mathematical induction ii

Mathematical Induction II

Lecture 21

Section 4.3

Mon, Feb 27, 2006

Example: Binary Search Trees

In a complete binary search tree of depth n - 1, (n rows) what is the average number of comparisons required to locate an element? (Assuming all positions are equally likely.)

Analysis Elements in row 1 require 1 comparison. Elements in row 2 require 2 comparisons. Elements in row 3 require 3 comparisons. In general, elements in row k require k

comparisons.

Example: Binary Search Trees

Further analysis Row 1 contains 1 element. Row 2 contains 2 elements. Row 3 contains 4 elements. In general, row k contains 2k - 1 elements.

Fact The average number of comparisons is the total

number of comparisons required for all elements divided by the number of elements.

Example: Binary Search Trees

The total number of elements is

1 + 2 + 4 + 8 + … + 2n – 1. The total number of comparisons is

11 + 22 + 34 + 48 + … + n2n – 1. Therefore, the average is

1

1

2421

2432211

n

nn

Example: Binary Search Trees

Prove by mathematical induction that

1 + 2 + 4 + 8 + … + 2n – 1 = 2n – 1. Prove by mathematical induction that

11 + 22 + 34 + 48 + … + n2n – 1

= (n – 1)2n + 1. Then, the average is

This is approximately equal to n – 1 (for large n).

12

12)1(

n

nn

Binary Search Trees

In reality, it often requires two comparisons at each node.Is the value less than the node?If not, is the value greater than the node?If not, then the value equals the node.

In other words, we must make a three-way decision at each node.

Binary Search Trees

So, exactly how many comparisons does it take to locate an element?

40

20

70503010

60

Binary Search Trees

So, exactly how many comparisons does it take to locate an element?

40

20

70503010

60

221

Binary Search Trees

So, exactly how many comparisons does it take to locate an element?

40

20

70503010

60

221

1 23

Binary Search Trees

So, exactly how many comparisons does it take to locate an element?

40

20

70503010

60

221

1 2

4

3

5

Binary Search Trees

So, exactly how many comparisons does it take to locate an element?

40

20

70503010

60

221

11 2 2

4

3

5

4

Binary Search Trees

So, exactly how many comparisons does it take to locate an element?

40

20

70503010

60

221

11 2 2

4

3

655

4

Binary Search Trees

It takes an average of 1.5 comparisons to move down one level.

Then it takes 2 comparisons to match a value at that level.

Level 1: 2 comparisons. Level 2: 1.5 + 2 = 3.5 comparisons. Level 3: 2(1.5) + 2 = 5 comparisons. Level n: (n – 1)(1.5) + 2 = 1.5n + 0.5.

Binary Search Trees

To find the average, we must compute

12 + 2(3.5) + 45 + … + 2n – 1(1.5n + 0.5). We could work this our from scratch… Or use what we have already worked out:

.1215.1

125.01215.1

25.025.15.05.121

1

1 1

11

n

nn

n

k

kn

k

n

k

kk

n

n

kk

Binary Search Trees

Therefore, the average is

For large n, this is approximately 1.5n – 1.

.

12

1215.1

n

nn

Binary Search Trees

What if we changed the order in which we did the comparisons at each node?Is the value equal to the node?If not, is the value less than the node?If not, then the value is greater than the

node. Will that be more or less efficient than the

previous order?

Binary Search Trees

Exactly how many comparisons does it take to locate an element?

40

20

70503010

60

Binary Search Trees

Exactly how many comparisons does it take to locate an element?

40

20

70503010

60

122

Binary Search Trees

Exactly how many comparisons does it take to locate an element?

40

20

70503010

60

122

2 23

Binary Search Trees

Exactly how many comparisons does it take to locate an element?

40

20

70503010

60

122

2 2

5

3

5

Binary Search Trees

Exactly how many comparisons does it take to locate an element?

40

20

70503010

60

122

22 2 2

5

3

5

3

Binary Search Trees

Exactly how many comparisons does it take to locate an element?

40

20

70503010

60

122

22 2 2

5

3

555

3

Binary Search Trees

To find the average, we must compute

11 + 23 + 45 + … + 2n – 1(2n – 1). Again, we can use what we have already

worked out:

.3232

121212

2221221

1

1

1

1

1

n

nn

n

k

kn

k

kn

k

k

n

n

kk

Binary Search Trees

Therefore, the average is

For large n, this is approximately 2n – 3. This is larger (worse) than 1.5n – 1. Therefore, it is better not to check for

equality first when searching a binary tree.

