Category: 计算机与 Internet


Maybe we can be put into a system,and be calculated to be a number。

This number describes what we shall do in next step。

So is there a rational number for a human?

In mathematics, irrational number means non-cycle limitless number.

Since a human is always of limit(his life is limited), he is always a rational number if it can be encoded a number.

(These are 2 different concepts, a human’s limited life and his encoded number’s length are different)

 

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很显然,需要对它的技术做修改。

所谓深度学习技术,其实是一种自下而上的技术,通过输入大量的数据,结合神经网络,总结出规律(总结出1个函数?函数拟合?)。

对于1个天才来说,它本身不需要学习。

深度学习技术,不过是把天才的行为函数化(自动化?系统化?)罢了,或者说解密天才的行为。

有意思的是,alphago团队本身并无1个围棋的职业选手,只有黄士杰是个业余n段兼围棋开发者;而且当5翻棋第4局出现bug之后,alphago的创造者回答提问也很有意思,“他们也不晓得alphago是如何学习的?”

alphago创造者虽然创造了alphago,但alphago的执行能力强于创造者,而且alphago的具体执行过程甚至创造者都未必能100%掌控。

 

remove 2345.com

This is an ugly spyware when u install something from untrusted source.

One key sympton is that when u open default browser, a default url is opened  http://www.hao601.com/?tn5_9_30=27711

Even if u set default home URL to about:blank, that URL is still opened.

In fact, it is implemented by appending that URL to ur browser’s quick start URL. So right click ur browser icon, property, remove “http://www.hao601.com/?tn5_9_30=27711” from target.

任给1棵树,都可以用表格来表示。

 

同样,任给1表格,也可以用树来表示。

http://news.163.com/special/tuling002/#!/scene-1

average

How many kinds of average number are there for a value depending on time series?

  1. simple moving average
  2. cumulative moving average
  3. weighted moving average
  4. exponential weighted moving average

Others?

EWMA is a very interesting one.

取石子游戏

Description

有两堆石子,数量任意,可以不同。游戏开始由两个人轮流取石子。游戏规定,每次有两种不同的取法,一是可以在任意的一堆中取走任意多的石子;二是可以在两堆中同时取走相同数量的石子。最后把石子全部取完者为胜者。现在给出初始的两堆石子的数目,如果轮到你先取,假设双方都采取最好的策略,问最后你是胜者还是败者。

Input

输入包含若干行,表示若干种石子的初始情况,其中每一行包含两个非负整数a和b,表示两堆石子的数目,a和b都不大于1,000,000,000。

Output

输出对应也有若干行,每行包含一个数字1或0,如果最后你是胜者,则为1,反之,则为0。

 

assuming input is (a, b), a is alway less than b, say a to be part0, say b to part1

  1. if a = b,print 1
  2. if (a,b) = (1,2) , print 0
  3. if (a,b) = (1,n), n>2
    firstly get n-2, then it can be reduced to scenario 2
    so print
    1
  4. if (a,b) = (2, n), n>2
    if I get 1from part0, it is reduced to (1,n), so I lose. excluding it
    if I get 2 from part0, it is reduced to (0,n), so I lose, excluding it
    So I can only get from part1. get n-1 from part1, reduce it to scenario 2
    so print
    1
  5. if (a,b) = (3, n), n>3
    can not get 1, or 2, or 3 from part0, it will lose   
        if n=4, (3,4) reduce to (3,3), lose
                    (3,4) reduce to (3,2), lose
                    (3,4) reduce to (3,1), lose
                    (3,4) reduce to (3,0), lose
              print 0 for (3,4)
        if n>4, reduce (3,n) to (3,4) by get n-4 from part1
              print 1 for (3,n) n>4
  6. if (a,b) = (4, n), n>4
    (4,n) reduce to (4,3) by get n-3 from part1
    print 1
  7. if(a,b) = (5, n), n>5
    can not get 1,2,3,4,5 from part0, will lose
  8.     if n=6, (5,6) reduce to (5,5), lose
                    (5,6) reduce to (5,4), lose
                    (5,6) reduce to (5,3), lose
                    (5,6) reduce to (5,2), lose
                    (5,6) reduce to (5,1), lose
                    (5,6) reduce to (5,0), lose
              print 0 for (5,6)
        if n>6, reduce (5,n) to (5,6) by get n-6 from part1
              print 1 for (5,n) n>6

  9. if(a,b) = (6, n), n>6
    (6,n) reduce to (6,5) by get n-5 from part1
    print 1

So we can see something here, summarize as such:

 

  • if a=b, ouput 1
  • sort the pair to be (a,b) to meet a<b
    if a=2n, output 1
    if a=2n+1, b=2n+2, output 0
    if a=2n+1, b>2n+2, output 1
0 n 1
1 2 0
1 n(>2) 1
2 n(>2) 1
3 4 0
3 n(>4) 1
4 n(>4) 1
5 6 0
5 n(>6) 1
6 n(>6) 1
7 8 0
7 n(n>8) 1

    Yellow means they can be changed as required.

