# Recursion

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Recursion

Lecture 14

Quiz

int hello(int n){

if (n==0)return 0;

elseprintf(“Hello World %d\n”,n);hello(n-1);

}

1. What would the program do if I call hello(10)?

2. What if I call hello(-1)?

3. What if the order of printf() and hello() is reversed?

Computing Sum of Arithmetic Progression

int AP(int n){

if (n==0)return 0;

elsereturn (n+AP(n-1));

}

Computing Exponential Function

int EX(int n){

if (n==0)return 1;

elsereturn (EX(n-1)+EX(n-1));

}

How many function calls if I run EX(n)?

Recursively Defined Sequences

We can also define a sequence by specifying its recurrence relation.

•Arithmetic sequence: (a, a+d, a+2d, a+3d, …, )

recursive definition: a0=a, ai+1=ai+d

•Geometric sequence: (a, ra, r2a, r3a, …, )

recursive definition: a0=a, ai+1=rai

•Harmonic sequence: (1, 1/2, 1/3, 1/4, …, )

recursive definition: a0=1, ai+1=iai/(i+1)

The Rabbit Population

• A mature boy/girl rabbit pair reproduces every month.

• Rabbits mature after one month.

wn::= # newborn pairs after n months

rn::= # reproducing pairs after n months

• Start with a newborn pair: w0 =1, r0 = 0

Rabbit Populations

How many rabbits after n months?

wn::= # newborn pairs after n months

rn::= # reproducing pairs after n months

r1 = 1

rn = rn-1 + wn-1

wn = rn-1 so

rn = rn-1 + rn-2

It was Fibonacci who was studying rabbit population growth.

Rabbit Populations

In-Class Exercise

How many n-bit string without the bit pattern 11?

Number of Partitions

How many ways to partition n elements into r non-empty groups?

S(4,4)=1 {x1} {x2} {x3} {x4}

{x1 x2} {x3 x4}{x1 x3} {x2 x4}{x1 x4} {x2 x3}{x1} {x2 x3 x4}{x2} {x1 x3 x4}{x3} {x1 x2 x4}{x4} {x1 x2 x3}

{x1 x2} {x3} {x4}{x2 x3} {x1} {x4}{x1 x3} {x2} {x4}{x2 x4} {x1} {x3}{x1 x4} {x2} {x3}{x3 x4} {x1} {x2}

S(4,2)=7

S(4,3)=6

Number of Partitions

How many ways to partition n elements into r non-empty groups?

Can you write a recurrence relation for S(n,r)?

(page 470-472 of the textbook)

Post #1 Post #2 Post #3

Tower of Hanoi

The goal is to move all the disks to post 3.

The rule is that a bigger disk cannot be placed on top of a smaller disk.

Tower of Hanoi

Can you write a program to solve this problem?

Think recursively!

Post #1 Post #2 Post #3

Move1,2(n)::= Move1,3(n-1);

biggest disk 12;

Move3,2(n-1)

Tower of Hanoi

http://www.mazeworks.com/hanoi/

Tower of Hanoi

Tower_of_Hanoi(int origin, int destination, int buffer, int number){

if (n==0)return;

Tower_of_Hanoi(origin, buffer, destination, number-1);printf(“Move Disk #%d from %d to %d\n”, number, origin, destination);Tower_of_Hanoi(buffer, destination, origin, number-1);

}

Solving Recurrence

a0=1, ak = ak-1 + 2

a1 = a0 + 2

a2 = a1 + 2 = (a0 + 2) + 2 = a0 + 4

a3 = a2 + 2 = (a0 + 4) + 2 = a0 + 6

a4 = a3 + 2 = (a0 + 6) + 2 = a0 + 8

See the pattern is ak = a0 + 2k = 2k+1

Verify by induction:

Solving Hanoi Sequence

a1=1, ak = 2ak-1 + 1

a2 = 2a1 + 1 = 3

a3 = 2a2 + 1 = 2(2a1 + 1) + 1 = 4a1 + 3 = 7

a4 = 2a3 + 1 = 2(4a1 + 3) + 1 = 8a1 + 7 = 15

a5 = 2a4 + 1 = 2(8a1 + 7) + 1 = 16a1 + 15 = 31

a6 = 2a5 + 1 = 2(16a1 + 15) + 1 = 32a1 + 31 = 63

Guess the pattern is ak = 2k-1

Verify by induction:

Solving Fibonacci Sequence

a0=0, a1=1, ak = ak-1 + ak-2

a2 = a1 + a0 = 1

a3 = a2 + a1 = 2a1 + a0 = 2

a4 = a3 + a2 = 2a2 + a1 = 3a1 + 2a0 = 3

a5 = a4 + a3 = 2a3 + a2 = 3a2 + 2a1 = 5a1 + 3a0 = 5

a6 = a5 + a4 = 2a4 + a3 = 3a3 + 2a2 = 5a2 + 3a1 = 8a1 + 5a0 = 8

a7 = a6 + a5 = 2a5 + a4 = 3a4 + 2a3 = 5a3 + 3a2 = 8a2 + 5a1 = 13a1 + 8a0 = 13

See the pattern an = an-kak+1 + an-k-1ak

but this does not give a formula for computing an

Second Order Recurrence Relation

In the book it is called “second-order linear homogeneous recurrence

relation with constant coefficients”.

ak = Aak-1 + Bak-2

A and B are real numbers and B≠0

For example, Fibonacci sequence is when A=B=1.

