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  • factorial

    2008-01-24 18:14:05

    Let P = {p1,p2, ... ,pX, ...} be infinite prime set.

    Factorial can be represent as the product of prime set P

    Let Z be the rounded integer of logpN, and p (- P

    Let q be the quotient of N/p

    we have,

    f(p) = Z[Z-1]/2

    In general,

    N! = p1^[q1 + f(p1)] * p2^[q2 + f(p2)] * ... * pN^[qN + f(pN)]

    for instance,

    7! = 2^(3+1) * 3^(2) * 5^(1) * 7^(1)

        = (2^4)(3^2)(5)(7)

    -> 6! = (2^4)(3^2)(5)

    28! = 2^(14+6) * 3^(9+3) * 5^(5+1) * 7^(4)

             * 11^(2) * 13^(2) * 17 * 19 * 23

          = (2^20)(3^12)(5^6)(7^4)(11^2)(13^2)

              (17)(19)(23)

     

    ------------------------------------------------------------

    Inverse factorial

    let !'(x) be the function that perform.

    the inverse of factorial such that

    !'(n!) = n

    !'(1) = 0 or 1

    !'(2) = 2

    !'(6) = 3

    !'(24) = 4

    Since, n(n-1)! = n!

    -> n = n!/(n-1)!

    -> !'(n!) = n!/(n-1)!

    !'(x)! = x

    for instance,

    !'(6)! = 3! = 6

    -> !'(x)! = !'(x!)

    or we can rewrite as

    x!!' = x!'!

    to be continue...

  • ..

    2007-11-05 22:03:45

    P versus NP

    The question is whether, for all problems for which a computer can verify a given solution quickly (that is, in polynomial time), it can also find that solution quickly. This is generally considered the most important open question in theoretical computer science.

    The official statement of the problem was given by Stephen Cook.

    The Hodge conjecture

    Main article: Hodge conjecture

    The Hodge conjecture is that for projective algebraic varieties, Hodge cycles are rational linear combinations of algebraic cycles.

    The official statement of the problem was given by Pierre Deligne.

    The Poincaré conjecture

    Main article: Poincaré conjecture

    In topology, a sphere with a two-dimensional surface is essentially characterized by the fact that it is simply connected. It is also true that every 2-dimensional surface which is both compact and simply connected is topologically a sphere. The Poincaré conjecture is that this is also true for spheres with three-dimensional surfaces. The question had long been solved for all dimensions above three. Solving it for three is central to the problem of classifying 3-manifolds. A proof of this conjecture was given by Grigori Perelman in 2003; its review was completed in August 2006, and Perelman was awarded the Fields Medal for his solution. Perelman declined the award.[1]

    The official statement of the problem was given by John Milnor.

    The Riemann hypothesis

    Main article: Riemann hypothesis

    The Riemann hypothesis is that all nontrivial zeros of the Riemann zeta function have a real part of 1/2. A proof or disproof of this would have far-reaching implications in number theory, especially for the distribution of prime numbers. This was Hilbert's eighth problem, and is still considered an important open problem a century later.

    The official statement of the problem was given by Enrico Bombieri.

    Yang-Mills existence and mass gap

    In physics, classical Yang-Mills theory is a generalization of the Maxwell theory of electromagnetism where the chromo-electromagnetic field itself carries charges. As a classical field theory it has solutions which travel at the speed of light so that its quantum version should describe massless particles (gluons). However, the deictic phenomenon of color confinement permits only bound states of gluons, forming massive particles. This is the mass gap. Another aspect of confinement is asymptotic freedom which makes it conceivable that quantum Yang-Mills theory exists without restriction to low energy scales. The problem is to establish rigorously the existence of the quantum Yang-Mills theory and a mass gap.

    The official statement of the problem was given by Arthur Jaffe and Edward Witten.

    Navier-Stokes existence and smoothness

    The Navier-Stokes equations describe the movement of liquids and gases. Although they were found in the 19th century, they still are not well understood. The problem is to make progress toward a mathematical theory that will give us insight into these equations.

    The official statement of the problem was given by Charles Fefferman.

    The Birch and Swinnerton-Dyer conjecture

    The Birch and Swinnerton-Dyer conjecture deals with a certain type of equation, those defining elliptic curves over the rational numbers. The conjecture is that there is a simple way to tell whether such equations have a finite or infinite number of rational solutions. Hilbert's tenth problem dealt with a more general type of equation, and in that case it was proven that there is no way to decide whether a given equation even has any solutions.

