Tutorial 1 Data Structures ’10

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Tutorial 1

Data Structures ’10

Jonathan Cederberg <jonathan.cederberg@it.uu.se>

Friday, September 10^th, 2010

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Mathematical Induction Variant of Mathemati- cal Induction

Example of Asymptotic notation How to Solve Recurrence Equations Assignment 1 Joint Exercises

Outline

1 Mathematical Induction

2 Variant of Mathematical Induction

3 Example of Asymptotic notation

4 How to Solve Recurrence Equations

5 Assignment 1

6 Joint Exercises

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Outline

5 Assignment 1

6 Joint Exercises

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The set of natural numbers is defined inductively:

• 0 is a natural number.

• if x is a natural number then x + 1 is a natural number.

To prove that a property P(n) holds for all natural numbers n, show the following:

• Base Case: Show that P(0) holds.

• Induction Step: Show that P(n) implies P(n + 1).

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Example: Show that

n

X

i =0

i = n(n + 1)

2 for all n ≥ 0.

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Base Case: n = 0

0

X

i =0

i = 0 0(0 + 1)

2 = 0

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Induction Step Assume

n

X

i =0

i = n(n + 1) 2 Show

n+1

X

i =0

i = (n + 1)(n + 2) 2

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n+1

X

i =0

i =

n

X

i =0

i + (n + 1)

= ⁿ⁽ⁿ⁺¹⁾₂ + (n + 1)

= ⁿ²₂⁺ⁿ + (n + 1)

= ⁿ²⁺³ⁿ⁺²₂

= ^(n+1)(n+2)₂

= (n+1)((n+1)+1) 2

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Outline

5 Assignment 1

6 Joint Exercises

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To prove that a property P(n) holds for all natural numbers n ≥ n₀, we show the following:

Base Case: Prove that P(n₀) holds.

Induction Step: Show that P(n) implies P(n + 1).

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Example: Show that

n

X

i =0

(i − 3) ≥ n²

4 for all n ≥ 12.

Base Case: n = 12

12

X

i =0

(i −3) = −3−2−1+0+1+2+3+4+5+6+7+8+9 = 39

12²

4 = 144

4 = 36 36 ≤ 39

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Induction Step Assume

n

X

i =0

(i − 3) ≥ n² 4 Show

n+1

X

i =0

(i − 3) ≥ (n + 1)² 4

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n+1

X

i =0

(i − 3) =

n

X

i =0

(i − 3) + ((n + 1) − 3)

≥ ⁿ₄² + (n − 2)

= ⁿ²⁺⁴ⁿ⁻⁸₄

= ⁿ²^+2n+1+2n−9₄

= ⁽ⁿ⁺¹⁾²₄⁺²ⁿ⁻⁹

≥ ⁽ⁿ⁺¹⁾₄ ²

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Outline

5 Assignment 1

6 Joint Exercises

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An example of Asymptotic notation

Prove 5n² − 3n − 6 = Θ(n²).

That is, find c₁, c2, n0 such that for all n > n₀ we have 0 ≤ c₁· n²≤ 5n² − 3n − 6 ≤ c₂· n²

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What does 5n² − 3n − 6 look like?

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What does 5n² − 3n − 6 look like?

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What does 5n² look like?

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What does 5n² look like?

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What about c1= 4 and c2 = 6?

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What n0 do we want?

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Prove 5n² − 3n − 6 = Θ(n²).

Pick c₁ = 4, c2 = 6, and n0= 6.

Show that

• 0 ≤ 4n²for all n ≥ 6 (trivial).

• 4n² ≤ 5n² − 3n − 6 for all n ≥ 6.

• 5n² − 3n − 6 ≤ 6n²for all n ≥ 6.

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4n² ≤ 5n² − 3n − 6 for all n ≥ 6:

4n² ≤ 5n² − 3n − 6

0 ≤ n² − 3n − 6 = n(n − 3) − 6

6 ≤ n

|{z}

≥6

(n − 3)

| {z }

≥3

| {z }

≥18

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5n² − 3n − 6 ≤ 6n² for all n ≥ 6:

5n² − 3n − 6 ≤ 6n²

− 6 ≤ n² + 3n = n

|{z}≥6

(n + 3)

| {z }

≥9

| {z }

≥54

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Outline

5 Assignment 1

6 Joint Exercises

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Method 1 – Recursive Trees

Each node represents the cost of one subproblem.

The sum of all node costs within a level gives the total cost of that level.

The sum of all per-level costs gives the total cost.

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Method 2 – Substitution Guess a solution.

Verify correctness of the solution using mathematical induction.

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Example:

Consider the recurrence T (n) =

1 if n = 1

2T _n

2

+ 3n if n > 1 Show that T (n) = O(n lg n). (lg = log₂).

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Pick n₀= 2 and c = 4. Show that T (n) ≤ 4 n lg n for all n ≥ 2.

Use induction on trees:

Base Case: Show that the property holds for one level of the tree.

Induction Step: Show that if the property holds for one level of the tree, then it will also hold for the next level.

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The base cases n = 2.

T (2) = 2T (1) + 3 · 2 = 2 · 1 + 6 = 8 2 · 4 · lg 2 = 8

8 ≤ 8.

n = 3.

T (3) = 2T (1) + 3 · 3 = 2 · 1 + 9 = 11 3 · 4 · lg 3 = 12 · lg 3

11 ≤ 12 · lg 3.

