4. 1 The Dutch National Flag in .NET

Development datamatrix 2d barcode in .NET 4. 1 The Dutch National Flag
1 4. 1 The Dutch National Flag
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We are required to construct a program, making use exclusively of the above operations together with simple arithmetic operations on indices, that will rearrange the stored values in such a way that, on termination, there are indices r and w such that
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A (Vi:r t^i<w:w hite.i) Note how the required postcondition has been formulated carefully to allow for the absence of stored values of each colour. Termination with M equal to r indicates that there are no red values, termination with r equal to w indicates no white values, and termination with w equal to N indicates no blue values.
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14.1.2 The Solution
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It is clear that a solution to the problem will involve an iterative process. Initially, all the colours are mixed and on termination all the colours should be sorted. We therefore seek an invariant property that has both the initial and final states as special cases. A reasonably straightforward idea is to strive for an invariant that partitions the array of values into four segments, three of the segments containing values all of the same colour (red, white or blue) and the fourth containing a mixture of colours. Four possible ways of arranging the boundaries between the segments are shown in Figure 14.1. In each case, the initial state is that the red, white and blue segments are empty and the 'mixed' segment is the entire array. Also, the final state is that the 'mixed' segment is empty. Any solutions based on the first and the last of these figures would be entirely symmetrical, as would solutions based on the two inner figures. The real choice is therefore between maintaining the 'mixed' section at one end of the array or in the interior of the array. The first figure corresponds to a program that uses a simple for statement with the control variable initialized to M and incremented at each repetition by one. The last possibility also corresponds to a for statement but where the control variable is initialized to N and continually decremented. The idea is that the values are processed one by one in order, and at each stage the set of values already processed is sorted. It is possible to construct a program of this nature, but it is not easy and the resulting program is not efficient! (Try to work out some of the details to see why.) This is the sort of solution that arises from an operational view of program construction, not taking sufficient time and effort to think about and formulate alternative strategies.
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14: Sorting and Searching Algorithms
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Figure 14.1 Possibilities for the invariant property.
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Maintaining the 'mixed1 segment in the interior of the array leads to a relatively simple and efficient solution. Adopting the second of the figures as the framework for a solution, we introduce the variables r, w and b with the following specifications.
(\/i:r^i<w:white.i)
blue.i) A (Vt Note that the colour of each element is specified for all indices, with the exception of indices i such that w ^ i < b. This is the 'mixed' segment of the array. Our goal is to design a simple loop (together with its initialization) that maintains this invariant whilst decreasing the size b-w of the 'mixed' segment. The chosen invariant property is depicted in Figure 14.2. Drawing a figure helps but you should never rely on figures. Figures are often ambiguous and do not properly capture the troublesome extreme cases. Always check your work against the formal specifications and nothing else. The initialization is easy. The red, white and blue segments are all empty. The assignment
r,w,b := M,M,JV guarantees that all three universal quantifications are vacuously true.
14.1 The Dutch National Flag
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Figure 14.2 Chosen invariant property.
The termination condition for the loop is w = b. When this state is reached, the 'mixed' segment is empty and the required postcondition is satisfied. In order to make progress towards the termination condition, it is reasonable to examine the colour of one of the elements at the boundary of the 'mixed' segment, that is, either the element with index w or the element with index b~l (or both). This way, we may hope to reduce the size of the 'mixed' segment by at least one at each iteration. With foresight, we choose to examine the colour of the element with index w. An easy case is if the colour of the element with index w is white. If so, the white segment can be extended by incrementing w. That is, execution of white.w - w := w + l is guaranteed to reduce b-w and maintain the invariant. Another relatively easy case is if the element with index w is blue. If so, the blue segment can be extended if the element with index w is swapped with the element with index b-l. That is, execution of