Newest Upgrade, Plan for Winter Break

It's been a long semester, and I have to drop a course because i am not doing well in all others. Well, frankly, I think I have learnt my lesson, that is study is not about concentration and resource, the more time you spend and the more resources you used, the better understanding you will have. So the plan for this winter is first, finish complex variable, and prof it at the beginning of the semester, also, check the am I still in the group, then study some part of 427,436,and of course if possible 470. This seems to be impractical, I will make time table for each in detail. Hope, I won't be lost again for the next semester, and also I have to start to search for summer research. So, the winter is clearly not a break for me. Living in busy time is kind of pleasure, if I can retrieve passion from it, no matter success or failure.

The step towards a subject, you should first sketch a rough blueprint for that, like a content table for a book, and then list some of the important theorem or methods or concepts for each topic, and then practice them for some amounts of problems (find a good problem book) to get a concrete understanding for each of the concept. THIS IS VERY CRUCIAL!!!!!!!!

语言决定世界

语言决定世界，那么在我掌握的语言之外会不会有什么是别的语言有的世界，但是中文却没有？

语言是个体所具有的世界的载体，虽然在国内没有办法将各个少数民族的文字进行信息化或者推广各族自己的文字（事实上我也不希望这样的东西产生在现在的时代，离心力过于强大），但是依然很渴望能够在另外的语言里面找寻一份不同。

有所启发

看了cuitianyi的博客，有所启发，看样子有规律的更新博客可以有效督促各种计划的完成，于是乎也在酝酿作一些受监督的计划。

Ural 1513的递推

做了Ural 1513,由于懒得写高精度，所以还是WA,不过这个递推是经过了思考的，思路如下，用一个二维数组来记录在长度为i时的可能情况数，这样的话，如果i<=k那么我们就不用顾忌，dp[i,'L']=dp[i-1,'L']+dp[i-1,'B'] and dp[i,'B']=dp[i-1,'L']+dp[i-1,'B'], 一旦i>k,那么dp[i,'L']=dp[i-1,'L']+dp[i-1,'B']-dp[i-k-1,'B'],也就是说，在累加所有上一个长度可能情况的同时，我们只需要排除从i到i-k都为L而i-k-1位置为B的情况(有待完善。。。)

很好玩

集合论告诉我们并不是所有能够描述的集合都能够构造出来，而维特根斯坦说对于我们不能说的，我们要保持沉默，很好玩的两句话


蛋白质的暴涨和宇宙的暴涨

Summer Plan

1.ACMs(Tire and Automata, Probability Problem Random Shooting SGU,SGU431，SGU 打印N个自身)
2.Self Study (Differential Equation, E&M II, Quantum II, Complex Analysis)
3.Crackme
4.Probability Challenge
5.Project Euler
6.Tensor
7.Harvard 100 Problem
8.CMU Puzzle
9.200 Physics Problem
10.Wittgenstein
11.Complex System
12.Mizar Proof(Prove Weak Form of Goldbach's Conjecture)
13.Numenta
14.Various Games(博弈)
15.felicity.iiit.ac.in
16.Different Metrcis(Godel Metric)
17.曙光 魔方 超级计算机(illinois supercomputing center)
18.AI-Code Robot Course
19. Develop a distributed project for my group

大创造

大创造需要大智慧，需要大问题，而不是拘泥于一个学科，一门学问，既然是宇宙的抽象模型，那么每一个问题的背后，每一个创造的背后都是一种互相关联的揭露，于是我知道如果我的想法是要改变世界的，那么他就不会平凡，不会拘泥于一道题，一个定理，而是一种理的追求。这才是正确的态度和心态。

控制论
信息论
随机数论
WolframAlpha

Randomness is Symmetric

Randomness is Symmetric

这回搞笑了，相声大师都不是汉族

这回搞笑了，相声大师都不是汉族的。。。

侯宝林 满族

马三立 回族

Beat Conway John

1 3
1 1 1 3
3 1 1 3
1 3 2 1 1 3
1 1 1 3 1 2 2 1 1 3
3 1 1 3 1 1 2 2 2 1 1 3
1 3 2 1 1 3 2 1 3 2 2 1 1 3
1 1 1 3 1 2 2 1 1 3 1 2 1 1 1 3 2 2 2 1 1 3

next line?

A nice course on rubik's cube

This course discusses the inner relationship between Group Theory and Rubik' Cube, which seems very fascinating to me.

http://akbar.marlboro.edu/~mahoney/courses/Spr00/rubik.html

Shity Recursion

This is a program that could be used to calculate rook polynomial which is putting rooks in a irregular chess board( more specifically, it can partially solve sgu269). Almost killed by recursion.

