J2EE之巅

 

2012年11月26日

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

Not like the previous solution here http://www.blogjava.net/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 超越巅峰 阅读(2606) | 评论 (0)编辑 收藏

2012年10月15日

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 超越巅峰 阅读(2591) | 评论 (0)编辑 收藏

2012年8月5日

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 超越巅峰 阅读(2271) | 评论 (0)编辑 收藏

2011年6月7日

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 超越巅峰 阅读(4118) | 评论 (3)编辑 收藏

2011年5月22日

发现自己是2010年下半年软考系统架构师总成绩第10名

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

posted @ 2011-05-22 13:50 超越巅峰 阅读(1959) | 评论 (1)编辑 收藏