vb.net code to generate barcode For the quick sort pivoting on median of three elements, the only changes needed are in the in Java

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For the quick sort pivoting on median of three elements, the only changes needed are in the
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partition() method: private static int partition(int[] a, int p, int q) { int m = (p + q)/2; m = indexOfMedian(a, p, m, q-1); swap(a, p, m); // The rest is the same as lines 20-32 in Example 14.9 on page 264 }
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This requires a method for locating the index of three array elements:
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private static int indexOfMedian(int[] a, int i, int j, int k) { // Returns the index of the median of {a[i], a[j], a[k]} if (a[i] <= a[j] && a[j] <= a[k]) return j; if (a[i] <= a[k] && a[k] <= a[j]) return k; if (a[j] <= a[i] && a[i] <= a[k]) return i; if (a[j] <= a[k] && a[k] <= a[i]) return k; if (a[k] <= a[i] && a[i] <= a[j]) return i; return j; }
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For the quick sort with reversion to the insertion sort on arrays of size < 8, the only changes needed are in the sort() method:
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private static void sort(int[] a, int p, int q) { if (q - p < 2) { return; } if (q - p < 8) { insertionSort(a, p, q); return; } int m = partition(a, p, q); sort(a, p, m); sort(a, m+1, q); } // steps 2 & 3 // step 4
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This requires a generalization of the insertion sort:
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public static void insertionSort(int[] a, int p, int q) { for (int i = p+1; i < q; i++) { int ai = a[i], j; for (j = i; j > 0 && a[j-1] > ai; j--) { a[j] = a[j-1]; } a[j] = ai; } }
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14.29 14.30 14.31
The heap sort is not always preferred over the quick sort because it is slower in the average case. The heap sort is not always preferred over the merge sort because it is not stable. The postcondition of heapify (Algorithm 14.9 on page 266) establishes the loop invariant in step 3. That guarantees that the root s0 is the maximum element of the subsequence. Step 5 inserts that maximum at the end of the subsequence. So when the loop at step 4 is finished, the sequence will be sorted. The heapify algorithm restores the heap property to the complete segment ss by applying the heapifyDown() method from its root. The Las Vegas sort has complexity O(n n ). There are n! different permutations of a deck of n cards. Shuffling them is equivalent to selecting one permutation at random. Only one of the n! permutations leaves the cards in order. So the expected number of random shuffles required before the correct one occurs is n!/2. Then each permutation takes n 1 comparisons to see if it is the correct one. So the total complexity is O(n n!/2). By Stirling s Formula (page 325), O(n n!/2) = O(n!) = O(2 n ).
Graphs
A graph is a nonlinear structure. Like linear data structures (lists), it can be implemented with either an indexed or a linked backing data structure. SIMPLE GRAPHS A (simple) graph is a pair G = (V, E), where V and E are finite sets and every element of E is a two-element subset of V (i.e., an unordered pair of distinct elements of V). The elements of V are called vertices (or nodes), and the elements of E are called edges (or arcs). If e E, then e = {a, b} for some a, b V. In this case, we can denote e more simply as e = ab = ba. We say that the edge e connects the two vertices a and b, that e is incident with a and b, that a and b are incident upon e, that a and b are the terminal points or end points of the edge e, and that a and b are adjacent. The size of a graph is the number of elements in its vertex set. EXAMPLE 15.1 A Simple Graph
Figure 15.1 shows a simple graph (V, E) of size 4. Its vertex set is V = {a, b, c, d}, and its edge set is E = {ab, ac, ad, bd, cd}. This graph has four vertices and five edges.
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