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CPSC 121: Models of Computation
2011 Winter Term 1

Revisiting Induction
Steve Wolfman, based on work by
Patrice Belleville and others

1
Outline
• Prereqs and Learning Goals
• Problems and Discussion
– Introductions
– Odd Numbers
– Horse Colours
– CS Induction: Duplicate Detection, Binary
Search, MergeSort
– More examples
• Next Lecture Notes
2
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.

3
Learning Goals: In-Class
By the end of this unit, you should be able to:
– Formally prove properties of the non-negative
integers (or a subset like integers larger than 3)
that have appropriate self-referential structure—
including both equalities and inequalities—using
either weak or strong induction as needed.
– Critique formal inductive proofs to determine
whether they are valid and where the error(s) lie if
they are invalid.

4
Online Quiz Notes:
New Problem
Imagine that your country's postal system
only issues 3 cent and 7 cent stamps. Prove
by induction that it is possible to pay for
postage using only these stamps for any
amount n cents, where n is at least 12.

We will use n=12 as one of our base cases.
(“We pay 12 cents with four 3 cent stamps”.)

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Extra Base Cases?
Our inductive step says that for “sufficiently
large” n, we can make n cents of postage by
making n-3 cents of postage and then using a
three cent stamp.
What additional base case (if any) do we need?
a. n = 6
b. n = 13
c. n = n-3
e. more than one additional base case
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n?
Our inductive step says that for “sufficiently large” n,
we can make n cents of postage by making n-3
cents of postage and then using a three cent stamp.
We use n=12, n=13, and n=14 as base cases.
How large must n be for the inductive step?
a. n  12
b. n  13
c. n  14
d. n  15
e. none of these

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Extra Base Cases?
Version 2
Our inductive step says that for “sufficiently large” n, we
can make n cents of postage by making n-1 cents of
postage and then doing one of: take out two 7c stamps
and add five 3c stamps or (if there are fewer than two 7c
stamps), take out two 3c stamps and add a 7c stamp.
What additional base case (if any) do we need?
a. n = 6
b. n = 13
c. n = n-1
e. more than one additional base case

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n?
Version 2
Our inductive step says that for “sufficiently large” n, we
can make n cents of postage by making n-1 cents of
postage and then doing one of: take out two 7c stamps
and add five 3c stamps or (if there are fewer than two 7c
stamps), take out two 3c stamps and add a 7c stamp.
We use n=12 as our only base case.
How large must n be for the inductive step?
a. n  12
b. n  13
c. n  14
d. n  15
e. none of these

9
Outline
• Prereqs and Learning Goals
• Problems and Discussion
– Introductions
– Odd Numbers
– Horse Colours
– CS Induction: Duplicate Detection, Binary
Search, MergeSort
– More examples
• Next Lecture Notes
10
Worked Problem:
How Many Introductions?
Problem: n people would like to introduce
themselves to each other. How many
introductions does it take?
For 2 people?
For 3 people?
For 4 people?
For 5 people?
…
For n people?
Sound familiar?
Let’s prove it.
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Worked Problem:
How Many Introductions?
Induction can feel very abstract. Let’s do
the first few steps concretely…

For 1 person?

Given the number for 1, for 2?

Given the number for 2, for 3?
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Worked Problem:
How Many Introductions?
Let’s let I(n) be the number of introductions
required for a group of n people…

I(1)? 0 introductions.                                   My turn!

Given I(k-1), what’s I(k)?
a. I(k-1) + 1
b. I(k-1) * (k-1)
c. I(k-1) + 2(k – 1)
d. I(k-1) + 2k
e. None of these
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Let’s work through the intuition and then use it to fill in the pattern!
Finishing the Intuition:
How Many Introductions?
I(k) =         I(k-1) + 2(k-1)
=         (k-1)((k-1)-1) + 2(k-1)
=         (k-1)(k-2) + 2(k-1)   My turn!
=         k2 – 3k + 2 + 2k – 2
=         k2 – k
=         k(k-1)

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Let’s use this to fill in our pattern!
Formally…
Definition: When two people greet each
other, we count that as two introductions.
Let I(k) be the number of introductions it
takes for a group of size k to introduce
themselves. (I(.) is not a predicate; it’s a
function!)
To prove: I(n) = n(n-1).

