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060310391

0560565

2

2016/10/3 13:00-14:45

@1 - 4

• 

• 

• 

– 

– 

– 

– 

• 

(2)

h/p://konicaminolta.jp/instruments/knowledge/color/part2/07.html

R 151

G 109

B 121

182 × 167 × 3

= 91182

pixel

256 0-255 3 256^3=16 777 216

•  RGB: Red, Green, Blue

–  3

•  CMY: Cyan, Magenta, Yellow

–  3

•  HSV: Hue SaturaRon Value

– 

•  HSL: Hue SaturaRon Luminance

–  100%

100% 0%

h/p://homepage2.niUy.com/studio_AURK/ccconv/Color/hue.html

(3)

•  RGB CMY

C = 255 – R

M = 255 – G

Y = 255 – B

•  RGB HSV

max,min RGB

V = max

S = 255 × (max – min) / max

R max

H = 60 × (B – G) / (max – min)

G max

H = 60 × (R – B) / (max – min)

B max

H = 60 × (G – R) / (max – min)

H H ← H + 360

• 

50×50

50 × 50 × 3 = 7500

0

255

1

(R, G, B) = (23, 14, 22)

(R, G, B) = (200, 198, 168)

(R, G, B) = (201, 200, 173)

(R, G, B) = (200, 199, 172)

50 × 50 × 3 = 7500

7500

Principal component analysis (PCA)

• 

50 × 50 × 3 = 7500

7500

• 

• 

(4)

2 1

u

1

x

1

, x

2

1 u

1

u

1

u

1

u

1

x

1

x

2

u

1

u

1

= u

1

u

1

x

1

x

2

ε

i

= y

i

− (α + βx

i

)

( 2

SSE

= ε

i2

i n

= (y

i

α − βx

i

)

2 i

n

:

∂SSE

∂β =−2 i (yiα − βxi)xi

n

= 0

∂SSE

∂α =−2 i (yiα − βxi)

n

= 0

a = yi

i

n n− b ixi

n n = y − bx

b =

xiyi

i

n ixi

n iyi

n n

xi2xi

i

n

( )

2 n

i

n

= i(xi− x )(yi− y )

n

(xi− x )2

i

n

n xi

i

n

xi

i

n xi 2 i

n

#

$

%

%

&

' ( ( a b

#

$ % & ' ( = iyi

n

xiyi

i

n

#

$

%

%

&

' (

-3 -2 -1 0 1 2 3 (

455055

x

y

yi ˆ

y i=

α

+

β

xi xi

ε

i

15

(a)  y

(b)

1

x,y

16

(5)

2 1

u

1 T

u

1

= u

11 2

+ u

12 2

= 1

1

n −1 z

1i

2

i=1 n

z1i

= x

i Tu

1

= x (

1i x2i

)

uu11

12

"

# $ %

&

' = u

11x1i

+ u

12x2i

(u1

x 0

u

1

x

i Z1i = u1 xi

Z1i

1 n−1

x1i2

i=1 n

x1ix2i i=1

n

x1ix2i i=1

n

x2i 2 i=1

n

$

%

&

&

&

&

'

( ) ) ) )

u11

u12

$

% & ' ( ) = λ

u11

u12

$

% & ' ( ) L(u

1,λ) =

1 n−1 (u11

x1i+ u12x2i)

2λ(u112+ u122 −1)

i=1 n

∂L

∂u11= 1

n−1i 2(u11x1i+ u12x2i)x1i− 2λu11= 0

=1 n

∂L

∂u12= 1

n−1 2(u11x1i+ u12x2i)x2i− 2λu12= 0

i=1 n

1 n−1u11 x1i

2 i=1

n

+ u12 x1ix2i i=1

n

$

% & '

( ) = λu11 1

n−1 u11 x1ix2i

i=1 n

+ u12 x2i 2 i=1

n

$

% & '

( ) = λu12

(V)

