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

Recursive Sequences A recurrence relation for a sequence {a n } 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 a 0, …, a n – 1.

Recursive Sequence Define a sequence {a k } by a 0 = 2, a 1 = 3, a k = a k – 1 + 2a k – 2, for all k  2. The next few terms are a 2 = 7, a 3 = 13, a 4 = 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.

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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?

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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. But how does one reassemble the top 8 disks on Peg 3?

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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. But how does one reassemble the top 7 disks on Peg 2? Etc.

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Ultimately, the question becomes, how does one reassemble the top 1 disk on Peg 2? That really is simple: just move it there.

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How many moves will it take? Let a n be the number of moves to reassemble n disks. Then a 1 = 1. a n = 2a n – 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 a k denote the value at the end of the k- th month. a 0 = d, a k = (1 + r)a k – 1 + d, for all k  1.

Future Value of an Annuity The first few terms are a 0 = d. a 1 = (1 + r)a 0 + d = 2d + rd. a 2 = (1 + r)a 1 + d = 3d + 3rd + r 2 d. a 3 = (1 + r)a 2 + d = 4d + 6rd + 4r 2 d + r 3 d. What is the pattern?

Future Value of an Annuity We might guess that the nonrecursive formula is a n = ((1 + r) n + 1 – 1)(d/r). We will verify this guess later.

Counting Strings Let  = {0, 1}. Let a k be the number of strings in  * of length k that do not contain 11. a 0 = 1, {  } a 1 = 2, {0, 1} a 2 = 3, {00, 01, 10} a 3 = 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, a k = a k – 1 + a k – 2, for all k  2. The next few terms are a 3 = a 2 + a 1 = 5, a 4 = a 3 + a 2 = 8, a 5 = a 4 + a 3 = 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 a n, r be the number of distinct r- partitions of A. Special cases a n, n = 1 for all n  1. a n, 1 = 1 for all n  1.

Counting r-Partitions Let A = {a, b, c, d}. a 4, 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}}

Counting r-Partitions 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}.

Counting r-Partitions 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}}

Counting r-Partitions 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}} Distinct 2-partitions of {b, c, d}

Counting r-Partitions 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 of same 3-partition of {b, c, d}

Counting r-Partitions 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, a n, r = a n – 1, r – 1 + r  a n – 1, r, for all n  1 and for all r, 1 < r < n.

Counting r-Partitions Compute a 4, 2. and a 5, 2. a 4, 2 = a 3, 1 + 2a 3, 2 = 1 + 2(a 2, 1 + 2a 2, 2 ) = 1 + 2(1 + 2  1) = 7. a 5, 2 = a 4, 1 + 2a 4, 2 =  7 = 15.

Taxicab Routes In a city, streets run either east-west or north-south. They form a grid, like graph paper. Let a m, 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 cases a 0, n = 1 for all n  0. a m, 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 a m,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 a m – 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, a m, n = a m – 1, n + a m, n – 1, for all m, n  1. Calculate the first few terms.