SOM-final

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Self-Organizing Map (SOM)

• The Self-Organizing Map was developed by professor Kohonen. The SOM has been proven useful in many applications

• One of the most popular neural network models. It belongs to the category of competitive learning networks.

• Based on unsupervised learning, which means that no human intervention is needed during the learning and that little needsto be known about the characteristics of the input data.

• Use the SOM for clustering data without knowing the class memberships of the input data. The SOM can be used to detect features inherent to the problem and thus has also been called SOFM, the Self-Organizing Feature Map.

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Self-Organizing Map (cont.)

• Provides a topology preserving mapping from the high dimensional space to map units. Map units, or neurons, usually form a two-dimensional lattice and thus the mapping is a mapping from high dimensional space onto a plane.

• The property of topology preserving means that the mapping preserves the relative distance between the points. Points that are near each other in the input space are mapped to nearby map units in the SOM. The SOM can thus serve as a cluster analyzing tool of high-dimensional data. Also, the SOM has the capability to generalize

• Generalization capability means that the network can recognize or characterize inputs it has never encountered before. A new input is assimilated with the map unit it is mapped to.

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Self Organizing Maps

• Often SOM’s are used with 2D topographies connecting the output units

• In this way, the final output can be interpreted spatially, i.e., as a map

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SOM Algorithm• Select output layer network topology

– Initialize current neighborhood distance, D(0), to a positive value• Initialize weights from inputs to outputs to small random values• Let t = 1• While computational bounds are not exceeded do

1) Select an input sample2) Compute the square of the Euclidean distance offrom weight vectors (wj) associated with each output node

3) Select output node j* that has weight vector with minimum value from step 2)

4) Update weights to all nodes within a topological distance given by D(t) from j*, using the weight update rule:

5) Increment t• Endwhile

From Mehotra et al. (1997), p. 189

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1 ,, ))((∑ =−

n

k kjkl twi

lili

))()(()()1( twittwtw jljj −+=+ η

1)1()(0 ≤−≤< tt ηηLearning rate generally decreases with time:

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Example Self-Organizing Map

• From Fausett (1994)• n = 4, m = 2

• Training samplesi1: (1, 1, 0, 0)i2: (0, 0, 0, 1)i3: (1, 0, 0, 0)i4: (0, 0, 1, 1)

Input units:

Output units: 1 2

What should we expect as outputs?

Network Architecture

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What are the Euclidean Distances Between the Data Samples?


0i40i3

0i20i1

i4i3i2i1

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Euclidean Distances Between Data Samples


0314i4021i3

03i20i1

i4i3i2i1

Input units:

Output units: 1 2 What might we expect from the SOM?

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Example Details• Training samples

i1: (1, 1, 0, 0)i2: (0, 0, 0, 1)i3: (1, 0, 0, 0)i4: (0, 0, 1, 1)

• Let neighborhood = 0– Only update weights associated with winning output unit (cluster) at each

iteration• Learning rate

η(t) = 0.6; 1 <= t <= 4η(t) = 0.5 η(1); 5 <= t <= 8η(t) = 0.5 η(5); 9 <= t <= 12etc.

• Initial weight matrix(random values between 0 and 1)

Input units:

Output units: 1 2

⎥⎦

⎤⎢⎣

⎡3.7.4.8.9.5.6.2.

2

1 ,, ))((∑ =−

n

k kjkl twi

))()(()()1( twittwtw jljj −+=+ η

d2 = (Euclidean distance)2 =

Weight update:

Unit 1:

Unit 2:

Problem: Calculate the weight updates for the first four steps

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First Weight Update

• Training sample: i1– Unit 1 weights

• d2 = (.2-1)2 + (.6-1)2 + (.5-0)2 + (.9-0)2 = 1.86– Unit 2 weights

• d2 = (.8-1)2 + (.4-1)2 + (.7-0)2 + (.3-0)2 = .98– Unit 2 wins– Weights on winning unit are updated

– Giving an updated weight matrix:

⎥⎦

⎤⎢⎣

⎡3.7.4.8.9.5.6.2.Unit 1:

Unit 2:

i1: (1, 1, 0, 0)i2: (0, 0, 0, 1)i3: (1, 0, 0, 0)i4: (0, 0, 1, 1)

])3.7.4.8.[-0] 0 1 [1(6.0]3.7.4.8.[2 =+=−− weightsunitnew

= [.92 .76 .28 .12]

⎥⎦

⎤⎢⎣

⎡12.9.

