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AS PER THE SYLLABUS OF BPUT FOR SEVENTH Lecture Notes | KISHORE KUMAR SAHU RIT, BERHAMPUR SOFT COMPUTING (PECS 3401)- NEURAL NETWORKS SEMESTER OF AE&IE BRANCH.

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Page 1: BERHAMPUR SOFT COMPUTING (PECS 3401)- NEURAL …kishoreinvssut.yolasite.com/resources/Soft...ON SOFT COMPU F INFORMATION 04 L the or uts They the fin weigh ii. A They not. T which

AS PER THE SYLLABUS OF BPUT FOR SEVENTH

Lecture Notes | KISHORE KUMAR SAHU

RIT, BERHAMPUR SOFT COMPUTING (PECS 3401)- NEURAL NETWORKS SEMESTER OF AE&IE BRANCH.

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INTRONeural nability toBIOLOThe basystemscomponas followi. DenThey acto the respons

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LECTURE NOTES

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CHAPTER-UNDAMENTAsystems that haveble for future use.

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are responsible fonal input to the acthed inputs i.e. =Activation Functioare responsible foThis is possible duh the inputs are alwput is propagated i.e. = ∅( ) =Outputs are responsible nding on the decisiHITECTURE OFtificial neural netwnumber of simpletecture inspired byNN structure can ls in always restrhs, where the verptic links. Broadly tSingle layer feed-ftype of networksrises of twos, namely theand output layer.input layersve the input signalthe output layersve the outputls. The synapticcarrying the

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LECTURE NOTES

R SAHU, DEPT O

n but not vice-vein nature. Despite tput layer alone tls to output layer.

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Supervised Learnnetwork is associdesired pattern. Aprocess, when a coand the correct exbe used to changeperformance. Gradient descendefined in terms it is required thdifferentiable, aserror E. Thus, if Δjth neuron of the leaning rate paraweight Stochastic learnfashion. An examechanism emplof NN systems.

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LECTURE NOTES

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al networks have es, handwritten chOptimization/ concomprises problemons. Examples of hortest possible tog out of industrialNNs. Forecasting and ral networks have s. They have, thorology, stock marControl systems: al networks have gms. Dozens of coporating NN technfor the control of c

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ACTIVTo gene∑functionA very csum is cthe outpknown aSignumAlso knoSigmoidThis fuasymptoslope pbetweenis an impHyperboThe funcThe othPiecewis

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KISHORE KUMA

ROPAGATIO

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that varies gragiven by: ∅(I) =ptness of the funidal functions ared can produce negaal Basis function, moidal and Bipola

LECTURE NOTES

R SAHU, DEPT O

CHAPTER-ON NETWOR

+ + ⋯ +tion function/ tran( ). g function. In this, is greater that , t is the step func . +1, I−1, I .

adually between . Where α is nction as it chan differentiable, whative outputs valueUnipolar Multimor Sigmoidal.

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SINGLEA singlea input lwe havex1, and xand w2 The oucalculateinputs aline, wrepresenproblemperceptr

E LAYER PERCe layer perceptronlayer and a outpute a perceptron witx2 as inputs with and a bias x0 withutput of the pered as the sum oand bias i.e. net= which makes cleanting problems thms that are linearlyron.

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CEPTRON n consists of t layer. Here h two input weights w1, h weight w0. rceptron is of weighted w0 + w1x1+ w2x2. Thar that single laat are linearly sepy separable, hence

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KISHORE KUMA

his represents the ayer perceptron arable. Since logice can be easily sim

LECTURE NOTES

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equation of a straare responsiblec gates correspondulated by single la

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ight of ds to ayer

Algor

Step 1

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x----

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rithm for training

1. Initialize tbetween 02. While stop3. For each tr4. Set the act5. Compute tActivation+1,−1,6. If o/p andfollows: ( ) an7. Test for sto

mple: Train araining set for AND

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g a single layer perthe weights and bi0-1. pping condition fairaining pair do stetivation of i/p. the net i/p to the function is use00 . d target are not eq( ) = ( ) nd ( ) = (opping condition.a single layer perceD gate is as follows

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ADALIADA lineuron single componsingle concludsame asupdate ( )Algorith

Step 1.

