242-535 ada: 1. induction1 objective o to introduce mathematical induction through examples...

242-535 ADA: 1. Induction 1

• Objectiveo to introduce mathematical induction through

examples

Algorithm Design and Analysis

(ADA)242-535, Semester 1 2014-2015

1. Mathematical

Induction


1. Motivation2. Induction Defined3. Maths Notation Reminder4. Four Examples5. More General Induction Proofs6. A Fun Tiling Problem

Overview


• Induction is used in mathematical proofs of many recursive algorithmso e.g. quicksort, binary search

• Induction is used to mathematically define recursive data structureso e.g. lists, trees, graphs

1. Motivation

continued


• Induction is often used to derive mathematical estimates of program running timeo timings based on the size of input data

• e.g. time increases linearly with the number of data items processed

o timings based on the number of times a loop executes


• Induction is used to solve problems such as:o is S(n) correct/true for all n values?

• usually for all n >= 0 or all n >=1

• Example:

o let S(n) be "n2 + 1 > 0"o is S(n) true for all n >= 1?

2. Induction Defined

continued

S(n) can be muchmore complicated,such as a programthat reads in an nvalue.


• How do we prove (disprove) S(n)?

• One approach is to try every value of n:o is S(1) true?o is S(2) true?o ...o is S(10,000) true?o ... forever!!!

Not very practical


• Induction is a technique for quickly proving S(n) true or false for all no we only have to do two things

• First show that S(1) is trueo do that by calculation as before

Induction to the Rescue

continued


• We prove that S(n) S(n+1) is true• Read as "if S(n) is true then S(n+1) is true"

• When S(1) is true and "S(n) S(n+1)" is true, then S(n) is true for all n>=1.

• Why?

continued

" stands for "implies"


• With S(1) and S(n) S(n+1)then S(2) is true o S(1) S(2) when n == 1



• and so on, for all n


• We prove the implication by:

• 1. Assume that S(n) is true; write it down

• 2. Write down some extra maths, which depends on the problemo e.g. (n+1)! = n! * (n+1)

• 3. Write down S(n+1)

• 4. Combine S(n) and the maths to obtain S(n+1)

Proving S(n) S(n+1)


• Prove S(n): "n2 + 1 > 0" for all n >= 1.

• First task: show S(1) is true by calculationo S(1) == 12 + 1 == 2, which is > 0o so S(1) is true

• Second task: prove S(n) S(n+1) is true

Let’s do it

continued


• Assume S(n) is trueo S(n): n2 + 1 > 0

• Extra maths: n > 0

• Write down S(n+1)o S(n+1) == (n+1)2 + 1

== n2 + 2n + 1 +1

== (n2 + 1) + 2n + 1

continued


• Use S(n) and the extra maths to show that S(n+1) is trueo S(n) means that n2 + 1 > 0, and the extra maths is n

> 0,

so (n2 + 1) + 2n + 1 > 0o this means that S(n+1) is trueo so S(n) S(n+1)

continued


• We have used induction to show two things:o S(1) is trueo S(n) S(n+1) is true

• From these it follows that S(n) is true for

all n >= 1


• Three pieces:o 1. A statement S(n) to be proved

• the statement must be about an integer n

o 2. A basis for the proof. This is the statement S(b) for some integer.

Often b = 0 or b = 1.

Induction More Formally

continued


• 3. An inductive step for the proof. We prove the statement “S(n) S(n+1)”

o The statement S(n), used in this proof, is called the inductive hypothesis

o We conclude that S(n) is true for all n >= b• S(n) might not be true for some n < b


• Summation: means 1+2+3+4+…+n

o e.g. means 4+9+16+…+m2

• Product: means 1*2*3*…*n

3. Maths Notation Reminder

n

i

i1

m

j

j2

2

n

i

i1


• Prove the statement S(n):for all n >= 1o e.g. 1+2+3+4 = (4*5)/2 = 10

• Basis. S(1), n = 1so 1 = (1*2)/2

4. Example 1

2

)1(

1

nni

n

i

1

1

1i

i

continued


• Inductive Step. Prove S(n) S(n+1)

• 1. Assume S(n) is true:

• 2. The extra maths involve summations:

• 3. We want to calculate S(n+1):

2

)1(

1

nni

n

i(1)

1

1 1

)1(n

i

n

i

nii (2)

2

)2)(1(

2

)11(11

1

nnnni

n

i

(3)

continued


• To prove S(n) S(n+1), combine S(n) (equation 1) and the extra maths (equation 2) to calculate S(n+1) (equation 3).

continued


• Substitute the right hand side (rhs) of (1) for the first operand of (2), to give:

