Concrete mathematics : a foundation for computer science

406 DISCRETE PROBABILITY
The mean value of A(hl , . . . , &), obtained by summing over all m” possibilities
(hl , . . . , &) and dividing by mn, will be the same as the mean value
we obtained before in (8.g4), But the variance of A(hl , , h,) is something
different; this is a variance of mn averages, not a variance of m” .n probe
counts. For example, if m == 1 (so that there is only one list), the “average”
value A(hl, . . . ,&) = A(l). . . , 1) is actually constant, so its variance VA is
zero; but the number of probes in a successful search is not constant, so the
variance VP is nonzero. But the VP is
We can illustrate this difference between variances by carrying out the nonzero “‘yin an
election year.
calculations for general m and n in the simplest case, when sk = l/n for
1 < k 6 n. In other words, we will assume temporarily that there is a uniform
distribution of search keys. Any given sequence of hash values (h, , . , h)
defines m lists that contain respectively (n, ,nz, . . . ,n,) entries for some
numbers ni, where
nl+n2+...+n, = n.
A successful search in which each of the n keys in the table is equally likely
will have an average running time of
(l+...+nl) + (l+...+nz) +...+ (l+...+n,)
A(h,,...,h,) = -
n
nl (nlfl) + nz(n2+1) + . . + n,(n,+l)
=-
2n
n:+n:+...+&+n
zz-
2n
probes. Our goal is to calculate the variance of this quantity A(hl , . . . , &),
over the probability space cionsisting of all m” sequences (hl , . . . , h,).
The calculations will be simpler, it turns out, if we compute the variance
of a slightly different quantity,
We have
B(h,,...,h,) =(?‘)+(T)+...+(y).
A(h,, . . . ,G = 1 +B(h,...,h,)/n,
hence the mean and varianc:e of A satisfy
EA = 1,;; VA = $. (8.99)