CS322Week 11 - Friday

Last time

What did we talk about last time? Graph isomorphism Trees

Questions?

Logical warmup

You are the despotic ruler of an ancient empire Tomorrow is the 25th anniversary of your reign You have 1,000 bottles of wine you were planning to open for the

celebration Your Grand Vizier uncovered a plot to murder you:

10 years ago, a rebel worked in the vineyard that makes your wine He poisoned one of the 1,000 bottles you are going to serve tonight and

has been waiting for revenge The rebel died an accidental death, and his papers revealed his plan, but

not which bottle was poisoned The poison exhibits no symptoms until death Death occurs within ten to twenty hours after consuming even the

tiniest amount of poison You have over a thousand slaves at your disposal and just under 24

hours to determine which single bottle is poisoned You have a few hundred prisoners, sentenced to death Can you use just the prisoners and risk no slaves? If so, what's the smallest number of prisoners you can risk and still

be sure to find the bottle?

Finishing Trees

Parse trees

A grammar for a formal language (such as we will discuss next week or the week after) is made up of rules that allow non-terminals to be turned into other non-terminals or terminals

For example:1. <sentence> <noun phrase><verb phrase>2. <noun phrase> <article><noun> |

<article><adjective><noun>3. <verb phrase> <verb><noun phrase>4. <article> a | an | the5. <adjective> funky6. <noun> DJ | beat7. <verb> plays | spins

Make a parse tree corresponding to the sentence, "The DJ plays a funky beat"

Describing trees

Any tree with more than one vertex has at least one vertex of degree 1

If a vertex in a tree has degree 1 it is called a terminal vertex (or leaf)

All vertices of degree greater than 1 in a tree are called internal vertices (or branch vertices)

A property of trees

For any positive integer n, a tree with n vertices must have n – 1 edges

Prove this by mathematical inductionHint: Any tree with 2 or more nodes

has a vertex of degree 1. What happens when you remove that vertex?

Rooted trees

In a rooted tree, one vertex is distinguished and called the root

The level of a vertex is the number of edges along the unique path between it and the root

The height of a rooted tree is the maximum level of any vertex of the tree

The children of any vertex v in a rooted tree are all those nodes that are adjacent to v and one level further away from the root than v

Two distinct vertices that are children of the same parent are called siblings

If w is a child of v, then v is the parent of w If v is on the unique path from w to the root, then v is

an ancestor of w and w is a descendant of v

Rooted tree example

Consider the following tree, rooted at 0

What is the level of 5? What is the level of 0? What is the height of this tree? What are the children of 3? What is the parent of 2? What are the siblings of 8? What are the descendants of 3?

3

0

5

21

7 9

6

4

8

Binary trees

A binary tree is a rooted tree in which every parent has at most two children

Each child is designated either the left child or the right child

In a full binary tree, each parent has exactly two children

Given a parent v in a binary tree, the left subtree of v is the binary tree whose root is the left child of v

Ditto for right subtree

Binary tree applications

As we all know from data structures, a binary tree can be used to make a data structure that is efficient for insertions, deletions, and searches

But, it doesn't stop there! We can represent binary arithmetic with a

binary tree Make a binary tree for the expression ((a –

b)∙c) + (d/e) The root of each subtree is an operator Each subtree is either a single operand or another

expression

Full Binary Tree Theorem 1

If k is a positive integer and T is a full binary tree with k internal vertices, then T has a total of 2k + 1 vertices and has k + 1 terminal vertices

Prove it!Hint: Induction isn't needed. We

just need to relate the number of non-terminal nodes to the number of terminal nodes

Full Binary Tree Theorem 2

If T is a perfect binary tree with height h, then it has 2h+1 – 1 vertices

Prove it using induction!

Generalizing that result

If T is a binary tree with t terminal vertices and height h, then t 2h

Prove it using strong induction on the height of the tree

Hint: Consider cases where the root of the tree has 1 child and 2 children separately

Graphing Functions

Graphing functions

I am assuming that you have been graphing functions since about 6th or 7th grade

Still, the formal definition of a graph of a function is: Let f be a real-valued function of a real

variable The graph of f is the set of all points (x,

y) in the Cartesian coordinate plane such that x is in the domain of f and y = f(x)

Power functions

In computer science, we are often interested in power functions of x written pa(x) where pa(x) = xa where x and a are nonnegative

Power functions are the building blocks of polynomial functions

Graph the following on the same coordinate axes p0

p1/2

p1

p2

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)

Upcoming

Next time…

More on asymptotic notation

Reminders

Keep reading Chapter 11 Keep working on Assignment 9

Due next Friday