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A radial basis function (RBF) neural network is a combination of Kohonen's LVQ-I and gradient descent. The architecture consists of three layers: an input layer, a hidden layer of J basis functions, and an output layer of K linear output units. The activation values of the hidden units are calculated as the closeness of the input vector zp to an I-dimensional parameter vector fij associated with hidden unit yj. The activation is given as n P -/*jii 2 2<T yj,p = e > (5.2) using a Gaussian basis function, or

We do not assume that there is at least one value of each colour. Swapping the pebbles in buckets i and j is effected by executing swap(ij). The effect of swap(ij) is specified formally by the Hoare triple: { M ^ i = I <N A M^ j =J <N A colour.i = X A colour.j = Y } swap(ij) { i = l A j =J A colour.i = Y A colour.j = X } , where the function colour is defined in the obvious way (colour.i = red = red.i, etc.). (Note the use of the ghost variables I, J, X and Y, in particular that it is not i and j that are swapped, but the colours at i and j.) It is convenient to assume that swap(i,i) is valid and has no effect on the state of the stored values (as predicted by the formal specification of swap}.

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which gives:

(c) Each value in the top-left heap is at most all values in the bottom-left heap Each value in the bottom-left heap is at most all values in the bottom-right heap And, each value in the bottom-right heap is at most all values in the top-right heap Note: the relationship is 'at most' It is incorrect to claim, for example, that all values in the top-left heap are lower than (or less than) all values in the bottom-left heap (d) If all cards in the pack have the same value, each iteration of step (2) adds just the borderline card to the top-left heap The repetition in step (2) will therefore take place 20 times.

The two remaining terms of equation (6.63) give:

(5.3)

If the pack is sorted and the borderline card is chosen to be the 20th card in the pack, one repetition of step (2) adds 20 cards to the bottom left deck whenever there are exactly 19 cards strictly lower than the 20th card If there are n cards strictly lower than the 20th, 19-n repetitions are needed (e) The borderline card is always either added to the top-left heap (in step (23)) or to the top-right heap (in step (24)) This means that between every repetition of (2) the size of the middle heap decreases by at least one If an arbitrary value is chosen for the borderline, termination is not guaranteed.

Again substituting r' for rT, then:

For example, if the borderline is chosen to be zero and all values in the pack are greater than zero, the algorithm will continually loop between a state in which all cards are in the bottom-right heap and one in which all cards are in the middle heap (f) The assumption is that M^N and (possibly) N * 0 The algorithm is obtained by replacing 100 by N, 20 by M, 21 by M+l, 80 by N-M and 81 by N-M+l When M is zero, the amalgamation in step (24) takes place Also the borderline card is moved to the top-right deck Thus, the algorithm terminates correctly after one repetition of step (2) When N equals M, the amalgamation in step (23) takes place and the borderline card is moved to the topleft deck.

cosh 0, then:

using a logistic function, where the norm is the Euclidean norm; aj and fij are respectively the standard deviation and mean of the basis function. The final output is calculated as j

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