Ravindra Ranwala's Blog

Exercise 9.3-1 asks you to show that the $SELECT$ algorithm still runs in linear time if the elements are divided into groups of 7. This problem asks about dividing into groups of 3. Show that $SELECT$ runs in linear time if you divide the elements into groups whose size is any odd constant greater than 3. Show that $SELECT$ runs in $O(n\; lg\; n)$ time if you divide the elements into groups of size 3 Because the bound in part (b) is just an upper bound, we do not know whether the groups-of-3 strategy actually runs in $O(n)$ time. But by repeating the groups-of-3 idea on the middle groups of medians, we can pick a pivot that guarantees $O(n)$ time. The $SELECT3$ algorithm on the next page determines the $i^{\;th}$ smallest of an input array of $n \gt 1$ distinct elements. Describe in English how the $SELECT3$ algorithm works.Include in your description one or more suitable diagrams. Show that $SELECT3$ runs in $O(n)$ time in the worst case. [1]   The worst-case linear-time median-findi

Problem statement Exercise 9.3-8 The $k$th quantiles of an $n$-element set are the $k$ - 1 order statistics that divide the sorted set into $k$ equal sized sets (to within 1). Give an $O(n \cdot lg k)$-time algorithm to list the $k$th quantiles of a set. [1] Definitions In statistics and probability, quantiles are cut points dividing the range of a probability distribution into continuous intervals with equal probabilities, or dividing the observations in a sample in the same way. There is one fewer quantile than the number of groups created. [2] Approach Let's write a divide and conquer algorithm using selection to solve this problem. What if we find the $l = \lfloor\frac{n \cdot \lfloor\frac{k}{2}\rfloor}{k}\rfloor$ th smallest number  and partition the array around it. Then, we can find the first $\lfloor\frac{k}{2}\rfloor$ cut points in the first half $l$ and the remaining $\lceil\frac{k}{2}\rceil$ cut points in the second half $n - l$. At each step, we keep on dividing $k$ by