= (n2 + n + 2n + 2) /2

= (n2+3n+2)/2

)1(2

)1(1

1

nnn

in

i

which is (3)2

)2)(1(1

1

nni

n

i

continued


• We have shown:o S(1) is trueo S(n) S(n+1) is true

• This means that S(n) is true for all n ≥ 1• Finished.

continued


• Prove the statement S(n):for all n >= 0o e.g. 1+2+4+8 = 16-1

• Basis. S(0), n = 0

so 20 = 21 -1

Example 2

122 1

0

nn

i

i

0

0

022i

i

continued






(1)

(2)

(3)

continued

122 1

0

nn

i

i

122 21

0

nn

i

i

1

0 0

1222n

i

n

i

nii



continued



= 2*(2n+1) - 1

which is (3)

111

0

2122

nnn

i

i

122 21

0

nn

i

i




continued


• Prove the statement S(n): n! >= 2n-1

for all n >= 1

o e.g. 5! >= 24, which is 120 >= 16

• Basis. S(1), n = 1: 1! >= 20

so 1 >= 1

Example 3

continued



• 1. Assume S(n) is true: n! >= 2n-1

• 2. The extra maths involve factorials: (n+1)! = n! * (n+1)

• 3. We want to calculate S(n+1): (n+1)! >= 2(n+1)-1

>= 2n

(1)

(2)

(3)

continued



continued



(n+1)! >= 2n-1 * (n+1)

>= 2n-1 * 2 since (n+1) >= 2

(n+1)! >= 2n which is (3)

why?




continued


• Prove the statement S(n):for all n >= 1o This proof can be used to

show that. the limit of the sum:is 1

• Basis. S(1), n = 1so 1/2 = 1/2

Example 4

1)1(

1

1

n

n

ii

n

i

1

1 11

1

)1(

1

i ii

continued

...4*3

1

3*2

1

2*1

1






(1)

(2)

(3)

continued

1)1(

1

1

n

n

ii

n

i

)2)(1(

1

)1(

1

)1(

11

1 1

nniiii

n

i

n

i

1)1(

)1(

)1(

11

1

n

n

ii

n

i



continued



=1)1(

)1(

n

nwhich is (3)

)2)(1(

1

1)1(

11

1

nnn

n

ii

n

i




continued


• There can be more than one basis case.

• We can use strong induction where the proof of S(n+1) may use any of S(b), S(b+1), …, S(n)o b is the lowest basis value

5. More General Inductive

Proofs


• Show:o S(1) is trueo S(2) is trueo S(3) is true

o S(n-2) S(n+1) is true

• This means that S(n) is true for all n ≥ 1• Why?

Strong Induction Example

3 base cases


• Every integer >= 24 can be written as 5a+7b for non-negative integers a and b.o note that some integers < 24 cannot be expressed this

way (e.g. 16, 23).

• Let S(n) be the statement (for n >= 24)“n = 5a + 7b, for a >= 0 and b >= 0”

Example

continued


• We will show:o S(24) is trueo S(25) is trueo S(26) is trueo S(27) is trueo S(28) is true

o S(n-4) S(n+1) is true

• This means that S(n) is true for all n ≥ 24• Why?

Using Strong Induction

5 base cases


• Basis. The 5 basis cases are 24 through 28.

o S(24) true because 24 = (5*2) + (7*2)o S(25) true because 25 = (5*5) + (7*0)o S(26) true because 26 = (5*1) + (7*3)o S(27) true because 27 = (5*4) + (7*1)o S(28) true because 28 = (5*0) + (7*4)

continued


• Inductive Step. Prove S(n-4) S(n+1)

• 1. Assume S(n-4) is true: n - 4 = 5a + 7b

• 2. No extra maths


n+1 = 5a' + 7b'

• To prove S(n-4) S(n+1), use S(n-4) (equation 1) to calculate S(n+1) (equation 2).

(1)

(2)

continued


• Use S(n-4): n-4 = 5a + 7b

• Add 5 to both sides of the equation:o n-4 + 5 = 5a + 7b + 5o n + 1 = 5(a + 1) + 7bo n + 1 = 5a' + 7b (a' is some new variable)o which means that S(n+1) is true


• We have shown:o S(24), S(25), S(26), S(27), S(28) are trueo S(n-4) S(n+1) is true


continued


• A right tromino is a “corner” shapemade of 3 squares:

• Use induction to prove that any number of right trominos can be used to tile (cover) any n*n size board of squareso but n must be a power of 2

• e.g. board sizes can be 2*2, 4*4, 8*8, 16*16, ...

• Show that a tiled board will always have 1 blank square.

6. A Tiling Problem

continued


• For example, a 4*4 board (22*22), with 1 blank square:

5 trominos used

242-535 ada: 1. induction1 objective o to introduce mathematical induction through examples...

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