28.5.

76.6.

92.2.Unit 1:

Unit 2:

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Second Weight Update


• d2 = (.2-0)2 + (.6-0)2 + (.5-0)2 + (.9-1)2 = .66– Unit 2 weights

• d2 = (.92-0)2 + (.76-0)2 + (.28-0)2 + (.12-1)2 = 2.28– Unit 1 wins– Weights on winning unit are updated


Unit 1:

Unit 2:

i1: (1, 1, 0, 0)i2: (0, 0, 0, 1)i3: (1, 0, 0, 0)i4: (0, 0, 1, 1)

])9.5.6.2.[-1] 0 0 [0(6.0]9.5.6.2.[1 =+=−− weightsunitnew

= [.08 .24 .20 .96]

Unit 1:

Unit 2:

⎥⎦

⎤⎢⎣

⎡12.9.

28.5.

76.6.

92.2.

⎥⎦

⎤⎢⎣

⎡12.96.

28.20.

76.24.

92.08.

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Third Weight Update


• d2 = (.08-1)2 + (.24-0)2 + (.2-0)2 + (.96-0)2 = 1.87– Unit 2 weights



Unit 1:

Unit 2:

i1: (1, 1, 0, 0)i2: (0, 0, 0, 1)i3: (1, 0, 0, 0)i4: (0, 0, 1, 1)

])12.28.76.92.[-0] 0 0 [1(6.0]12.28.76.92.[2 =+=−− weightsunitnew

= [.97 .30 .11 .05]

Unit 1:

Unit 2:

⎥⎦

⎤⎢⎣

⎡12.96.

28.20.

76.24.

92.08.

⎥⎦

⎤⎢⎣

⎡05.96.

11.20.

30.24.

97.08.

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Fourth Weight Update


• d2 = (.08-0)2 + (.24-0)2 + (.2-1)2 + (.96-1)2 = .71– Unit 2 weights



Unit 1:

Unit 2:

i1: (1, 1, 0, 0)i2: (0, 0, 0, 1)i3: (1, 0, 0, 0)i4: (0, 0, 1, 1)

])96.20.24.08.[-1] 1 0 [0(6.0]96.20.24.08.[1 =+=−− weightsunitnew

= [.03 .10 .68 .98]

Unit 1:

Unit 2: ⎥⎦

⎤⎢⎣

⎡05.98.

11.68.

30.10.

97.03.

⎥⎦

⎤⎢⎣

⎡05.96.

11.20.

30.24.

97.08.

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Applying the SOM Algorithm

0.60Unit 14

0.60Unit 23

0.60Unit 12

0.60Unit 21

η(t)D(t)4321time (t)

Data sample utilized

‘winning’ output unit

Unit 1:

Unit 2: ⎥⎦

⎤⎢⎣

⎡00.1

05.

5.0

0.10

After many iterations (epochs) through the data set:

Did we get the clustering that we expected?

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What clusters do thedata samples fall into?

Unit 1:

Unit 2: ⎥⎦

⎤⎢⎣

⎡00.1

05.

5.0

0.10

WeightsInput units:

Output units: 1 2

Training samplesi1: (1, 1, 0, 0)i2: (0, 0, 0, 1)i3: (1, 0, 0, 0)i4: (0, 0, 1, 1)

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Solution

• Sample: i1– Distance from unit1 weights

• (1-0)2 + (1-0)2 + (0-.5)2 + (0-1.0)2 = 1+1+.25+1=3.25

– Distance from unit2 weights• (1-1)2 + (1-.5)2 + (0-0)2 + (0-0)2 = 0+.25+0+0=.25 (winner)


• (0-0)2 + (0-0)2 + (0-.5)2 + (1-1.0)2 = 0+0+.25+0 (winner)

– Distance from unit2 weights• (0-1)2 + (0-.5)2 + (0-0)2 + (1-0)2 =1+.25+0+1=2.25

Unit 1:

Unit 2: ⎥⎦

⎤⎢⎣

⎡00.1

05.

