CS322Week 12 - Monday

Last time

What did we talk about last time? Trees Graphing functions

Questions?

Logical warmup

Two vegetarians and two cannibals are on one bank of a river

They have a boat that can hold at most two people

Come up with a sequence of boat loads that will convey all four people safely to the other side of the river

The cannibals on any given bank cannot outnumber the vegetarians…. or else!

Spanning TreesWhich we somehow seemed to skip earlier…

Turning graphs into trees

Consider the following graph that shows all the routes an airline has between citiesMinneapolis

Milwaukee

Chicago

St. Louis

Detroit

Cincinnati

Louisville

Nashville

Turning graphs into trees What if we want to remove routes (to save

money)? How can we keep all cities connected?

MinneapolisMilwaukee

Chicago

St. Louis

Detroit

Cincinnati

Louisville

Nashville

The best tree?

Does this tree have the smallest number of routes?

Why?Minneapolis

Milwaukee

Chicago

St. Louis

Detroit

Cincinnati

Louisville

Nashville

Spanning trees

A spanning tree for a graph G is a subgraph of G that contains every vertex of G and is a tree

Some properties: Every connected graph has a spanning

tree▪ Why?

Any two spanning trees for a graph have the same number of edges▪ Why?

Weighted graphs

In computer science, we often talk about weighted graphs when tackling practical applications

A weighted graph is a graph for which each edge has a real number weight

The sum of the weights of all the edges is the total weight of the graph

Notation: If e is an edge in graph G, then w(e) is the weight of e and w(G) is the total weight of G

A minimum spanning tree (MST) is a spanning tree of lowest possible total weight

Weighted graphs example

Here is the graph from before, with labeled weights

MinneapolisMilwaukee

Chicago

St. Louis

Detroit

Cincinnati

Louisville

Nashville

355

695

262

74

348

269

242

151

230306

83

Finding a minimum spanning tree

Kruskal's algorithm gives an easy to follow technique for finding an MST on a weighted, connected graph

Informally, go through the edges, adding the smallest one, unless it forms a circuit

Algorithm: Input: Graph G with n vertices Create a subgraph T with all the vertices of G (but no edges) Let E be the set of all edges in G Set m = 0 While m < n – 1▪ Find an edge e in E of least weight▪ Delete e from E▪ If adding e to T doesn't make a circuit▪ Add e to T ▪ Set m = m + 1

Output: T

Minimum spanning tree example

Run Kruskal's algorithm on the city graph:

MinneapolisMilwaukee

Chicago

St. Louis

Detroit

Cincinnati

Louisville

Nashville

355

695

262

74

348

269

242

151

230306

83

Minimum spanning tree output

Output:

MinneapolisMilwaukee

Chicago

St. Louis

Detroit

Cincinnati

Louisville

Nashville

355

262

74

242

151

230

83

Prim's algorithm

Prim's algorithm gives another way to find an MST Informally, start at a vertex and add the next closest

node not already in the MST Algorithm:

Input: Graph G with n vertices Let subgraph T contain a single vertex v from G Let V be the set of all vertices in G except for v For i from 1 to n – 1▪ Find an edge e in G such that: ▪ e connects T to one of the vertices in V▪ e has the lowest weight of all such edges

▪ Let w be the endpoint of e in V▪ Add e and w to T▪ Delete w from V

Output: T

Prim fights Kruskal

Apply Kruskal's algorithm to the graph below

Now, apply Prim's algorithm to the graph below

Is there any other MST we could make? 2

3

3

4 11

1

Back to Functions

Discontinuous functions

Recall the definition of the floor of x: x = the largest integer that is less than

or equal to x Graph f(x) = x Defining functions on integers

instead of real values affects their graphs a great deal

Graph p1(x) = x, x R Graph f(n) = n, n N

Multiples of functions

There is a strong visual (and of course mathematical) correlation a function that is the multiple of another function

Examples: g(x) = x + 2 2g(x) = 2x + 4

Given f graphed below, sketch 2f

- 6 - 5 - 4 - 3 - 2 - 1 1 2 3 4 5 6

21

-1-2

Absolute value

Consider the absolute value function f(x) = |x|

Left of the origin it is constantly decreasing

Right of the origin it is constantly increasing

- 6 - 5 - 4 - 3 - 2 - 1

1 2 3 4 5 6

21

-1-2

Increasing and decreasing functions

We say that f is decreasing on the set S iff for all real numbers x1 and x2 in S, if x1 < x2, then f(x1) > f(x2)

