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    J2EE之巔

     

    2010年6月18日

    The Clojure Program To solve N Queens Problem (Without back tracing)

    Not like the previous solution here http://m.tkk7.com/chaocai/archive/2012/08/05/384844.html
    The following solution not using the back tracing way is more concise and readable, but for the searching space becomes huger, the performance is much worser then the previous one.

    (ns SICP.unit3)
    (defn conflictInCol? [s col]
      (some #(= col %) s)
    )

    (defn conflictInDia? [s col]
      (let [dia (count s)
            n1 (fn [c
    ] (Math/abs (- dia (.indexOf s c))))
            n2 (fn [c] (Math/abs (- col c)))]
        (some #(= (n1 %) (n2 %)) s)
      )
    )

    (defn safe? [s col] 
      (not (or (conflictInCol? s col) (conflictInDia? s col)))
    )
      
    (defn next-level-queens [solutions-for-prev-level board-size current-level]
      (let [solutions (atom [])]
        (doseq [s solutions-for-prev-level]
          (doseq [col (range 0 board-size)]
            (if (safe? s col)
              (reset! solutions (cons (conj s col) @solutions))
         
            )
           )
       
        )
       
          (if (< current-level (dec board-size))
            (recur @solutions board-size (inc current-level))
            (count @solutions)
          )
       )
    )

    (defn queens [board-size]
      (next-level-queens  (apply vector (map #(vector %) (range 0 board-size))) board-size 1)
    )

    Chao Cai (蔡超)
    Sr. SDE
    Amazon


     

    posted @ 2012-11-26 12:21 超越巔峰 閱讀(2837) | 評論 (0)編輯 收藏

    Clojure XPath

    The functions to support using XPath in Clojure.

    Source Code

     1 ;The code was implemented by caichao@amazon.com
     2 ;You could use the code anyway, but should keep the comments
     3 ;Created 2012.10
     4 (ns clojure.ccsoft.xml
     5   (:require [clojure.xml :as xml]))
     6  
     7 (import '(java.io StringReader)
     8         '(java.io ByteArrayInputStream))
     9  
    10 (defn xml-structure [xml-txt] 
    11    [ (xml/parse (-> xml-txt
    12               (.getBytes)
    13               (ByteArrayInputStream.)
    14      )
    15     )]
    16 )
    17  
    18 (defn node [tag xmlStruct]
    19  
    20   (first (filter #(= (:tag %) tag) (:content xmlStruct)))
    21 )
    22  
    23 (defn node [path xml-txt]
    24    (loop [path path 
    25           xml-content (xml-structure xml-txt) 
    26           ]
    27       (let [current-tag (first path) current-elem (first xml-content)]
    28         (if (= (:tag current-elem ) current-tag)
    29  
    30           (if (= (count path) 1)
    31             current-elem 
    32             (recur  (rest path) (:content current-elem ))
    33           )
    34           (if (> (count  xml-content) 1)
    35            (recur path  (rest xml-content))
    36           )
    37         )
    38      )
    39     )
    40  )

    How to Use

    (def cmd-example "<command>
                       <header>
                         
    <type>script</type>
                         
    <transaction_id>12345</transaction_id>
                       
    </header>
                       
    <body>
                          println 
    3+4;
                       
    </body>
                      
    </command>")
     
     
    (node [:command :header :transaction_id] cmd
    -example)


    posted @ 2012-10-15 10:15 超越巔峰 閱讀(2871) | 評論 (0)編輯 收藏

    The Clojure Program To solve N Queens Problem

    The following program is about solving N-Queens problem (http://en.wikipedia.org/wiki/Eight_queens_puzzle) by Clojure. If you have the better solution in Clojure or Haskell, welcome to provide your solution.
    (ns queens)
    (defn conflictInRow? [queens newqueen]
      (some #(= newqueen %) queens)
    )
    (defn conflictInDia? [queens newqueen]
      (let [dia (count queens) 
            n1 (fn [queen] (Math/abs (- dia (.indexOf queens queen))))
            n2 (fn [queen] (Math/abs (- newqueen queen)))]
        (some #(= (n1 %) (n2 %)) queens)
       )
     )
    (defn conflict? [queens newqueen]
      (or (conflictInRow? queens newqueen) (conflictInDia? queens newqueen))
     )
    (def cnt (atom 0))
    (defn put-queens [queens newqueen boardSize ]
      (if (= (count queens) boardSize)  
        (do
          (println queens)
          (reset! cnt (inc @cnt))
        )
        (do 
          ;(println queens)
          (if (> newqueen boardSize)
         
