Recursively Defined Sequences Lecture 35 Section 8.1 Wed, Mar 23, 2005.

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Recursively Defined Sequences

Lecture 35

Section 8.1

Wed, Mar 23, 2005

Recursive Sequences

A recurrence relation for a sequence {an} is an equation that defines each term of the sequence as a function of previous terms, from some point on.

The initial conditions are equations that specify the values of the first several terms a0, …, an – 1.

Recursive Sequence

Define a sequence {ak} bya0 = 2,a1 = 3,ak = ak – 1 + 2ak – 2, for all k 2.

The next few terms area2 = 7,a3 = 13,a4 = 27.

The Towers of Hanoi

The game board has three pegs, Peg 1, Peg 2, Peg 3, and 10 disks.

Initially, the 10 disks are stacked on Peg 1, each disk smaller than the disk below it.

By moving one disk at a time from peg to peg, reassemble the disks on Peg 3 in the original order.

At no point may a larger disk be placed on a smaller disk.

The Towers of Hanoi

Finish

The Towers of Hanoi

There is a very simple recursive solution.Reassemble the top 9 disks on Peg 2.Move Disk 10 from Peg 1 to Peg 3.Reassemble the top 9 disks on Peg 3.

But how does one reassemble the top 9 disks on Peg 2?

The Towers of Hanoi

It is very simple to reassemble the top 9 disks on Peg 2.Reassemble the top 8 disks on Peg 3.Move Disk 9 from Peg 1 to Peg 2.Reassemble the top 8 disks on Peg 2.

The Towers of Hanoi

It is very simple to reassemble the top 8 disks on Peg 3.Reassemble the top 7 disks on Peg 2.Move Disk 8 from Peg 1 to Peg 3.Reassemble the top 7 disks on Peg 3.

The Towers of Hanoi

Ultimately, the question becomes, how does one reassemble the top 1 disk on Peg 2?

That really is simple: just move it there.

The Towers of Hanoi

How many moves will it take? Let an be the number of moves to

reassemble n disks. Then

a1 = 1.

an = 2an – 1 + 1, for all n 2.

Future Value of an Annuity

Begin with an initial deposit of $d. At the end of each month

Add interest at a monthly interest rate r.Deposit an additional $d.

Let ak denote the value at the end of the k-th month.a0 = d,ak = (1 + r)ak – 1 + d, for all k 1.

The first few terms area0 = d.

a1 = (1 + r)a0 + d = 2d + rd.

a2 = (1 + r)a1 + d = 3d + 3rd + r2d.

a3 = (1 + r)a2 + d = 4d + 6rd + 4r2d + r3d.

What is the pattern?

We might guess that the nonrecursive formula is

an = ((1 + r)n + 1 – 1)(d/r). We will verify this guess later.

Counting Strings

Let = {0, 1}. Let ak be the number of strings in * of

length k that do not contain 11.a0 = 1, {}a1 = 2, {0, 1}a2 = 3, {00, 01, 10}a3 = 5, {000, 001, 010, 100, 101}

What is the pattern?

Counting Strings

Consider strings of length k, for some k 2, that do not contain 11.

If the first character is 0, then the remainder of the string is a string of length k – 1 which does not contain 11.

If the first character is 1, then the next character must be 0 and the remainder is a string that does not contain 11.

Counting Strings

k = 1: {0, 1}

Counting Strings

k = 1: {0, 1} k = 2: {00, 01, 10}

Counting Strings

k = 1: {0, 1} k = 2: {00, 01, 10} k = 3: {000, 001, 010} {100, 101}

Counting Strings

k = 1: {0, 1} k = 2: {00, 01, 10} k = 3: {000, 001, 010, 100, 101}

Counting Strings

k = 1: {0, 1} k = 2: {00, 01, 10} k = 3: {000, 001, 010, 100, 101} k = 4: {0000, 0001, 0010, 0100, 0101}

{1000, 1001, 1010}

Counting Strings

k = 1: {0, 1} k = 2: {00, 01, 10} k = 3: {000, 001, 010, 100, 101} k = 4: {0000, 0001, 0010, 0100, 0101,

1000, 1001, 1010}

Counting Strings

Therefore,ak = ak – 1 + ak – 2, for all k 2.