We say that f is increasing on the set S iff for all real numbers x1 and x2 in S, if x1 < x2, then f(x1) < f(x2)

We say that f is an increasing (or decreasing) function iff f is increasing (or decreasing) on its entire domain

Clearly, a positive multiple of an increasing function is increasing

Virtually all running time functions are increasing functions

Asymptotic Notation (Big Oh)Student Lecture

Asymptotic Notations

Growth of functions

Mathematicians worry about the growth of various functions

They usually express such things in terms of limits, maybe with derivatives

We are focused primarily on functions that bound running time spent and memory consumed

We just need a rough guide We want to know the order of the

growth

Definitions

Let f and g be real-valued functions defined on the same set of nonnegative real numbers

f is of order at least g, written f(x) is (g(x)), iff there is a positive A R and a nonnegative a R such that A|g(x)| ≤ |f(x)| for all x > a

f is of order at most g, written f(x) is O(g(x)), iff there is a positive B R and a nonnegative b R such that |f(x)| ≤ B|g(x)| for all x > b

f is of order g, written f(x) is (g(x)), iff there are positive A, B R and a nonnegative k R such that A|g(x)| ≤ |f(x)| ≤ B|g(x)| for all x > k

Using the notation

Express the following statements using appropriate notation: 10|x6| ≤ |17x6 – 45x3 + 2x + 8| ≤ 30|x6|, for x > 2

Justify the following:

is (x)

0,1

)92(1515

xx

xxx

6,151

)92(15

xxx

xx

1)92(15

xxx

Properties of -, O-, and - notation

Let f and g be real-valued functions defined on the same set of nonnegative real numbers

1. f(x) is (g(x)) and f(x) is O(g(x)) iff f(x) is (g(x))2. f(x) is (g(x)) iff g(x) is O(f(x))3. f(x) is (f(x)), f(x) is O(f(x)), and f(x) is (f(x))4. If f(x) is O(g(x)) and g(x) is O(h(x)) then f(x) is

O(h(x))5. If f(x) is O(g(x)) and c is a positive real, then cf(x)

is O(g(x))6. If f(x) is O(h(x)) and g(x) is O(k(x)) then f(x) + g(x)

is O(G(x)) where G(x) = max(|h(x)|,|k(x)|) for all x7. If f(x) is O(h(x)) and g(x) is O(k(x)) then f(x)g(x) is

O(h(x)k(x))

Orders of functions

If 1 < x, then x < x2

x2 < x3

… So, for r, s R, where r < s and x >

1, xr < xs

By extension, xr is O(xs)

Proving bounds

Prove a bound for g(x) = (1/4)(x – 1)(x + 1) for x R

Prove that x2 is not O(x) Hint: Proof by contradiction

Polynomials

Let f(x) be a polynomial with degree n f(x) = anxn + an-1xn-1 + an-2xn-2 … + a1x + a0

By extension from the previous results, if an is a positive real, then f(x) is O(xs) for all integers s n f(x) is (xr) for all integers r ≤ n f(x) is (xn)

Furthermore, let g(x) be a polynomial with degree m g(x) = bmxm + bm-1xm-1 + bm-2xm-2 … + b1x + b0

If an and bm are positive reals, then f(x)/g(x) is O(xc) for real numbers c > n - m f(x)/g(x) is not O(xc) for all integers c > n - m f(x)/g(x) is (xn - m)

Algorithm Efficiency

Extending notation to algorithms

We can easily extend our -, O-, and - notations to analyzing the running time of algorithms

Imagine that an algorithm A is composed of some number of elementary operations (usually arithmetic, storing variables, etc.)

We can imagine that the running time is tied purely to the number of operations

This is, of course, a lie Not all operations take the same amount of time Even the same operation takes different amounts of

time depending on caching, disk access, etc.

