four kinds of design for adding operation of two rationals

有理數加法的幾種設計: 結果回傳或覆蓋 vs 類別或物件方法
  (1) public static Rational add(Ratinoal r1, Rational r2)
      // 用法:  Rational r3 = Rational.add(r1,r2);  // r3=r1+r2

  (2) public static void add(Ratinoal r1, Rational r2)
      // 用法:  Rational.add(r1,r2);  // r1=r1+r2

  (3) public Rational add(Rational r2)
      // 用法:  Rational r3 = r1.add(r2);  // r3=r1+r2

  (4) public void add(Rational r2)
      // 用法:  r1.add(r2);  // r1=r1+r2

Canonical Weighted Sum (CWS) Gates for Simplifying Bayesian Network Inference

decampos-10-ijar-combining content-based and collaborative recommendations- a hybrid approach based on bayesian networks

The following will explain the examples in pp793-794 of the paper.

Assume that I and F nodes in the bayesian network are binary random variables.
If the set of parent nodes of node I6 is Pa(I6) = { F6, F7, F8},
then the 8 possible configurations of conditional joint probabilities are
     pa1(I6) = (f6,1 & f7,1 & f8,1)
     pa2(I6) = (f6,0 & f7,1 & f8,1)
     pa3(I6) = (f6,1 & f7,0 & f8,1)
     pa4(I6) = (f6,1 & f7,1 & f8,0)
     pa5(I6) = (f6,0 & f7,0 & f8,1)
     pa6(I6) = (f6,0 & f7,1 & f8,0)
     pa7(I6) = (f6,1 & f7,0 & f8,0)
     pa8(I6) = (f6,0 & f7,0 & f8,0)

By use of Canonical Weighted Sum (CWS) Gates, which consist of 2x2x3=12 weights,
computation of 16 conditional joint probabilities (having 8 degrees of freedom)
can be decomposed into the summation of 12 weights (having 3 degrees of freedom):
 w(f6,1&i6,1) = w(f6,0&i6,0) = 0.3
 w(f7,1&i6,1) = w(f7,0&i6,0) = 0.4
 w(f8,1&i6,1) = w(f8,0&i6,0) = 0.3
and
 w(f6,1&i6,0) = w(f6,0&i6,1) = 0
 w(f7,1&i6,0) = w(f7,0&i6,1) = 0
 w(f8,1&i6,0) = w(f8,0&i6,1) = 0

Pr(i6,0 | pa1(I6))= Pr(i6,0 | f6,1 & f7,1 & f8,1) = w(f6,1&i6,0) + w(f7,1&i6,0) + w(f8,1&i6,0) = 0
Pr(i6,1 | pa1(I6))= Pr(i6,1 | f6,1 & f7,1 & f8,1) = w(f6,1&i6,1) + w(f7,1&i6,1) + w(f8,1&i6,1) = 0.3 + 0.4 + 0.3 = 1

Pr(i6,0 | pa2(I6))= Pr(i6,0 | f6,0 & f7,1 & f8,1) = w(f6,0&i6,0) + w(f7,1&i6,0) + w(f8,1&i6,0) = 0.3
Pr(i6,1 | pa2(I6))= Pr(i6,1 | f6,0 & f7,1 & f8,1) = w(f6,0&i6,1) + w(f7,1&i6,1) + w(f8,1&i6,1) = 0 + 0.4 + 0.3 = 0.7

Pr(i6,0 | pa3(I6))= Pr(i6,0 | f6,1 & f7,0 & f8,1) = w(f6,1&i6,0) + w(f7,0&i6,0) + w(f8,1&i6,0) = 0.4
Pr(i6,1 | pa3(I6))= Pr(i6,1 | f6,1 & f7,0 & f8,1) = w(f6,1&i6,1) + w(f7,0&i6,1) + w(f8,1&i6,1) = 0.3 + 0 + 0.3 = 0.6

