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problem that makes union easy but find hard-the quick-union algorithm Even so, the running time for any sequence of at most M find and N - 1 union operations is only negligibly more than O(M + N ) time, and moreover, only a single array of integers is used
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A tree is represented by an array of integers representing parent nodesThe set name of any node in a tree is the root of a tree
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The union operation is constant time
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The cost of a find depends on the depth of the accessed node and could be linear
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Recall that the unionlfind problem does not require a find operation to return any specific name, just that finds on two elements return the same answer if and only if they are in the same set One possibility might be to use a tree to represent a set, as each element in a tree has the same root and the root can be used to name the set Each set is represented by a tree (recall that a collection of trees is called a forest) The name of a set is ~ i v e n the node at the root Our trees are not bv " necessarily binary trees, but their representation is easy because the only information we need is the parent Thus we need only an array of integers: Each entry p [ iI in the array represents the parent of element i,and we can use -1 as a parent to indicate a root Figure 2412shows a forest and the array that represents it To perform a union of two sets, we merge the two trees by making the root of one tree a child of the root of the other This operation clearly takes constant time Figures 2413-2415 represent the forest after each of union(4, 5) union(6, 7 ,and union(4,6), where we have adopted the con) vention that the new root after union@, y) is x A find operation on element x is performed by returning the root of the tree containing x The time for performing this operation is proportional to the number of nodes on the path from x to the root The union strategy outlined previously enables us to create a tree whose every node is on the path
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Figure 2412 A forest and its eight elements, initially in different sets
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Figure 2413 The forest after the union of trees with roots 4 and 5
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Figure 2414 The forest after the union of trees with roots 6 and 7
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Figure 2415 The forest after the union of trees with roots 4 and 6
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to x, resulting in a worst-case running time of O(N) per f i n d Typically (as shown in the preceding applications), the running time is computed for a sequence of M intermixed instructions In the worst case, M consecutive operations could take O(MN) time
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Quadratic running time for a sequence of operations is generally unacceptable Fortunately, there are several ways to easily ensure that this running time does not occur
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union-by-size guarantees logarithmic finds
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Instead of -1 being for the negative of the size is stored
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We performed the previous unions rather arbitrarily by making the second tree a subtree of the first A simple improvement is always to make the smaller tree a subtree of the larger, breaking ties by any method, an approach called union-by-size The preceding three u n i o n operations were all ties, so we can consider that they-were peLformed by size If the next operation is union(3, 4), the forest shown in Figure 2416 forms Had the size heuristic not been used, a deeper forest would have been formed (three nodes rather than one would have been one level deeper) If the u n i o n operation is done by size, the depth of any node is never more than log N A node is initially at depth 0, and when its depth increases as a result of a u n i o n , it is placed in a tree that is at least twice as large as before Thus its depth can be increased at most log N times (We used this argument in the quick-find algorithm in Section 243) This outcome implies that the running time for a f i n d operation is O(log N) and that a sequence of M operations takes at most O(M log N ) time The tree shown in Figure 2417 illustrates the worst tree possible after 15 u n i o n operations and is obtained if all the u n i o n s are between trees of equal size (The worst-case tree is called a binomial tree Binomial trees have other applications in advanced data structures) To implement this strategy, we need to keep track of the size of each tree Since we are just using an array, we can have the array entry of the root contain the negative of the size of the tree, as shown in Figure 2416 Thus the initial representation of the tree with all -1s is reasonable When a
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