Step 2. Step 3. Step 4. Step 5.

Step 6.

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MADAThe MAhandle problemwhich several networkperceptrcalled areason t

INE NETWOKSne networks (anetworks) are simlayer perceptrnents are similar layer perceptrone the ADA line s single layer percequation are ) = ( ) + (hm for training a Initialize thebetween 0-1While stoppiFor each traiSet the activaCompute theActivation f+1, 0−1, 0If o/p and tafollows: ( )Test for stop

ALINE NETWORDA line networks linearly separam. This is a netwis composed constituent ADA lk or single laron. This networkas MADA line duethat is has a numbeBY: K

adaptive linear milar to that of ron. All the to that of a n. So we can networks are eptron, except theas follows: − ).i.e. Tk is repsingle layer percee weights and bias. ing condition fails ining pair do step-ation of i/p. e net i/p to the actfunction is used . arget are not equa ( ) = el= ( ) and (pping condition.

RKS can able ork of line ayer k is e to er of hidden layers

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KISHORE KUMA

e in the training al( ) = ( )placed with the ereptron s to zero. Also set do steps-3 to step4 to step-6. tivation function to compute thal then change the( ) = ( )= ( ) + (lse( ) = ( ).

s.

LECTURE NOTES

R SAHU, DEPT O

lgorithm we have + ( − )ror ( − ). the learning rate -7.

= + (∑he o/p = (e weights and bia) + ( − )− ).

ON SOFT COMPU

F INFORMATION

the and in ). ) =

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Step 1

Step 2Step 3Step 4Step 5Step 6

Step 7

Step 8

Step 9

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x1  x

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rithm for MADA lin

1. Initialize trandom va0.5. 2. While stop3. For each tr4. Set the act5. Compute t6. Activation= (7. Calculate apply activ8. If o/p andfollows:( ) = (9. Test for sto

mple: performT

x2  w1  w2  b

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1  ‐0.77  0.43  0

1  0.43  0.43  0

‐1  0.43  0.43  0

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ne the i/p and hiddenalues and hidden pping condition fairaining pair do stetivation of i/p unitthe net i/p to the h function is us( ) . the net i/p for vation function ford target are not eq = ( ) + ( −) = ( ) and opping condition.m the training proTRAINING X‐OR BY MADb1  yin  y  T

0.5  ‐0.36  ‐1  ‐

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0.5  0.5  1  1

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n neuron layer welayer and o/p layils do steps-3 to steep-4 to step-8. s. hidden units ( )sed to compute o/p neuron (r each yi, =qual then change t ( ) = (− ). else ( ) = ( )cess with MADA liDALINE ALGORITHM T  ∆w1  ∆w2  ∆b

1  0  0  0

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T  ∆w1  ∆w2  ∆b

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eights and bias to yer weights and bep-9. = + (∑ )hidden unit o) = + (∑( ) . the weights and b) + ( − )).

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small bias to ). output, ) and bias as and X-OR.

b1n 

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x1  x2  w1  w2  b1  yin  y  T  ∆w1  ∆w2  ∆b1  w1n  w2n  b1n 

