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(AdaBoost)
•               Naive Bayes      Yes, No              ○   ×


    •                            ○   ×        1   0
•       3
•                     (Weighted Voting)


    •
        •
            •
    •
        •
•
•        1   0
•                   N
•                (Xi , ci )(i = 1, . . . , N )   C

    XC = (No, Yes, Yes, Yes), cC = 1
•   R:
    •
•
    •
                           1
         i                wi
    •
                  N
             wi = 1/N (i = 1, . . . , N )
              1

    •                    10
             wi = 1/10
              1
•        t=1,...,R
1.             : t                                                      i
                                            t
          pt
           i                   pt =
                                           wi
                                i          N      t
                                           i=1   wi
                           N

     •                                                p1 = wi = 1/10
                                                            1
         t=1                      t
                                 wi   =1               i
                           i=1


2.                                          WeakLearner
           WeakLearner                                (   t   < 1/2 )        ht
                                              t
                                 N
                       t   =          pt |ht (Xi ) − ci | < 1/2
                                       i
                               i=1


                                            0, ht (Xi )        ci
               ht (Xi ) − ci         =
                                            1,

                                                                    Step 3
•   t=1   WeakLearner




               ID     A F                     h1(Xi) = 1
                E,F             h1(Xi)=ci   D,J ci=0
               ID     G,H,I,J                  h1(Xi) = 0


                                WeakLearner
               10                2
                p1 = 1/10
                 i
                 1 = 1/10 × 2 = 1/5 < 1/2
                                 T2
               WeakLearner
t+1
3.           wi                                                βt
                            t
                    βt =                  0≤        < 1/2,   0 ≤ βt < 1
                           1−    t
                                                t

     •   β
                                     1−|ht (Xi )−ci |
                  wi = wi βt
                   t+1  t


         •          εt               βt
         •   εt            βt
         •                      WeakLearner             ht
•                                t=1
           1 = 0.2
                   1
          β1 =             = 0.2/0.8 = 0.25
                1−     1
    •          A D, G J            ht         E, F

     2
    wA   = wB = wC = wD = wG = wH = wI = wJ
            2     2     2     2 2    2    2

         = 1/10 × β1 = 0.025
                   1
     2
    wE   = wF = 1/10 × β1 = 0.1
            2            0
WeakLearner
•   t=1               WeakLearner       T2
•               WeakLearner
    •                                        WeakLearner
    •
        WeakLearner
•   t=1
    •      T1   Yes       ε1   ε1(T1=Yes)
p1 = 1/10(i ∈ {A, B, ..., J})
 i
                            1
        1 (T1   = Yes) =        × 4 = 0.4
                           10
                            1
        1 (T2   = Yes) =        × 2 = 0.2
                           10
                            1
        1 (T3   = Yes) =        × 3 = 0.3
                           10
                            1
        1 (T4   = Yes) =        × 2 = 0.2
                           10
                                T2=Yes      T4=Yes
                                    T2
•   t=2
     wA2
              = wB = wC = wD = wG = wH = wI = wJ
                 2     2     2     2 2    2    2

              = 1/10 × β1 = 0.025
                        1
      2
     wE       = wF = 1/10 × β1 = 0.1
                 2            0



                  wi = 0.400
                   2

    i∈{A,...,J}

     p2 = p2
      A    B        =   p2 = p2 = p2 = p2 = p2 = p2
                         C     D    G     H  I    J
                    =   0.025/0.400 = 0.0625
     p2 = p2
      E    F        =   0.1/0.400 = 0.25

•                               WeakLearner
      2 (T1   = Yes)    =   0.0625 × 2 + 0.25 × 2 = 0.625
                             2 (T1 = No) = 0.375
      2 (T2 = Yes)      =   0.25 + 0.25 = 0.5
      2 (T3 = Yes)      =   0.0625 × 2 + 0.25 × 1 = 0.3125
      2 (T4 = Yes)      =   0.0625 × 2 = 0.125
                                                       T4=Yes
β2 =   2 /(1   −   2)   = 0.143
R
                          1   if   t=1 (− log10   βt )ht (X) ≥   (− log10 βt ) 1
hf (X)      =                                                                  2
                          0   othrewise
                t
     βt =
            1−       t



 •                                                                     1,
                               0
     •   Log
 •                  εt    0               βt                 -Log βt
                         WeakLearner             ht(X)
 •                                     -Log βt               ht(X)
β1 = 0.25, β2 = 0.143
 2
                1
    (− log βt )
t=1
                2
                       1                1
     =   − log10 0.25 × − log10 0.143 ×
                       2                2
     =   0.723

