CSCI 212 - Class Notes
02/10/09
Select Problem
Given: A list Aof nelements a1, a2, . . . , anin an arbitrary order and an integer k, 1 ≤k≤n.
Objective: Find the kth smallest element in O(n) time.
We find the desired element by recursively partitioning the list around the median-of-medians. To
find the median-of-medians, we divide the list into groups of size 7, create a list of the medians
from each group, and find the median of the list of medians recursively.
The elements are grouped as shown below. When the total number of elements is not evenly di-
visible by 7, i.e., n6= 7bn
7c, the remaining elements a7bn
7c+1, . . . , anare not included in any of the
groups. Note that we used groups of size 5 when previously discussing the algorithm.
a1, a2, . . . , a7
| {z }
g1
, a8, a9, . . . , a14
| {z }
g2
, . . . , a7(i−1)+1, . . . , a7i
| {z }
gi
, . . . , a7(bn
7c−1)+1, . . . , a7bn
7c
| {z }
gbn
7c
Our algorithm appears below. We denote with mithe median of group gi.
Select(A,k)
1. Compute M, the list of the medians from each group. M← {m1, m2, . . . , mbn
7c}.
2. Compute m∗, the median of the list of medians. m∗←Select(M,db n
7c/2e).
3. Partition the elements of Ainto three groups L,E, and Rsuch that (i) ∀a∈L, a < m∗; (ii)
∀a∈E, a =m∗; and (iii) ∀a∈R, a > m∗.
4. If |L|< k ≤ |L|+|E|, return m∗.
If k < |L|, return Select(L,k).
Otherwise, return Select(R,k− |L|−|E|).
Let T(n) denote the time complexity of calling Select on a list with nelements. From the four steps
of the algorithm, T(n) = O(n) + T(bn
7c) + O(n) + T(max {|L|,|R|}). Combining terms and forming
an upper bound, we have T(n)≤cn +T(n
7) + T(max {|L|,|R|}). To obtain an upper bound for
|L|and |R|, consider each group to be a sorted column of elements in a matrix containing exactly
one column for each group. Now, arrange the columns so that the medians, i.e., middle elements
of the columns, are in a non-decreasing order. The number of elements in Athat are ≤m∗(≥m∗)
is ≥4§bn
7/2c¨≥2bn
7c. This implies that |R|(|L|), the number of elements in Athat are greater
(less) than m∗, is ≤n−2bn
7c ≤ n−2¡n−6
7¢=5
7n+12
7.
We now find two constants αand βsuch that |L| ≤ 5
7n+12
7≤βn,|R| ≤ 5
7n+12
7≤βn, and
T(n)≤αcn. By substitution, we have T(n)≤cn +T(n
7) + T(βn). This implies that we need
cn +αcn
7+αcβn ≤αcn. Simplifying, we have 1 + α
7+αβ ≤α⇒1 + αβ ≤6
7α⇒β < 6
7. For β,
the previously-obtained upper bound combined with the lower bound implied from 5
7n+12
7≤βn
yields 5
7< β < 6
7. We choose β=3
4. Substituting for β, we have 1 + 3
4α≤6
7α⇒α≥28
3. We
choose α= 10. We derived our choices of αand βbased on 5
7n+12
7≤βn, which can be shown, by
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