-Artificial Neural Network- Chapter 9 Self Organization Map(SOM)

朝陽科技大學資訊管理系李麗華 教授

朝陽科技大學 李麗華 教授 2

Introduction• It’s proposed by Kohonen in 1980.• SOM is an unsupervised two layered network that can o

rganize a topological map from a random starting point.

SOM is also called Kohnen’s self organizating feature map.

The resulting map shows the natural relationships among the patterns that are given to the network.

• The application is good for clustering analysis.

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Network Structure– One input layer– One competitive layer which is usually a 2-Dim grid

Input layer ： f(x) ： xOutput layer ： competitive layer with topological map relationship Weights ： randomly assigned

X1X2

Yk1 Yk2Ykj

j

Ykj

Y1kY11 Y12

Yjk

（ Xij,Yij ）

W1jk Wijk

k

朝陽科技大學 李麗華 教授 4

Concept of NeighborhoodCenter ： the winning node C is the center.Distance ：

R Factor （鄰近係數）： RFj ： f （ rj,R ） =e(- rj /R)

e(-rj/R)→ 1 when rj = ∮ e(-rj/R)→ when r∮ j= ∞

e(-rj/R)→ 0.368 when rj = R The longer the distance, the smaller the neighborhood area.

R Factor Adjustment ： Rn=R-rate Rn-1 , R-rate<1.0

22yyxxj CNCNr N is the node is output N

to C rj is the distance from N to C

R ： the radius of the neighborhood rj ： distance from N to C

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Learning1. Setup network2. Randomly assign weights to W3. Set the coordinate value of the output layer N （ x,y ）4. Input a training vector X5. Compute the winning node6. Update weight W with R factor△7. ηn=η-rate ηn-1

Rn=R-rate ηn-1

8. Repeat from 4 to 7 until converge

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Reuse the network

1. Setup the network

2. Read the weight matrix

3. Set the coordriate value of the output layer N （ x,y ）

4. Read input vector

5. Compute the winning node

6. Output the clustering result Y.

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Computation process

X1~Xn （ Input Vector ） Njk （ Output ）

jkkjkjnetnet

min**

i

ijkijk WXnet 2

1 j=j*&k=k* ifφ others

3. Yjh =

2. Compute winning node

1. Setup network

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Computation process (cont.)

△Wijk=η （ Xi － Wijk ）． RF

jk

When Rr

JKjkeRF /

Wijk= W△ ijk+Wijk

rjk= 222** *)(*)( kkjjNN kjjk

4. Update Weights

7. Repeat until converge

ηn=η-rate‧ηn-1 Rn= R-rate R‧ n-16.

5. Repeat 1-4 for all input

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Example• Let there be one 2-Dim clustering problem.

• The vector space include 5 different clusters and each has 2 sample sets.

X1 X2

1 -0.9 -0.8

2 -0.8 0.6

3 0.9 0.6

4 0.7 -0.4

5 -0.2 0.2

6 -0.7 -0.6

7 -0.9 0.8

8 0.7 0.6

9 0.8 -0.8

10 0.1 -0.2

-1

-1

1

1

1

6

4

9

5

10

83

72

朝陽科技大學 李麗華 教授 10

SolutionSetup an 2X9 network

Randomly assign weights

Let R=2.0 η=1.0

代入第一個 pattern[-0.9, -0.8]

RFjk= R

rj

e

X1 X2

Wi00 -0.2 -0.8

Wi01 0.2 -0.4

Wi02 0.3 0.6

Wi10 -0.4 0.6

Wi11 -0.3 0.2

Wi12 -0.6 -0.2

Wi20 0.7 0.2

Wi21 0.8 -0.6

Wi22 -0.8 -0.6

Wi10

Wi11

Wi02

Wi20

Wi12

Wi22 Wi00

Wi01Wi21

-1

1

-1

1

• net00=[(-0.9+0.2)2+(-0.8+0.8) 2]=0.49

• net01=[(-0.9-0.2) 2+(-0.8+0.4) 2]=1.37

• net02=[(-0.9+0.3) 2+(-0.8-0.6) 2]=2.32

• net10=[(-0.9+0.4) 2+(-0.8-0.6) 2]=1.71

• net11=[(-0.9+0.3) 2+(-0.8-0.2) 2]=1.36

• net12=[(-0.9+0.6) 2+(-0.8+0.2) 2]=0.45

• net20=[(-0.9+0.7) 2+(-0.8-0.2) 2]=1.04

• net21=[(-0.9-0.8) 2+(-0.8+0.6) 2]=2.93

• net22=[(-0.9+0.8) 2+(-0.8+0.6) 2]=0.05

kjjknet

,

]min[ net

22

【MIN】Winning Node

朝陽科技大學 李麗華 教授 11

Solution (cont.)Update weight

282.2)20()20( 22

j*=2 k*=2

r00= RF00=0.243

22 )21()20(

22 )22()20(

r01=RF01=

r02 =

r22= 22 )22()22( RF22= 1

W00=η (X‧ 1-W00) RF‧ 00

1.0 × （ -0.9+0.2 ） × （ 0.243 ） = － 0.17

-0.17

Wi10

Wi11

Wi02

Wi20

Wi12

Wi22

Wi00

Wi01

Wi21

-1

1

-1

1