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2.5 Exercises
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think that 367 is the twentieth smallest value. All we need to do is to go through the pack, putting all the cards lower than the borderline value on a left-hand heap, and all cards higher than it on a righthand heap, and so on. At the end, we expect that the left-hand heap will contain exactly the required twenty values. This process requires only one scan of the entire pack, and will take just over one and a half minutes in our small manual example. Returning to the computer problem, it could carry out the whole process on a hundred thousand observations in one second very much better than the five hours required if the previous method had been used. But it seems that this gain in speed can be achieved only if we have prior knowledge of the correct borderline value, and this knowledge we do not have. But now suppose we make a guess of the correct borderline value, and carry out the partitioning process as before. I suggest that we choose the borderline as the value of an actual card in the pack, since this will ensure that we never choose a ridiculously high or low value. Now if our guess, say 367, was too high, the left-hand heap ends up too large, containing more than twenty cards; and the right-hand heap is too small, containing less than eighty cards. Similarly, if our guess was too low, the left-hand heap is too small and the right-hand heap is too large. Thus we always know afterwards whether the first guess was too high or too low, and perhaps we can use this knowledge to make a better guess next time. As before, it is a good idea to select as next guess an actual card of the pack, which is known to be better than the previously guessed wrong borderline. This can be done very easily by selecting a card from the appropriate heap, for example, the left-hand heap if the original guess was too high; for it is known that all cards in this heap are smaller that the previous (too high) borderline. So we can repeat the process with this new borderline. But now consider the right-hand heap which was too small. This heap contains only cards which should be there, in the sense that they are already in the same heap as they will be in when the correct borderline is found. There is no point in scanning the cards of this heap again. 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,
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2: A Searching Problem and Its Solution
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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.
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