Pr(i6,0 | pa4(I6))= Pr(i6,0 | f6,1 & f7,1 & f8,0) = w(f6,1&i6,0) + w(f7,1&i6,0) + w(f8,0&i6,0) = 0.3
Pr(i6,1 | pa4(I6))= Pr(i6,1 | f6,1 & f7,1 & f8,0) = w(f6,1&i6,1) + w(f7,1&i6,1) + w(f8,0&i6,1) = 0.3 + 0.4 + 0 = 0.7

Pr(i6,0 | pa5(I6))= Pr(i6,0 | f6,0 & f7,0 & f8,1) = w(f6,0&i6,0) + w(f7,0&i6,0) + w(f8,1&i6,0) = 0.3 + 0.4 = 0.7
Pr(i6,1 | pa5(I6))= Pr(i6,1 | f6,0 & f7,0 & f8,1) = w(f6,0&i6,1) + w(f7,0&i6,1) + w(f8,1&i6,1) = 0 + 0 + 0.3 = 0.3

Pr(i6,0 | pa6(I6))= Pr(i6,0 | f6,0 & f7,1 & f8,0) = w(f6,0&i6,0) + w(f7,1&i6,0) + w(f8,0&i6,0) = 0.3 + 0 + 0.3 = 0.6
Pr(i6,1 | pa6(I6))= Pr(i6,1 | f6,0 & f7,1 & f8,0) = w(f6,0&i6,1) + w(f7,1&i6,1) + w(f8,0&i6,1) = 0 + 0.4 + 0 = 0.4

Pr(i6,0 | pa7(I6))= Pr(i6,0 | f6,1 & f7,0 & f8,0) = w(f6,1&i6,0) + w(f7,0&i6,0) + w(f8,0&i6,0) = 0 + 0.4 + 0.3 = 0.7
Pr(i6,1 | pa7(I6))= Pr(i6,1 | f6,1 & f7,0 & f8,0) = w(f6,1&i6,1) + w(f7,0&i6,1) + w(f8,0&i6,1) = 0.3 + 0 + 0 = 0.3

Pr(i6,0 | pa8(I6))= Pr(i6,0 | f6,0 & f7,0 & f8,0) = w(f6,0&i6,0) + w(f7,0&i6,0) + w(f8,0&i6,0) = 0.3 + 0.4 + 0.3 = 1
Pr(i6,1 | pa8(I6))= Pr(i6,1 | f6,0 & f7,0 & f8,0) = w(f6,0&i6,1) + w(f7,0&i6,1) + w(f8,0&i6,1) = 0

1 = Pr(i6,0 | pa(I6)) + Pr(i6,1 | pa(I6))
  = w(f6,1&i6,0) + w(f7,1&i6,0) + w(f8,1&i6,0) + w(f6,1&i6,1) + w(f7,1&i6,1) + w(f8,1&i6,1) = 1

============================

Given Pa(U4) = { I1, I3, I6, I7, I8, I10 } with six-value (0,1,~,5) random variable U,
the network consists of 2x6x6=72 weights.

By the following 12 nonzero weights having 6 degrees of freedom (with all other weights being 0):
       w(i1,1&u4,1) = w(i3,1&u4,1) = w(i6,1&u4,2) = w(i7,1&u4,1) = w(i8,1&u4,3) = w(i10,1&u4,5) = 0.166
and
       w(i1,0&u4,0) = w(i3,0&u4,0) = w(i6,0&u4,0) = w(i7,0&u4,0) = w(i8,0&u4,0) = w(i10,0&u4,0) = 0.166
we can compute any Pr(U4 | pa(U4)) conditional probabilities by summation of the 12 nonzero weights.