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Finishing the Intuition:
How Many Introductions?
I(k) =   I(k-1) + 2(k-1)
=   (k-1)((k-1)-1) + 2(k-1)                  My turn!
=   (k-1)(k-2) + 2(k-1)
=   k2 – 3k + 2 + 2k – 2
=   k2 – k
=   k(k-1)

Def’ns: Two people greet: two intros. I(k) is the #
of intros for a group of size k to intro themselves.
To prove: I(n) = n(n-1) for all groups of at least
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one person.
A Pattern For Induction

P(n) is ______________.
Theorem: P(n) is true for all n  _______.
Proof: We proceed by induction on n.
Base Case(s) (P(.) is true for _______):
Prove each base case via your other techniques.
Inductive Step (if P(.) is true for ________________, then P(n) is true,
for n > ____________):

WLOG, let n be greater than ____________.
Assume P(.) is true for __________________.
Break P(n) down in terms of the smaller case(s).
The smaller cases are true, by assumption.
Build back up to show that P(n) is true.
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This completes our induction proof. QED
Introductions Proof

P(n) is ______________.

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Introductions Proof

P(n) is I(n) = n(n-1).
Theorem: P(n) is true for all n  _______.

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Introductions Proof

P(n) is I(n) = n(n-1).
Theorem: P(n) is true for all n  1.
Proof: We proceed by induction on n.
Base Case(s) (P(.) is true for _______):
Prove each base case via your other techniques.

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Introductions Proof

P(n) is I(n) = n(n-1).
Theorem: P(n) is true for all n  1.
Proof: We proceed by induction on n.
Base Case(s) : a “group” of one person takes no introductions .
I(1) = 0 = 1(1-1). 
Inductive Step (if P(.) is true for ________________, then P(n) is true,
for n > ____________):

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Introductions Proof

P(n) is I(n) = n(n-1).
Theorem: P(n) is true for all n  1.
Proof: We proceed by induction on n.
Base Case(s) : a “group” of one person takes no introductions .
I(1) = 0 = 1(1-1). 
Inductive Step:
WLOG, let n be an integer greater than 1.
Assume I(k-1) = (k-1)((k-1)-1).
Break P(n) down in terms of the smaller case(s).
The smaller cases are true, by assumption.
Build back up to show that P(n) is true.

This completes our induction proof. QED
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Introductions Proof Complete

P(n) is I(n) = n(n-1).
Theorem: P(n) is true for all n  1.
Proof: We proceed by induction on n.
Base Case(s) : a “group” of one person takes no introductions .
I(1) = 0 = 1(1-1). 
Inductive Step:
WLOG, let n be an integer greater than 1.
Assume I(k-1) = (k-1)((k-1)-1) = (k-1)(k-2).
I(k) = I(k-1) + 2(k-1)
= (k-1)(k-2) + 2(k-1)
= k2 – 3k + 2 + 2k – 2
= k2 – k
= k(k-1)
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This completes our induction proof. QED
Outline
• Prereqs and Learning Goals
• Induction as a Formal Argument Form
• Problems and Discussion
– Introductions
– Odd Numbers
– Horse Colours
– CS Induction: Duplicate Detection, Binary Search,
MergeSort
– More examples
• Next Lecture Notes

24
Historical Problem:
Sum of Odd Numbers
Problem: What is the sum of the first n odd
numbers?

First, find the pattern. Then, prove it’s correct.
The first 1 odd number?
The first 2 odd numbers?
The first 3 odd numbers?
The first n odd numbers?
Historical note: Francesco Maurolico made the first
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recorded use of induction in 1575 to prove this theorem!
Sum of Odd Numbers:
Insight
Problem: Prove that the sum of the first n
odd numbers is n2.

How can we break the sum of the first,
second, …, nth odd number up in terms of
a simpler sum of odd numbers?

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Sum of Odd Numbers:
Insight
Problem: Prove that the sum of the first n
odd numbers is n2.

The sum of the first n odd numbers is the
sum of the first n-1 odd numbers plus the
nth odd number.

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Historical Problem:
Sum of Odd Numbers
Now, how do we prove it? We need...
Property: For all positive integers n... ?
Base Case: Establish for n=?
Inductive Hypothesis: Assume... ?
Inductive Step: To prove... ?

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Historical Problem:
Sum of Odd Numbers
Property:
For all positive integers n, the sum of the
first n odd natural numbers is n2.
Basis Case: Establish for n=?
Inductive Hypothesis: Assume... ?
Inductive Step: To prove... ?