 •                          1.
•   h1: T2=Yes         C=1, h2: T4=Yes   C=1

 2   •             K
      (− log βt )ht (XK )
t=1
     = − log10 0.25 × 0 − log10 0.143 × 1
     = 0.845
                   0.723                            ○


 2
     •             L
      (− log βt )ht (XL )
t=1
     =   − log10 0.25 × 1 − log10 0.143 × 0
     =   0.602        0.723                         ×
AdaBoost
•              hf                          ε
                    =                      D(i)
                        {i|hf (Xi )=yi }

    •   D(i)
•                   R
                                                  εt   t

               ≤          2     t (1   − t)
                    t=1
    •   t                              εt<1/2

               2     t (1     − t) < 1

    •                                           WeakLearner
                          ε
•           1
     •   R                                  R+1

                       N                    R         1/2
                             R+1
                            wi   ≥               βt
                      i=1                  t=1
 •
     •           N
                      R+1                               R+1
                     wi       ≥                        wi
             i=1                    {i|hf (Xi )=yi }



                                   t 1−|hf (Xi )−yi |
                      t+1
                     wi     =     wi βi
                                                             R
                                                                    1−|hf (Xi )−yi |
                      t+1
                     wi     =                         D(i)         βt
{i|hf (Xi )=yi }                  {i|hf (Xi )=yi }           t=1
R                    hf                               (hf (Xi ) = yi )
            1−|hf (Xi )−yi |
           βt                        hf (Xi ) = 1       yi = 0                 hf (Xi ) = 0
     t=1
                    2
             hf                             2
              hf (Xi ) = 1           yi = 0                              5.1
       hf (Xi ) = 1                yi = 0,          hf(Xi)=1
                     (hf (Xi ) = R i )
                                 y                               R
                                                                              1
Xi ) = 1          yi = 0              h(− log βt )hf (Xi ) y≥ = 1 (− log βt )
                                        f (Xi ) = 0         i
                                    t=1                       t=1
                                                                              2

                        R
 0                      t=1 (log5.1
                                βt )
                           R        R                                R
                                     1                                          1
− log βt )hf (Xi ) ≥     (− log(log βt )(1 − hf (Xi )) ≥
                                βt )                                  (log βt )
                                                                                2
                     t=1    t=1      2                            t=1

            1 − hf (Xi ) = 1− | hf (Xi ) − yi |

                                        R                        R          1/2
                               R
                                     1t f (Xi )−yi | ≥
                                       1−|h
g βt )(1 − hf (Xi )) ≥     (log βt ) β                                 βt
                                t=1 2                            t=1
                       t=1
t                              t
                           t=1                             t=1

      hf (Xi )hf (Xi ) = 0 yi = i1= 1
               =0              y                                         5.1
                          R                               R
                                                                 1
                           (− log βt )ht (Xi ) <     (− log βt )
                       t=1                       t=1
                                                                 2
174                  −1                  hf (Xi ) = 1− | ht (Xi ) 5 yi |
                                                                  −
                                                                         1
                          R                                   R          2
                                 1−|ht (Xi )−yi |
                                βt                    >             βt
                          t=1                                 t=1

                                  hf (Xi ) = 1                yi = 0           hf (Xi ) =
yi = 1
                                                                         1
                          R                                   R          2
                                 1−|ht (Xi )−yi |
                                βt                    ≥             βt
                          t=1                                 t=1
5.2
                                                                                                              1   
                          R                                                                    R               2
                                 1−|hf (Xi )−yi |                               D(i)                              
                   D(i)         βt                      ≥                                            βt
{i|hf (Xi )=yi }          t=1                                {i|hf (Xi )=yi }                  t=1
                                                                                                                 1
                                                                                                   R               2

                                                        =                          D(i)                 βt
                                                                 {i|hf (Xi )=yi }               t=1



                                                                                         1
                                                                                 R         2
                                 R+1
                                wi   ≥                            D(i)              βt
             {i|hf (Xi )=yi }                  {i|hf (Xi )=yi }                 t=1
                                                               1
                                                    R          2

                                       =     ·          βt
                                                  t=1




      5.2
= ·
                              t=1            βt
                                       t=1