{

Vu

1

= λu

1

}

u

1

= 0

[ ]

var(x) = 1

n−1 (xi− x

i=1 n

)2

cov(x, y) = 1

n−1i=1(xi− x

n

)(yi− y ) 2

r = cov(x, y) var(x) var(y)=

(xi− x

i=1 n

)(yi− y )

(xi− x

i=1 n

)2 (yi− y i=1

n

)2

-3 -2 -1 0 1 2 3

-2-10123

length

width

(xi− x )(yi− y )

x y

2 4

×(-1)

2

V

=

6.00 2.03

2.03 2.98

"

# $ %

&

' =

6 2

2 3

"

# $ %

&

'

7, 2 u1=

2 / 5 1/ 5

"

#

$ %

& ' u2=

1/ 5

−2 / 5

#

$

% & ' (

1 2.5 -0.2

2 3.4 3.4

3 1.0 1.9

4 -2.4 -0.8

5 -3.0 -2.4

6 0.1 -1.6

7 -2.5 -0.5

8 -2.9 1.2

9 1.8 0.1

10 1.6 -1.0

-3 -2 -1 0 1 2 3

-2-10123

length

width

(6)

Au = λ u

A 0 u A λ

A n×n A n λ1, λ2, …, λn

u1, u2, …, un

{u1, u2, …, un}

(A − λI)u = 0

u=0 0

0

A − λI = 0

V

=

6 2

2 3

"

# $ %

&

'

6 2

2 3

"

# $ %

&

'

u1

u2

"

# $ %

&

' = λ

u1

u2

"

# $ %

&

'

6 λ 2

2 3 λ

$

% & '

( )

u1

u2

$

% & '

( ) =

0

0

$

% & '

( )

6 λ 2

2 3 λ = 0

(6 −λ)(3 −λ) − 4 = (λ− 2)(λ− 7) = 0

λ=2 2u21+ u22= 0 u212 + u222 = 1 u21 u22

"

# $ %

& ' = 1/ 5

−2 / 5

"

# $ %

& ' λ=7 u11− 2u12= 0 u112+ u122 = 1 u11

u12

"

# $ %

& ' = 2 / 5

1/ 5

"

# $ %

& '

2, 7

u1= u11 u12

"

# $ %

& ' =

2 / 5 1/ 5

"

# $ %

&

' u2=

u11 u12

"

# $ %

& ' =

1/ 5

−2 / 5

"

# $ %

& '

k 2 B

50 × 50 × 3 = 7500

7500

z1, z2, …

• 

• 

z

ij

= u

j1

x

1i

++ u

jM

x

Mi

= x

iT

u

j

j i principal component score

1. 

2.  -3 -2 -1 0 1 2 3

-2-10123

length

width

z11= 2.5( −0.2)2 / 5

1/ 5

#

$

% & ' ( = 2.2

z12=

(

2.5 −0.2

)

1 / 5

−2 / 5

= 1.3

PC1 PC2

1 2.5 -0.2 2.2 1.3

2 3.4 3.4 4.6 -1.5

3 1.0 1.9 1.8 -1.2

4 -2.4 -0.8 -2.5 -0.3

5 -3.0 -2.4 -3.7 0.8

6 0.1 -1.6 -0.6 1.5

7 -2.5 -0.5 -2.4 -0.6

8 -2.9 1.2 -2.0 -2.3

9 1.8 0.1 1.7 0.7

10 1.6 -1.0 1.0 1.6

0

(7)

1 N − 1 zji

2 i N

= 1 N − 1

zj Tz

j= 1

N − 1(Xuj)

T(Xuj)

