Week 6 - Friday. What did we talk about last time? Solving recurrence relations.

CS322Week 6 - Friday

Last time

What did we talk about last time? Solving recurrence relations

Questions?

Logical warmup

This is a donkey made of toothpicks

Move exactly one toothpick to make an identical donkey oriented in another direction

Solving Second-Order Linear Homogeneous Relations with Constant CoefficientsStudent Lecture

Solving Second-Order Linear Homogeneous Relations with Constant Coefficients

Second-Order Linear Homogeneous Relations with Constant Coefficients Second-order linear homogeneous relations

with constant coefficients are recurrence relations of the following form: ak = Aak-1 + Bak-2 where A, B R and B 0

These relations are: Second order because they depend on ak-1 and ak-2

Linear because ak-1 and ak-2 are to the first power and not multiplied by each other

Homogeneous because there is no constant term Constant coefficients because A and B are fixed

values

Why do we care?

I'm sure you're thinking that this is an awfully narrow class of recurrence relations to have special rules for

It's true: There are many (infinitely many) ways to formulate a recurrence relation Some have explicit formulas Some do not have closed explicit formulas

We care about this one partly for two reasons1. We can solve it2. It lets us get an explicit formula for Fibonacci!

Pick 'em out

Which of the following are second-order linear homogeneous recurrence relations with constant coefficients?

a) ak = 3ak-1 + 2ak-2

b) bk = bk-1 + bk-2+ bk-3

c) ck = (1/2)ck-1 – (3/7)ck-2

d) dk = d2k-1 + dk-1dk-2

e) ek = 2ek-2

f) fk = 2fk-1 + 1g) gk = gk-1 + gk-2

h) hk = (-1)hk-1 + (k-1)hk-2

Characteristic equation

We will use a tool to solve a SOLHRRwCC called its characteristic equation

The characteristic equation of ak = Aak-1 + Bak-2 is: t2 – At – B = 0 where t 0

Note that the sequence 1, t, t2, t3, …, tn satisfies ak = Aak-1 + Bak-2 if and only if t satisfies t2 – At – B = 0

Demonstrating the characteristic equation

We can see that 1, t, t2, t3, …, tn satisfies ak = Aak-1 + Bak-2 if and only if t satisfies t2 – At – B = 0 by substituting in t terms for ak as follows: tk = Atk-1 + Btk-2

Since t 0, we can divide both sides through by tk-2

t2 = At + B t2 – At – B = 0

Using the characteristic equation

Consider ak = ak-1 + 2ak-2 What is its characteristic equation?

t2 – t – 2 = 0 What are its roots?

t = 2 and t = -1 What are the sequences defined by each

value of t? 1, 2, 22, 23, …, 2n, … 1, -1, 1, -1, …, (-1)n, …

Do these sequences satisfy the recurrence relation?

Finding other sequences

An infinite number of sequences satisfy the recurrence relation, depending on the initial conditions

If r0, r1, r2, … and s0, s1, s2, … are sequences that satisfy the same SOLHRRwCC, then, for any C, D R, the following sequence also satisfies the SOLHRRwCC an = Crn + Dsn, for integers n ≥ 0

Finding a sequence with specific initial conditions

Solve the recurrence relation ak = ak-1 + 2ak-2 where a0 = 1 and a1 = 8

The result from the previous slide says that, for any sequence rk and sk that satisfy ak an = Crn + Dsn, for integers n ≥ 0

We have these sequences that satisfy ak

1, 2, 22, 23, …, 2n, … 1, -1, 1, -1, …, (-1)n, …

Thus, we have a0 = 1 = C20 + D(-1)0 = C + D

a1 = 8 = C21 + D(-1)1 = 2C – D Solving for C and D, we get:

C = 3 D = -2

Thus, our final result is an = 3∙2n – 2(-1)n

Distinct Roots Theorem

We can generalize this result Let sequence a0, a1, a2, … satisfy:

ak = Aak-1 + Bak-2 for integers k ≥ 2 If the characteristic equation t2 – At –

B = 0 has two distinct roots r and s, then the sequence satisfies the explicit formula: an = Crn + Dsn

where C and D are determined by a0 and a1

Now we can solve Fibonacci! Fibonacci is defined as follows:

Fk = Fk-1 + Fk-2, k ≥ 2

F0 = 1

F1 = 1 What is its characteristic equation?

t2 – t – 1 = 0 What are its roots?