•         2
5.2 •   5.2        N           N
                       t+1
                      wi N ≥       t
                                  wi   N×     2 i
                  i=1
                             t+1
                           wi i=1≥           wi × 2
                                               t
                                                      i
 •
: α≥0         r = {0, 1}
                        i=1            i=1

        : α≥0       r = {0, 1}
                     αr ≤ 1 − (1 − α)r
                           αr ≤ 1 − (1 − α)r
5.6.                                                                                     175
       N              N
                                1−|ht (Xi )−yi |
              t+1
             wi   =          t
                            wi βt
       i=1            i=1
                       N
                 ≤          wi (1 − (1 − βt )(1 − |ht (Xi ) − yi |))
                             t

                      i=1
                       N                           N          N
                 =          wi − (1 − βt )
                             t                          t
                                                       wi −        wi |ht (Xi ) − yi |
                                                                    t

                      i=1                      i=1            i=1
                       N                        N                N
                 =          wi − (1 − βt )
                             t                          t
                                                       wi −   t
                                                                         t
                                                                        wi
                      i=1                      i=1                i=1
                       N                        N
                 =          wi − (1 − βt )
                             t                          t
                                                       wi   (1 − t )
                      i=1                      i=1
                        N
                 =             t
                              wi    × (1 − (1 − βt )(1 − t ))
                       i=1
                       βt =     t /(1   − t)
                          N
                 =             t
                              wi    ×2    t
                       i=1
•
5.3                 t                                      WeakLearner
          •
          t
               εt        t                                WeakLearner
                                          hf                                         hf                 ε
                                                     R
                                            ≤             2        t (1   − t)
      •                      1/2
                                                 t=1
              R                             N                        N                                 N          R
:         5.1 βt                     ≤ 5.2 wi
                                            R+1
                                                ≤                          R
                                                                          wi          ×2   t   ≤              1
                                                                                                             wi         2   t
              t=1                                                                                                 t=1
                         R           1/2 i=1 N                      i=1                                i=1
                                   N

                              βt         wi = 1
                                          1
                                            ≤              R+1
                                                          wi   (                     5.1           )
                        t=1        i=1
                                                 i=1
                                            R
                                                   N
                                     =           2
                                            ≤
                                           t=1
                                                     t          R
                                                               wi         × 2 t(               5.2            )
                                                         i=1
          βt =          t /(1 − t )                  5.2           t = R − 1, R − 2, . . . , 1
                         R                                     R
                                           −1/2 N                         R
              ≤                2 t×       βt    =                12           t (1   − t)
                                           ≤                   wi              2     t
                        t=1                                    t=1
•       3
    •
        •
    •              (Weighted Voting)
•
•   WeakLearner                        WeakLearner
                         NaiveBayes                    K-NN,
    SVM
    •                                                Ensemble
        Learning
4   5   2   A   9   3   1   C

    1                                   8   6   7   B
            2           4   5   2   A
4
                                        4   5   2   A   8   6   7   B
                3       8   6   7   B
8       6
                    5                   9   3   1   C
    9           7
                        9   3   1   C
                                        8   6   7   B   4   5   2   A

                                        9   3   1   C
1                   1   2   3   4   5   6   7   8   9
            2
4                       1   2   3   4   5   6   7   9   8
                3
8       6               1   2   3   4   5   6   8   9   7
                    5
    9           7       1   2   3   4   5   7   8   9   6

                        1   2   3   4   6   7   8   9   5

                        1   2   3   5   6   7   8   9   4

                        1   2   4   5   6   7   8   9   3

                        1   3   4   5   6   7   8   9   2

                        2   3   4   5   6   7   8   9   1
AdaBoost Classification Algorithm

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AdaBoost Classification Algorithm