(1) pa(U4) = { i1,1 & i3,1 & i6,1 & i7,1 & i8,1 & i10,1 }

Pr(u4,0 | pa(U4)) = w(i1,1&u4,0) + w(i3,1&u4,0) + w(i6,1&u4,0) + w(i7,1&u4,0) + w(i8,1&u4,0) + w(i10,1&u4,0)
    = 0            + 0            + 0            + 0            + 0            + 0
    = 0

Pr(u4,1 | pa(U4)) = w(i1,1&u4,1) + w(i3,1&u4,1) + w(i6,1&u4,1) + w(i7,1&u4,1) + w(i8,1&u4,1) + w(i10,1&u4,1)
    = 0.166        + 0.166        + 0            + 0.166        + 0            + 0
    = 0.5

Pr(u4,2 | pa(U4)) = w(i1,1&u4,2) + w(i3,1&u4,2) + w(i6,1&u4,2) + w(i7,1&u4,2) + w(i8,1&u4,2) + w(i10,1&u4,2)
    = 0            + 0            + 0.166        + 0            + 0            + 0
    = 0.166

Pr(u4,3 | pa(U4)) = w(i1,1&u4,3) + w(i3,1&u4,3) + w(i6,1&u4,3) + w(i7,1&u4,3) + w(i8,1&u4,3) + w(i10,1&u4,3)
    = 0            + 0            + 0            + 0            + 0.166        + 0
    = 0.166

Pr(u4,4 | pa(U4)) = w(i1,1&u4,4) + w(i3,1&u4,4) + w(i6,1&u4,4) + w(i7,1&u4,4) + w(i8,1&u4,4) + w(i10,1&u4,4)
    = 0            + 0            + 0            + 0            + 0            + 0
    = 0

Pr(u4,5 | pa(U4)) = w(i1,1&u4,5) + w(i3,1&u4,5) + w(i6,1&u4,5) + w(i7,1&u4,5) + w(i8,1&u4,5) + w(i10,1&u4,5)
    = 0            + 0            + 0            + 0            + 0            + 0.166
    = 0.166

1 = Pr(u4,1 | pa(U4)) + Pr(u4,2 | pa(U4)) +　Pr(u4,3 | pa(U4)) + Pr(u4,4 | pa(U4)) + Pr(u4,5 | pa(U4))
  = 0.5 + 0.166 + 0.166 + 0 + 0.166

(2) pa(U4) = { i1,0 & i3,0 & i6,1 & i7,1 & i8,0 & i10,0 }

Pr(u4,0 | pa(U4)) = w(i1,0&u4,0) + w(i3,0&u4,0) + w(i6,1&u4,0) + w(i7,1&u4,0) + w(i8,0&u4,0) + w(i10,0&u4,0)
    = 0.166        + 0.166        + 0            + 0            + 0.166        + 0.166
    = 0.666

Pr(u4,1 | pa(U4)) = w(i1,0&u4,1) + w(i3,0&u4,1) + w(i6,1&u4,1) + w(i7,1&u4,1) + w(i8,0&u4,1) + w(i10,0&u4,1)
    = 0            + 0            + 0            + 0.166        + 0            + 0
    = 0.166

Pr(u4,2 | pa(U4)) = w(i1,0&u4,2) + w(i3,0&u4,2) + w(i6,1&u4,2) + w(i7,1&u4,2) + w(i8,0&u4,2) + w(i10,0&u4,2)
    = 0            + 0            + 0.166        + 0            + 0            + 0
    = 0.166

Pr(u4,3 | pa(U4)) = w(i1,0&u4,3) + w(i3,0&u4,3) + w(i6,1&u4,3) + w(i7,1&u4,3) + w(i8,0&u4,3) + w(i10,0&u4,3)
    = 0            + 0            + 0            + 0            + 0            + 0
    = 0

Pr(u4,4 | pa(U4)) = w(i1,0&u4,4) + w(i3,0&u4,4) + w(i6,1&u4,4) + w(i7,1&u4,4) + w(i8,0&u4,4) + w(i10,0&u4,4)
    = 0            + 0            + 0            + 0            + 0            + 0
    = 0

Pr(u4,5 | pa(U4)) = w(i1,0&u4,5) + w(i3,0&u4,5) + w(i6,1&u4,5) + w(i7,1&u4,5) + w(i8,0&u4,5) + w(i10,0&u4,5)
    = 0            + 0            + 0            + 0            + 0            + 0
    = 0

1 = Pr(u4,1 | pa(U4)) + Pr(u4,2 | pa(U4)) +　Pr(u4,3 | pa(U4)) + Pr(u4,4 | pa(U4)) + Pr(u4,5 | pa(U4))
  = 0.666 + 0.166 + 0.166 + 0 + 0

============================

Suppose that Acf = A4 or user 4.
Given Pa(Acf) = { Ux, Uy, Uz } with with six-value (0,1,~,5) random variable Acf,
the network consists of 6x6x3=108 weights.