‐1  ‐1  0.43  0.43  0.5  ‐0.36  ‐1  ‐1  0  0  0  0.43  0.43  0.5 

1  1  ‐1.37  0.43  0.5  ‐0.44  ‐1  ‐1  0  0  0  ‐1.37  0.43  0.5 

‐1  1  0.43  0.43  0.5  0.5  1  1  0  0  0  0.43  0.43  0.5 

1  ‐1  0.43  0.43  0.5  0.5  1  1  0  0  0  0.43  0.43  0.5  BACK PROPAGATION NEURAL NETWORKS A multilayer feed forward back propagation neural network with one layer of z hidden units is shown in the figure. The layer having the bias from w01, w02,…,w0m and hidden neuron having the bias v01, v02,…,v0m. The figure only feed forward network is shown but during the back propagation phase of learning the signals are sent in the reverse direction, i.e. from o/p to i/p layer. The training algorithm of back propagation involves four steps: Initialization of weights and bias- The bias and the weights are chosen as small numbers. Feedforward-Each i/p pair receives the i/p and transmits to the hidden layer. Each hidden unit then calculates the o/p and transmits this signal to the o/p units. Back propagation of error-Each o/p is compared with target o/p and the determines the error. Based on the error the factor δk is computed and is used to distribute the error at o/p unit yk back to all units in the previous layer. Similarly the factor δj is computed for each hidden unit. Update the weights and the bias- The weights and the bias are updated using the δ-factor and activation. δ=(tk-yk)f ’ (tk-yk)(Generalized δ rule of learning). Algorithm for Back propagation

Step 1. Initialize weights and bias to small random values. Step 2. While stopping condition fails do steps-3 to step-10. Step 3. For each training pair do step-4 to step-9. Step 4. Each i/p unit receives the signal xi and transmits this signal to all other units. Step 5. Compute the net i/p to the hidden units ( ) = + ∑ . . Activation function is used to compute the o/p of hidden units, = ( ) .

Step 6. Calculate the net i/p for o/p neuron ( ) = + ∑ . and apply activation function for each yk, = ( ) . Step 7. For each yk calculate the δk=(tk-yk)f '(yin(k)) = (tk-yk).f (yin(k)).(1- f (yin(k))). Step 8. Each hidden unit sums its δ i/p from the previous layer and o/p is given as: ( ) = ∑ . . The error information is calculated as ∆ = ( ). ( ( )). Step 9. Each of the o/p unit update its bias and weights as given below. ( ) = ( ) + ∆ where ∆ = and bias ( ) = ( ) + ∆ where ∆ = . Each hidden unit update its weights and bias as follows ( ) = ( ) +∆ where ∆ = ∆ and bias ( ) = ( ) + ∆ where ∆ = ∆ . Step 10. Test for stopping condition. Merits of Back propagation Algorithm 1. The mathematical formulation is so compatible for any kind of networks. 2. Multilayer neural network trained with back propagation algorithm has got a greater representation capability. Any non linear activation function. 3. It requires good set of training data. It can tolerate noise and missing data in training sample. 4. Easy to implement. 5. The computation time is reduced if the weights chosen are small at the beginning stage. 6. Wider application. 7. Can be used to store a huge amount of pattern. 8. The batch update of weights exists, which provides a smoothing effect on weight corrections. Demerits of Back propagation Algorithm 1. Learning often takes longer time to converge. 2. Complex functions requires more iterations. 3. Gradient descent method used in back propagation algorithm gives guarantee to minimize error at local minima. 4. The network may be trapped in a local minima, though a better solution is available nearby. 5. The training may sometimes cause temporal instability to the system.

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LECTURE NOTES ON SOFT COMPUTING P a g e | 9

BY: KISHORE KUMAR SAHU, DEPT OF INFORMATION TECHNOLOGY, RIT, BERHAMPUR.

Example: Find the new weights for a network with i/p pattern [0.6 0.8 0] and target o/p is 0.9. The i/p hidden weights are = 2 1 01 2 20 3 1 and hidden o/p weights are w=[-1 1 2]T. v03=-1 and w01=-1. The learning rate = 0.3 and use binary sigmoid activation function. Solution: ( ) = ; ( ) = + ∑ . and = ( ) = ( ).