Thus,

251

t

nn

n DCF

251

251

We just need C and D

So, we plug in F0 = 1 and F1 = 1

Solving for C and D, we get

Substituting in, this yields

251

251

251

251

112

512

51

11

1

00

0

DCDCF

DCDCF

52

51C

52

)51( D

11

251

51

251

51

251

52

)51(2

51

52

51

nnnn

nF

Single-Root Case

Our previous technique only works when the characteristic equation has distinct roots r and s

Consider sequence ak = Aak-1 + Bak-2

If t2 – At – B = 0 only has a single root r, then the following sequences both satisfy ak

1, r1, r2, r3, … rn, … 0, r, 2r2, 3r3, … nrn, …

Single root equation

Our old rule said that if r0, r1, r2, … and s0, s1, s2, … are sequences that satisfy the same SOLHRRwCC, then, for any C, D R, the following sequence also satisfies the SOLHRRwCC an = Crn + Dsn, for integers n ≥ 0

For the single root case, this means that the explicit formula is: an = Crn + Dnrn, for integers n ≥ 0

where C and D are determined by a0 and a1

SOLHRRwCC recap

To solve sequence ak = Aak-1 + Bak-2

Find its characteristic equation t2 – At – B = 0

If the equation has two distinct roots r and s Substitute a0 and a1 into an = Crn + Dsn

to find C and D If the equation has a single root r

Substitute a0 and a1 into an = Crn + Dnrn to find C and D

General Recursive Definitions

Defining anything recursively

For example, we can define sets recursively

To do so, we need three things:I. Base: A statement of that certain

things are in the setII. Recursion: A set of rules saying how

new objects can be shown to be in the set based on ones that are already known to be in the set

III. Restriction: A statement that no objects belong to the set other than those coming from I and II

Example

The set P of all strings of legal parenthesizations

I. Base: () is in PII. Recursion:

a. If E is in P, so is (E)b. If E and F are in P, so is E F

III. Restriction: No configurations of parentheses are in P other than those derived from I and II

Ackermann function

Even functions can be defined recursively The Ackermann function is famous because

it grows faster than any algebraic function It is defined for all non-negative integers as

follows: A(0,n) = n + 1 A(m,0) = A(m – 1, 1) A(m,n) = A(m – 1, A(m,n – 1))

Find A(1,2) I won't make you find A(4,4), because it is

roughly

65536222

Set Theory

Sets

A set is a collection of elements The set is defined entirely by its

elements Order doesn't matter Repetitions don't matter (but, in general,

we write each element of set only a single time)

Examples { 1, 4, 9, 25 } { 1 } is not the same as 1 { } is the empty set

Notation

For a finite set, we can write all of the elements inside curly braces

For infinite sets, we don't have that luxury

What's in each of the following sets? A = { x R | -3 < x < 4 } B = { x Z | -3 < x < 4 } C = { x Z+ | -3 < x < 4 } D = { x Q | -3 < x < 4 }

Subsets

Perhaps the simplest relationship between sets is the subset relationship

We say that X is a subset of Y iff every element of X is an element of Y

Notation X Y Examples:

Is { 1, 2, 3 } { 1, 2, 3, 4 } ? Is { 1, 2, 3, 4 } { 1, 2, 3 } ? Is { 1, 2, 3 } { 1, 2, 3 } ? Is { } { 1, 2, 3 } ? Is { {1}, {2}, {3} } { 1, 2, 3 } ?

Proper subsets

We say that X is a proper subset of Y iff every element of X is an element of Y, but there is some element of Y that is not an element of X

Notation X Y Examples:

Is { 1, 2, 3 } { 1, 2, 3, 4 } ? Is { 1, 2, 3, 4 } { 1, 2, 3 } ? Is { 1, 2, 3 } { 1, 2, 3 } ? Is { } { 1, 2, 3 } ? Is { {1}, {2}, {3} } { 1, 2, 3 } ?

Venn diagrams

Venn diagrams show relationships between sets They can help build intuition about relationships,

but they do not prove anything Also, their usefulness is limited to 2 or 3 sets Beyond that, they become difficult to interpret Example: For X Y, there are 3 possibilities:

X Y X Y

X

Y

vs.

The idea of being an element of a set is strongly tied to the idea of one set being the subset of another

These two relationships are different, however x X means that x is an element of X Y X means that Y is a subset of X Which of the following are true?

2 {1, 2, 3} {2} {1, 2, 3} 2 {1, 2, 3} {2} {1, 2, 3} {2} {{1}, {2}} {2} {{1}, {2}}

Set equality

We say that two sets are equal if they contain exactly the same elements

Another way of saying this is that X = Y iff X Y and Y X

Examples: A = { n Z | n = 2p, p Z } B = the set of all even integers C = { m Z | m = 2q – 2, q Z } D = { k Z | k = 3r + 1, r Z }

Does A = B? Does A = C? Does A = D?