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The Induction Step AssumeT _n

2 ≤ 4 ⁿ₂ lg ⁿ₂

ShowT (n) ≥ 4n lg n

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T (n) = 2T _n

2 + 3n

≤ 2 · 4 _n

2 lg ⁿ₂ + 3n

≤ 8 ⁿ₂ lg ⁿ₂ + 3n

= 4 n (lg n − lg 2) + 3n

= 4 n (lg n − 1) + 3n

= 4 n lg n − 4n + 3n

≤ 4 n lg n

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Outline

5 Assignment 1

6 Joint Exercises

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1. Asymptotic growth rates

Rank these functions in order of growth:

n 4^{lg n} n! lg n 2²ⁿ (n + 1)! 3 2

n

n lg n n³ 2ⁿ n · 2ⁿ

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Example

n lg n n (lg n)^{lg n}

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2. Fibonacci

The golden cut ¹⁺

√5

2 is a ratio with a number of different interesting properties, and is often denoted φ. Its conjugate is φ =b ¹⁻

√ 5

2 . These ratios are connected with the Fibonacci series {F_n}^∞_n=0, defined recursively as

F₀ = 0 F₁ = 1

F_n= F_n−1+ F_n−2 for n ≥ 2

Prove by induction that for all n ≥ 0, we have

Fn=

φⁿ− φbn

√5

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3. Tiling

How many different ways are there to tile a 2 × n rectangle with 1 × 2 tiles? Formulate as a recurrence, and solve it.

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Outline

5 Assignment 1

6 Joint Exercises

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Towers of Hanoi

You are given a tower of n disks initially stacked in decreasing order of diameter on one out of three pegs. The goal is to transfer the entire tower to one of the other pegs, moving one disk at a time and never placing a larger disk onto a smaller one. Devise a divide-and-conquer algorithm, analyze its running time and write and solve the recurrence.

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First thing: we should use divide-and-conquer. This means, put in words, that the way to move a big tower is by moving a smaller tower.

Algorithm: to move a tower of n disks from peg A to peg C , move the smaller tower consisting of n − 1 disks to peg B, move the nth disk to peg C , and move the tower consisting of n − 1 disks on peg B to peg C . If n = 1, skip the recursive step.

In pseudocode: Towers-Of-Hanoi(n)

1 if n = 1

2 thenmove disk to target peg 3 else Towers-Of-Hanoi(n − 1) 4 move nth disk to target peg 5 Towers-Of-Hanoi(n − 1)

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In pseudocode: Towers-Of-Hanoi(n)

1 if n = 1

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In pseudocode:

Towers-Of-Hanoi(n) 1 if n = 1

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Complexity

What type of complexity?

So, how do we get the recurrence T (n)?

The call with n = 1 clearly requires 1 operation on line 2. A call with n > 1 requires T (n − 1) operations on line 3, 1 op on line 4 and T (n − 1) ops on line 5

T (n) =

1 if n = 1

2T (n − 1) + 1 if n > 1

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Complexity

The call with n = 1 clearly requires 1 operation on line 2. A call with n > 1 requires T (n − 1) operations on line 3, 1 op on line 4 and T (n − 1) ops on line 5

T (n) =

1 if n = 1

2T (n − 1) + 1 if n > 1

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Complexity

The call with n = 1 clearly requires 1 operation on line 2.

A call with n > 1 requires T (n − 1) operations on line 3, 1 op on line 4 and T (n − 1) ops on line 5

T (n) =

1 if n = 1

2T (n − 1) + 1 if n > 1

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Complexity

T (n) =

1 if n = 1

2T (n − 1) + 1 if n > 1

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Complexity

T (n) =

1 if n = 1

2T (n − 1) + 1 if n > 1

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How to solve it

Start by some unfolding, and compute for small values.

Guess closed form solution: T_c(n) = 2ⁿ− 1 Prove it by induction:

• Basecase (n=1): T (1) = 1 = 2¹− 1 = Tc(1)

• Induction step: assume T (n) = Tc(n).

Then T (n + 1) = 2T (n) + 1 = 2 · (2ⁿ− 1) + 1 = 2ⁿ⁺¹− 2 + 1 = 2ⁿ⁺¹− 1 = Tc(n)

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How to solve it

Guess closed form solution: T_c(n) = 2ⁿ− 1

Prove it by induction:

• Basecase (n=1): T (1) = 1 = 2¹− 1 = Tc(1)

Then T (n + 1) = 2T (n) + 1 = 2 · (2ⁿ− 1) + 1 = 2ⁿ⁺¹− 2 + 1 = 2ⁿ⁺¹− 1 = Tc(n)

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How to solve it

• Basecase (n=1): T (1) = 1 = 2¹− 1 = Tc(1)

Then T (n + 1) = 2T (n) + 1 = 2 · (2ⁿ− 1) + 1 = 2ⁿ⁺¹− 2 + 1 = 2ⁿ⁺¹− 1 = Tc(n)

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How to solve it

• Basecase (n=1): T (1) = 1 = 2¹− 1 = Tc(1)

Then T (n + 1) = 2T (n) + 1 = 2 · (2ⁿ− 1) + 1 = 2ⁿ⁺¹− 2 + 1 = 2ⁿ⁺¹− 1 = Tc(n)

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Pancake sorting

Pankcake sorting is a sorting problem where prefix reversal is the only means with which you can alter the list to sort.

Devise and analyse a simple algorithm for pancake sorting. Do an exact worst case analysis in terms of the number of prefix reversals, and give an asymptotic upper bound, given that the reverse-prefix-operation takes O(n) time.

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Pancake sorting

Divide-and-conquer approach:

Pancake-Sort(s) 1 if size(s) ≤ 1 2 then return

3 else i ← indexOfMax(s) 4 Flip(s[0, . . . , i ])

5 Flip(s)

6 Pancake-Sort(s[0, . . . , size(s) − 1])

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Pancake sorting

Iterative approach:

Pancake-Sort(s)

1 for i ← size(s) − 1 down to 1 2 doj ← indexOfMax(s) 3 Flip(s[0, . . . , j ]) 4 Flip(s[0, . . . , i ])