CSCI2110 – Discrete Mathematics Tutorial 6 Recursion

CSCI2110 – Discrete MathematicsTutorial 6Recursion

Wong Chung Hoi (Hollis)12-7-2011

Agenda

• How to solve recurrence• Counting Rabbits• Double Tower of Hanoi• Tower of Hanoi with adjacency requirement• Counting strings• Triangulated polygon

How to solve recurrence• Recurrence Equation and Boundary Values

– E.g. Rn = Rn-1 + Rn-2 + n, T0 = 0, T1 = 1

• Solving recurrence problem:1. Define a recurrence variable, e.g. Rn

2. Try for a few base cases, e.g. R1, R2,…3. Express Rn in terms of Rn-1, Rn-2,…

• If it is not possible, try another approach

4. Find its closed form if necessary, e.g. calculating large n• By observation, guessing or iteration

5. Verify the closed form• By base cases or induction

Agenda


Counting Rabbits - Problem• Problem

– Start with 1 pair of baby rabbit.– Rabbits take 2 months to become

an adult.– Adult rabbits gives birth to 3 pairs

of baby rabbits every month.– Rabbits NEVER DIE!– Number of pairs of rabbits in the end of the year?

1. Denote Rn as the number of rabbit after n month.2. R0= 1, R1 = 1, R2 = 1, R3 = 3+1=4, R4 = 4+3 = 7 …

Counting Rabbits - Solution

3. Observation: Rabbit born at nth month = 3Rn-3

– Rn = 3Rn-3 + Rn-1 – By Repeatedly substitution

R5 = 10R6 = 22R7= 43R8 = 73R9 = 139R10 = 268R11 = 487R12 = 904

Agenda


Double tower of Hanoi – Problem

• How many moves is needed to move the whole tower?

1. Denote Tn as the number of moves for moving 2n discs from one pole to another.

A B C

2n discs

Double tower of Hanoi – Solution

2. T1 = ?


2. T1 = 1


2. T1 = 2


2. T1 = 2, T2 = ?


2. T1 = 2, T2 = 1


2. T1 = 2, T2 = 2


2. T1 = 2, T2 = 3


2. T1 = 2, T2 = 4


2. T1 = 2, T2 = 5


2. T1 = 2, T2 = 6, T3 = ?3. What is the recurrence equation?


3. Observation: – We must first move the top 2(n-1) discs


3. Observation: – We must first move the top 2(n-1) discs– Then move the 2 largest discs


3. Observation: – We must first move the top 2(n-1) discs– Then move the 2 largest discs


3. Observation: – We must first move the top 2(n-1) discs– Then move the 2 largest discs– Move the 2(n-1) discs again


3. Observation: – We must first move the top 2(n-1) discs– Then move the 2 largest discs– Move the 2(n-1) discs again– Tn = Tn-1 + 2 + Tn-1 = 2Tn-1 + 2


4. Finding the closed form of Tn = 2Tn-1 + 2– By iteration:

Tn = 2Tn-1 + 2= 2(2Tn-2 + 2) + 2= 22Tn-2+22+2= 22(2Tn-3+2)+22+2= 23Tn-3+23+22+2…= 2n-1T1+2n-1+2n-2+…+2= 2n+2n-1+…+2= 2(1-2n) / (1-2)= 2n+1-2


5. Verifying Tn = 2Tn-1+2, T1 = 2 is equivalent to Tn=2n+1-2– By induction:• Case n = 1, T1 = 21+1-2 = 2• Assume when n = k-1, they are equivalent• Case n = k,

Tk = 2Tk-1+2= 2(2(k-1)+1-2) + 2= 2k+1 -2

Double tower of Hanoi – Extension

• What about triple, quadruple tower of Hanoi?

• What if we have to preserve the order of disc?








1. Let Sn be the number of moves for 2n discs that preserves order.

2. S1 = ?



2. S1 = 1



2. S1 = 2



2. S1 = 3



2. S1 = 3, S2 = ?



2. S1 = 3, S2 = 1



2. S1 = 3, S2 = 2



2. S1 = 3, S2 = 3



2. S1 = 3, S2 = 4



2. S1 = 3, S2 = 5



2. S1 = 3, S2 = 6



2. S1 = 3, S2 = 7



2. S1 = 3, S2 = 8



2. S1 = 3, S2 = 9



2. S1 = 3, S2 = 10



2. S1 = 3, S2 = 11


3. Observation:– Sn =


3. Observation:– Sn = Sn-1


3. Observation:– Sn = Sn-1 + 1


3. Observation:– Sn = Sn-1 + 1 + Sn-1


3. Observation:– Sn = Sn-1 + 1 + Sn-1 + 1


3. Observation:– Sn = Sn-1 + 1 + Sn-1 + 1 + Sn-1


3. Observation:– Sn = Sn-1 + 1 + Sn-1 + 1 + Sn-1 + 1


3. Observation:– Sn = Sn-1 + 1 + Sn-1 + 1 + Sn-1 + 1 + Sn-1

– In previous example, Tn-1 moves actually move 2(n-1) discs tower with only bottom layer flipped

