.

12

3232

n

nn

Example: Recursive Sequences

Define a sequence a1 = 2, an = 2an – 1 – 1 for n 2.

Find a non-recursive formula for an and prove that it is correct.

Analysis{an} = {2, 3, 5, 9, 17, 33, …}.{an – 1} = {1, 2, 4, 8, 16, 32, …}.Conjecture that an = 2n – 1 + 1.

Example: Recursive Sequences

Proof: Basic Step

P(1) is true since a1 = 2 = 20 + 1.

Inductive StepSuppose that ak = 2k – 1 + 1 for some k 1.

Example: Recursive Sequences

Then ak + 1 = 2ak – 1 (by def.)

= 2(2k – 1 + 1) – 1 (by ind. hyp.)

= 2k + 2 – 1

= 2k + 1.Therefore, P(k + 1) is true.

Therefore, an = 2n + 1 for all n 1.

Example: Contest Problem

Let n be a positive integer. Suppose we have n red dots and n blue

dots in the plane such that no three dots are collinear.

Prove that it is possible to connect the red dots to the blue dots in distinct pairs such that none of the line segments intersect.

Example: Contest Problem

Example: Contest Problem

Example: Contest Problem

Proof: Basic Step

P(1) is obviously true since there is only one pair of dots and only one segment.

Example: Contest Problem

Proof: Basic Step

P(1) is obviously true since there is only one pair of dots and only one segment.

Example: Contest Problem

Inductive StepSuppose P(1), …, P(k) are true for some k

1.Now suppose we have a collection of k + 1

red dots and k + 1 blue dots.Consider the convex hull of this set.

Example: Contest Problem

There are two possibilities. Case 1: The dots on the convex hull are not all

the same color. Case 2: The dots on the convex hull are all the

same color.

Example: Contest Problem

There are two possibilities. Case 1: The dots on the convex hull are not all

the same color. Case 2: The dots on the convex hull are all the

same color.

Example: Contest Problem

Case 1: Suppose they are not all the same color.Choose two adjacent dots of different

colors on the convex hull and connect them with a segment.

Example: Contest Problem

Example: Contest Problem

Example: Contest Problem

Example: Contest Problem

By induction, the remaining collection of k red dots and k blue dots may be connected so that no two segments intersect.

Nor will any of those segments intersect the first segment because it is on the convex hull and no three dots are collinear.

Therefore, the entire set of dots may be connected with non-intersecting segments.

Example: Contest Problem

Example: Contest Problem

Example: Contest Problem

Case 2: Suppose they are all the same color, say red.Choose a line that is not parallel to any

segment connecting any two dots.Move the line across the set of dots, from

right to left.Keep count of the number of red dots and

the number of blue dots to the right of the line.

Example: Contest Problem

0 red dots, so far (r = 0). 0 blue dot, so far (b = 0).

Example: Contest Problem

1 red dots, so far (r = 1). 0 blue dot, so far (b = 0).

Example: Contest Problem

1 red dots, so far (r = 1). 1 blue dot, so far (b = 1).

Example: Contest Problem

2 red dots, so far (r = 2). 1 blue dot, so far (b = 1).

Example: Contest Problem

3 red dots, so far (r = 3). 1 blue dot, so far (b = 1).

Example: Contest Problem

Initially, r – b = 0. At the end, r – b = 0. Just after passing the first dot, r – b = 1. Just before passing the last dot, r – b = –1. The line meets only one dot at a time. Therefore, somewhere in between, r – b

must be 0.

Example: Contest Problem

3 red dots (r = 3). 1 blue dot (b = 1). r – b = 2.

Example: Contest Problem

3 red dots (r = 3). 2 blue dot (b = 2). r – b = 1.

Example: Contest Problem

3 red dots (r = 3). 3 blue dot (b = 3). r – b = 0.

Example: Contest Problem

Apply the induction hypothesis.The dots on the right may be connected

with non-intersecting segments.The dots on the left may be connected with

non-intersecting segments.Therefore, the entire set of dots may be

connected with non-intersecting segments.

Example: Contest Problem

Therefore, P(n) is true for all n 1.

Example: Contest Problem

Therefore, P(n) is true for all n 1.

Example: Contest Problem

Therefore, P(n) is true for all n 1.