  • 1.
  • 2. (AdaBoost) • Naive Bayes Yes, No ○ × • ○ × 1 0 • 3 • (Weighted Voting) • • • • • •
  • 3. 1 0 • N • (Xi , ci )(i = 1, . . . , N ) C XC = (No, Yes, Yes, Yes), cC = 1
  • 4. R: • • • 1 i wi • N wi = 1/N (i = 1, . . . , N ) 1 • 10 wi = 1/10 1
  • 5. t=1,...,R 1. : t i t pt i pt = wi i N t i=1 wi N • p1 = wi = 1/10 1 t=1 t wi =1 i i=1 2. WeakLearner WeakLearner ( t < 1/2 ) ht t N t = pt |ht (Xi ) − ci | < 1/2 i i=1 0, ht (Xi ) ci ht (Xi ) − ci = 1, Step 3
  • 6. t=1 WeakLearner ID A F h1(Xi) = 1 E,F h1(Xi)=ci D,J ci=0 ID G,H,I,J h1(Xi) = 0 WeakLearner 10 2 p1 = 1/10 i 1 = 1/10 × 2 = 1/5 < 1/2 T2 WeakLearner
  • 7. t+1 3. wi βt t βt = 0≤ < 1/2, 0 ≤ βt < 1 1− t t • β 1−|ht (Xi )−ci | wi = wi βt t+1 t • εt βt • εt βt • WeakLearner ht
  • 8. t=1 1 = 0.2 1 β1 = = 0.2/0.8 = 0.25 1− 1 • A D, G J ht E, F 2 wA = wB = wC = wD = wG = wH = wI = wJ 2 2 2 2 2 2 2 = 1/10 × β1 = 0.025 1 2 wE = wF = 1/10 × β1 = 0.1 2 0
  • 9. WeakLearner • t=1 WeakLearner T2 • WeakLearner • WeakLearner • WeakLearner • t=1 • T1 Yes ε1 ε1(T1=Yes)
  • 10. p1 = 1/10(i ∈ {A, B, ..., J}) i 1 1 (T1 = Yes) = × 4 = 0.4 10 1 1 (T2 = Yes) = × 2 = 0.2 10 1 1 (T3 = Yes) = × 3 = 0.3 10 1 1 (T4 = Yes) = × 2 = 0.2 10 T2=Yes T4=Yes T2
  • 11. t=2 wA2 = wB = wC = wD = wG = wH = wI = wJ 2 2 2 2 2 2 2 = 1/10 × β1 = 0.025 1 2 wE = wF = 1/10 × β1 = 0.1 2 0 wi = 0.400 2 i∈{A,...,J} p2 = p2 A B = p2 = p2 = p2 = p2 = p2 = p2 C D G H I J = 0.025/0.400 = 0.0625 p2 = p2 E F = 0.1/0.400 = 0.25 • WeakLearner 2 (T1 = Yes) = 0.0625 × 2 + 0.25 × 2 = 0.625 2 (T1 = No) = 0.375 2 (T2 = Yes) = 0.25 + 0.25 = 0.5 2 (T3 = Yes) = 0.0625 × 2 + 0.25 × 1 = 0.3125 2 (T4 = Yes) = 0.0625 × 2 = 0.125 T4=Yes
  • 12. β2 = 2 /(1 − 2) = 0.143
  • 13. R 1 if t=1 (− log10 βt )ht (X) ≥ (− log10 βt ) 1 hf (X) = 2 0 othrewise t βt = 1− t • 1, 0 • Log • εt 0 βt -Log βt WeakLearner ht(X) • -Log βt ht(X)
  • 14. β1 = 0.25, β2 = 0.143 2 1 (− log βt ) t=1 2 1 1 = − log10 0.25 × − log10 0.143 × 2 2 = 0.723 • 1.
  • 15. h1: T2=Yes C=1, h2: T4=Yes C=1 2 • K (− log βt )ht (XK ) t=1 = − log10 0.25 × 0 − log10 0.143 × 1 = 0.845 0.723 ○ 2 • L (− log βt )ht (XL ) t=1 = − log10 0.25 × 1 − log10 0.143 × 0 = 0.602 0.723 ×
  • 16. AdaBoost • hf ε = D(i) {i|hf (Xi )=yi } • D(i) • R εt t ≤ 2 t (1 − t) t=1 • t εt<1/2 2 t (1 − t) < 1 • WeakLearner ε
