Algorithms and Data Structures for External Memory

120 Dynamic and Kinetic Data Structures
involved. In this section, we look at a technique that is based upon
the properties of the problem itself rather than upon that of the data
structure.
We call a problem decomposable if we can answer a query by querying
individual subsets of the problem data and then computing the
final result from the solutions to each subset. Dictionary search and
range searching are obvious examples of decomposable problems. Bentley
developed the logarithmic method [83, 278] to convert efficient static
data structures for decomposable problems into general dynamic ones.
In the internal memory setting, the logarithmic method consists of
maintaining a series of static substructures, at most one each of sizes
1, 2, 4, 8, . . . . When a new item is inserted, it is initialized in a substructure
of size 1. If a substructure of size 1 already exists, the two
substructures are combined into a single substructure of size 2. If there
is already a substructure of size 2, they in turn are combined into a
single substructure of size 4, and so on. For the current value of N, itis
easy to see that the kth substructure (i.e., of size 2 k ) is present exactly
when the kth bit in the binary representation of N is 1. Since there are
at most log N substructures, the search time bound is logN times the
search time per substructure. As the number of items increases from 1
to N, the kth structure is built a total of N/2 k times (assuming N is a
power of 2). If it can be built in O(2 k ) time, the total time for all insertions
and all substructures is thus O(N logN), making the amortized
insertion time O(logN). If we use up to three substructures of size 2 k
at a time, we can do the reconstructions in advance and convert the
amortized update bounds to worst-case [278].
In the EM setting, in order to eliminate the dependence upon the
binary logarithm in the I/O bounds, the number of substructures must
be reduced from logN to log B N, and thus the maximum size of the kth
substructure must be increased from 2 k to B k . As the number of items
increases from 1 to N, the kth substructure has to be built NB/B k
times (when N isapowerofB), each time taking O � B k (log B N)/B �
I/Os. The key point is that the extra factor of B in the numerator of
the first term is cancelled by the factor of B in the denominator of
the second term, and thus the resulting total insertion time over all N
insertions and all log B N structures is O � N(log B N) 2� I/Os, which is