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•               (
    Classification, Pattern Recognition)

    •
    •
    •      A
(3/3)
•                       (Clustering)
    •
    •   A                 B
    •           A            B
•                   (Association Rules)
    •
    •       A                  B
    •
•
•




    9
•




        K


    L


            10
•
    •
    •
•
    •


                    T                T’
              Yes       No     Yes        No




        (A)                  (B)               11
•   T     v∈T                   cost(v)           v       T
                 .
           
               {cost(x) | x ∈ T is leaf}
5.4.                                                                 133

                                  X1


                         X2                X3


                     1        0        1            X4
                : 2           2        2
                                                1            0

                                                3            3        12
13
5.1
                EXACT COVER BY 3-SET
 •          NP                                 NP
      •     NP             NP            EXACT EXACT COVER BY 3-SET
                                               COVER BY 3-SET
                                                        EXACT COVER BY 3-SET

 •     EXACT COVER BY 3-SET
            5.2 3                                        X     3                X
                     S = {T 1, T 2, ...}                                         S1 ⊂ S
                                                NP
         (1) ∪{T |T ∈ S1 } = X
      5.4.                           X                  S1                              135
134                                                 5
          (2)         i=j                  Ti ∩ Tj = φ S 1

          5.4     EXACT COVER BY 3-SET
                         1     2     3                        {{1, 4, 7}, {2, 3, 5}, {6, 8, 9}}
                                    X                                   EXACT COVER
      BY 3-SET             4         5              6        EXACT COVER
           :                      NP                          BY 3-SET
                           7               8        9

                                                                                                  14
BY 3-SET
    •:                                 NP
            •
            •
    X                                            1                               |X|
            •       X Y = {y1 , y2 , ..., y|X| }                    |X|                        0   Y
        t           X            Y             Y = {y1 , y2 , . . . X |X| }
                                                                    ,y                    1            Y
            •       t 0X               Y                                     X            1    Y           0

                                                         1    t∈X
                                           t[A] =
                                                         0    t∈Y

X               3                                         T1 , T 2 , . . .
    1                              0                                                      yi
                           1

                                           1    t ∈ Ti                       1   t = yi
0 t∈Y
•X X     3
                  3
                                                      T1 , T 2 , . . . T1 , T 2 , . . .
    1                    1        0                  0                                    yi
•   yi                  1                  1
                                          1 t ∈ Ti                   1 t = yi
                       t[Ti ] =                      , t[yi ] =
                                          0 t ∈ Ti                   0 t = yi

•
    Ti       yi
•
•                 2
    •                                                    Ti
         9                            3                                    5.5(A)
    •                 5.5(B)
•
    •                      |X|
                                   |X| |X|
                     1 + 2 + ··· +    +
                                    3   3
        •
        •         EXACT COVER BY 3-SET
            136    EXACT COVER BY 3-SET5
                                           :              :
Y                     T1         T2   T3
                                               yi                       yi
                                                     T1
            |Y | = |X|                                                  9
            5.6(A)                             1          T2




                                                     1         T3




                     1 + 2 + · · · + |X| + |X|1                     0




                     (A)                       (B)
5.6(A)
•                            Ti             yi        yi             yi
          |Y| = |X|
•                1 + 2 + · · · + |X| + |X|

•     2                                 1 + 2 + ··· +
                                                      |X| |X|
                                                         +
                                                       3   3
•                 EXACT COVER BY 3-SET
                                       EXACT COVER BY 3-SET
                             1 + 2 + · · · + |X|/3 + |X|/3
                                                                                              NP
          5.4.                                                                     137

                                                           y1




                      5.4.2                      0              y2


                      y1          y2   y3                                                NP
                                                           0




                                                                          y9



                                                                     0         1


                       (A)                           (B)
19
T                T’
      Yes       No     Yes        No




(A)                  (B)
                                       20
S = {(x1 , c1 ), (x2 , c2 ), . . . , (xN , cN )}


     H(C) = −p log2 p − p× log2 p×

     p   p×
p
                                                             21
•                     4        6
              p   =    , p× =
                     10        10
• H(C)   = −p log2 p − p× log2 p×
             4       4   6       6
         = − log2      −   log2     = 0.971
            10      10 10        10




                                              22
•




             30
                  YES   NO
    C:             2    2    4
         ×         2    4    6
                   4    6    10   23
T1: 30
                         YES       NO
   C:                     2         2       4
                 ×        2         4       6
                          4         6      10
                           2       2 2    2
  H(C | T1 = Yes)    =    − log2    − log2 = 1.0
                           4       4 4    4
                           2       2 4    4
   H(C | T1 = No)    =    − log2    − log2 = 0.918
                           6       6 6    6
  •
              4                   6
H(C | T1 ) =    H(C | T1 = Yes) + H(C | T1 = No) = 0.951
             10                  10

                                                      24
•        T                                 I(T)

             I(T ) = H(C) − H(C | T )
•
    I(T1 ) = H(C) − H(C | T1 ) = 0.971 − 0.951 = 0.020
•
    I(T2 ) = 0.420, I(T3 ) = 0.091, I(T4 ) = 0.420
•                                         T2
                       T2
    •   T4
        T2

                                                     25
T2:

        Yes         No




•
    •
    •
•
•    Yes


            4    4 2    2
    H(C) = − log2 − log2 = 0.918
            6    6 6    6



                                                           
           T1                      4     2    2 2       2
                      H(C | T1 ) =     − log2 − log2
                                   6 4       4 4      4
                                   2     2    2
                                       − log2
                                   6     2    2
                                 = 0.667
                         I(T1 ) = 0.918 − 0.667 = 0.251
                          I(T3 ) = 0, I(T4 ) = 0.918
                                               T4           27
•
•
    •
        •   naive bayes
    •
        •
    •
•
    •
    •
        •
    •                     29
•
•
    •
    •   ID3
                                               2


