By B. Yegnanarayana

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**Example text**

Using the Matlab command newff([zeros(4,1) ones(4,1)],[8 3],{’logsig’, ’logsig’}) the feed-forward network has been created. 2284]T Using the Matlab command train(net,InputData,DesiredOutput), the network has been trained. 2366]T function[]=prob4_9d() %Next set of lines is to generate the input and output data according to the given condition p=[0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1; 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1; ... 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1; 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1]; t=[0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1; 0 0 1 1 0 1 1 0 1 1 0 0 1 0 0 1; ...

Solution to problem 10: Let s = [s1 s2 s3 ]T represent state vector. Then for the weights matrix W = [wij ], the energy of 53 the feedback network is given by 1 V = − sT W s 2 Therefore, gradient of V is given by ∇V = −W s = − 0 −1 −1 s1 −1 0 1 −1 1 0 The Hessian matrix is given by ∇2 V = − W = − s2 s3 = −s2 − s3 −s1 + s3 −s1 + s2 0 −1 −1 −1 0 −1 1 1 0 Solution to problem 11: Consider the representation of each of the ten digits (0, 1, · · · , 9) by a matrix of 10x10 elements, where each element is either 1 or -1.

Irrespective of sold k , ∆V = 0. For Case C, Thus we have ∆V ≤ 0. Solution to problem 3: Consider a 3-unit modelwith bipolar {−1, +1} units. 69) ⇒ where we consider the energy terms with the threshold terms θi . 108), e−Eab /T = e−F a Zclamped = a /T b we have, F a = −T log e−Eab /T b ∂F a − ∂wij T = e −Ean /T − b 1 −Eab /T ab e (−sab i sj ) T n = b e−Eab /T ab ab si sj e−Ean /T n 49 where n is a dummy summation index. But e−Eab /T = P (Hb |Va ) e−Ean /T n Therefore, − ∂F a = ∂wij ab P (Hb |Va )sab i sj b a ∂F is given by, ∂wij ∂ F¯ a − = P + (Va ) ∂wij a Hence the average value of − ab P (Hb |Va )sab i sj b + ab P (Va ∧ Hb )sab = p+ i sj ij = ab Solution to problem 6: (See Hertz, 1991, p.