may include searching for interleaving parts and run merge algorithm for them only. It could save up much time, when sizes of merged arrays differ in scores of times.

Complexity analysis

Merge algorithm's time complexity is O(n + m). Additionally, it requires O(n + m) additional space to store resulting array.

Code snippets

Java implementation

// size of C array must be equal or greater than

// sum of A and B arrays' sizes

public void merge(int[] A, int[] B, int[] C) {

      int i,j,k ;

i = 0;

j=0;

k=0;

m = A.length;

n = B.length;

while(i < m && j < n){

if(A[i]<= B[j]){

C[k] = A[i];

i++;

}else{

C[k] = B[j];

j++;

}

k++;

while(i

C[k] = A[i]

i++;

k++;

}

while(j

C[k] = B[j]

j++;

k++;

}

Python implementation

def merege(left,right):

result = []

i,j = 0

while i< len(left) and j < len(right):

if left[i]<= right[j]:

result.append(left[i])

i = i + 1

else:

result.append(right[j])

j = j + 1

while i< len(left):

result.append(left[i])

i = i + 1

while　j< len(right):

result.append(right[j])

j = j + 1

return result


MergSort:

import operator

def mergeSort(L, compare = operator.lt):
     if len(L) < 2:
          return L[:]
     else:
          middle = int(len(L)/2)
          left = mergeSort(L[:middle], compare)
          right= mergeSort(L[middle:], compare)
          return merge(left, right, compare)

def merge(left, right, compare):
     result = []
     i, j = 0, 0

     while i < len(left) and j < len(right):
          if compare(left[i], right[j]):
               result.append(left[i])
               i += 1
          else:
                result.append(right[j])
                j += 1
     while i < len(left):
          result.append(left[i])
          i += 1
     while j < len(right):
          result.append(right[j])
          j += 1
     return result

ytl 2011-05-06 16:55 发表评论

Sorting algorithms --Selection Sort

ytl — Fri, 06 May 2011 08:16:00 GMT

Selection Sort

Selection sort is one of the O(n²) sorting algorithms, which makes it quite inefficient for sorting large data volumes. Selection sort is notable for its programming simplicity and it can over perform other sorts in certain situations (see complexity analysis for more details).

Algorithm

The idea of algorithm is quite simple. Array is imaginary divided into two parts - sorted one and unsorted one. At the beginning, sorted part is empty, while unsorted one contains whole array. At every step, algorithm finds minimal element in the unsorted part and adds it to the end of the sorted one. When unsorted part becomes empty, algorithmstops.

When algorithm sorts an array, it swaps first element of unsorted part with minimal element and then it is included to the sorted part. This implementation of selection sort in not stable. In case of linked list is sorted, and, instead of swaps, minimal element is linked to the unsorted part, selection sort is stable.

Let us see an example of sorting an array to make the idea of selection sort clearer.

Example. Sort {5, 1, 12, -5, 16, 2, 12, 14} using selection sort.

Complexity analysis

Selection sort stops, when unsorted part becomes empty. As we know, on every step number of unsorted elements decreased by one. Therefore, selection sort makes n steps (n is number of elements in array) of outer loop, before stop. Every step of outer loop requires finding minimum in unsorted part. Summing up, n + (n - 1) + (n - 2) + ... + 1, results in O(n²) number of comparisons. Number of swaps may vary from zero (in case of sorted array) to n - 1 (in case array was sorted in reversed order), which results in O(n) number of swaps. Overall algorithm complexity is O(n²).

Fact, that selection sort requires n - 1 number of swaps at most, makes it very efficient in situations, when write operation is significantly more expensive, than read operation.

Code snippets

Java

public void selectionSort(int[] arr) {

      int i, j, minIndex, tmp;

      int n = arr.length;

      for (i = 0; i < n - 1; i++) {

            minIndex = i;

            for (j = i + 1; j < n; j++)

                  if (arr[j] < arr[minIndex])

                        minIndex = j;

            if (minIndex != i) {

                  tmp = arr[i];

                  arr[i] = arr[minIndex];

                  arr[minIndex] = tmp;

            }

      }

}

Python

for i in range(len(L)-1):
minIndex = i
minValue = L[i]
j = i + 1
while j< len(L):
if minValue > L[j]:
minIndex = j
minValue = L[j]
j += 1
if minIndex != i:
temp = L[i]
L[i] = L[minIndex]
L[minIndex] = temp

ytl 2011-05-06 16:16 发表评论