               (if (and (= (peek queens) boardSize) (= (count queens) 1))
                   (throw (Exception. (str "That's all " @cnt)))
                   (recur (pop queens) (inc (peek queens)) boardSize )
               )
         
            (if (conflict? queens newqueen)
                
                 (recur queens (inc newqueen) boardSize )
                 
              (do
                 (put-queens (conj queens newqueen) 1 boardSize )
                 (recur queens (inc newqueen) boardSize )
               )
            )
           )
          )
        )
        
    )
    (defn queens [boardSize] 
        (put-queens [] 1 boardSize)
     )


    Chao Cai (蔡超)

    Sr. Software Dev Engineer 
    Amazon.com

     

    posted @ 2012-08-05 23:26 超越巔峰 閱讀(2473) | 評論 (0)編輯 收藏

    Spring AOP on Annotation

    1 The annotation:
    @Retention(RetentionPolicy.RUNTIME)
    @Target(ElementType.METHOD)
    @Inherited
    public @interface NeedToRetry {
        Class<?>[] recoverableExceptions();
        int retryTime();
        int intervalIncrementalFactor() default 0;
        long retryInterval() default 0L;
    }

    2 The Aspect
    @Aspect
    public class InvokingRetryInterceptor {
        private static Logger log = Logger.getLogger(InvokingRetryInterceptor.class);
        private boolean isNeedToRetry(Throwable t,Class<?>[] recoverableExceptions){
            String exceptionName= t.getClass().getName();
            for (Class<?> exp:recoverableExceptions){            
                if (exp.isInstance(t)){
                    return true;
                }
            }
            log.warn("The exception doesn't need recover!"+exceptionName);
            return false;
        }

        private long getRetryInterval(int tryTimes,long interval,int incrementalFactor){
            return interval+(tryTimes*incrementalFactor);
        }
        
        @Around(value="@annotation(amazon.internal.dropship.common.NeedToRetry)&&@annotation(retryParam)",argNames="retryParam")
        public Object process(ProceedingJoinPoint pjp,NeedToRetry retryParam ) throws Throwable{
            boolean isNeedTry=true;
            int count=0;
            Throwable fault;            
            Class<?>[] recoverableExceptions=retryParam.recoverableExceptions();
            int retryTime=retryParam.retryTime();
            long retryInterval=retryParam.retryInterval();
            int incrementalFactor=retryParam.intervalIncrementalFactor();
            do{
                try{                
                    return pjp.proceed();            
                }catch(Throwable t){
                    fault=t;
                    if (!isNeedToRetry(t,recoverableExceptions)){
                        break;
                    }
                    Thread.sleep(getRetryInterval(retryTime,retryInterval,incrementalFactor));
                }
                count++;
            }while(count<(retryTime+1));
            throw fault;
            
        }
    }

    posted @ 2011-06-07 11:34 超越巔峰 閱讀(4426) | 評論 (3)編輯 收藏

    發現自己是2010年下半年軟考系統架構師總成績第10名

    http://www.rkb.gov.cn/jsj/cms/s_contents/download/s_dt201103170102.html

    posted @ 2011-05-22 13:50 超越巔峰 閱讀(2127) | 評論 (1)編輯 收藏

    JBehave in practice

    ATDD (Acceptance Test Driven Development) is the extension of TDD, which helps us deliver exactly what the customer wants. Now ATDD has already been the hot spot in the software development world. There are several variations of ATDD including BDD, EDD and etc, also more and more frameworks have been created to help us develop with ATDD, for example  FIT and JBehave.
    The followings will introduce how to use the JBehave in your real project effectively.


     

    Figure 1 Test Code Structure

    Each test implementation could be divided into four layers, this structure could help us improve the codes reusability and maintainability, So, it will make us implement the tests quickly and easily.