Geometric-Sequence Solution

ak = Aak-1 + Bak-2

Find solutions of the form (1, t, t2, t3, t4, …, tn, …)

tk = Atk-1 + Btk-2

That is, suppose ak=tk

t2 = At + B

t2 - At – B = 0

So t is a root of the quadratic equation t2 - At – B = 0.

Example

ak = ak-1 + 2ak-2

Find solutions of the form (1, t, t2, t3, t4, …, tn, …)

So t must be a root of the quadratic equation t2 - t – 2 = 0.

This implies that t=2 or t=-1.

So solutions of the form (1, t, t2, t3, t4, …, tn, …) are:

(i) (1,2,4,8,16,32,64,…)

(ii) (1,-1,1,-1,1,-1,…)

Example

ak = ak-1 + 2ak-2

So solutions of the form (1, t, t2, t3, t4, …, tn, …) are:

(i) (1,2,4,8,16,32,64,…)

(ii) (1,-1,1,-1,1,-1,1,…)

Are there other solutions?

Try (2,1,5,7,17,31,65,…)

(0,3,3,9,15,33,63,…)

(4,5,13,23,49,95,193,…)

How to obtain these solutions?

Linear Combination of Two Solutions

If (r0,r1,r2,r3,…) and (s0,s1,s2,s3,…) are solutions to ak = Aak-1 + Bak-2,

then the sequence (a0,a1,a2,a3,…) defined by the formula

ak = Crk + Dsk

also satisfies the same recurrence relation for any C and D.

(page 490 of the textbook)

Distinct-Roots Theorem

Suppose a sequence (a0,a1,a2,a3,…) satisfies a recurrence relation

ak = Aak-1 + Bak-2

If t2 - At – B = 0 has two distinct roots r and s,

then an = Crn + Dsn for some C and D.

(page 491-493 of the textbook)

If we are given a0 and a1, then C and D are uniquely determined.

The theorem says that all the solutions of the recurrence relationare a linear combination of the two series (1,r,r2,r3,r4,…,rn,…) and (1,s,s2,s3,s4,…,sn,…) defined by the distinct roots of t2 - At – B = 0.

Solving Fibonacci Sequence

a0=0, a1=1, ak = ak-1 + ak-2

First solve the quadratic equation t2 - t – 1 = 0.

So the distinct roots are:

Solving Fibonacci Sequence

a0=0, a1=1, ak = ak-1 + ak-2

By the distinct-roots theorem, the solutions satisfies the formula:

To figure out C and D, we substitute the value of a0 and a1:

Multinomial Theorem

Solving these two equations, we get:

Therefore:

Single-Root Case

ak = Aak-1 + Bak-2

Find solutions of the form (1, t, t2, t3, t4, …, tn, …)

So t is a root of the quadratic equation t2 - At – B = 0.

Suppose this quadratic equation has only one root r,

then we know that (1, r, r2, r3, r4, …, rn, …) satisfies the recurrence relation.

Are there other solutions?

ak = Aak-1 + Bak-2

So t is a root of the quadratic equation t2 - At – B = 0.

Another Solution of the Single-Root Case

(0, r, 2r2, 3r3, 4r4, …, nrn, …) also satisfies the recurrence relation.

ak = Aak-1 + Bak-2

Let r be the single root of the quadratic equation t2 - At – B = 0.

Since r is the single root, A=2r and B=-r2.

ak = 2rak-1 - r2ak-2Therefore we just need to verify that

The right hand side is:

which is equal to the left hand side!

Single-Root Theorem

Suppose a sequence (a0,a1,a2,a3,…) satisfies a recurrence relation

ak = Aak-1 + Bak-2

If t2 - At – B = 0 has only one root r,

then an = Crn + Dnrn for some C and D.

If we are given a0 and a1, then C and D are uniquely determined.

The theorem says that all the solutions of the recurrence relationare a linear combination of the two series (1,r,r2,r3,r4,…,rn,…) and (0,r,2r2,3r3,4r4,…,nrn,…) defined by the only root of t2 - At – B = 0.

In-Class Exercise

a0=1, a1=3, ak = 4ak-1 - 4ak-2

Gambler’s Ruin

Initially a gambler has n dollars.

He repeatedly bets 1 dollar for the coin to come up head.

If the coin comes up head, then he wins 1 dollar; otherwise he loses 1 dollar.

The gambler will stop if he wins M dollars or when he loses all his money.

Suppose the coin is fair, what is the probability that he can win M dollars?

(page 497 of the textbook)