Snick snack CPSC 121: Models of Computation 2013W2 Introduction to Induction Steve Wolfman 1

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CPSC 121: Models of Computation2013W2

Introduction to Induction

Steve Wolfman

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Outline

• Prereqs, Learning Goals, and Quiz Notes

• Bonus Prelude

• A Pattern for Induction

• Problems and Discussion– Single-Elimination Tournaments– Induction on Numbers

• Next Lecture Notes

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Learning Goals: Pre-Class

By the start of class, you should be able to:– Convert sequences to and from explicit formulas

that describe the sequence. – Convert sums to and from summation/“sigma”

notation.– Convert products to and from product/“pi”

notation.– Manipulate formulas in summation/product

notation by adjusting their bounds, merging or splitting summations/products, and factoring out values.

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Learning Goals: In-Class

By the end of this unit, you should be able to:– Given a theorem to prove via induction and

the insight into how the problem breaks down into one or more smaller problem(s), write out a complete proof strategy.

– Prove naturally self-referential properties by induction. (That is, identify the “insight” and flesh out the strategy into a complete proof.)

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Where We Are inThe Big Stories

Theory

How do we model computational systems?

Now: Developing a new proof technique that’s perfect for algorithms involving recursion (or iteration)... which is almost all interesting algorithms!

Hardware

How do we build devices to compute?

Now: Taking a break in lecture. In lab, nearly at

complete, working computer!

(Although lab’s taking a break too, for regular expressions.)

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Outline


• Bonus Prelude




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Prelude: A Binary Tree with n Nodes contains n+1 Empty Trees

Note: We’ll define a binary tree as one of an empty tree or a “node” with two subtrees (each itself a binary tree). (What a “node” is depends on what we want to do with the tree but doesn’t concern us here!)

Translate the definition and the theorem to predicate logic; use the proofs sheet to think about how to prove it.

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Induction: Regular Old ProofPlus One Powerful Step

A binary tree is: an empty tree or a node with two subtrees (each itself a binary tree).

Theorem: A binary tree with n nodes contains n+1 empty trees.

10Induction gives us one extra tool: assuming our theorem holds for recursive sub-cases.

“Binary Trees”

A “binary tree” is either an empty tree or a node with two children, each of which is a binary tree.

11(They let us do some cool CS things... see CS110, 210, 221.)

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New Proof Technique: (Structural) Induction

1. Identify the key recursive structure in the problem.

For example, binary trees.

A binary tree is one of:

• An empty tree

• A node with two subtrees, each of which is a binary tree

12The book discusses “weak” and “strong” induction.As far as we are concerned, there can be only one induction: structural.


2. Circle each recursive appearance of the structure inside its definition.

A binary tree is one of:

• An empty tree

• A node with two subtrees, each of which is a binary tree

13So, the two subtrees are recursive appearances of binary trees.


3. Divide the cases into: those without recursive appearances (“base cases”) and those with (“recursive” or “inductive” cases)

A binary tree is one of:• An empty tree• A node with two subtrees, each of which is

a binary tree

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Base case

Inductive case

Reuse all of these steps whenever you reuse the recursive structure.


4. Write a predicate describing what you want to say about the structure.

For example:

P(t) ≡ the number of empty trees in the binary tree t is one more than the number of nodes in t

15Always in terms of one of the recursive structure.


16This step is easy!


6. Write out your proof template:

P(a) ≡ [fill in predicate definition]

Theorem: For all [recursive structure] a, P(a) holds.

Proof by structural induction:

Base case: [For each base case you ID’d, write the structure of that case and prove the theorem holds for that case. If you don’t END by showing your theorem true for the base case structure, you probably messed up!]

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Example

P(t) ≡ the number of empty trees in the binary tree t is one more than the number of nodes in t

Theorem: For all binary trees t, P(t) holds.


Base case: t is an empty tree.

In that case, t has no nodes and one empty tree.

1 = 0 + 1 as expected.

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6. Proof template continued:

Inductive Step: [For each recursive case write out an inductive step…]

Consider an arbitrary [recursive case structure] a.

Induction Hypothesis: Assume the theorem holds for [each recursive appearance of the structure in this case].

We now show that the theorem holds for a itself.

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Example

Inductive Step:

Consider an arbitrary non-empty binary tree t (i.e., a node with two subtrees, each of which is a binary tree).

Induction Hypothesis: Assume the theorem holds for the two subtrees of t.

We now show that the theorem holds for t itself.

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6. Proof template completed:

[Show that theorem holds! Three signs that something has probably gone wrong:

1) You don’t END by showing that your theorem holds for a.

2) You don’t USE the “Induction Hypothesis” assumption(s) you made.

3) You EVER take a recursive appearance and use the recursive definition to break it down further (Example BAD thing to say: “Each of the subtrees is also a binary tree; so, it can have subtrees and so on until we reach an empty tree.)]

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Example

Consider an arbitrary binary tree t.