  1. Next State
    image
    Input State Input Symbol Output State Output Symbol Action
    S0 i0 S1 i0 0
  2. If then else
    image
    Input State Input Symbol Output State Output Symbol Action
    S0 i0 S1 i0 0
    S0 i1 S2 i1 0
  3. empty Loop
    S0: initial
    S1: loop body start
    S2: loop body end 
    image

    Turing machine Tape Initials:        image

    Input State Input Symbol Output State Output Symbol Action
    S0 0 S1 1 R
    S0 1 S1 2 R
    S0 S1 R
    S0 n-1 S1 n R
    S0 n Sout    
    S1 !FlagS1 S1 FlagS1 0
    S1 FlagS1 S2   R
    S2 any S2   L
    S2 FlagS1 S0 !FlagS1 L
  4. Loop with something between S1 and S2
    Take a look at the red entry at above table, modify S2 to others, and make sure there is a new entry of returning to S2 after ur loop body logic
    Input State Input Symbol Output State Output Symbol Action
    S1 FlagS1 Others   R
      S2   R
    S2 any S2   L
    S2 FlagS1 S0 !FlagS1 L
  5. Copy of A String 111…111
    Input State Input Symbol Output State Output Symbol Action
    S0 0 Sout    
    S0 1 S1 Flag R
    S1 1 S1   R
    S1 0 S2   R
    S2 1 S2   R
    S2 0 S3 1 L
    S3 any-Flag S3   L
    S3 Flag S0 1 R
  6. Append a String 111…111 to a String 111…111||null 
    identical to 5,here is its flow graph
    image
  7. Multiplication
    sample:
     image

    Input State Input Symbol Output State Output Symbol Action Comment
    S0 0 Sout     exception
    S0 1 S1   R  
    S1 0 S6   R S6=start moving to the end and add the tailing 1
    S1 1 S2 Flag R  
    S2 1 S2   R  
    S2 0 S3   R complete reading 1st number
    S3 0 Sout     exception
    S3 1 S4   R  
    S4   S5     append following 1s to subsequent 1s
    S5 any-Flag S5   L  
    S5 Flag S1 1 R  
    S6 1 S7   R  
    S6 0 Sout     exception
    S7 1 S7   R  
    S7 0 S8   R complete reading 2nd number
    S8 1 S8   R  
    S8 0 Sout 1   normal

Turing Machines

About Turing Machines, this article is a must.

http://plato.stanford.edu/entries/turing-machine/

 

Simply say, A Truing Machine is composed of 5 elements.

  1. Initial State                                  s0
  2. Final State                                   f0
  3. A finite set of state                      S
  4. A finite set of input symbols        I 
  5. A series of functions                    S x I –> S x I x {L, R, 0}

Of course there are others such as tape, read|write head etc.

In this article, it discusses a method to store number by Turing machine, which is different from normal binary system used by current computers. That is, 0 represented by one 1. 1 by two 1s. 2 by three 1s. … …

Using this representation, it gives a Turing machine to define f(n)=n+1.
Further, it gives a Turing machine of f(m,n)=m+n.

Turing machines can be encoded as sequences of 0|1. Further its arguments and it can be set as input of a UTM(Universal Turing Machine) to simulate its behavior. UTM can be thought of as a programmable computer. And Turing machines can be thought of as a program.

Further, UTM can also be encoded as 0|1 sequences and set as input to itself to simulate its behavior.

 

The number of Turing machine is countable. And computable functions can be represented by a Turing machine.

The number of functions over Natural numbers is uncountable.(https://proofwiki.org/wiki/Natural_Number_Functions_are_Uncountable)

So there exists functions which can not be represented by a Turing machine.

 

Let us take a look at the way to prove Functions over Natural Number is uncountable(F: NxN).

Firstly, for each n, let us construct the function Fn(x): N—>n, Fn(x) is indefinite and bijectioin to N. So the set of Fn(x) is countable. 

Secondly, assuming F is countable, so there exists a bijection M: N—>Qn(x),Qn(x) belong to F. Now let us construct another a bijection P: N—>Qn(x)+1,

    for each m belongiing to N, P(m)=Qn(m)+1

    Qn(m)+1 is a Natural number,

so P is a member of F.

On the other side, P is different from each of Qn(x), so F is uncountable.

 

The above method to prove F is uncountable is similar to the way of proving the sequence of 0|1 is uncountable by Cantor.

Let us see what is the common way.

Assuming the set S is countable, then construct an element E from some or all of the set’s elements. On one side, E is not identical to any of the set’s element. On the other side, from the set’s definition, E also belong to S. Contradiction happens here. So S is not countable.

win8 tips

  1. win8下应用字体模糊blurry
  2. 参见这个http://productforums.google.com/forum/#!topic/chrome/9DnjIpD3xoE

    The point is to enable a option of the app exe.

    image

  3. remove skydrive/onedrive from win8
    http://www.thewindowsclub.com/disable-skydrive-integration-windows-8-1
    using gpedit.msc