= 1

N − 1 uj

TXTXu j= uj

TVu j= λjuj

Tu j= λj

Vuj=λjuj

V eigenvalue λ1≥λ2≥  ≥λM≥ 0 ui

zj j

λ

j

/ λ

i

i=1 M

λi

i=1 j

/ λi

i=1 M

j contribuRon cumulaRve

1 M −1X

TX= V

x 0

i λi

x

0.77 0.77

0.33 1.00

var(z

1

) = 7.0

PC1 PC2

1 2.5 -0.2 2.2 1.3

2 3.4 3.4 4.6 -1.5

3 1.0 1.9 1.8 -1.2

4 -2.4 -0.8 -2.5 -0.3

5 -3.0 -2.4 -3.7 0.8

6 0.1 -1.6 -0.6 1.5

7 -2.5 -0.5 -2.4 -0.6

8 -2.9 1.2 -2.0 -2.3

9 1.8 0.1 1.7 0.7

10 1.6 -1.0 1.0 1.6

var(l) = 6.0

V

=

6 2

2 3

"

# $ %

&

'

var(w) = 3.0

var(z

2

) = 2.0

• 

• 

1

•  1

• 

2

R =

1 r

r 1

⎝⎜

⎠⎟

R − λI = 0 ⇔ (1− λ)2

− r

2

= 0

∴ λ

1

= 1+ r, λ

2

= 1− r

Ra1

= λ

1a1

y1= 1

2(x1+ x2), y2= 1

2(x1− x2)

∴a

11

= a

12

= 1

2

a2

a11+ ra12= (1+ r)a11 ra11+ a12= (1+ r)a12

(8)

-2s.d. 平均 +2s.d. PC1

PC2

PC3

PC4

*** 24.05

*** 57.28

*** 119.71

*** 48.27 品種効果

自由度174, 875

主成分

寄与率

(43.5%)

(15.0%)

(9.3%)

(4.5%)

z

iq

= x

iT

u

q

(z

i1

,..., z

iq

) = x

iT

(u

1

,..., u

q

) z

iT

= x

iT

U ⇔ z

i

= U

T

x

i

z

i

1

= x

iT

u

1

UTU = u1

Tu

1 u1

Tu q

  

uqTu

1 uq

Tu q

⎢⎢

⎥⎥

= I

U

T

= U

1

⇔ UU

T

= I

∴x

i

= Uz

i

= 2 x T 2

xi = U k λ1

0

 0

⎜⎜

⎜⎜

⎟⎟

⎟⎟

2

2

eigenfaces

Murphy KP (2012) "Machine Learning: A ProbabilisRc PerspecRve" The MIT Press.

(9)

102

30

30

• 

•  A B

–  100 100

A B 100/100 = 1

–  100 0

A B 0/100 = 0

–  100 65

A B 65/100 = 0.65

Principal co-ordinate analysis (PCO)

• 

h/p://aoki2.si.gunma-u.ac.jp/lecture/misc/princo-ex.html

(10)

B = XX

T

•  q n

•  n q n×q X i

j ij

b

ij

= x

ik

x

jk

k=1

q

•  x bij i j

d

ij2

= (x

ik

− x

jk

)

2

k=1

q

= x

ik

2 k=1

q

+ x

ik 2 k=1

q

− 2

k=1

x

ik

x

jk

q

= b

ii

+ b

jj

− 2b

ij

•  i j dij

b

ij

i=1

n

=

k=1

x

ik

x

jk

q

=

i=1

n

x

jk i=1

x

ik

k

= 0

k =1

q

{

•  dij

2 i=1

n = i=1(bii+ bjj− 2bij)

n = i=1bii

n + nbjj

dij 2 j=1

n = j=1(bii+ bjj− 2bij)

n = j=1bjj

n + nbii= i=1bii

n + nbii

dij2

j=1

n i=1

n = j=1(bii+ bjj− 2bij)

n i=1

n = n i=1bii

n + nn j=1bjj

n = 2n i=1bii

n

• 

di.2=1 n dij

2 j=1

n =1 n i=1bii

n + bii

di.2=1 n dij

2 i=1

n =1 n i=1bii

n + bjj

d.. 2= 1

n2 dij

2 j=1

n i=1

n =2 n i=1bii

n

bij= −1 2(dij

2− di. 2− d.j

2+ d.. 2)

X 2 2

X ,

A(u1

,..., u

q

) = (u

1

,..., u

q

)