Running time

First, assume that the number of operations performed by A on input size n is dependent only on n, not the values of the data If f(n) is (g(n)), we say that A is (g(n)) or that A is of

order g(n) If the number of operations depends not only on n

but also on the values of the data Let b(n) be the minimum number of operations where

b(n) is (g(n)), then we say that in the best case, A is (g(n)) or that A has a best case order of g(n)

Let w(n) be the maximum number of operations where w(n) is (g(n)), then we say that in the worst case, A is (g(n)) or that A has a worst case order of g(n)

Time

Approximate Time for f(n) Operations Assuming One Operation Per Nanosecond

f(n) n = 10 n = 1,000 n = 100,000 n = 10,000,000

log2 n3.3 x 10-

9 s 10-8 s 1.7 x 10-8 s 2.3 x 10-8 s

n 10-8 s 10-6 s 0.0001 s 0.01 s

n log2 n

3.3 x 10-

8 s 10-5 s 0.0017 s 0.23 s

n2 10-7 s 0.001 s 10 s 27.8 hours

n3 10-6 s 1 s 11.6 days 31,668 years

2n 10-6 s 3.4 x 10284 years

3.1 x 1030086 years

2.9 x 103010283 years

Computing running time

With a single for (or other) loop, we simply count the number of operations that must be performed:int p = 0;int x = 2;for( int i = 2; i <= n; i++ )p = (p + i)*x;

Counting multiplies and adds, (n – 1) iterations times 2 operations = 2n – 2

As a polynomial, 2n – 2 is (n)

Nested loops

When loops do not depend on each other, we can simply multiply their iterations (and asymptotic bounds)int p = 0;for( int i = 2; i <= n; i++ )

for( int j = 3; j <= n; j++ )p++;

Clearly (n – 1)(n -2) is (n2)

Trickier nested loops

When loops depend on each other, we have to do more analysisint s = 0;for( int i = 1; i <= n; i++ )

for( int j = 1; j <= i; j++ )s = s + j*(i – j + 1);

What's the running time here? Arithmetic sequence saves the day

(for the millionth time)

Iterations with floor

When loops depend on floor, what happens to the running time?int a = 0;for( int i = n/2; i <= n; i++ )

a = n - i; Floor is used implicitly here, because

we are using integer division What's the running time? Hint:

Consider n as odd or as even separately

Sequential search

Consider a basic sequential search algorithm:int search( int[] array, int n, int value)

{for( int i = 0; i < n; i++ )

if( array[i] == value )return i;

return -1;} What's its best case running time? What's its worst case running time? What's its average case running time?

Insertion sort algorithm

Insertion sort is a common introductory sort

It is suboptimal, but it is one of the fastest ways to sort a small list (10 elements or fewer)

The idea is to sort initial segments of an array, insert new elements in the right place as they are found

So, for each new item, keep moving it up until the element above it is too small (or we hit the top)

Insertion sort in code

public static void sort( int[] array, int n){for( int i = 1; i < n; i++ ){

int next = array[i];int j = i - 1;while( j != 0 && array[j] > next ){array[j+1] = array[j];j--;}array[j] = next;

}}

Best case analysis of insertion sort

What is the best case analysis of insertion sort?

Hint: Imagine the array is already sorted

Worst case analysis of insertion sort

What is the worst case analysis of insertion sort?

Hint: Imagine the array is sorted in reverse order

Average case analysis of insertion sort

What is the average case analysis of insertion sort?

Much harder than the previous two! Let's look at it recursively Let Ek be the average number of comparisons

needed to sort k elements Ek can be computed as the sum of the average

number of comparisons needed to sort k – 1 elements plus the average number of comparisons (x) needed to insert the kth element in the right place Ek = Ek-1 + x

Finding x

We can employ the idea of expected value from probability

There are k possible locations for the element to go

We assume that any of these k locations is equally likely

For each turn of the loop, there are 2 comparisons to do

There could be 1, 2, 3, … up to k turns of the loop

Thus, weighting each possible number of iterations evenly gives us 1

2)1(2

21

1

kkk

kj

kx

k

j

Finishing the analysis

Having found x, our recurrence relation is: Ek = Ek-1 + k + 1 Sorting one element takes no time, so E1 =

0 Solve this recurrence relation! Well, if you really banged away at it, you

might find: En = (1/2)(n2 + 3n – 4)

By the polynomial rules, this is (n2) and so the average case running time is the same as the worst case

Upcoming

Next time…

Finish algorithmic efficiency Exponential and logarithmic

functions

Reminders

Keep reading Chapter 11 Keep working on Assignment 9

Due Friday before midnight Exam 3 is next Monday

Review on Friday