Extensible Networking Platform 1 1 - CSE 240 – Logic and Discrete Mathematics Review: Mathematical Induction Use induction to prove that the sum of the

Extensible Networking Platform 1 1 - CSE 240 – Logic and Discrete Mathematics

Review: Mathematical Induction

Use induction to prove that the sum of the first n odd integers is n2.

Base case (n=1): the sum of the first 1 odd integer is 12. Yes, 1 = 12.

Assume P(k): the sum of the first k odd ints is k2. 1 + 3 + … + (2k - 1) = k2

Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2

1 + 3 + … + (2k-1) + (2k+1) =

k2 + (2k + 1)= (k+1)2


Mathematical Induction - a cool example

Deficient Tiling

A 2n x 2n sized grid is deficient if all but one cell is tiled.

2n

2n



• We want to show that all 2n x 2n sized deficient grids can be tiled with tiles, called triominoes, shaped like:



• Is it true for all 21 x 21 grids?



Inductive Hypothesis:We can tile any 2k x 2k

deficient board using our fancy designer tiles.

Use this to prove:We can tile any 2k+1 x 2k+1

deficient board using our fancy designer tiles.



2k

2k 2k

2k

2k+1

OK!! (by IH)

?

?

?



2k

2k 2k

2k

2k+1

OK!! (by IH)

OK!! (by IH)

OK!! (by IH)

OK!! (by IH)




Mathematical Induction - why does it work?

Definition:A set S is “well-ordered” if every non-

empty subset of S has a least element.

Given (we take as an axiom): the set of natural numbers (N) is well-ordered.

Is the set of integers (Z) well ordered?

No. { x Z : x < 0 }

has no least element.



Is the set of non-negative reals (R) well ordered?



Proof of Mathematical Induction:

We prove that (P(0) (k P(k) P(k+1))) (n P(n))

Assume1. P(0)2. k P(k) P(k+1)3. n P(n) n P(n)



Assume1. P(0)2. n P(n) P(n+1)3. n P(n) n P(n)

Let S = { n : P(n) }

What do we know? -P(k) is false because it’s in S. -k 0 because P(0) is true. -P(k-1) is true because P(k) is the least

element in S.


Strong Mathematical Induction

If P(0) and n0 (P(0) P(1) … P(n)) P(n+1)

Thenn0 P(n) In our proofs, to show P(k+1), our

inductive hypothesis assumes that ALL of P(0), P(1), … P(k)

are true, so we can use ANY of them to make the inference.


Game with Matches

• Two players take turns removing any number of matches from one of two piles of matches. The player who removes the last match wins

• Show that if two piles contain the same number of matches initially, then the second player is guaranteed a win


Strategy for Second Player

• Let P(n) denote the statement “the second player wins when they are initially n matches in each pile”

• Basis step: P(1) is true, because only 1 match in each pile, first player must remove one match from one pile. Second player removes other match and wins

• Inductive step: suppose P(j) is True for all j 1<=j <= k.

• Prove that P(k+1) is true, that is the second player wins when each piles contains k+1 matches


Strategy for Second Player

• Suppose that the first player removes r matches from one pile, leaving k+1 –r matches there

• By removing the same number of matches from the other pile the second player creates the situation of two piles with k+1-r matches in each. Apply the inductive hypothesis and the second player wins each time.


Postage Stamp Example

• Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps

• P(n) : Postage of n cents can be formed using 4-cent and 5-cent stamps

• All n >= 12, P(n) is true


Postage Stamp Proof• Base Case: n = 12, n = 13, n = 14, n = 15

– We can form postage of 12 cents using 3, 4-cent stamps

– We can form postage of 13 cents using 2, 4- cent stamps and 1 5-cent stamp

– We can form postage of 14 cents using 1, 4-cent stamp and 2 5-cent stamps

– We can form postage of 15 cents using 3, 5-cent stamps

• Induction Step– Let n >= 15– Assume P(k) is true for 12 <= k <= n, that is postage

of k cents can be formed with 4-cent and 5-cent stamps (Inductive Hypothesis)

– Prove P(n+1)– To form postage of n +1 cents, use the stamps that

form postage of n-3 cents (from I.H) with a 4-cent stamp


Recursive Definitions

We completely understand the function f(n) = n!, right?

As a reminder, here’s the definition:n! = 1 · 2 · 3 · … · (n-1) · n, n 1

But equivalently, we could define it like this:

0 n if 1

1n if )!1(!

nnn


Recursive Definitions

Another VERY common example:

Fibonacci Numbers

1 if )2()1(

1 if 1

0 if 0

)(

nnfnf

n

n

nf

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Extensible Networking Platform 1 1 - CSE 240 – Logic and Discrete Mathematics Review: Mathematical Induction Use induction to prove that the sum of the