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Historical Problem:
Sum of Odd Numbers
Theorem: For all positive integers n, the sum of
the first n odd natural numbers is n2.
Base Case:

Establish for n=1. The sum of the “first 1 odd
natural numbers” is 1, which equals 12. 

Inductive Hypothesis: Assume... ?
Inductive Step: To prove... ?

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Historical Problem:
Sum of Odd Numbers
Theorem: For all positive integers n, the sum
of the first n odd natural numbers is n2.
Base Case for n = 1 
Inductive Hypothesis: For some arbitrary
k > 1, assume the sum of the first k-1 odd
natural numbers is (k-1)2
Inductive Step: To prove... ?

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Historical Problem:
Sum of Odd Numbers

(by the IH)

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Outline
• Prereqs and Learning Goals
• Induction as a Formal Argument Form
• Problems and Discussion
– Introductions
– Odd Numbers
– Horse Colours
– CS Induction: Duplicate Detection, Binary Search,
MergeSort
– More examples
• Next Lecture Notes

33
Problem: Proof Critique
Theorem: All horses are the same colour.
See handout.

Problem: Critique the proof.

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Problem: Proof Critique
Theorem: All horses are the same colour.
See handout.

Proof critique: Is the proof valid?
a. Yes, because each step follows irrefutably
from the previous steps.
b. Yes, because the premises are false.
c. Yes, but not for the reasons listed here.
d. No, because the conclusion is false.
e. No, but not for the reasons listed here.
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Problem: Proof Critique
Theorem: All horses are the same colour.
See handout.

Proof critique: Where is the error in the proof?
a. The base case.
b. The inductive hypothesis.
c. The inductive step.
d. None of these, but there must be an error.
e. None of these, and there may be no error.

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Problem: Proof Critique
Theorem: All horses are the same colour.
See handout.

Proof critique: Can the proof be fixed?

37
Outline
• Prereqs and Learning Goals
• Induction as a Formal Argument Form
• Problems and Discussion
– Introductions
– Odd Numbers
– Horse Colours
– CS Induction: Duplicate Detection, Binary Search,
MergeSort
– More examples
• Next Lecture Notes

38
Induction and Computer Science
How important is induction? Is it just for proving

Induction forms the basis of…
– proofs of correctness and efficiency for many
algorithms: e.g., How long does it take to detect
duplicates in a list of n inputs? Does binary search
work?
– “recursive” algorithms: e.g., “merge sort”
– “recursive data structures”: e.g., How big can a “tree”
of height n be?
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Problem: Detecting Duplicates
Problem: How long does it take to detect
duplicates in a list of n Inputs?
Let’s assume the amount of “time” taken is
proportional to the number of comparisons
we make between numbers.
We could compare each element to each
other element and see if they’re the same.
Sound familiar? We already know how long
this takes!
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Note: this isn’t the fastest algorithm!
Problem: Binary Search Works
Problem: Prove that binary search works.
Binary search is when we search a sorted
list by checking the middle element.
If it’s what we’re looking for, we’re done.
Otherwise, we “throw out” the part of the list
we don’t need (e.g., the left half if the
element we looked at was too small) and
start over on what remains.
41
Sorting by “Merging”
Problem: sort a list of names.
Algorithm:
1. If the list is of length 1, it’s sorted.
2. Otherwise:
a) Divide the list in half (or as close as possible).
b) Sort each half using this algorithm.
c) Merge the sorted lists back together.

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The Merge Step
Problem: given two lists of names in sorted
order, merge them into a single sorted list.
Merge(a,b):
1. If a is empty, return b
2. If b is empty, return a
3. Otherwise, if the first element of b comes before
the first element of a, return a list with the first
element of b at the front and the result of
Merge(a, rest of b) as the rest
4. Otherwise, return a list with the first element of
a at the front and the result of
Merge(rest of a, b) as the rest.                     43
Familiar?
Problem: Prove Merge works

44
Outline
• Prereqs and Learning Goals
• Induction as a Formal Argument Form
• Problems and Discussion
– Introductions
– Odd Numbers
– Horse Colours
– CS Induction: Duplicate Detection, Binary Search,
MergeSort
– More examples
• Next Lecture Notes

45
Problem:
Prove that 2n < n!
Problem: prove that 2n < n!?

Note: is 2n < 46
n!?
Outline
• Prereqs and Learning Goals
• Induction as a Formal Argument Form
• Problems and Discussion
– Introductions
– Odd Numbers
– Horse Colours
– CS Induction: Duplicate Detection, Binary Search,
MergeSort
– More examples
• Next Lecture Notes

47
Learning Goals: In-Class
By the end of this unit, you should be able to:
– Formally prove properties of the non-negative
integers (or a subset like integers larger than 3)
that have appropriate self-referential structure—
including both equalities and inequalities—using
either weak or strong induction as needed.
– Critique formal inductive proofs to determine
whether they are valid and where the error(s) lie if
they are invalid.