( ) = + (∑ . ) = 0 + (0.6*2 + 0.8*1 + 0*0) = 2; ( ) = + (∑ . ) = 0 + (0.6*1 + 0.8*2 + 0*3) = 2.2; ( ) = + (∑ . ) = -1 + (0.6*0 + 0.8*2 + 0*1) = 0.6; = ( ) = ( ) = . ∗ = . = 0.645; = ( ) = ( ) = . ∗ . = . = 0.659; = ( ) = ( ) = . ∗ . = . = 0.5448; ( ) = + ∑ . and = ( ) = ( ). ( ) = + ∑ . = -1 + (0.645*-1 + 0.659*1 + 0.5448*2) = 0.1; = ( ) = ( ) = . ∗ . = . = 0.50749 ≈ 0.5075;

δk=(tk-yk)f '(yin(k)) = (tk-yk).f (yin(k)).(1- f (yin(k))) δ1 = (0.9 – 0.5075) * 0.5075 * (1-0.5075)=0.0977. Error at hidden layers ( ) = ∑ . ;

( ) = ∑ . = 0.0977*-1 = -0.0977; ( ) = ∑ . = 0.0977*1 = 0.0977;

( ) = ∑ . = 0.0977*2 = 0.1954; Error term at hidden layer: ∆ = ( ). ( ) = ( ). ( ) 1 − ( ) ; ∆ = ( ). ( ) 1 − ( ) = -0.09777*0.645*(1-0.645) = -0.01802; ∆ = ( ). ( ) 1 − ( ) = 0.09777*0.659*(1-0.659) = 0.01769; ∆ = ( ). ( ) 1 − ( ) = 0.1954*0.5448*(1-0.5448) = 0.03906; Weight Updating Change in weight (i/p-hidden layer) ∆ = ∆ ∆ = ∆ =0.3* -0.02238*0.6= -0.00324; ∆ = ∆ =0.3* -0.01769*0.6= 0.00318; ∆ = ∆ =0.3* 0.03906*0.6= 0.000703; ∆ = ∆ =0.3* -0.02238*0.8= 0.00432; ∆ = ∆ =0.3*0.01769*0.8= 0.00424; ∆ = ∆ =0.3*0.03906*0.8= 0.000937; ∆ = ∆ =0.3*-0.02238*0= 0; ∆ = ∆ =0.3*0.01769*0= 0; ∆ = ∆ =0.3*0.03906*0= 0; New weights are as follows ( ) = ( ) + ∆

= 2 − 0.00324 1 + 0.00318 0 + 0.0007031 + 0.00432 2 + 0.00424 2 + 0.0009370 + 0 3 + 0 1 + 0 = 0 0 00 0 00 0 0

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Change ∆ =∆ =∆ =New bia= 0Change ∆ =∆ =∆ =New wew = [-1+Change ∆ =New bia(

in bias (hidden lay∆ =0.3* -0.02238∆ =0.3* -0.01769∆ =0.3* 0.03906as are ( ) =− 0.0054 0 + 0.00in the weights for = 0.3 ∗ 0.09= 0.3 ∗ 0.09= 0.3 ∗ 0.09eights for the hidde+0.01526 1+0.015in bias for o/p laye= 0.3 ∗ 0.097as for output layer) = ( ) + ∆

BY: K

yer): ∆ = ∆ 8= -0.0054; 9= 0.0053; = 0.00117; ( ) + ∆ ; 053 − 1 + 0.0011hidden-o/p layer:977 ∗ 0.645 = 0.01977 ∗ 0.659 = 0.01977 ∗ 0.5448 = 0.0en-o/p layer: (58 2+0.01287] = er: ∆ = 7 = 0.023; ( ) = (∆ = −1 + 0.023

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KISHORE KUMA

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ECTURE NOTES O

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LECTURE NOTES ON SOFT COMPUTING P a g e | 11

BY: KISHORE KUMAR SAHU, DEPT OF INFORMATION TECHNOLOGY, RIT, BERHAMPUR.

information necessary so that the algorithm can move on to the next time-step, t+1. Pseudo-code Pseudo-code for BPTT: Back_Propagation_Through_Time(a, y) // a[t] is the input at time t. y[t] is the output Unfold the network to contain k instances of f do until stopping criteria is met: x = the zero-magnitude vector;// x is the current context for t from 0 to n - 1 // t is time. n is the length of the training sequence Set the network inputs to x, a[t], a[t+1], ..., a[t+k-1] p = forward-propagate the inputs over the whole unfolded network e = y[t+k] - p; // error = target - prediction Back-propagate the error, e, back across the whole unfolded network Update all the weights in the network Average the weights in each instance of f together, so that each f is identical x = f(x); // compute the context for the next time-step

