peer to peer network properties

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 TilingA 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 10 if 0

)(nnfnfnn

nf

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