#include <iostream>

#include <fstream>

#include <memory>

using namespace std;

fstream fin;

int n,k,num[5],rk[5],ans[5][8],ma,mi;

void ready()

{

    fin>>n>>k;

    for (int i=1;i<=n;i++)

     {

       fin>>num[i];

       if (num[i]>ma) ma=num[i];

       rk[i]=(1 << num[i])-1;

     }

     if (ma>n)

      mi=n;

      else

      mi=ma;

}

bool chk(int rook[])

{

    for (int i=1;i<=n;i++)

      if (rook[i]!=0)

        return false;

    return true;

}

void poly(int dep,int lev,int pow)

{  

   int tmp[5];

  if (chk(rk))  //如果检查是空棋盘

   {

ans[0][dep]=1;   

     return;

   }

           

             memcpy(tmp,rk,sizeof(tmp));

             while (rk[lev]==0) lev++;

while ((rk[lev] & (1 << pow))!=(1 << pow))

  pow++;

rk[lev]=rk[lev]- (1 << pow);

             if (rk[lev]==0)

poly(dep+1,lev+1,0);

else poly(dep+1,lev,pow+1);

            for (int l=0;l<=mi;l++)

{ans[l][dep]+=ans[l][dep+1];

ans[l][dep+1]=0;}

memcpy(rk,tmp,sizeof(rk));

                rk[lev]=0;

             for (int k=lev+1;k<=n;k++)

              if ((rk[k])>=(1 << pow))

rk[k]=rk[k]- (1 << pow);

              poly(dep+1,lev+1,0);

            for (int l=1;l<=mi;l++)

{ans[l][dep]+=ans[l-1][dep+1];

ans[l-1][dep+1]=0;}

memcpy(rk,tmp,sizeof(rk));

         

}

int main()

{   fin.open("c:\\test.txt");

    ready();

    poly(1,1,0);

cout<<ans[k][1];

    fin.close();

}

POJ 2449 A* search K shortest path

First real C++ program

#include<iostream>

#include<fstream>

using namespace std;

int n,m,s,t,k,size;

int l[1001],l2[1001],her[1001],time[1001];

bool used[1001];

fstream fin;

int a,b,r;

struct edge{

    int v,l,t;

};

struct key{

int v,value;

};

edge map[1001][1001],map2[1001][1001];

key data[1001],tmp;

void ready()

{

fin>>n>>m;

   for (int i=1;i<=m;i++)

   {  

  fin>>a>>b>>r;

  l[a]++;

      map[a][l[a]].v=b;

      map[a][l[a]].l=r;

      l2[b]++;

      map2[b][l2[b]].v=a;

      map2[b][l2[b]].l=r;

   }

   fin>>s>>t>>k;

   if (s==t) k++;

}

void dijkstra()

{ int z;

   for (int i=1;i<=n;i++)

her[i]=1000000;

   her[t]=0;

   for (int i=1;i<=n;i++)

   {

  int min=1000000;

  for (int j=1;j<=n;j++)

    if ((!used[j]) && (her[j]<min))

{

min=her[j];

z=j;

  }

used[z]=true;

for (int j=1;j<=l2[z];j++)

{

   if (her[z]+map2[z][j].l<her[map2[z][j].v])

            her[map2[z][j].v]=her[z]+map2[z][j].l;

}

   }

}

void push(int v,int i)

{    int tmp1;

     key temp;

    size++;

data[size].v=map[v][i].v;

data[size].value=tmp.value+map[v][i].l;

tmp1=size;

while (tmp1 > 1)

{

if (data[tmp1 >> 1].value+her[data[tmp1>>1].v]<data[tmp1].value+her[map[v][i].v]) break;

temp=data[tmp1];

data[tmp1]=data[tmp1 >> 1];

data[tmp1 >> 1]=temp;

        tmp1= tmp1 >> 1;

}

}

void del()

{   key temp;

temp=data[size];

data[size]=data[1];

data[1]=temp;

size--;

int i = 1;

while ((i << 1) <= size)

{

int j;

j = i << 1;

if ((j < size) && (data[j + 1].value+her[data[j+1].v]< data[j].value+her[data[j].v])) j++;

if (data[i].value+her[data[i].v]< data[j].value+her[data[j].v]) break;

temp=data[i];

data[i]=data[j];

data[j]=temp;

i = j;

}

}

int a_star()

{ 

if( her[s]==1000000 ) return -1;

data[1].v=s;

data[1].value=0;

  time[s]=0;

  size=1;

  

  while (size>0) {

    time[data[1].v]++;

 if (time[t]==k)

 return data[1].value;

 

      tmp=data[1];

 del();

 if (time[data[1].v]>k) continue;

 for (int e=1;e<=l[tmp.v];e++)

 {push(tmp.v,e);                           

  }

  }

return -1;

}

int main()

{

 fin.open("c:\\test.txt");

ready(); 

 dijkstra();

 int ans;

 ans=a_star();

 cout<<ans;  

 fin.close();

}