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2: A Searching Problem and Its Solution card is chosen to be the 20th in the pack. Is it the case that step (2) will be repeated exactly once If not, give a necessary and sufficient condition for this to be the case. (e) Why is it important that the borderline is a card in the middle heap Show by example what may go wrong if the chosen borderline value is not the value of one of the cards. (Hint: think about why the algorithm is guaranteed to terminate.) (f) Generalize your answers to (a)-(c) for the case that the problem is to find the M lowest values in a pack of N cards, where M and N are input parameters to the algorithm. Modify the algorithm accordingly. Check that your algorithm works for the extreme cases when M and/or N is zero, and when M equals the number of cards in the pack. D The exercises below are aimed at getting you to think in terms of invariant properties. The invariant properties are very simple even 'obvious' once pointed out. Finding them can be difficult, however, because thinking in terms of invariants is a skill that is rarely taught properly, if at all. The solutions are provided at the back of the book, but try not to give up too early! Exercise 2.2 (The Domino Problem). A chess board has had its top-right and bottom-left squares removed so that there are 62 squares remaining. (See Figure 2.4.)

can also be used, where P is the total number of patterns in the data set considered. However, the SSE is not a good measure when the performance on different data set sizes are compared. An additional error measure is required for classification problems, since the MSE alone is not a good descriptor of accuracy. In the case of classification problems, the percentage correctly classified (or incorrectly classified) patterns is used as a measure of accuracy. The reason why the MSE is not a good measure, is that the network may have a good accuracy in terms of the number of correct classifications, while having a relatively large MSE. If just the MSE is used to indicate when training should stop, it can happen that the network is trained too long in order to reach the low MSE. Hence, wasting time and increasing the chances of overfitting the training data (with reference to the number of correct classifications). But when is a pattern classified as correct When the output class of the NN is the same as the target class which is not a problem to determine when the ramp or step function is used as the activation function in the output layer. In the case of continuous activation functions, a pattern p is usually considered as being correctly classified if for each output unit OK, ((ok,p, > 0.5 + 0 and tk,p = 1) or (ok,p < 0.5 - 9 and tk,p>p = 0)), where 0 6 [0, 0.5] - of course, assuming that the target classes are binary encoded. An additional measure of accuracy is to calculate the correlation between the output and target values for all patterns. This measure, referred to as the correlation coefficient, is calculated as

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Using the specific forms forw1srsandw2sras above, then:

A {Vij : ( K i < s A s^j <N : a [ i ] < a [ j ] ) A (Vij :Q^i<l A A ( V i j :s^i<l A It is at this point that the development differs from Hoare's algorithm. The key insight is that it is possible to use the Dutch National Flag program to sort the elements in the 'medium' segment into values less than X (the 'red* values), values equal to X (the 'white' values) and values greater than X (the 'blue' values). Let us suppose this is done as the first step of maintaining the invariant. Let us also assume that the Dutch National Flag program returns two indices, ra and n, delimiting the segment containing values equal to X. This segment is known to

where I/A' = l/Ai s + l/A 2S .Writing this as L(a) = L\ aL-2, then obtain:

0 5 m n 'small' 'medium < X' 'medium = X'

Vji)

The remaining functions, F(a), G(a), J(a) and K(a), can all be derived by using similar procedures to the 1s case, and after lengthy manipulation are given by:

'medium > X'

In addition, G(a) = G\ + G where:

'large'

V""*'/

and:

0 5 m 'small' 'medium < X'

Continuing for the remaining functions:

and Zitp / 0. Then, netyjp = ep>>p cos(7r<^)p) The output value for each output unit is then calculated as

'medium = X'

and finally:

'medium > X'

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