By the following ? nonzero weights having ? degrees of freedom (with all other weights being 0):
       w(ux,5&acf,4) = 0.54
       w(uy,1&acf,4) = 0.15
       w(uz,4&acf,4) = 0.09
and
       w(ux,0&acf,0) = 0.6
       w(uy,0&acf,0) = 0.3
       w(uz,0&acf,0) = 0.1
we can compute any Pr(Acf | pa(Acf)) conditional probabilities by summation of the ? nonzero weights.


(1) pa(Acf) = { ux,5 & uy,1 & uz,4 }

Pr(acf,4 | pa(U4)) = w(ux,5&acf,4)               + w(uy,1&acf,4)               + w(uz,4&acf,4)
     = RSim(Ux,Acf) * Pr(A=4|Ux=5) + RSim(Uy,Acf) * Pr(A=4|Uy=1) + RSim(Uz,Acf) * Pr(A=4|Uz=4)
     = 0.6          * 0.9          + 0.3          * 0.5          + 0.1          * 0.9
                   = 0.78

(2) pa(Acf) = { ux,0 & uy,1 & uz,4 }

Pr(acf,4 | pa(U4)) = w(ux,0&acf,4)               + w(uy,1&acf,4)               + w(uz,4&acf,4)
     = 0                           + RSim(Uy,Acf) * Pr(A=4|Uy=1) + RSim(Uz,Acf) * Pr(A=4|Uz=4)
     = 0                           + 0.3          * 0.5          + 0.1          * 0.9
                   = 0.24

(3) pa(Acf) = { ux,0 & uy,1 & uz,4 }

Pr(acf,0 | pa(U4)) = w(ux,0&acf,0)               + w(uy,1&acf,0)               + w(uz,4&acf,0)
     = RSim(Ux,Acf)                + 0                           + 0
     = 0.6                         + 0                           + 0
     = 0.6

why object-oriented programming provides encapsulation, inheritance, and polymorphism

一般完整的物件導向語言提供封裝、繼承、多型三大功能，允許以合成或繼承方式建構新物件。

其中，封裝(encapsulation)旨在保護私密資料的安全、完整性，其存取唯有透過公開的屬性存取(窗口)或運算方法達成之。

合成(composition)是一種利用物件建構物件的方法，物件之間的擁有(has-a)關係可形成階層樹狀圖。

繼承(inheritance)旨在提供新用舊的機制，讓後代需要用到親代原來的屬性及方法時，直接沿用，不必重造車輪。
繼承也是一種利用物件建構物件的方法，物件之間的屬於(is-a)關係可形成階層樹狀圖。

多型(polymorphism)旨在繼承架構下提供舊用新的機制，讓親代也有機會用到後代創新的新行為、新功能。

find the unique second smallest number in excel

一般 Excel 找最小或最大值可以使用 MIN(range) 或 MAX(range)函數，其中，range表示儲存格範圍。 如果想要找第2小或第2大也可以使用 SMALL(range, 2) 或 LARGE(range, 2)，其中，2表示順序第2。 但是以找最小為例，有時候遇到第1小及第2小值一樣，就看不出值分布情形。若希望能看到不同值的第2小數字，可以使用如下函數:

     = SMALL(range, COUNTIF(range, MIN(range)) + 1)

其中， SMALL(range,  n) 可回傳 range 範圍內順序第 n 小的數字。 COUNTIF(range, MIN(range)) 可回傳 range 範圍內符合最小值的儲存格個數。 此個數加1之後，即為不同值第2小數字之順序。 因此可餵給 SMALL 函數，回傳不同值第2小數字出來。

至於不同值第3或第n小數字，可能就需要更複雜的 FREQUENCY 陣列函數或自己寫巨集計算了。