    The official statement of the problem was given by Andrew Wiles.

  • 問題集

    2007-09-16 07:21:19

    7^2222 / 9

    remainder = ?

    -1=-1

    1/-1 = -1/1

    (1/-1)^(1/2) = (-1/1)^(1/2)

    1^(1/2) / (-1)^(1/2) = (-1)^(1/2) / 1^(1/2)

    1^(1/2) * 1^(1/2) = (-1)^(1/2) * (-1)^(1/2)

    1 = -1

    -1 = -1 -> 1 = -1 ? why?

    one day, A, B and C get the same amount of money to buy food in McDonald. A use all of the money to buy chicken nugget($7/each)and remain $2. B use all of the money to buy burger($9/each) and remain $8. C use all of the money to buy fillet O'fish meal($13/each), and remain $3. How much they get at the beginning? (Show steps by steps)

     有一日,小明,小強和小李拿着相同數值的錢去麥當勞. 小明用了所有的錢買麥樂雞(每份$7), 剩餘$2. 小強用了所有的錢買漢堡包(每份$9), 剩餘$8. 小李用了所有的錢買魚柳包餐(每份$13), 剩餘$3. 他們一開始拿了多少錢?

    甲、乙兩項工作,單獨做,張需10小時做完甲工作,15小時做完乙工作,李需8小時做完甲工作,20小時做完乙工作,兩人合作完成兩項工作,最少需要多少小時呢?

    () 5 () 5 () 1 () 5 () = 24

    在()中填上 + - x / ( ) 使算式成立.

    在圖形W 內試加上三直線,製造出九個三角形.(三角形內不可以有另一個三角形) 完成後請回報所花時間

    大家知唔知道點解水渠蓋係圓形?

    第一個問題:

    1/3 = 0.33333333...

    1/3 * 3 = 0.33333333... * 3

    1 = 0.99999999...?

    第二個問題:

    "這個句子是錯的!"

    那麼上面的句子本身是錯的還是對的?

    本人想到一個3進2制法, 也可以理解為1.5進制一般情況, 加數進位只進1, 但係3進2制必會進2!

    例: 10進制 -> 3進2制

    1 -> 1

    2 -> 2

    3 -> 20 (正如在10進制下, 10要進1位. 而在3進2制下,3要進2位)

    4 -> 21

    5 -> 22

    6 -> 23 -> 40 -> 210

    7 -> 24 -> 41 -> 211

    8 -> 25 -> 42 -> 212

    9 -> 26 -> 43 -> 60 -> 230 -> 400 -> 2100

    10 -> 27 -> 44 -> 61 -> 231 -> 401 -> 2101

    3進2制 -> 10進制

    211

    = 2 * (1.5)^2 + 1 * (1.5)^1 + 1 * (1.5)^0

    = 2 * 2.25 + 1.5 + 1

    = 4.5 + 1.5 + 1

    = 7

    雖然唔知有冇實際用途, 純粹分享!

    如何用5個數字或數學符號來製造出最大的數?

    例一: 9^9^9^9^9

    例二: 9!!!!

    從前,有一個愛酒如命的國王. 他為了他的生辰宴會而在他最珍貴的收藏系列中選出100樽酒.

    不幸地, 其中一樽酒給刺客下了毒藥, 而該刺客已自殺身亡. 那是一種無色無味的慢性毒藥, 滲進酒後, 只需一小滴便可殺人於無形.它的藥性要1個月後才會發生作用.

    可是, 1個月後便是他的生辰宴會. 因此, 他決定找人來試酒出毒酒!

    究竟國王最少要用多少人和用了什麼方法來試出哪一樽才是毒酒? (即每人只可喝一次酒, 藥的不穏定性難以以時間差來分出)

    在一埸龜兔賽跑中, 因為白兔比烏龜跑得快很多, 為了讓比賽更公平的進行, 烏龜可以先前進一段.

    但過程中出現了些少問題:

    1)白兔超越烏龜前必須趕上烏龜.

    2)白兔雖然跑得很快,但不管跑得多快,在牠趕到烏龜前所在的那一點時, 必定會花費一些時間.