The next few terms area3 = a2 + a1 = 5,

a4 = a3 + a2 = 8,

a5 = a4 + a3 = 13.

Counting r-Partitions

An r-partition of a set is a partition of the set into r nonempty subsets.

Let A be a set of size n. Let an, r be the number of distinct r-

partitions of A. Special cases

an, n = 1 for all n 1.an, 1 = 1 for all n 1.

Let A = {a, b, c, d}. a4, 2 = 7 since the 2-partitions are

{{a}, {b}, {c, d}}{{a}, {c}, {b, d}}{{a}, {d}, {b, c}}{{b}, {c}, {a, d}}{{b}, {d}, {a, c}}{{c}, {d}, {a, b}}

Consider an r-partition of a set A. Let x A. Either x is in a set {x} by itself or it isn’t. If it is, then the remaining sets form an

(r – 1)-partition of A – {x}. If it isn’t, then if we remove x, we have an r-

partition of A – {x}.

The 3-partitions that contain {a}.{{a}, {b}, {c, d}}{{a}, {c}, {b, d}}{{a}, {d}, {b, c}}

The 3-partitions that do not contain {a}.{{b}, {c}, {a, d}}{{b}, {d}, {a, c}}{{c}, {d}, {a, b}}

The 3-partitions that contain {a}.{{a}, {b}, {c, d}} {{b}, {c, d}}{{a}, {c}, {b, d}} {{c}, {b, d}} {{a}, {d}, {b, c}} {{d}, {b, c}}

The 3-partitions that do not contain {a}.{{b}, {c}, {a, d}}{{b}, {d}, {a, c}}{{c}, {d}, {a, b}}

Distinct2-partitionsof {b, c, d}

The 3-partitions that contain {a}.{{a}, {b}, {c, d}} {{b}, {c, d}}{{a}, {c}, {b, d}} {{c}, {b, d}} {{a}, {d}, {b, c}} {{d}, {b, c}}

The 3-partitions that do not contain {a}.{{b}, {c}, {a, d}} {{b}, {c}, {d}} {{b}, {d}, {a, c}} {{b}, {d}, {c}} {{c}, {d}, {a, b}} {{c}, {d}, {b}}

Three copies ofsame 3-partition

of {b, c, d}

In fact, we get the same r-partition of A – {x} that we would get had x been a member of any other set in the partition.

Thus, each r-partition of A – {x} gives rise to r r-partitions of A.

Therefore,

an, r = an – 1, r – 1 + r an – 1, r , for all n 1 and for all r, 1 < r < n.

Compute a4, 2. and a5, 2.a4, 2 = a3, 1 + 2a3, 2

= 1 + 2(a2, 1 + 2a2, 2)= 1 + 2(1 + 2 1)= 7.

a5, 2 = a4, 1 + 2a4, 2

= 1 + 2 7= 15.

Taxicab Routes

In a city, streets run either east-west or north-south.

They form a grid, like graph paper. Let am, n denote the number of routes that a

taxicab may follow to reach a point that is m blocks east and n blocks north of its current location.

Assume the cab travels only east and north.

Taxicab Routes

Special casesa0, n = 1 for all n 0.

am, 0 = 1 for all m 0.

One block from the destination, he is either one block south or one block west.

One Block South

4 6 Grid

One block south

One Block West

4 6 Grid

One block north

Taxicab Routes

If he is one block south of the destination, then there are am,n – 1 routes to get to that point, followed by one route (north) to get to the destination.

One Block South

4 6 Grid

One Block South

3 6 Grid

Taxicab Routes

If he is one block west of the destination, there there are am – 1,n routes to get to that point, followed by one route (east) to get to the destination.

One Block West

4 6 Grid

One Block West

4 5 Grid

Taxicab Routes

Therefore,

am, n = am – 1, n + am, n – 1, for all m, n 1. Calculate the first few terms.

Recursively Defined Sequences Lecture 35 Section 8.1 Wed, Mar 23, 2005.

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