•
    •
    •   CART (Classification And Regression Tree)   C4.5


                                                          30
•   CART
    •           2
    •
    •
•   C4.5
    •
    •
•
    •
        •
    •
            Forest
                     31
(10/21)

•              sesejun+dm10@sel.is.ocha.ac.jp
•
•            11/2(   )
•
    •   http://togodb.sel.is.ocha.ac.jp/        22
                                       2010



                                                     32
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  • 3.
  • 4. (1/3) • • • • •
  • 5. (2/3) • ( Classification, Pattern Recognition) • • • A
  • 6. (3/3) • (Clustering) • • A B • A B • (Association Rules) • • A B •
  • 7.
  • 8.
  • 10. K L 10
  • 11. • • • • T T’ Yes No Yes No (A) (B) 11
  • 12. T v∈T cost(v) v T . {cost(x) | x ∈ T is leaf} 5.4. 133 X1 X2 X3 1 0 1 X4 : 2 2 2 1 0 3 3 12
  • 13. 13
  • 14. 5.1 EXACT COVER BY 3-SET • NP NP • NP NP EXACT EXACT COVER BY 3-SET COVER BY 3-SET EXACT COVER BY 3-SET • EXACT COVER BY 3-SET 5.2 3 X 3 X S = {T 1, T 2, ...} S1 ⊂ S NP (1) ∪{T |T ∈ S1 } = X 5.4. X S1 135 134 5 (2) i=j Ti ∩ Tj = φ S 1 5.4 EXACT COVER BY 3-SET 1 2 3 {{1, 4, 7}, {2, 3, 5}, {6, 8, 9}} X EXACT COVER BY 3-SET 4 5 6 EXACT COVER : NP BY 3-SET 7 8 9 14
  • 15. BY 3-SET •: NP • • X 1 |X| • X Y = {y1 , y2 , ..., y|X| } |X| 0 Y t X Y Y = {y1 , y2 , . . . X |X| } ,y 1 Y • t 0X Y X 1 Y 0 1 t∈X t[A] = 0 t∈Y X 3 T1 , T 2 , . . . 1 0 yi 1 1 t ∈ Ti 1 t = yi
  • 16. 0 t∈Y •X X 3 3 T1 , T 2 , . . . T1 , T 2 , . . . 1 1 0 0 yi • yi 1 1 1 t ∈ Ti 1 t = yi t[Ti ] = , t[yi ] = 0 t ∈ Ti 0 t = yi • Ti yi • • 2 • Ti 9 3 5.5(A) • 5.5(B)
  • 17. • |X| |X| |X| 1 + 2 + ··· + + 3 3 • • EXACT COVER BY 3-SET 136 EXACT COVER BY 3-SET5 : : Y T1 T2 T3 yi yi T1 |Y | = |X| 9 5.6(A) 1 T2 1 T3 1 + 2 + · · · + |X| + |X|1 0 (A) (B)
  • 18. 5.6(A) • Ti yi yi yi |Y| = |X| • 1 + 2 + · · · + |X| + |X| • 2 1 + 2 + ··· + |X| |X| + 3 3 • EXACT COVER BY 3-SET EXACT COVER BY 3-SET 1 + 2 + · · · + |X|/3 + |X|/3 NP 5.4. 137 y1 5.4.2 0 y2 y1 y2 y3 NP 0 y9 0 1 (A) (B)
  • 19. 19
  • 20. T T’ Yes No Yes No (A) (B) 20
  • 21. S = {(x1 , c1 ), (x2 , c2 ), . . . , (xN , cN )} H(C) = −p log2 p − p× log2 p× p p× p 21
  • 22. 4 6 p = , p× = 10 10 • H(C) = −p log2 p − p× log2 p× 4 4 6 6 = − log2 − log2 = 0.971 10 10 10 10 22
  • 23. 30 YES NO C: 2 2 4 × 2 4 6 4 6 10 23
  • 24. T1: 30 YES NO C: 2 2 4 × 2 4 6 4 6 10 2 2 2 2 H(C | T1 = Yes) = − log2 − log2 = 1.0 4 4 4 4 2 2 4 4 H(C | T1 = No) = − log2 − log2 = 0.918 6 6 6 6 • 4 6 H(C | T1 ) = H(C | T1 = Yes) + H(C | T1 = No) = 0.951 10 10 24
  • 25. T I(T) I(T ) = H(C) − H(C | T ) • I(T1 ) = H(C) − H(C | T1 ) = 0.971 − 0.951 = 0.020 • I(T2 ) = 0.420, I(T3 ) = 0.091, I(T4 ) = 0.420 • T2 T2 • T4 T2 25
  • 26. T2: Yes No • • • •
  • 27. Yes 4 4 2 2 H(C) = − log2 − log2 = 0.918 6 6 6 6 T1 4 2 2 2 2 H(C | T1 ) = − log2 − log2 6 4 4 4 4 2 2 2 − log2 6 2 2 = 0.667 I(T1 ) = 0.918 − 0.667 = 0.251 I(T3 ) = 0, I(T4 ) = 0.918 T4 27
  • 28.
  • 29. • • naive bayes • • • • • • • • 29
  • 30. • • • • ID3 2 • • • CART (Classification And Regression Tree) C4.5 30
  • 31. CART • 2 • • • C4.5 • • • • • • Forest 31
  • 32. (10/21) • sesejun+dm10@sel.is.ocha.ac.jp • • 11/2( ) • • http://togodb.sel.is.ocha.ac.jp/ 22 2010 32
  • 33. 33