    3)烏龜雖然很慢, 但不管跑得多慢, 白兔花費的時間一定足以讓烏龜往前進一點點.

    4)接下來也一樣, 當白兔又趕到烏龜先前所在那一點之後, 又必須往前繼續趕上烏龜, 而烏龜這時又往前進了一點點. 當然, 這次和烏龜的距離較前一次更近, 但仍然有距離. 白兔會越來越近, 卻永遠無法趕上烏龜.

    在沒有時間限制的比賽中, 白兔當然能趕上烏龜. 但是, 以邏輯上而言, 為何白兔永遠無法超越烏龜?

  • 2-3 system

    2007-08-27 19:28:41

    #include <cstdlib>
    #include <iostream>

    using namespace std;

    void convert(long x){
         long a[30000] = {0};
         long n = 0; 
         while (x != 0){
              a[n] = x % 3;
              x = (x/3) * 2;
              n++;
         }
        
         n--;
         while (n != -1){
              cout << a[n];
              n--;
              }
    }

    int main(int argc, char *argv[])
    {
        long start, end;
        cout << "Convert decimal number to 3-2 system!\nYou can print value in range!\n";
        cout << "Start printing value from?";
        cin >> start;
        cout << "End printing value at?";
        cin >> end;
        for (long i = start; i <= end; i++){
            cout << i << " -> ";
            convert(i);
            cout << "\n";
            }
        system("PAUSE");
        return EXIT_SUCCESS;
    }

  • collatz generator

    2007-08-27 19:25:33

    #include <cstdlib>
    #include <iostream>

    using namespace std;

    bool even(int a){
         return ((a%2) == 0);
    }

    long even_function(long &a){
           a = a/2;
           cout << a << "\n";
           return a;
    }

    long odd_function(long &a){
           a = ((a+a+a) + 1);
           cout << a << "\n";
           return a;
    }

    void output_generator(long &a, long y){
           if (even(a))
              even_function(a);
           else
              odd_function(a);
    }          

    int main(int argc, char *argv[])
    {
        long x;
        long limit;
        long y = 0;
        cout << "limit? ";
        cin >> limit;
        cout << "Calculate...\n";
        x =limit;
        while ( x > 1 ){
          y++;   
          output_generator(x,y);
        }
        cout << "success\n";
        system("PAUSE");
        return EXIT_SUCCESS;
    }

  • [論壇] 黑幫的密碼(轉貼修改)

    2007-08-27 19:17:52

    一天, 某黑幫的一個頭頭 收到一封由老大寄來的信 ,內容如下 :

       TelL MuM and dad thAt maXI lOVe mARy anD JeSS oK ? WiSH UR FD LiVE bAD    (文法可能有錯,但不礙事)

    他收到信後不明所以, " 說什麼東東呀?"

    他發現信中付著一首詩, 加以解讀後發現了一些東西

    某人將會被殺


    ------------------------------------------------------------------------


    the poet

    Mr. little looks like a LINE , Mr. Capital looks like a DOT

    They have a race

    The referee is Mr. MORSE

    After the race they EXCHANGE with their OPPOSING MEMBERS

    ----------------------------Answer-------------------------------


    TelL MuM and dad thAt maXI lOVe mARy anD JeSS oK ? WiSH UR FD LiVE bAD

    大階代表點, 細階代表線, 經轉換後成為以下:

    .--. .-. --- --- --.- --.. -..- .--. --. .-.. -. .-.. .. .. .-.. -..

    以上為摩斯密碼, 轉換成為英文字母後為:

    PROOQZXPGLNLIILD

    英文字母順序倒轉置換:

    a -> z
    b -> y
    c -> x
    d -> w
    e -> v
    f -> u
    g -> t
    h -> s
    i -> r
    j -> q
    k -> p
    l -> o
    m -> n
    n -> m
    o -> l
    p -> k
    q -> j
    r -> i
    s -> h
    t -> g
    u -> f
    v -> e
    w -> d
    x -> c
    y -> b
    z -> a

    prooqzxpglnliild -> killjacktomorrow

    jack將會被殺!


    [ 本帖最後由 hks_sing 於 2007-8-26 07:52 編輯 ]

數據統計

  • 訪問量: 1414
  • 日誌數: 8
  • 圖片數: 5
  • 文件數: 1
  • 建立時間: 2007-04-24
  • 更新時間: 2008-04-20

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