# CSC 301.01, Class 05: Divide-and-conquer algorithms and structures

Overview

• Preliminaries
• Notes and news
• Upcoming work
• Extra credit
• Questions
• Quick followup on asymptotic analysis
• Divide and conquer, in theory
• Review of divide-and-conquer algorithms and structures
• Finding the median value in an array
• Exponentiation with integer exponents
• Analyzing divide-and-conquer algorithms

### News / Etc.

• Happy Labor day.
• Piazza is useful.

### Upcoming work

• Rosenfield symposium, this week. (Lots of different events)
• CS Table on Doxxing.

### Extra credit (Peer)

• Coffee Hour with German House, Tuesday, Above the Grill, 4:15 pm

### Extra Credit (Misc)

• Community Hour (Dialogues Across Difference), Tuesday at 11 a.m. in JRC 209.
• CLS Kick-Off Event, Tuesday at 11 a.m. in “North Campus Grove”.
• Host a Prospie!

## Quick followup on asymptotic analysis

We have a formal definition of big-O and an informal understanding of what the parts of the definition represent. What changes about the definition when we do Big-Theta and Big-Omega.

• How does the big O definition change for Theta(n) or Omega(n)
• Yay! The folks called on seem fairly confident.

How do you figure out the big-O (or other) bound on an algorithm?

• For an iterative algorithm, you count the loops.
• For a recursive algorithm, think about it.

Sample iterative algorithm

i = n
while (i > 1)
for (j = 0; j < i; j++)
some-constant-time-operation
end for
i = i/2;
end while

• The outer loop runs log_2(n)-ish.
• The inner loop runs O(n)-ish, at worst.
• So, we multiply the two and say this is an O(nlogn) algorithm.

Is there a tighter bound?

• n + n/2 + n/4 + … + 1
• n
• n + n/2 = 3/2 n
• n + n/2 + n/4 = 7/4 n
• n + n/2 + n/4 + n/8 = 15/8 n
• < 2*n
• O(n)

Moral: We may have to think more carefully about how the loops are interacting.

Another approach: Run the program on lots of data points and interpolate the graph.

## Divide and conquer, in theory

An approach to algorithm design that involves splitting the input in half, solving the problem (perhaps modified slightly) on one or both halves, and then combining those solutions into a solution to the overall problem.

• Example: Find Occifer Motta

## Review of divide-and-conquer algorithms and structures

What are some algorithms and data structures that you have encountered that involve divide and conquer?

• Trees - Divide and conquer data structure (heaps, binary search trees)
• Binary search
• Quicksort
• Merge sort

The divide should be reasonably good for the behavior we want from divide-and-conquer algorithms. For example, Quicksort works poorly if the pivot is always the second-largest value.

Complexities:

• Choosing what to divide.
• Choosng how to divide.

## Finding the median value in an array of values

Given an unsorted array, [a1 … an], where n is odd and no two values are the same, find the value a such that (n-1)/2 elements are smaller than a and (n-1)/2 elements are larger. You may not sort the array, but you may rearrange the values.

### Idea zero: Ignore the requirements

1. Sort
2. Look in the middle

### Idea one: Just plow ahead

1. Find the median of the left half (not including the middle element?)
2. Find the median of the right half (not including the middle element?)
3. Do something with those two medians

Example input: 1,10,100,1000,2,20,200

### Idea two: Change the problem slightly - Put the median in the middle.

1. Pick a pivot randomly from the elements of the array
2. Rearrange the elements so that everything less than the pivot is left of the pivot, everything greater than the pivot is to the right.
3. Repeat on the subarray.
• n steps to partition the first time
• n/2 steps to partition the second time

### Idea three: Change the problem significantly - Find the kth largest

element in an arbitrary subarray.

## Exponentiation with integer exponents

In a language that provides arbitrary precision numbers, compute x^n for non-negative integer n.

• If n is even, x^n = square(x^(n/2))
• If n is odd, x^n = x*(square(x^((n-1)/2)))

On Wednesday.