Module 10: Proof Techniques (part 2) Mathematical Induction

CPSC 121: Models of Computation

Module 10: Proof Techniques (part 2)Mathematical Induction

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Module 10: Announcements

Pre-class quiz #11 is due Monday November 30th at 19:00.

Textbook sections:

Epp, 5th or 4th edition: remainder of 6.1, 7.1

Epp, 3rd edition: remainder of 5.1, 6.1

Rosen, 6th edition: remainder of 2.1, 2.3 up to the top of page 136.

Rosen, 7th edition: remainder of 2.1, 2.3 down to the bottom of page 141.

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Module 10: Announcements

Assignment #5 is due Wednesday December 2nd at 19:00.

There is no pre-class quiz #12. ☺☺☺

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Module 10: Mathematical Induction

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/Σ notation.

Convert products to and from product/π notation.

Manipulate formulas in summation/product notation by adjusting their bounds, merging or splitting summations/products, and factoring out values.

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Pre-class quiz #10

Well done overall.

You had a bit more trouble on the questions about the coin-changing induction proof.

We will discuss induction proofs at length and do a lot of examples.

Being comfortable with summations is important not only for CPSC 121, but for following courses too!

As usual, we will revisit the open-ended question shortly.

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If you haven’t already done so,

read the rest of the chapters on mathematical induction (Epp4, 5.2 to 5.4, etc).

After this reading, you will be able to:

Given a theorem to prove stated in terms of its induction variable (i.e., usually, in terms of n), 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.

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By the end of this module, you should be able to:

Establish properties of self-referential structures using inductive proofs that naturally build on those self-references.

Critique formal inductive proofs to determine whether they are valid and where the error(s) lie if they are invalid.

Prove properties of the non-negative integers (or a subset) with appropriate self-referential structure using weak or strong induction as needed.

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CPSC 121: the BIG questions:

1. How can we convince ourselves that an algorithm does what it's supposed to do?

2. How do we determine whether or not one algorithm is better than another one?

Mathematical induction is a veryvery useful tool when proving the correctness or efficiency of an algorithm.

We will see several examples of this.

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Module Summary

Example: single-elimination tournaments.

Defining and validating mathematical induction.

More examples where we can use induction.

A slightly different type of induction.

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Module 10.1: Single-elimination tournaments

Problem: single-elimination tournament

Teams play one another in pairs

The winner of each pair advances to the next roundLos Angeles

Montréal

Toronto

New York Isl.

Vancouver

St. Louis

Pittsburgh

Buffalo

Round 2 Round 3 Round 4

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The winner of each pair advances to the next round

Montréal

Los Angeles

Montréal

Toronto

New York Isl.

Vancouver

St. Louis

Pittsburgh

Buffalo

Los AngelesToronto

New York Isl.


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Montréal

Montréal

Los Angeles

Montréal

Toronto

New York Isl.

Vancouver

St. Louis

Pittsburgh

Buffalo

Los Angeles

Los Angeles

Toronto

New York Isl.


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Montréal

Montréal

Los Angeles

Montréal

Toronto

New York Isl.

Vancouver

St. Louis

Pittsburgh

Buffalo

Los Angeles

Los Angeles

Toronto

New York Isl.Montréal


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Question: if we have n rounds of playoffs, how many teams can participate?

a) n

b) 2n

c) n2

d) 2n

e) None of the above.

▷

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How can we prove this result formally?

We will use a technique called mathematical induction.

We will convince ourselves that it is a valid proof technique.

Many theorems involve summations, so it is important to be comfortable manipulating them.

Hence the inclusion of section 5.1 in the reading for pre-class quiz #10.

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First we look at the following question instead:

If t teams can participate in a playoff with n rounds, how many teams can participate in a playoff with n+1 rounds?

a) t+1

b) 2t

c) t2

d) 2t


▷

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Proof:

Consider an unspecified playoff with n rounds. Assume that t teams can participate in that playoff.

We can think of a playoff with n+1 rounds as follows:

Two playoffs with n rounds proceed in parallel.

The two winners then play each other in the n+1st round.

Since each playoff with n rounds has n teams, we thus have 2n teams in total.

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How do we use this to prove what we want?

Recall Universal Modus Ponens:P(a)∀x D, P(x) → Q(x)∈\ Q(a)

If we define

P(t, n): Up to t teams can participate in a playoff with n rounds

then we have proved

∀t Z∈ + n Z∀ ∈ + , P(t, n)→P(2t, n+1)

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So

We know that only 2 teams can play with 1 round.

This is P(2, 1).

