Last lecture summary

SOM

• supervised x unsupervised• regression x classification• Topology?• Main features?• Codebook vector?• Output from the neuron?

Compettive learning

• BMU• Scaling• Which neuron gets updated?• How it will be updated?• Topology preservation• Neighborhood

• Variable parameters– NS (neighborhood strength)– Neighborhood size– Learning rate

1 NS , 1j j jt t t d t t t w w x w

Multidimensional data

• IRIS (attributes: sepal length, sepal widht, petal length, petal width)

• Since we have class labels, we can assess the classification accuracy of the map.

• So first we train the map using all 150 patterns.

• And then we present input patterns individually again and note the winning neuron.– The class to which the input belongs is the class

associated with this BMU codebook vector (see previous slide, Class panel).

– Only the winner decides classification.

Vers (2) – 100% accuracySet (1) – 86%Virg (3) – 88%Overall accuracy = 91.3%

Vers (2) – 100% accuracySet (1) – 90%Virg (3) – 94%Overall accuracy = 94.7%

Sandhya Samarasinghe, Neural Networks for Applied Sciences and Engineering, 2006

Only winner decides the classification

Neighborhood of size 2 decides the classification

U-matrix• Distances between the neighboring codebook

vectors can highlight different cluster regions in the map and can be a useful visualization tool

• Two neurons: w1 = {w11, w21, … wn1}, w2 = {w12, w22, … wn2}

• Euclidean distance between them

• The average of the distance to the nearest neighbors – unified distance, U -matrix

2 2 2

12 11 12 21 22 1 2n nd w w w w w w

The larger the distance between neurons, the larger (i.e., lighter color) is the U value.

Large distance between this cluster (Iris versicolor) and the middle cluster (Iris setosa). Large distances between codebook vectors indicate a sharp boundary between the clusters.

Surface graphThe height represents the distance.

3rd row – large height = separation

Other two clusters are not separated.

Quantization error• Measure of the distance between codebook

vectors and inputs.• If for input vector x the winner is wc, then

distortion error e can be calculated as

• Comput e for all input vectors and get average – quantization error, average map distortion error E.

, NS ,cii

e d e d c ix w x w

1 1, NS ,ci

n n i

E d E dN N

n ni ix w x w

Iris quantization error

High distortion error indicates areas where the codebook vector is relatively far from the inputs. Such information can be used to refine the map to obtain a more uniform distortion error measure if a more faithful reproduction of the input distribution from the map is desired.

Genetic algorithms(new stuff)

Optimization in DM

f’(x) = 0f’’(x) > 0 … minimumf’’(x) < 0 … maximum

Optimization in DM

• traditional methods (exact)– e.g. gradient based methods

• heuristics (approximate)– deterministic– stochastic (chance)• e.g. genetic algorithms, simulated annealing, ant colony

optimization, particle swarm optimization

Optimization in DM

• Applications of optimization techniques in DM are numerous.

• Optimize parameters to obtain the best performance.

• Optimize weights in NN• From many features, find the best (small)

subset giving the best performance (feature selection).

• …

Biology Inspiration• Every organism has a set of rules describing how that

organism is built up from the tiny building blocks of life. These rules are encoded in genes.

• Genes are connected together into long strings called chromosomes.

locus

• gene for color of teeth• allele for blue teeth

• Genes + alleles = genotype.• Physical expression of the

genotype = phenotype.

http://biology.unm.edu/ccouncil/Biology_124/Images/chromosome.gif

• When two organisms mate they share their genes. The resultant offspring may end up having half the genes from one parent and half from the other. This process is called recombination (crossover).

• Very occasionally a gene may be mutated.

http://members.cox.net/amgough/Chromosome_recombination-01_05_04.jpg

• Life on earth has evolved through the processes of natural selection, recombination and mutation.

• The individuals with better traits will survive longer and produce more offsprings.– Their survivability is given by their fitness.

• This continues to happen, with the individuals becoming more suited to their environment every generation.

• It was this continuous improvement that inspired John Holland in 1970’s to create genetic algorithms.

GA step by step

• Objective: find the maximum of the function O(x1, x2) = x1

2 + x22

– This function is called objective function.– And it will be use to evaluate the fitness.

