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Solution 4.3. (a) This is not the case. For example, 0 < 1, but 0^2 = 0 = 1-^2. (b) This is not the case either; 1^2 ^ 0^2 but it is not the case that 1^0. (c) The difference is that addition is invertible. (That is, adding m can be undone by subtracting m.) Integer division is not invertible. (d) Multiplication by a negative number is anti-monotonic. (That is, if n < 0,

The transition can be made from Cartesian (x, y, z) to spherical polar coordinates, in effect just r, in the same way as detailed in Section 8.5. Using equation (8.45), each

(Note the reversal of the ordering.) Division inverts multiplication. So any implementation of integer division should also be anti-monotonic. That is, if n < 0 ,

of the three Cartesian axes gives an equation of the following form:

LEARNING)

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Solution 4.4. The assignment k := Us clearly correct, and the assignment k := r clearly incorrect it will generate an array bound error the very first time the loop body is executed whatever the size of the array. The assignment k := (l+r)+2 is correct. The property I < k is satisfied because I ^ (i+r-l)-r2, as proved above, l+r-1 ^l+r, and integer division by 2 is monotonic. The proof that (l+r)-r2 <r is the only time that we need to use the fact that integer division rounds towards zero.

Therefore, the complete V 2 w ( r ) is given by:

arithmetic } If integer division is defined to round away from zero, the assignment k := (l+r-\)+2 is correct, the assignment fe := (i+r)-r2 is not. To show that the latter is incorrect consider an array of size 1. Then I and r are initialized to 0 and 1, respectively. So (l+r)/2 rounds up to 1. If this value is assigned to fe, an array bound error will occur. A general conclusion is that the assignment k := (l+r-1) +2 is safer than the assignment k := (l+r)+2 because its correctness is independent of whether integer division is implemented by rounding down or up. D Solution 4.5. First, if card[r-l] = card[l], a division-by-zero error will occur. Assuming this not to be the case, the right side of the assignment to k gives a value in the range I up to r-1 if

(as, for example, in [4], p. 188). Substituting into the Schrodinger equation then:

Decomposition (SVD) to analyze the hidden unit activation covariance matrix to determine the optimal number of hidden units Based on the assumption that outputs are linearly activated, the rank of the covariance matrix is the optimal number of hidden units (also see [Fujita 1992]) SVD of this information matrix results in an eigenvalue and eigenvector decomposition where low eigenvalues correspond to irrelevant hidden units The rank is the number of non-zero eigenvalues Fletcher, Katkovnik, Steffens and Engelbrecht use the SVD of the conditional Fisher information matrix, as given in equation (722), together with likelihood-ratio tests to determine irrelevant hidden units [Fletcher et al 1998] In this case the conditional Fisher information matrix is restricted to weights between the hidden and output layer only, whereas previous techniques are based on all network weights Each iteration of the pruning algorithm identifies exactly which hidden units to prune.

Solution 4.6. (a) In the scenario given, hi and 1 o are initialized to 2 and 0, respectively. The first iteration of the loop sets centre to 1 the fact that there are 'On! y two i terns 1 eft' is not observed and resets 1 o to 1. In the second iteration, centre is again assigned the value 1. This time, the method erroneously concludes 'Only two items left'. The test v[ cent re] .equal s(o) fails and then an array bound error occurs when the test v [cent re+1] . equal s (o) is executed. Reading the comment 'Only two items left', it would appear that the program implicitly assumes that v. length is at least 2. Indeed, the program will always give an array bound error if its length is zero, and it will also do so if its length is one and the entry being sought is greater than the entry in the array. The clumsy code preceded by the comment 'Only two items 1 eft' was possibly inserted to fix a bug that had been found. But, as we have just seen, the fix is not just clumsy, it does not work! (It is very common for additional tests to be added

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