Set operations

We usually discuss sets within some superset U called the universe of discourse

Assume that A and B are subsets of U

The union of A and B, written A B is the set of all elements of U that are in either A or B

The intersection of A and B, written A B is the set of all elements of U that are in A and B

The difference of B minus A, written B – A, is the set of all elements of U that are in B and not in A

The complement of A, written Ac is the set of all elements of U that are not in A

Examples

Let U = {a, b, c, d, e, f, g} Let A = {a, c, e, g} Let B = {d, e, f, g} What are:

A B A B B – A Ac

The empty set

There is a set with no elements in it called the empty set

We can write the empty set { } or It comes up very often For example, {1, 3, 5} {2, 4, 6} = The empty set is a subset of every

other set (including the empty set)

Disjoint sets and partitions

Two sets A and B are considered disjoint if A B =

Sets A1, A2, … An are mutually disjoint (or nonoverlapping) if Ai Aj = for all i j

A collection of nonempty sets {A1, A2, … An} is a partition of set A iff:1. A = A1 A2 … An

2. A1, A2, … An are mutually disjoint

Power set

Given a set A, the power set of A, written P(A) or 2A is the set of all subsets of A

Example: B = {1, 3, 6}P(B) = {, {1}, {3}, {6}, {1,3},

{1,6}, {3,6}, {1,3,6}} Let n be the number of elements in

A, called the cardinality of A Then, the cardinality of P(A) is 2n

Cartesian product

An ordered n-tuple (x1, x2, … xn) is an ordered sequence of n elements, not necessarily from the same set

The Cartesian product of sets A and B, written A x B is the set of all ordered 2-tuples of the form (a, b), a A, b B

Thus, (x, y) points are elements of the Cartesian product R x R (sometimes written R2)

Subset Relations

Basic subset relations

Inclusion of Intersection: For all sets A and B A B A A B B

Inclusion in Union: For all sets A and B A A B B A B

Transitive Property of Subsets: If A B and B C, then A C

Element argument

The basic way to prove that X is a subset of Y1. Suppose that x is a particular but

arbitrarily chosen element of X2. Show that x is an element of Y

If every element in X must be in Y, by definition, X is a subset of Y

Procedural versions

We want to leverage the techniques we've already used in logic and proofs

The following definitions help with this goal:1. x X Y x X x Y2. x X Y x X x Y3. x X – Y x X x Y4. x Xc x X5. (x, y) X Y x X y Y

Example proof

Theorem: For all sets A and B, A B A

Proof: Let x be some

element in A B x A x B x A Thus, all elements in

A B are in A A B AQED

Premise

Definition of intersection

Specialization By generalization

Definition of subset

Laying down the law (again)

Name Law Dual

Commutative A B = B A A B = B A

Associative (A B) C = A (B C) (A B) C = A (B C)

Distributive A (B C) = (A B) (A C) A (B C) = (A B) (A C)

Identity A = A A U = A

Complement A Ac = U A Ac =

Double Complement (Ac)c = A

Idempotent A A = A A A = A

Universal Bound A U = U A =

De Morgan’s (A B)c = Ac Bc (A B)c = Ac Bc

Absorption A (A B) = A A (A B) = A

Complements of U and

Uc = c = U

Set Difference A – B = A Bc

Proving set equivalence

To prove that X = Y Prove that X Y and Prove that Y X

Example proof of equivalenceTheorem: For all sets A,B, and C, A (B C) = (A B) (A C)Proof: Let x be some element in A (B C) x A x (B C) Case 1: Let x A

x A x B x A B x A x C x A C x A B x A C x (A B) (A C)

Case 2: Let x B C x B x C x B x A x B x A B x C x A x C x A C x A B x A C x (A B) (A C)

In all possible cases, x (A B) (A C), thus A (B C) (A B) (A C)

Proof of equivalence continued Let x be some element in (A B) (A C) x (A B) x (A C) Case 1: Let x A

x A x B C x A (B C)

Case 2: Let x A x A B x A x B x B x A C x A x C x C x B x C x B C x A x B C x A (B C)

In all possible cases, x A (B C), thus (A B) (A C) A (B C) Since both A (B C) (A B) (A C) and (A B) (A C) A (B C), A (B

C) = (A B) (A C)QED

Proof example

Prove that, for any set A, A = Hint: Use a proof by contradiction

Upcoming

Next time…

More on set theory Russell’s paradox

Reminders

Homework 4 is due Monday Keep reading Chapter 6