Adopted from Genetic Algorithms – A step by step tutorial, Max Moorkap, Barcelona, 29th November 2005

Encoding

• A model parameters (x1, x2) are encoded into binary strings.

• How to encode (and decode back) a real number as a binary string?– For each real valued variable x we need to know:• the domain of the variable x ϵ [xL,xU]• length of the gene k

5-bit x1 x2xL -1 0xU 1 3.1

Stepsize 0.0645 0.1

x1 ϵ [-1, 1] x2 ϵ [0, 3.1]

c1 = (0101110011) → (01011) = -1 + 11 * 0.0645 = -0.29

(10011) = 0 + 19 * 0.1 = 1.9chromosome

gene

stepsize=𝑥𝑈−𝑥 𝐿

2𝑛−1

number=𝑥 𝐿+stepsize×(decoded value of string )

• At the start a population of N random models is generated

• c1 = (0101110011) → (01011) = -1 + 11 * 0.0645 = -0.29

(10011) = 0 + 19 * 0.1 = 1.9• c2 = (1111010110) → (11110) = -1 + 30 * 0.0645 = 0.935

(10110) = 0 + 22 * 0.1 = 2.2• c3 = (1001010001) → (10010) = -1 + 18 * 0.0645 = 0.161

(10001) = 0 + 17 * 0.1 = 1.7• c4 = (0110100001) → (01101) = -1 + 13 * 0.0645 = -0.161

(00001) = 0 + 1 * 0.1 = 0.1

• For each member of the population calculate the value of the objective function O(x1, x2) = x1

2 + x22

O1 = O(-0.29, 1.9) = 3.69

O2 = O(0.935, 2.2) = 5.71

O3 = O(0.161, 1.7) = 2.92

O4 = O(-0.161, 0.1) = 0.04

genotype phenotype

• Chromosome with bigger fitness has higher probability to be selected for breeding.

• We will use the following formula

O1 = 3.69O2 = 5.71O3 = 2.92O4 = 0.04

∑Oj = 12.36

P1 = 0.30P2 = 0.46P3 = 0.24P4 = 0.003

𝑃 𝑖=𝑂𝑖

∑𝑗=1

𝑛

𝑂 𝑗

Roulette wheel

p2 (46%)

p3 (24%)

p1(30%)

P4 (0.3%)

• Now select two chromosomes according to roulette wheel.– Allow the same chromosome to be selected more

than once for breeding.• These two chromosomes will:

1. cross over2. mutate

• Let’s say c2 = (1111010110) and c3 = (1001010001) chromosomes were selected.

• With probability Pc these two chromosomes will exchange their parts at the randomly selected locus (crossover point).

1111010110

1001010001

1111010001

1001010110

Pc

1111010001

1001010110

Pm

1111010110

1001010001

1111010001

1001010110

Pc

1110010001

1001010110

Pm

1111010110

1001010001

1111010001

1001010110

Pc

1110010001

1001011110

Pm

• Crossover point is selected randomly.• Pc generally should be high, about 80%-95%– If the crossover is not performed, just clone two parents

into new generation.

• Pm should be low, about 0.5%-1%– Perform mutation on each of the two offsprings at each

locus.

• Very big population size usually does not improve performance of GA.– Good size: 20-30, sometimes 50-100 reported as

best– Depends on size of encoded string

• Repeat previous steps till the size of new population reaches N. – The new population replaces the old one.

• Each cycle throught this algorithm is called generation.

• Check whether termination criteria have been met.– Change in the mean fitness from generation to

generation.– Preset the number of generation.

1. [Start] Generate random population of N chromosomes (suitable solutions for the problem)

2. [Fitness] Evaluate the fitness f(x) of each chromosome x in the population 3. [New population] Create a new population by repeating following steps

until the new population is complete 1. [Selection] Select two parent chromosomes from a population

according to their fitness (the better fitness, the bigger chance to be selected)

2. [Crossover] With a crossover probability cross over the parents to form a new offspring (children). If no crossover was performed, offspring is an exact copy of parents.

3. [Mutation] With a mutation probability mutate new offspring at each locus (position in chromosome).

4. [Accepting] Place new offspring in a new population 4. [Replace] Use new generated population for a further run of algorithm 5. [Test] If the end condition is satisfied, stop, and return the best solution in

current population 6. [Loop] Go to step 2.