Advantages BPTT tends to be significantly faster for training recurrent neural networks than general-purpose optimization techniques such as evolutionary optimization. Disadvantages BPTT has difficulty with local optima. With recurrent neural networks, local optima is a much more significant problem than it is with feed-forward neural

networks. The recurrent feedback in such networks tends to create chaotic responses in the error surface which cause local optima to occur frequently, and in very poor locations on the error surface. RADIAL BASIS FUNCTION NETWORK (RBFN) Let us look at some regression models: 1. Polynomial regression with one variable ( , ) = + + + ⋯ = ∑ 2. Simple linear regression with D variable ( , ) = + + ⋯ + = , in one-dimensional case ( , ) = + , which is a straight line. 3. Linear regression with Basis function ∅ ( ) ( , ) = + ∑ ∅ ( ) = ∅( ) there are now M parameters instead of D parameters For this basis function we use Radial Basis functions. A radial basis function depends only on the radial distance (typically Euclidean) from the origin, i.e. ∅( ) = ∅(‖ ‖). If the basis function is centered at then ∅ ( ) = ℎ(‖ − ‖). We would look at radial basis functions centered at the data points , n=1,...,N.

Typically h(x) is a Gaussian ∅ ( ) = − ‖ ‖22 or a logistic function ∅ ( ) = 11+ ‖ ‖2 . The RBFN is characterised by two types of weights i.e. hypothetical (fixed weights of input-hidden layer) and adjustable weights (weights that can be adjusted, hidden-output layer weights i.e. wi). Algorithm for Radial Basis Function N/W

Step 1. Initialize weights to small random values. Step 2. While stopping conditions fails do step-3 to step-9. Step 3. Activate the inputs neurons by applying a set of inputs.

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LECTURE NOTES ON SOFT COMPUTING P a g e | 12

BY: KISHORE KUMAR SAHU, DEPT OF INFORMATION TECHNOLOGY, RIT, BERHAMPUR.

Step 4. For each input do step-5 to step-8. Step 5. Calculate RBF. Step 6. Choose centre for each radial basis function. Step 7. Calculate output for each hidden neuron as ℎ ( ) = − ∑ ‖ ‖22=0 where xi is the applied input, it the centre of Gaussian function and is the smoothing parameter or width of the Gaussian function. Step 8. Calculate the output of the network as ( ) = ∑ ℎ ( ) +=1 where weights of the adjustable connections, ℎ ( )is the response of the RB neurons, and is the bias for output neurons. Step 9. Test for the stopping condition. = ∑ ∑ ( ) − , where E is the error and where n is the number of input patterns, k is the sum of the values for each output node k. ( ) is the achieved output for node k given input ( ) and is the desired output for k node given input n. Comparison of RBFs and BP-MLP

RBF N/W BP_MLPAlways consists of three layers, i.e. input, output and hidden layer. Consists of one input and output layer and a number of hidden layer. The input-hidden layer weights are hypothetical (cannot be changed). All the connections are adjustable.The response of the hidden layers follows the Gaussian function. The response of the hidden layers is a linear connection of all the inputs of that neuron Constructs local approximation to non-linear input-output mapping. Constructs global approximation to non-linear input-output mapping. Has non-linear activation function It has linear activation function.

KOHONEN SELF ORGANIZING FEATURE MAP (SOM) The key principle for map formation is that training should be taken place over an extended region of the network, which uses the concept of neighbourhood neurons. The competitive networks is similar to a single layer network, except there exist an interconnection between the output neurons because of which all the output neurons compete with each other and the winner will be selected. There are two methods to select a winner. 1. Squared Euclidean method: Here we will find the square of the distance between input vector xi and weight vector wij i.e. ( ) = ∑ ( − )2=1 . The winner is that neuron which is having small distance. 2. Dot Product method: Using this method will find the dot product of input vector and weight vector which is given by ( ) = ∑ ∗=1 . That neuron is treated as winner that is having the large amount of dot products. Algorithm for Kohonen Self Organizing Feature Map Initially the weights and learning rates (α) are set to a small values. The inputs vector to be clustered presented to the network. Once the input vector is given and based on the initial weights all the output neurons will compete with each other and the winner will be selected. Based on the winner selection, weights are up to date for a particular unit.

