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CSCI2110 – Discrete Mathematics
Tutorial 6
Recursion
Wong Chung Hoi (Hollis)
12-7-2011
Agenda
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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
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By observation, guessing or iteration
5. Verify the closed form
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By base cases or induction
Agenda
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•
•
•
•
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How to solve recurrence
Counting Rabbits
Double Tower of Hanoi
Tower of Hanoi with adjacency requirement
Counting strings
Triangulated polygon
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 = 10
R6 = 22
R7= 43
R8 = 73
R9 = 139
R10 = 268
R11 = 487
R12 = 904
Agenda
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How to solve recurrence
Counting Rabbits
Double Tower of Hanoi
Tower of Hanoi with adjacency requirement
Counting strings
Triangulated polygon
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
2n discs
B
C
Double tower of Hanoi – Solution
2. T1 = ?
Double tower of Hanoi – Solution
2. T1 = 1
Double tower of Hanoi – Solution
2. T1 = 2
Double tower of Hanoi – Solution
2. T1 = 2, T2 = ?
Double tower of Hanoi – Solution
2. T1 = 2, T2 = 1
Double tower of Hanoi – Solution
2. T1 = 2, T2 = 2
Double tower of Hanoi – Solution
2. T1 = 2, T2 = 3
Double tower of Hanoi – Solution
2. T1 = 2, T2 = 4
Double tower of Hanoi – Solution
2. T1 = 2, T2 = 5
Double tower of Hanoi – Solution
2. T1 = 2, T2 = 6, T3 = ?
3. What is the recurrence equation?
Double tower of Hanoi – Solution
3. Observation:
– We must first move the top 2(n-1) discs
Double tower of Hanoi – Solution
3. Observation:
– We must first move the top 2(n-1) discs
– Then move the 2 largest discs
Double tower of Hanoi – Solution
3. Observation:
– We must first move the top 2(n-1) discs
– Then move the 2 largest discs
Double tower of Hanoi – Solution
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
Double tower of Hanoi – Solution
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
Double tower of Hanoi – Solution
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
Double tower of Hanoi – Solution
5. Verifying Tn = 2Tn-1+2, T1 = 2 is equivalent to
Tn=2n+1-2
– By induction:
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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?
Double tower of Hanoi – Extension
• What about triple, quadruple tower of
Hanoi?
• What if we have to preserve the order of
disc?
Double tower of Hanoi – Extension
• What about triple, quadruple tower of
Hanoi?
• What if we have to preserve the order of
disc?
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = ?
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 1
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 2
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = ?
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 1
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 2
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 3
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 4
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 5
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 6
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 7
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 8
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 9
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 10
Double tower of Hanoi – Extension
1. Let Sn be the number of moves for 2n discs
that preserves order.
2. S1 = 3, S2 = 11
Double tower of Hanoi – Extension
3. Observation:
– Sn =
Double tower of Hanoi – Extension
3. Observation:
– Sn = Sn-1
Double tower of Hanoi – Extension
3. Observation:
– Sn = Sn-1 + 1
Double tower of Hanoi – Extension
3. Observation:
– Sn = Sn-1 + 1 + Sn-1
Double tower of Hanoi – Extension
3. Observation:
– Sn = Sn-1 + 1 + Sn-1 + 1
Double tower of Hanoi – Extension
3. Observation:
– Sn = Sn-1 + 1 + Sn-1 + 1 + Sn-1
Double tower of Hanoi – Extension
3. Observation:
– Sn = Sn-1 + 1 + Sn-1 + 1 + Sn-1 + 1
Double tower of Hanoi – Extension
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
Double tower of Hanoi – Extension
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
Double tower of Hanoi – Extension
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
Double tower of Hanoi – Extension
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
Double tower of Hanoi – Extension
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
Double tower of Hanoi – Extension
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
Double tower of Hanoi – Extension
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
Double tower of Hanoi – Extension
4. Exercise: Show that Sn = 2n+2-5
5. 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 C
1. Let Tn be the number of move from A to C
A
B
C
Agenda
•
•
•
•
•
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How to solve recurrence
Counting Rabbits
Double Tower of Hanoi
Tower of Hanoi with adjacency requirement
Counting strings
Triangulated polygon
Tower of Hanoi with Adjacency
Requirement
2. T1 = ?
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 1
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = ?
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = 1
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = 2
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = 3
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = 4
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = 5
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = 6
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = 7
A
B
C
Tower of Hanoi with Adjacency
Requirement
2. T1 = 2, T2 = 8, T3 = ?
A
B
C
Tower of Hanoi with Adjacency
Requirement
3. Observation:
– Tn = ?
A
B
C
Tower of Hanoi with Adjacency
Requirement
3. Observation:
– Tn = Tn-1
A
B
C
Tower of Hanoi with Adjacency
Requirement
3. Observation:
– Tn = Tn-1 + 1
A
B
C
Tower of Hanoi with Adjacency
Requirement
3. Observation:
– Tn = Tn-1 + 1 + Tn-1
A
B
C
Tower of Hanoi with Adjacency
Requirement
3. Observation:
– Tn = Tn-1 + 1 + Tn-1 + 1
A
B
C
Tower of Hanoi with Adjacency
Requirement
3. Observation:
– Tn = Tn-1 + 1 + Tn-1 + 1 + Tn-1 = 3Tn-1 + 2
4. Exercise: Show that Sn = 3n – 1
5. Exercise: Verify by induction
A
B
C
Agenda
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How to solve recurrence
Counting Rabbits
Double Tower of Hanoi
Tower of Hanoi with adjacency requirement
Counting strings
Triangulated polygon
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
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How to solve recurrence
Counting Rabbits
Double Tower of Hanoi
Tower of Hanoi with adjacency requirement
Counting strings
Triangulated polygon
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.
Triangulated Polygon
2. T0 = 1, T1 = 1, T2 = 2, T3 = 5
Triangulated Polygon
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
n=4
A
B
Triangulated Polygon
3. Observation:
– Tn = T0 Tn-1
n=4
Tn-1
T0
A
B
Triangulated Polygon
3. Observation:
– Tn = T0 Tn-1 + T1Tn-2
n=4
Tn-2
A
T1
B
Triangulated Polygon
3. Observation:
– Tn = T0 Tn-1 + T1Tn-2 + T2Tn-3 + …
n=4
Tn-3
A
T2
B
Triangulated Polygon
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?
n=4
Tn-1
T0
A
B
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

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