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This suggests that in any subsequent scan we can put these cards aside, say at the top of the card table The importance of this suggestion arises from the fact that subsequent scans will be shorter than earlier ones, so eventually we will get down to a single card, which must then be the right borderline So having put to the top the right-hand heap which was too small, we move the other heap to the middle, select a new borderline, say 196,.

which in analogy to three dimensions, could also be written:

One problem that may occur in LVQ networks is that one cluster unit may dominate as the winning cluster unit. The danger of such a scenario is that most patterns will be in one cluster. To prevent one output unit from dominating, a "conscience" factor is incorporated in a function to determine the winning output unit. The conscience factor penalizes an output for winning too many times. The activation value of output units is calculated using _ J 1 for minVfc{d;fc,p(zp,Mfc) - bk(t)} ~ \ 0 otherwise where M*)= 7(7 - (*))

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and proceed with the split. At the end of the second split, we will have a borderline value and three heaps: 1. A top right heap, with cards higher than the first borderline 367. 2. A bottom right heap, with cards lying between the two borderlines 196 and 367. 3. A bottom left heap, with cards lower than the second smaller borderline 196. It may happen now that it is the left of the two bottom heaps which is too small; it will therefore contain only cards which properly belong to the left heap; and as before, we can put it on the card table, and omit it from future examination. Then we place the borderline value on that heap. Next we move the remaining bottom heap up to the middle and repeat the process, selecting again an arbitrary trial borderline (say 229), and splitting the middle heap into a bottom left heap and a bottom right heap. Then we have a picture as shown in Figure 2.2.

where the two-dimensional effective density of states:

Moreover, if statement 5 has been constructed to satisfy (133), we can use the rule of sequential composition to infer that { P } S ; do ^done T od { Q } (136) Loops and Induction The validity of (135) is proved by a case analysis on bf ^ 0 and bf > 0 The first case involves combining (132) and (131) The case bf > 0 is proved by (strong) induction on the value of the ghost variable C in (134) Formally, the inductive hypothesis is (Vn :: { inv A bf = n } do^done^Tod { Q }> , (137) where the type of n is the set of all natural numbers (We assume, of course, that n is not one of the program variables) Property (135) is then a consequence, because of (132).

and the Fermi energy would be equivalent to that in equation (5.151) but with , instead of Nc and the donor density Nd would be taken as a sheet density. Figure 5.37 shows the corresponding curve to Fig. 5.36 for a two-dimensional system. The generic behaviour is the same as in bulk, i.e. that the higher the temperature the greater proportion of donors are ionised and the greater the doping density the lower the proportion of ionised donors. Again the number of free carriers generated by doping might seem surprisingly small. Although this view has been recently reinforced [125] it is actually too simplistic and the proportion of carriers ionised is too small. The reason is that the model assumes that the impurity ionisation energy (which in this section would be Ec Ed) is left as a constant, whereas it was found that screening should see it reduce as the impurity concentration increases [126]. Pearson and Bardeen [126] argued that the ionisation energy of an impurity would be reduced by an energy inversely proportional to the distance between the impurities, hence the ionisation energy should be equal

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