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grows even slower than log*N For instance, A(3, 1) = A(2, 2) = 222= 16 Thus 3 for N < 2 ' 6 , a(M, N) I Further, because A(4, 1) = A(3, 2) = A (2, A(3, 1)) = A(2, 16), which is 2 raised to a power of 16 stacked 2s, in practice, a(M, N) 14However, a(M, N) is not a constant when M is slightly more than N, so the running time is not linears In the remainder of this section, we prove a slightly weaker result We show that any sequence of M = R(N) union and find operations takes a total of O(M log*N) time The same bound holds if we replace union-byrank with union-by-size This analysis is probably the most complex in this text and is one of the first truly complex analyses ever performed for an algorithm that is essentially trivial to implement By extending this technique, we can show the stronger bound claimed previously

= Q2(1 - x q }.

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In this section, we establish a fairly tight bound on the running time of a sequence of M = R(N) union and find operations The union and find operations may occur in any order, but union is done by rank and find is done with path compression We begin with some theorems concerning the number of nodes of rank r Intuitively, because of the union-by-rank rule, there are many more nodes of small rank than of large rank In particular, there can be at most one node of rank log N What we want to do is to produce as precise a bound as possible on the number of nodes of any particular rank r Because ranks change only when union operations are performed (and then only when the two trees have the same rank), we can prove this bound by ignoring path compression We do so in Theorem 241

In the absence of path compression, when a sequence of uni on instructions is being executed, a node qf rank r must have 2' descendants (including itself}

123 109 62 104 57 37 44 100 16 28 138 105 159 75 88 164 169 167 149 167

(14.72)

The proof is by induction The basis r = 0 is clearly true Let T be the tree of rank r with the,fewest number of descendants a n d x be T's root Suppose that the last union with which x was invohved was between T , and T2 Suppose that TI's root was x If T , had rank I; then T I would be a

5 Note however, that if M = N log",

then a(M, N) is at most 2 Thus, so long as M is slightly more than linear the running time is linear in M

<I>(x) ==

tree of rank r with fewer descendants than 7: This condition contradicts the assumption that T is the tree ~ : i tthe smallest number of descendants l~ Hence the rank of T , is at most r - I The rank of T2 is at most the rank of T I because of union-by-rank As T has rank rand the rank could only increase because of Tz, it follow~sthat the rank of T2 is r - 1 Then the rank of T I is also r - I By the induction hypothesis, each tree has at least 2' 1 descendants, giving a total of 2' and establishing the theorem

Source: Daniel and Wood (I980, p. 46).

Proof (continued)

!1 (

Theorem 241 says that if no path compression is performed, any node of rank r must have at least 2 r descendants Path compression can change this condition, of course, because it can remove descendants from a node However, when u n i o n operations are performed-even with path compressionwe are using ranks, or estimated heights These ranks behave as if there is no path compression Thus when the number of nodes of rank r are being bounded, path compression can be ignored, as in Theorem 242

The number of nodes of rank r is at ~ n o s tN / 2 '

Without path compression, each node of rank r is the root of a subtree of at least 2' nodes No other node in the subtree can have rank K Thus all subtrees of nodes of rank rare disjoint Therefore there are at most N / 2 ' disjoint subtrees and hence N / 2 ' nodes of rank K

90 80 70

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