Induction Hypothesis: Assume the theorem holds for the two subtrees of t.

t’s left subtree tl has some number of nodes nl and some number of empty trees el. By assumption, el = nl + 1. Similarly for t’s right subtree tr, er = nr + 1.

t’s number of nodes n includes all of tl’s and tr’s nodes plus its own node. t’s number of empty trees e includes all and only the empty trees in tl and tr.

Thus n = nl + nr + 1 and e = el + er. Substituting from above: e = el + er = n/ + 1 + nr + 1= nl + nr + 1 + 1 = n + 1.

Thus, e = n + 1, as we wanted to prove.

QED22

Outline


• Bonus Prelude




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A Pattern for Induction

1. Identify the recursive structure in the problem.

2. Circle each recursive appearance of the structure inside its definition.

3. Divide the cases into: those without recursive appearances (“base cases”) and those with (“recursive” or “inductive” cases)

4. Write a predicate P(a) describing what you want to say about the structure.

5. Write your theorem a?,P(a)

6. Complete the proof template. 24

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Base case: [For each non-recursive case, prove the theorem holds for that case. End by showing your theorem true for the base case structure!]

Inductive Step: [For each recursive case…]



We now show the theorem holds for a. [And then do so! You should:

1)END by showing that your theorem holds for a.

2)USE the “Induction Hypothesis” assumption(s) you made.

3)NEVER take a recursive appearance and use the recursive definition to break it down further, like this BAD example: “each subtree is a binary tree that can have subtrees and so on until we reach an empty tree.”]

It helps to write out what you want to prove rather than just “show the theorem holds for a”.(Even though neither one is strictly necessary.)

Outline


• Bonus Prelude




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Single-Elimination Tournaments

In each round teams play in pairs. Losing teams are eliminated. The tournament ends when only one team remains.

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What is a Tournament?

Let’s think like CPSC 110ers…

A tournament is one of:

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The “insight into how the problem breaks down”

A tournament of n rounds is made up of two tournaments of (n-1) rounds, like how the Stanley Cup Finals is set “on top of” the “east half” and “west half” tournaments.

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One extra round:east winner vs. west winner

Single-Elimination Tournament Problem

What’s the maximum number of teams in a 0-round single-elimination tournament?

a.0 teams

b.1 team

c.2 teams

d.3 teams

e.None of these

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Think: what does this correspond to in the data definition?



a.0 teams

b.1 team

c.2 teams

d.3 teams

e.None of these

31A “one round tournament” has only one set of games.A “two round tournament” has two sets of games (finals and semi-finals).



a.0 teams

b.1 team

c.2 teams

d.3 teams

e.None of these

32Would there be any “byes” in the tournament?A team with a “bye” gets to advance to the next round “for free”.


What’s the maximum number of teams in an n-round single-elimination tournament?

a.n teams

b.2n teams

c.n2 teams

d.2n teams

e.None of these

33(Induction is for proving our theorem;

it’s not for figuring out what we want to prove!)

A Pattern for Induction

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Coincidentally, t’s a good variable name again!

Side Note: How Many Rounds Does a Tournament Have?

How many rounds does a tournament that’s just a single winner have?

How many rounds does a tournament that’s a final game between the winners of two smaller tournaments have?

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Side Note: How Many Rounds Does a Tournament Have?

A tournament has:

•0 rounds if it’s just a single “winner”

•1 more round than the largest number of rounds of either of its subtournaments, otherwise

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P(a) ≡ there are at most 2n teams in tournament t with n rounds











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Theorem: For all tournaments t, P(t) holds.




Consider an arbitrary [recursive case structure] t.


We now show the theorem holds for t. [And then do so! You should:

1)END by showing that your theorem holds for t.



I changed all the variables to t just because t’s a good name for a tournament.

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Base case: A tournament that is just a single “winner”… [TODO: finish!]








We’ll leave the proof of the base case until we’ve finished the structural pieces.

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Inductive Step:







There’s just the one recursive case.

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Inductive Step:

Consider an arbitrary tournament t that is a final game between the winners of two subtournaments (which we’ll call s1 and s2).






It’s often handy to name those recursive appearances! E.g., subtournaments s1 and s2.

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Inductive Step:

Consider an arbitrary tournament t that is a final game between the winners of two subtournaments s1 and s2.

Induction Hypothesis: Assume the theorem holds for s1 and s2.





44It helps to write out what you want to prove rather than just “show the theorem holds for a”.

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The “insight into how the problem breaks down”

The n-round tournament with the maximum number of teams is made from two (n-1)-round tournaments with the maximum number of teams.

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Step-by-Step?

Here’s the logical structure of our theorems:MT(0,20).

nZ0, (n > 0) MT(n-1,2n-1) MT(n,2n).

Do these prove MT(1,21)?

a.Yes.

b.No.

c.I don’t know.

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Step-by-Step?


nZ0, (n > 0) MT(n-1,2n-1) MT(n,2n).