Step 1. Initialize weights and learning rate. Step 2. While stopping condition fails do step-3 to step-9. Step 3. For each input vector do step-4 to step-6. Step 4. For each j compute the squared Euclidean distance ( ) = ∑ ( − )2=1 for 1 ≤ ≤ 1 ≤ ≤ . Step 5. Find the index j for minimum Dj. Step 6. For all units j for a specified neighbourhood of j and for all i update the weights. ( ) = + ( − ). Step 7. Update the learning rule. Step 8. Reduce the radius of topological neighbourhood. Step 9. Test for stopping condition.

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LECTURE NOTES ON SOFT COMPUTING P a g e | 13

BY: KISHORE KUMAR SAHU, DEPT OF INFORMATION TECHNOLOGY, RIT, BERHAMPUR.

Example: = 0.2 0.6 0.4 0.9 0.20.3 0.5 0.7 0.6 0.8 , = 0.3 0.4 , α=0.3. Solution: ( ) = ∑ ( − )2=1

D(1)=(x1-w11)2+(x2-w21)2=(0.3-0.2)2+(0.4-0.3)2=0.01+0.01=0.02, D(2)=(x1-w12)2+(x2-w22)2=(0.3-0.6)2+(0.4-0.5)2=0.09+0.01=0.1, D(3)=(x1-w13)2+(x2-w23)2=(0.3-0.4)2+(0.4-0.7)2=0.01+0.09=0.1, D(4)=(x1-w14)2+(x2-w24)2=(0.3-0.9)2+(0.4-0.6)2=0.36+0.04=0.4, D(5)=(x1-w15)2+(x2-w25)2=(0.3-0.2)2+(0.4-0.8)2=0.01+0.16=0.17, ( ) = + ( − )

w11n=w11o+α(x1-w11)=0.2+0.3(0.3-0.2)=0.2+0.03=0.23;

w21n=w21o+α(x2-w21)=0.3+0.3(0.4-0.3)=0.3+0.03=0.33;

= 0.23 0.6 0.4 0.9 0.20.33 0.5 0.7 0.6 0.8 . LEARNING VECTOR QUANTIZATION (LVQ) The architecture of LVQ is similar to SOM architecture. In LVQ networks each output unit has a known class. Hence it uses supervised learning method. The method for initializing the reference vector. 1. Take first m training data and use them as weight vectors and remaining vectors used for training. 2. Initialize the reference vector randomly and assign initial weights and classes randomly. Training Algorithm of LVQ The algorithm for the LVQ network is to find the output unit that has a matched pattern with the input vector. At the end of the process if x (input vector) and w (weight vector) belongs to the same class, then weights are moved towards the new input vector.

In this method winner neuron is identified. The winner neuron index is compared with the target and based on the comparison results weights are updated. Step 1. Initialize the weights and learning rates. Step 2. If stopping condition fails do step-3 to step-7. Step 3. For each training input perform step-4 to step-5. Step 4. Compute j using square Euclidean distance method. ( ) = ∑ − 2=1 , find j when D(j) is minimum. Step 5. Update wij as follows: if T=Cj then ( ) = ( ) + − ( ) else ( ) = ( ) − − ( ) Step 6. Reduce the learning rate α. Step 7. Test for the stopping condition.