    Specification/Scenario layer:

    This layer describes system’s behaviors and functionalities by the scenarios.  For using JBehave, we can use the natural language describe the scenarios and just need to follow the JBehave ‘Given-When-Then’ rule.

    Parser layer:

    We don’t need to implement this layer , this layer has been implemented by JBehave. What exactly JBehave do is to relate the steps of the scenario to the methods of the test codes.

    Step Logic Layer:

    The layer implements test logics associating with every step of the scenarios. Every step are implemented by a Java method.

    Action/Utils layer

    This the very important layer to improve the reusability of our codes. This layer provides the utility methods to help you implement step logics. These utility methods usually involved the system state checking, mock requests sending and so on.

    For example, we can provide the methods to check the data in database/file or check the state of the middleware, also so frameworks are very useful to implement the logic simulating the client browser’s requests.

     


    Chao Cai

    Working for Amazon.com

    chaocai2001@yahoo.com.cn

     

    posted @ 2011-02-26 13:34 超越巔峰 閱讀(3274) | 評論 (0)編輯 收藏

    TDD Tips

    How to design the testable software? You may always find some best practices about designing for scalable, extensible or maintainable. To be testable, the best way should be TDD. Followings are some tips from my real practices on TDD.

    1 TDD is design process; it let you design for testing, naturally

    Write the test firstly, it does not only help you find the bugs; but the most important point is to let you design for test naturally.

    Also you should keep in mind, tests not only help you find bugs, but also protect your codes; when some changes impact on your existing codes, the tests will be broken.

     

    2 Keep the implementation simple

    Keep your implementation simple, just let the test pass. The complex implementation may introduce the logics or codes not covered by the tests, even leads some codes not testable.

     

    3 TDD in each scope.

    You may get to know the concept ATDD (acceptance test driven development). TDD could be used in every phase of the development and by the different granularity.

    To ATDD, you could consider on using some existing framework such as FIT, these frameworks will be bridge between business logic and implementation logic.

    Recently, the concept BDD (behavior driven development) is introduced to the ATDD process, so the BDD frameworks such as JBehave is also the good choice.

     


    Different TDD process could be nested and should be nested don’t let your step too large.

     

    4 keep each step small enough

    Always keep each step small to avoid introducing the untestable codes or logics and pass each test quickly.

     

     

    6 Always refactor

    This step is always overlooked in TDD process; however it is the very important step. Also, never forget refactor should involve all your tests.

     

    Why can't write test firstly?

     1.not think how to meature the codes

    2. The current step maybe too large, should separate into small ones

    3. The codes with ugly dependencies


     http://blog.csdn.net/chaocai2004/archive/2011/01/09/6125479.aspx


    Chao Cai (蔡超)

    Sr. SDE

    Amazon.com

     

    posted @ 2011-01-09 16:55 超越巔峰 閱讀(3417) | 評論 (1)編輯 收藏

    算法的時間復雜度

     

    相信大家對于算法的時間復雜度O都不會陌生,不過你知道一個算法的時間復雜度是如何計算出來的嗎?

    以前在學習算法和數據結構的時候,對于每種算法的復雜度都是死記的并沒有真正的去研究他們是如何計算出來,最近突然對算法產生了興趣,迫使自己研究了一下算法復雜度的計算方法。

    概念

    O表示法表示時間復雜性,注意它是某一個算法的時間復雜性。大O表示只是說有上界,由定義如果f(n)=O(n),那顯然成立f(n)=O(n^2),它給你一個上界,但并不是上確界,但人們在表示的時候一般都習慣表示前者

    另外除了這個官方概念,個人認為大O表示的是問題規模n和算法中語句執行次數的關系。

    以二分查找為例,我們求解它的時間復雜度

    1 設規模為n個元素時,要執行T(n)次

    T(n)=T(n/2)+1

    T(n)=[T(n/4)+1]+1

    T(n)=T(n/2^m)+m

    n=2^m

    T(n)=T(1)+log2n

    T(1)=1

    所以其算法復雜度為O(log2n)

    posted @ 2010-06-18 15:26 超越巔峰 閱讀(3583) | 評論 (1)編輯 收藏

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