48
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Extra Slides

49
Problem: Proof Critique
Theorem: All integers ≥ 2 are even.

Problem: Critique the proof. Is it valid? If
not, why not? Can it be fixed, and how?

50
Problem:
Sum of a Geometric Series
Problem: What is the sum of the first n
terms of the form ai, if a is a real number
between 0 and 1? (Note: we mean terms
0 through n.)
n 1
a      1
Note: we’re looking for
a 1

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Can we use induction over the real numbers between 0 and 1?
Induction: a Domino Effect
A proof by induction is like toppling a chain
of dominoes.
Let’s say I tell you:
– the first domino in the chain toppled
– for every domino in the chain, if the domino
before it toppled, then the domino itself
toppled

Did all the dominoes in the chain topple?
52
Formal (Weak) Induction
A formal induction proof follows the same pattern. To
prove some property P(.) applies to all positive integers
n, we prove:

The first domino topples.   P(0) is true.
If one domino topples…      [For arbitrary k] If P(k) is true...
Then the next one           Then P(k+1) is also true.
topples.

Typically, we prove the second step by
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antecedent assumption/direct proof.
Form of an Induction Proof
Theorem: some property that depends on n

Basis step: show that a small case works

holds for n=k where k ≥ your basis step.

Inductive Step: Under this assumption, you
need to prove that your property holds for
n=k+1.
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Side Note: Practical Induction
That’s how you formally write out the proof,
but how do you figure out the proof?
Start at the inductive step!

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Side Note: Practical Induction
Look at a “big” problem (of size n).
Figure out how to break it down into smaller
pieces.
Assume those smaller pieces work. That
will end up as your inductive hypothesis.
Figure out which problems cannot be broken
down (usually small ones!). Those will
end up as your basis step(s).
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Proof
Base case: A group of size 1 takes 0 intros.
(Because there are no pairs.)

Note: we talked about induction to prove properties for
57
the non-negative integers, but it can be more general.
Proof (Continued)
Inductive step, to prove: If a group of size k-1
takes (k-1)((k-1)-1) intros, then a group
of size k takes k(k-1) intros.
Assume size (k-1) takes (k-1)((k-1)-1)
or (k-1)(k-2) intros (the Inductive
Hypothesis)
...

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Proof (Continued)
Inductive step, continued: A group of size k, where
k > 0, is a group of size k-1 plus one person.
Imagine size k-1 group has completed introductions.
That took (k-1)(k-2) intros by the IH.
Add one more person to the group, and that person
must greet all k-1 other people, for a total of
(k-1)(k-2) + 2(k-1) introductions.
(k-1)(k-2) + 2(k-1) = (k-1)[k-2 + 2]
= (k-1)k

QED: if k-1 takes (k-1)(k-2), k takes k(k-1).
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Proof (Completed)
Base case: A group of size 1 takes 0 intros.
(See proof above.)

Inductive step: If a group of size k-1 takes
(k-1)(k-2) intros, then a group of size k takes
k(k-1) intros. (See proof above.)

QED.

From these two we could (but don’t have to) crank out
a proof for any n that a group of size n requires
n(n-1) introductions.
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Form of a Strong Induction
Proof
Theorem: some property that depends on n

Basis step: show that one or more small cases
(as needed) work

Inductive Hypothesis: Assume your property holds
for any n = i where your basis step ≤ i < k and k
is an integer > your basis step.

Inductive Step: Under this assumption, you need to
prove that your property holds for n = k.
Hint: there’s really no difference. Just assume what you need in
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order to break the problem into smaller pieces.
Side Note: “Strong Induction”
We’ve used weak induction: basis step
(n = a) plus “if it works for n then it works
for n+1”.

In strong induction, we use: basis step plus
“if it works for all a  i < k then it works
for k”.

Can we still build our individual proofs (for n62
= 1, n = 2, n = 3, ...) from that?

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