5.0

0.10

Weights

Input units:

Output units: 1 2


2

1 ,, ))((∑ =−

n

k kjkl twid2 = (Euclidean distance)2 =

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Solution


• (1-0)2 + (0-0)2 + (0-.5)2 + (0-1.0)2 = 1+0+.25+1=2.25

– Distance from unit2 weights• (1-1)2 + (0-.5)2 + (0-0)2 + (0-0)2 = 0+.25+0+0=.25 (winner)


• (0-0)2 + (0-0)2 + (1-.5)2 + (1-1.0)2 = 0+0+.25+0 (winner)

– Distance from unit2 weights• (0-1)2 + (0-.5)2 + (1-0)2 + (1-0)2 = 1+.25+1+1=3.25

Unit 1:

Unit 2: ⎥⎦

⎤⎢⎣

⎡00.1

05.

5.0

0.10

Weights

Input units:

Output units: 1 2


2

1 ,, ))((∑ =−

n

k kjkl twid2 = (Euclidean distance)2 =

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Conclusion

• Samples i1, i3 cluster with unit 2• Samples i2, i4 cluster with unit 1

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What aboutgeneralization?

• New data samplei5: (1, 1, 1, 0)

• What unit should this cluster with?• What unit does this cluster with?


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Example 5.7• pp. 191-194 of Mehotra et al. (1997)

111I60.50.50.5I5

001I41.50.50I3

000I21.81.71.1I1

Input Data Samples

• What do we expect as outputs from this example?

n = m = 3

B A C

Input units:

Output units

3100.7521.531.14I6

00.751.250.753.49I5

03.516.14I4

02.52.74I3

07.34I2

0I1

I6I5I4I3I2I1

Squared Euclidean Distancesof One Input Value to Another

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111I6

0.50.50.5I5

001I4

1.50.50I3

000I2

1.81.71.1I1

ZYX

I2

I1I3

I4

I5

I6I1

I2

I3

I4

I5

I6

Data Samples Plotted as X,Y,Z points in 3D Space

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Example Details:Neighborhood distance &

Learning Rate• Neighborhood distance

D(t) gives output unit neighborhood as a function of time0 <= t <= 6, D(t) = 1t > 6, D(t) = 0

• Learning rate also varies with time0 <= t <= 5, η(t) = 0.66 <= t <= 12, η(t) = .25t > 12, η(t) = 0.1

http://www.cprince.com/courses/cs5541/lectures/SOM/SOM.xls

Initial weightsWa 0.2 0.7 0.3Wb 0.1 0.1 0.9Wc 1 1 1

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First Iteration

• Use input data in order I1, I2, …, I6– Start with I1: 1.1 1.7 1.8

• 1) Compute Euclidean distance of data from current weight vectors for output units

• 2) Compute weight updates

Initial weightsWa 0.2 0.7 0.3Wb 0.1 0.1 0.9Wc 1 1 1

B A C

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Applying the SOM Algorithm

A0.10A18

B0.10B17

B0.10B16

C0.10C15

B0.10B14

C0.10C13

A0.250A12

B0.250B11

B0.250B10

C0.250C9

B0.250B8

C0.250C7

C, A0.51C6

A, B, C0.51A5

B, A0.51B4

A, B, C0.51A3

B, A0.51B2

C, A0.51C1

Weights Updated

η(t)D(t)654321time (t)

Data sample utilized

‘winning’ output node

B A C

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Results: Classification & Weights

1, 3C

2, 4, 5B6A

Data sampleOutput node

Classification

1.340.950.61Wc0.30.230.47Wb

0.810.770.83Wa

Weights after 15 time steps

Weights after 21 time steps

1.3861.0250.659Wc0.26370.212670.46863Wb

0.8290.7930.847Wa

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I1

I2

I3

I4

I5

I6

1.340.950.61Wc

0.30.230.47Wb

0.810.770.83Wa

111I6

0.50.50.5I5

001I4

1.50.50I3

000I2

1.81.71.1I1

ZYX

Data Samples & Weights Plotted as X,Y,Z points in 3D Space

I2

I1I3

I4

I5

I6

Wc

Wb

Wa

Wa

Wb

Wc

SOM-final

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