Given

P(2, 1)

∀t Z∈ + n Z∀ ∈ +, P(t, n) → P(2t, n+1)

We can deduce:

P(4, 2)

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Given

P(4, 2)∀t Z∈ + n Z∀ ∈ +, P(t, n) → P(2t, n+1)

We can deduce:

P(8, 3)

Given

P(8, 3)∀t Z∈ + n Z∀ ∈ +, P(t, n) → P(2t, n+1)

We can deduce:

P(16, 4)

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Given

P(16, 4)∀t Z∈ + n Z∀ ∈ +, P(t, n) → P(2t, n+1)

We can deduce:

P(32, 5)

Etc...

Therefore we can show that

∀n Z∈ +, P(2n, n)

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Another way to think about this:

Given an unspecified positive integer n.

To prove P(2n+1, n+1):

First we prove P(2n, n).

Then we use the fact that∀t Z∈ + n Z∀ ∈ +, P(t, n) → P(2t, n+1)

Hence induction is more or less the same as recursion!

A fun example involving induction: http://www.youtube.com/watch?v=daRS98L9rHs&feature=related

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Module Summary





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Module 10.2: Defining and validating M.I.

Define Q(n) by:

Q(n): P(2n,n)

In the previous example, we proved

Q(1)

∀n Z∈ +, Q(n) → Q(n+1)

and deduced that

∀n Z∈ +, Q(n)

Why is this a valid proof technique?

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Theorem:

Q(1) ^ ( n Z∀ ∈ +, Q(n) → Q(n+1)) → n Z∀ ∈ +, Q(n)

Proof:

We use a proof by contradiction.

Suppose that the premises hold, but that n Z∀ ∈ +, Q(n) is false. There is at least one value of n for which Q(n) does not hold.

Let y be the smallest such value. Clearly y > 1, so y – 1 Z∈ +.

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Proof (continued):

Because y is the smallest value for which Q(n) is false, and y – 1 Z∈ +, Q(y - 1) is true.

But then the 2nd premise implies that Q(y) is true, a contradiction (since we assumed Q(y) was false).

Therefore the conclusion must be true, that is n ∀ ∈Z+, Q(n) holds.

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So a proof by mathematical induction has

One (or more) base case(s): Q(1)

Usually very straightforward to prove.

An induction step:

∀n Z∈ +, Q(n) → Q(n+1) OR n Z∀ ∈ +, Q(n-1) → Q(n)

We can use any proof technique we know for it.

Usually a direct proof will work well.

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Module Summary





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Module 10.3: More induction examples

Example: recall the exercises from our first class:

Make group of 5 students.

Then order each group from in order of day of birth by swapping pairs of adjacent students.

We claimed the maximum number of swaps for n students is n(n-1)/2.

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How many swaps do we need?

Suppose we place students from left to right.

The students already placed are ordered by day of birth.

We swap each new student with his/her neighbour until he/she is at the right place.

The ith student may be swapped with all previous i-1 students.

So the total number of swaps is at most

Hence we need to prove that

∑i=0

n−1

i

∑i=0

n−1

i=n(n−1)

2

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Which fact(s) do we need to prove?

a)

b)

c)

d) Both (a) and (b)

e) Both (a) and (c)

∑i=0

0

i=0

∀ n∈ℤ+ (∑

i=0

n−1

i=n(n−1)

2 )→(∑i=0

n

i=(n+1) n

2 )(∀n∈ℤ+ ,∑

i=0

n−1

i=n(n−1)

2 )→(∀ n∈ℤ+ ,∑i=0

n

i=(n+1)n

2 )

▷

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Proof:

Base case: n = 1

Clearly, with only 1 student, no swap is needed: which is equal to 1 (1-1)/2

Induction step:

Pick an unspecified n ≥ 1. Assume that when we have n students, we need at most n(n-1)/2 swaps. Equivalently we assume that

This is called the Induction Hypothesis.

∑i=0

0

i=0

∑i=0

n−1

i=n(n−1)2

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Proof (continued)

Induction step (continued)

For n+1 students, we need at most swaps.

By the induction hypothesis, this is equal to

swaps, as required for the induction step.

Hence by the principle of M.I., the theorem holds. QED

∑i=0

n

i=(∑i=0

n−1

i)+ n

n(n−1)2

+ n=(n+ 1)n2

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Worksheet problems 1 and 2

Theorem: for every value of a ≠0, 1:

These summations occur frequently when we need to determine the running time of divide-and-conquer algorithms.

Theorem: n ≥ 4, 2n < n!

∑i=0

t

a i=

a t+ 1−1

a−1

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Rules for inequalities:

Start from one side (say the left side)

Work step by step towards the other.

When dealing with <, you are allowed to make the expression on the RHS larger, but never smaller.

Example: if I am smaller than you, then I am still smaller than you when you stand on a bench.