Plus, we know MT(1,21). Do all of these prove MT(2,22)?

a.Yes.

b.No.

c.I don’t know.51

Step-by-Step?


nZ0, (n > 0) MT(n-1,2n-1) MT(n,2n).

Plus, we know MT(1,21) and MT(2,22).Do all of these prove MT(3,23)?

a.Yes.

b.No.

c.I don’t know.52

Step-by-Step?


nZ0, (n > 0) MT(n-1,2n-1) MT(n,2n).

From this, can we prove MT(n,2n)for any particular integer n?

a.Yes.

b.No.

c.I don’t know.53

The Magic of Induction!

Our “proof” is really a recursive program.

Plug in a tournament of a particular number of rounds n, and it produces a proof that that tournament cannot have more than 2n teams in it.

The magic of induction is that we agree that writing this program is the same as writing all the proofs the program can create.

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Outline


• Bonus Prelude




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Natural Numbers and Induction

Can we prove things inductively about the natural numbers without a recursive data definition? Are they “self-referential”?

Here’s one answer:

A natural number is:

- 0

- The next number after (successor of) a natural number

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Natural Numbers and Induction

Can we prove things inductively about the natural numbers without a recursive data definition? Are they “self-referential”?

But, let’s just try one!

Problem: What is the sum of the natural numbers 0..n?

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Partly Worked Problem: Sum of 0..n

Partly Worked Problem: What is the sum of the natural numbers 0..n?

Induction doesn’t help with insight...

But spoilers do: n(n+1)/2.

Now, how do we prove it?

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Partly Worked Problem: Sum of 0..n

Partly Worked Problem: Prove that the sum of the natural numbers 0..n is n(n+1)/2.

Is this self-referential? If we knew the sum of the numbers 0..(n-1), would it tell us anything about the sum of the numbers 0..n?

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Sigma, Pi, Factorial, Powers:All Self-Referential

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So? So, you can do inductive proofs on them!

Outline


• Bonus Prelude




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Learning Goals: In-Class

By the end of this unit, you should be able to:– Establish properties of self-referential

structures using inductive proofs that naturally build on those self-references.

Note: this learning goal is not yet looking for formal inductive proofs. Instead, we’re just exploring how we

work with things that are defined in terms of themselves.62

Next Lecture Learning Goals: Pre-Class

By the start of class, you should be able to:– Given a theorem to prove and the insight into

how to break the problem down in terms of smaller problems, write out the skeleton of an inductive proof including: the base case(s) that need to be proven, the induction hypothesis, and the inductive step that needs to be proven.

So, take what we did in this lecture and use the textbook readings to formalize your understanding.

We will have much more practice on inductive proofs.63

Next Lecture Prerequisites

See the Mathematical Induction Textbook Sections at the bottom of the “Textbook and References” page.

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Extra Slides

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One Extra Step We Sometimes Do

Really, we are done. But just to be thorough, we’ll add:

Termination: the number of rounds n is a non-negative integer, and each application of the inductive step reduces it by at least 1. Therefore, it must reach our base case (0) in a finite number of steps.

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Binary Tree Data Definition

A binary-tree is one of:-An empty tree-(make-node binary-tree

binary-tree number)

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Binary Trees’ Height

A binary tree’s “height” is the maximum number of steps it takes to get from the root down to a node with only empty trees as children.

This one’s height is:a. 0b. 1c. 2d. 3e. 4

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Problem 1: What’s the maximum number of nodes in a binary tree of height n?

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Problem 2: Prove your result.

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Binary Trees, Take Two

Problem 1: Give an algorithm to determine the height of a binary tree.

Problem 2: Prove that your algorithm is correct.

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Worked: Algorithm for Height

Height(t):

Is t an empty tree?

Yes: evaluate to -1

No: recursively find the height of each subtree hl and hr and evaluate to max(hl, hr) + 1.

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How big are the subtrees?We need to assume the algorithm works on trees of “those sizes”.

Worked: Proof of Algorithm’s Correctness

Proof: We proceed by induction.Base case: On an empty tree, our algorithm yields -1. We simply make this correct by definition. (Or, you can imagine we go “up one level” to get to a node.)Induction Hypothesis: For an arbitrary (finite-sized) non-empty tree t, assume our algorithm works correctly on all trees smaller than t.Inductive step: t is not an empty tree; so, our algorithm finds the height of each subtree hl and hr. These heights are correct by assumption, since the subtrees must be smaller than t (lacking at least t itself!). To reach a node with only leaves as children, we go into either the left or right subtree. The length of the path through the left subtree is therefore hl+1, with the +1 for the step down into that subtree. Similarly, the right path is hr +1. The height is the larger of these possible paths; so, it’s max(hl+1, hr+1) = max(hl, hr) + 1, which is what our algorithm evaluates to. Termination: We call this only on finite trees, and we reduce the size of the tree by at least one node at each inductive step. Thus, we eventually reach an empty tree.QED

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Snick snack CPSC 121: Models of Computation 2013W2 Introduction to Induction Steve Wolfman 1