Example: 1 0 1 00 0 1 111 10 00 01 is the vector and class is given by 1212 and α=0.1 to 0.05. Solution: Let us consider 1st two vectors as reference w1=[1 0 1 0] and w2=[0 0 1 1] and other two vectors used as training data. α=0.1, x1=[1 1 0 0] and T=1. Calculate ( ) = ∑ − 2=1 D(1) = (1-1)2 + (0-1)2 + (1-0)2 + (0-0)2 = 2. D(2) = (0-1)2 + (0-1)2 + (1-0)2 + (1-0)2 = 4. Therefore D(1) is minimum i.e. j=1 and Cj = 1. Now ( ) = ( ) − − ( ) w1(new) = [1 0 1 0] + 0.1*([1 1 0 0] – [1 0 1 0]) = [1 0 1 0] + 0.1*([0 1 -1 0]) = [1 0 1 0] + [0 0.1 -0.1 0] = [1 0.1 0.9 0]

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LECTURE NOTES ON SOFT COMPUTING P a g e | 14

BY: KISHORE KUMAR SAHU, DEPT OF INFORMATION TECHNOLOGY, RIT, BERHAMPUR.

w1 = [1 0.1 0.9 0] w2 = [0 0 1 1] x2 = [1 0 0 1] α=0.1 and T=2. D(1) = (1 - 0)2 + (0.1 - 0)2 + (0.9 - 1)2 + (0 - 1)2 = 1.82 D(2) = (0 - 1)2 + (0 - 0)2 + (1 - 0)2 + (1 - 1)2 = 2 D(1) is minimum and j=1 and Cj = 1 and T!=Cj w1(new) = [1 0.1 0.9 0] + 0.1*([1 0 0 1] – [1 0.1 0.9 0]) = [1 0.1 0.9 0] + 0.1*([0 -0.9 -0.1 1]) = [1 0.1 0.9 0] + [0 -0.09 -0.01 0.1] = [1 0.11 0.81 -0.1] αn = 0.5 x0.1 = 0.05. SIMULATED ANNEALING NEURAL NETWORKS Simulated annealing was developed in the mid 1970's by Scott Kirkpatric, along with a few other researchers. Simulated annealing was original developed to better optimized the design of integrated circuit (IC) chips. Annealing is the metallurgical process of heating up a solid and then cooling slowly until it crystallizes. If this cooling process is carried out too quickly many irregularities and defects will be seen in the crystal structure. Ideally the temperature should be deceased at a slower rate. A slower fall to the lower energy rates will allow a more consistent crystal structure to form. This more stable crystal form will allow the metal to be much more durable. Simulated annealing seeks to emulate this process. Simulated annealing begins at a very high temperature where the input values are allowed to assume a great range of random values. As the training progresses the temperature is allowed to fall. This restricts the degree to which the inputs are allowed to vary. This often leads the simulated annealing algorithm to a better solution, just as a metal achieves a better crystal structure through the actual annealing process. Simulated annealing can be used to find the minimum of an arbitrary equation that has a specified number of inputs. In the case of a neural network, as we will learn in Chapter 10, this equation is the error function of the neural network. he weight matrix of a neural network makes for an excellent set of inputs for the simulated annealing algorithm to minimize for. Different sets of weights are used for the

neural network, until one is found that produces a sufficiently low return from the error function. First, for each temperature the simulated annealing algorithm runs through a number of cycles. This number of cycles is predetermined by the programmer. As the cycle runs the inputs are randomized. Only randomizations which produce a better suited set of inputs will be kept. Once the specified number of training cycles has been completed, the temperature can be lowered. Once the temperature is lowered, it is determined of the temperature has reached the lowest allowed temperature. If the temperature is not lower than the lowest allowed temperature, then the temperature is lowered and another cycle of randomizations will take place. If the temperature is lower than the minimum temperature allowed, the simulated annealing algorithm is complete. A neural network's weight matrix can be thought of as a linear array of floating point numbers. Each weight is independent of the others. It does not matter if two weights contain the same value. The only major constraint is that there are ranges that all weights must fall within. Because of this the process generally used to randomize the weight matrix of a neural network is relatively simple. Using the temperature, a random ratio is applied to all of the weights in the matrix. This ratio is calculated using the temperature and a random number. The higher the

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