λ

1

0

0 λ

q

⎢ ⎢

⎥ ⎥

Au

1

= λ

1

u

1

Au

q

= λ

q

u

q

AU = UΛ

u

i T

u

j

= 0

u

i

T

u

i

= 1, U

T

U = I ⇔ U

T

= U

1

⇔ UU

T

= I

∴A = UΛU

T

u

*i

= λ

i

u

i

A = U

*

U

*T U*

= (u

1

*

,..., u

*q

)

B = XX

T X

X = UΛ

1/2

Λ

1/2

= diag( λ

11/2

,…, λ

q1/2

)

B

(11)

感覚距離 PC2

PC4 主成分得点の差 0123

0.2 0.4 0.6 0.8 1.0

012345

1

( )

PC1 PC2 PC3 PC4 PC5 PC6 PC7 PC8 PC9 PC10

PCO1 -1.0-0.50.00.51.0

PC1 PC2 PC3 PC4 PC5 PC6 PC7 PC8 PC9 PC10

PCO2 -1.0-0.50.00.51.0

u

2

3

4 1

x x

(12)

• 

–  3 4

• 

– 

h/p://konicaminolta.jp/instruments/knowledge/color/part2/06.html

II. :

(m) 10 2

1

10 -2

10 -4

10 -6

10 -8

10 -10

10 -12

10 -14







 





 FM



780

700

600

500

400

380

(nm)

780

700

600

520

440

360

(nm)

< >

(13)

< >

Brassica rapa L.

Brassica napus L.

rapa

< >

Brassica rapa L.

Brassica napus L.

: Nikon D70

: Ultra AchromaRc Takumar

• 

Syafaruddin et al. (2006) Breed Sci 56:75-79

f (x)

g( x) = 0

L(x, λ ) ≡ f (x) + λ g(x)

x

(Lagrangian)

(Lagrange mulRplier)

∇L = ∇f + λ ∇g = 0

(14)

h/p://www.isigas.com/LagrangeMP.html

g(x

1

, x

2

) = 0

f (x

1

, x

2

)

∇g ∇f

g(x)=0 f(x)

∇f − λ ∇g = 0

λ

g(x

1

, x

2

) = x

1

+ x

2

−1 = 0

f (x

1

, x

2

) = 1− x

12

− x

22

L (x, λ ) = 1− x

1

2

− x

22

+ λ (x

1

+ x

2

−1)

x

L

x

1

= −2x

1

+ λ = 0

L

x

2

= −2x

2

+ λ = 0

L

∂λ = x

1

+ x

2

−1 = 0

x

1

=

1

2 ,

x

2

=

1

2 , λ = 1

3

z

1i

= u

11

x

1i

+ + u

1m

x

mi

= u

1

T

x

i

V (z

1

) = 1

n −1 (z

1i

− z

1

)

2 i=1

n

= s

jk

u

1j

u

1k

= u

1 T

Vu

1 k =1

m

j =1 m

sjk xj xk

u

T

u = u

112

+  + u

1M2

= 1

z1

L(u

1

,λ) = u

1 T

Vu

1

λ(u

1 T

u

1

−1)

∂L

∂u

1

= 2Vu

1

− 2λu

1

= 0 (V − λ I)u

1

= 0

V(z2) = u2TVu2

L(u

2

,λ,ν ) = u

2 T

Vu

2

λ(u

2 T

u

2

−1) − ν (u

2 T

u

1

)

3 ...

x

1 1

z

2i

= u

21

x

1i

+ + u

2m

x

mi

= u

2

T

x

i

u

2T

u

2

= u

212

+  + u

2M2

= 1

u

2

T

u

1

= u

21

u

11

+  + u

2M

u

1M

= 0

=0 →

∂L

∂u

2

= 2Vu

2

− 2λu

2

νu

1

= 0 u

2

T

u

1

= u

1 T

u

2

= 0, u

1 T

u

1

= 1

ν = 0

(V − λI)u = 0

2

(15)

1,2

Q.

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