  • 17. 1 • R R+1 N R 1/2 R+1 wi ≥ βt i=1 t=1 • • N R+1 R+1 wi ≥ wi i=1 {i|hf (Xi )=yi } t 1−|hf (Xi )−yi | t+1 wi = wi βi R 1−|hf (Xi )−yi | t+1 wi = D(i) βt {i|hf (Xi )=yi } {i|hf (Xi )=yi } t=1
  • 18. R hf (hf (Xi ) = yi ) 1−|hf (Xi )−yi | βt hf (Xi ) = 1 yi = 0 hf (Xi ) = 0 t=1 2 hf 2 hf (Xi ) = 1 yi = 0 5.1 hf (Xi ) = 1 yi = 0, hf(Xi)=1 (hf (Xi ) = R i ) y R 1 Xi ) = 1 yi = 0 h(− log βt )hf (Xi ) y≥ = 1 (− log βt ) f (Xi ) = 0 i t=1 t=1 2 R 0 t=1 (log5.1 βt ) R R R 1 1 − log βt )hf (Xi ) ≥ (− log(log βt )(1 − hf (Xi )) ≥ βt ) (log βt ) 2 t=1 t=1 2 t=1 1 − hf (Xi ) = 1− | hf (Xi ) − yi | R R 1/2 R 1t f (Xi )−yi | ≥ 1−|h g βt )(1 − hf (Xi )) ≥ (log βt ) β βt t=1 2 t=1 t=1
  • 19. t t t=1 t=1 hf (Xi )hf (Xi ) = 0 yi = i1= 1 =0 y 5.1 R R 1 (− log βt )ht (Xi ) < (− log βt ) t=1 t=1 2 174 −1 hf (Xi ) = 1− | ht (Xi ) 5 yi | − 1 R R 2 1−|ht (Xi )−yi | βt > βt t=1 t=1 hf (Xi ) = 1 yi = 0 hf (Xi ) = yi = 1 1 R R 2 1−|ht (Xi )−yi | βt ≥ βt t=1 t=1
  • 20. 5.2  1  R R 2 1−|hf (Xi )−yi | D(i)  D(i) βt ≥ βt {i|hf (Xi )=yi } t=1 {i|hf (Xi )=yi } t=1   1 R 2 = D(i) βt {i|hf (Xi )=yi } t=1   1 R 2 R+1 wi ≥ D(i) βt {i|hf (Xi )=yi } {i|hf (Xi )=yi } t=1 1 R 2 = · βt t=1 5.2
  • 21. = · t=1 βt t=1 • 2 5.2 • 5.2 N N t+1 wi N ≥ t wi N× 2 i i=1 t+1 wi i=1≥ wi × 2 t i • : α≥0 r = {0, 1} i=1 i=1 : α≥0 r = {0, 1} αr ≤ 1 − (1 − α)r αr ≤ 1 − (1 − α)r
  • 22. 5.6. 175 N N 1−|ht (Xi )−yi | t+1 wi = t wi βt i=1 i=1 N ≤ wi (1 − (1 − βt )(1 − |ht (Xi ) − yi |)) t i=1 N N N = wi − (1 − βt ) t t wi − wi |ht (Xi ) − yi | t i=1 i=1 i=1 N N N = wi − (1 − βt ) t t wi − t t wi i=1 i=1 i=1 N N = wi − (1 − βt ) t t wi (1 − t ) i=1 i=1 N = t wi × (1 − (1 − βt )(1 − t )) i=1 βt = t /(1 − t) N = t wi ×2 t i=1
  • 23. • 5.3 t WeakLearner • t εt t WeakLearner hf hf ε R ≤ 2 t (1 − t) • 1/2 t=1 R N N N R : 5.1 βt ≤ 5.2 wi R+1 ≤ R wi ×2 t ≤ 1 wi 2 t t=1 t=1 R 1/2 i=1 N i=1 i=1 N βt wi = 1 1 ≤ R+1 wi ( 5.1 ) t=1 i=1 i=1 R N = 2 ≤ t=1 t R wi × 2 t( 5.2 ) i=1 βt = t /(1 − t ) 5.2 t = R − 1, R − 2, . . . , 1 R R −1/2 N R ≤ 2 t× βt = 12 t (1 − t) ≤ wi 2 t t=1 t=1
  • 24. 3 • • • (Weighted Voting) • • WeakLearner WeakLearner NaiveBayes K-NN, SVM • Ensemble Learning
  • 25.
  • 26.
  • 27.
  • 28. 4 5 2 A 9 3 1 C 1 8 6 7 B 2 4 5 2 A 4 4 5 2 A 8 6 7 B 3 8 6 7 B 8 6 5 9 3 1 C 9 7 9 3 1 C 8 6 7 B 4 5 2 A 9 3 1 C
  • 29. 1 1 2 3 4 5 6 7 8 9 2 4 1 2 3 4 5 6 7 9 8 3 8 6 1 2 3 4 5 6 8 9 7 5 9 7 1 2 3 4 5 7 8 9 6 1 2 3 4 6 7 8 9 5 1 2 3 5 6 7 8 9 4 1 2 4 5 6 7 8 9 3 1 3 4 5 6 7 8 9 2 2 3 4 5 6 7 8 9 1