CSCI2110 Tutorial 8: Solving Recurrence

CSCI2110 Tutorial 8:Solving Recurrence

Chow Chi Wang (cwchow ‘at’ cse.cuhk.edu.hk)

Solving Recurrence

How to solve recurrence?• By iteration• By guessing / observation• For second order linear homogenous recurrence with

constant coefficients• By Distinct-Roots Theorem• By Single-Root Theorem

Solving recurrence by iteration• Consider this recurrence relation:

, and

• How do we solve this?

Solving recurrence by iteration

Second Order Recurrence• You only need to know how to solve second order linear

homogeneous recurrence relation with constant coefficients for this course.

• It looks like: , where are constants and .

• Concrete examples:• (the Fibonacci recurrence)

Second Order Recurrencesecond order linear homogeneous recurrence relation with

constant coefficients

• Second Order means that depends only on previous two terms . And it must depends on (i.e. ).

Second Order Recurrencesecond order linear homogeneous recurrence relation with

constant coefficients

• Linear means that the sequence terms in the recurrence relation must be of degree 1. So we are not allowed to have something like etc. in our recurrence relation.

Second Order Recurrencesecond order linear homogeneous recurrence relation with

constant coefficients

• Homogeneous means that the space of solutions is a linear space. In other words, the sum of any set of solutions or multiples of solutions is also a solution.

• In our case, this simply means we don’t have the degree-0 term. So we are not allowed to have something this:

, where .

Second Order Recurrencesecond order linear homogeneous recurrence relation with

constant coefficients

• Constant coefficients means that both and are constants.

Second Order Recurrence• Which of following recurrence relations are second order

linear homogeneous recurrence with constant coefficients?1. 2. 3. 4. 5. 6.

Solving Second Order Recurrence• To solve :

Solve the corresponding characteristic equation:

By Distinct-Roots Theorem By Single-Root Theorem

Solve using

distinct roots ? single root ?

Step 1

Step 2

Solving Second Order Recurrence (Step 1)

• We have to solve the characteristic equation

• You may use the quadratic formula:• The roots of the quadratic equation is given by

Solving Second Order Recurrence (Step 2)

• Depending on the root(s) we calculated in step 1:• if are distinct roots of the characteristic equation.• if is the repeated root of the characteristic equation.

• We will use the initial conditions to determine the values of .

Solving Second Order Recurrence (Step 2)

• By substituting these conditions into the recurrence relation, we have:

for the distinct root case. We can then determine the values of by solving this system of linear equations.

Counting Strings - Revisit• The Problem

• A string made of 3 types of alphabets “a”, “b”, “c”• Count the number of string with length and without the pattern “aa”

• Let be the number of strings of length without the pattern “aa”. Find a recurrence relation and then solve it.

Counting Strings - Revisit• The recurrence relation is given by: .• And the initial conditions are .

• The characteristic equation is .• The roots of the equation are .• By the distinct roots theorem, we have:

for some constants .

Counting Strings - Revisit• The recurrence relation is given by: .• And the initial conditions are .

• Using the initial conditions, we have:

• So we have .

Counting Strings - Revisit• The recurrence relation is given by: .• And the initial conditions are .

• Therefore:

• Tip: After finding the general formulae of the sequence, always remember to test it’s correctness by trying some base cases.

Gambler’s Ruin• Problem:

• Betting $1 on the outcome of a die until he has $300.• +$1 if he won, -$1 if he lost.• Probability that the gambler is ruined when he begins with $n?

• Let be the probability that the gambler is ruined when he begins with $. Find a recurrence relation and then solve it.

Gambler’s Ruin• The gambler guesses the correct outcome of a die with

probability , so we have:

for

• And the boundary conditions are .

Gambler’s Ruin• The recurrence relation is given by: • And the boundary conditions are.

• The characteristic equation is .• The roots of the equation are or .• By the distinct roots theorem, we have:

for some constants .

Gambler’s Ruin• The recurrence relation is given by: • And the boundary conditions are.

• Using the boundary conditions, we have:

• So we have .

Gambler’s Ruin• The recurrence relation is given by: • And the boundary conditions are.

• Therefore:

Gambler’s Ruin 2• Problem:

• Betting $1 on the outcome of a coin until he has $300.• +$1 if he won, -$1 if he lost.• Probability that the gambler is ruined when he begin with $n?

• Let be the probability that the gambler is ruined when he begins with $. Find a recurrence relation and then solve it.

• Answer: .

Exercise• Solve the following recurrence:

1. for .2. for .3. for .

Summary• You may solve simple recurrence by iteration, guessing /

observing.

• How to solve second order linear homogeneous recurrence with constant coefficients? (slide 12)

• Tip: After finding the general formulae of the sequence, always remember to test it’s correctness by trying some base cases.

Thank You!