3. Observation:– Sn = Tn-1 + 1 + Tn-1 + 1 + Tn-1 + 1 + Tn-1 = 4Tn-1 + 3– In previous example, Tn-1 moves actually move

2(n-1) discs tower with only bottom layer flipped


4. Exercise: Show that Sn = 2n+2-55. Exercise: Verify by induction

Tower of Hanoi with Adjacency Requirement

• From A to C, from C to A is now allowed• We want to move the tower from A to C1. Let Tn be the number of move from A to C

A B C

Agenda



2. T1 = ?

A B C


2. T1 = 1

A B C


2. T1 = 2

A B C


2. T1 = 2, T2 = ?

A B C


2. T1 = 2, T2 = 1

A B C


2. T1 = 2, T2 = 2

A B C


2. T1 = 2, T2 = 3

A B C


2. T1 = 2, T2 = 4

A B C


2. T1 = 2, T2 = 5

A B C


2. T1 = 2, T2 = 6

A B C


2. T1 = 2, T2 = 7

A B C


2. T1 = 2, T2 = 8, T3 = ?

A B C


3. Observation:– Tn = ?

A B C


3. Observation:– Tn = Tn-1

A B C


3. Observation:– Tn = Tn-1 + 1

A B C


3. Observation:– Tn = Tn-1 + 1 + Tn-1

A B C


3. Observation:– Tn = Tn-1 + 1 + Tn-1 + 1

A B C


3. Observation:– Tn = Tn-1 + 1 + Tn-1 + 1 + Tn-1 = 3Tn-1 + 2

4. Exercise: Show that Sn = 3n – 15. Exercise: Verify by induction

A B C

Agenda


Counting Strings• Problem– A string make of 3 types of alphabets “a”, “b”, “c”– Count the number of string with length n and without the pattern

“aa”1. Let Ln be the number of string with length n and without

pattern “aa”2. L1 = 3, {a,b,c}

L2 = 8, {ab,ac,ba,bb,bc,ca,cb,cc}L3 = 22, {aba,abb,abc,aca,acb,acc,bab,bac,

bba,bbb,bbc,bca,bcb,bcc,cab,cac,cba,cbb,cbc,cca,ccb,ccc}

Counting Strings

3. Observation:– If the length n string starts with “b” or “c”, its safe

to append all n-1 length string to it.– Else, the length n string must start with “a”.• The second character must be either “b” or “c”,• Therefore, its safe to append all n-2 length string after

“ab” or “ac”

– Ln = 2Ln-1 + 2Ln-2

4. Exercise: Show that L6 = 448

Agenda


Catalan Number

• Recall that recurrence equation of the form

has a closed form

which is called the Catalan number

Triangulated Polygon

• Problem– How many ways to cut a convex polygons into

triangles?1. Let Tn be the number of ways to cut an n+2

sides polygon into triangles.


2. T0 = 1, T1 = 1, T2 = 2, T3 = 5


3. Observation:– If we fix an edge AB on a polygon, in the final

cutting, this edge must form a triangle with other n vertex.

– This triangle partition the polygon into 3 regions

A B

n = 4


3. Observation:– Tn = T0 Tn-1

A B

n = 4

Tn-1

T0


3. Observation:– Tn = T0 Tn-1 + T1Tn-2

A B

n = 4

T1

Tn-2


3. Observation:– Tn = T0 Tn-1 + T1Tn-2 + T2Tn-3 + …

A B

n = 4

T2Tn-3


3. Observation:– Tn = T0 Tn-1 + T1Tn-2 + T2Tn-3 + … + Tn-1T0 =

4. The closed form is given by Catalan number Cn. How many way to triangulate an octagon?

A B

n = 4

T0

Tn-1

Conclusion

• Be familiar with the 5 steps in solving recurrence– Defining recurrence variable– Check for boundary cases– Come up with recurrence equation– Find the closed form– Verify the closed form

Documents

CSCI2110 – Discrete Mathematics Tutorial 6 Recursion