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Worksheet problem 3

Theorem: prove that for every n ≥ 1,

Hint:

Note (this doesn’t help with the proof; it’s simply an interesting mathematical fact):

∑i=1

n1i2≤2−

1n

∑i=1

∞ 1

i2=π

2

6

1

n2<

1n (n−1)

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Example: consider the following DFA:

3

20

4

1

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This DFA accepts the empty string, and which other strings?

a) Strings whose length is divisible by 5.

b) Unsigned binary integers that are prime numbers.

c) Strings where every 1 is followed by 01.

d) Unsigned binary integers that are divisible by 5.


▷

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Worksheet problem 4

Theorem: After reading the string s representing the unsigned binary integer xs, the DFA is in state xs mod 5

Hint 1: use induction on the length n of s.

Hint 2: if bn-1

bn-2

...b1b

0 represents the integer x,

then bn-1

bn-2

...b1b

0b represents the integer 2x+b.

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Mathematical induction is how we analyze recursive algorithms:

Example:

(define (sum n)

(if (= n 0)

0

(+ n (sum (- n 1)))))

The type of induction we discussed so far works when the recursive call is with argument (- n 1)

∑i=0

n

i=n+ (∑i=0

n−1

i)

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Module Summary





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Module 10.4: A slightly different type of induction

Some recursions have recursive calls where the argument is not (- n 1):

(function1 (- n 2))

(function2 (quotient n 2))

(function3 x) where x is some integer in the range 0 < x < n

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How do we handle these?

A recursive function/method has

One or more base casesWhen it can compute the answer directly

An induction stepIt computes recursively the solutions to one or more “smaller” sub-problems, and combines them to obtain its answer.

Our induction proofs mimic this structure.

It is similar to the templates you used in CPSC 110.

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We use a slightly different induction hypothesis.

Instead of proving that

∀x Z∈ +, Q(x) → Q(x+1)

We prove that

∀x Z∈ +, (Q(1) ^ Q(2) ^ ... ^ Q(x)) → Q(x+1)

We can also show that this type of induction is a valid proof technique.

The proof is the same proof by contradiction that we looked at earlier.

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Example: Wilma Flinstone asks Fred to buy dinosaurs eggs to make a large omelet for a party. Eggs are sold in packs of 4 and 7 eggs. Prove:

F(x): Fred can buy exactly x dinosaur eggs for any integer x ≥ 18.

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How do we decide which base cases we need?

Look at the induction step. Decide for which values of n it can be used to prove F(x).

The induction step might not be usable for some x:

because it needs Q(some k < x) and k would be smaller than smallest x for which Q holds.

because some mathematical steps only work for large enough values of x.

If the induction step is not usable for x, then x needs to be a base case!

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In the simplest induction step possible, which fact will we use to prove F(x) ?

a) F(x-1)

b) F(x-2)

c) F(x-3)

d) F(x-4)

e) F(y) where y is a value other than x-1, x-2, x-3, x-4.

▷

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Induction step:

Consider an unspecified integer x ≥

Assume F(k) holds when k is in the range 18 ≤ k < x.

Then Fred can buy x eggs as follows:Fred buys x - eggs.Then Fred buys a pack of eggs.

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Base case: what case(s) do we need to prove?

a) x = 18

b) x = 18, x = 19

c) x = 18, x = 19, x = 20, x = 21

d) x = 18 up to x = 26


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Base cases:

18 = 2 x 7 + 1 x 4

19 = 1 x 7 + 3 x 4

20 = 0 x 7 + 5 x 4

21 = 3 x 7 + 0 x 4

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Worksheet problems 5 and 6

Theorem: we define the Gibbonacci numbers as follows: G(1) = G(2) = 1, G(n) = 2G(n-1) + 3G(n-2) for n ≥ 3. Prove that G(n) ≤ 3n.

Theorem: every positive integer n greater than 1 that is either a prime or can be written as a product of primes.

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Problem: find the ith smallest element in a list

Algorithm:

Pick a random element x of the list.

Divide the list into three sublists:

list-smaller: elements smaller than x

list-equal: elements equal to x

list-larger: elements larger than x

Then searchlist-smaller if i ≤ length of list-smallerlist-larger if i > length of list smaller + length of list-equalotherwise return x

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This algorithm is called randomized-quick-select.

Racket code will be posted on the course web site.

A student shows the expected number of steps S(n) of the algorithm on a list with n elements is:

S(1) = 4c

S(2) = 12c

S(3) = 20c

S(n) ≤ 2cn + S(ë3n/4û) when n ≥ 4

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Worksheet problem 7

Prove that for every n ≥ 1, S(n) ≤ 8cn.

Documents

Module 10: Proof Techniques (part 2) Mathematical Induction