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

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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 first n odd integers is n 2. Prove a base case (n=1) Base case (n=1): the sum of the first 1 odd integer is 1 2. Yes, 1 = 1 2. Prove P(k) P(k+1) Assume P(k): the sum of the first k odd ints is k 2. 1 + 3 + … + (2k - 1) = k 2 Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1) 2 Inductive hypothesis 1 + 3 + … + (2k-1) + (2k+1) =k 2 + (2k + 1) By inductive hypothesis = (k+1) 2 By arithmetic

Extensible Networking Platform 2 2 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - a cool example Deficient Tiling A 2 n x 2 n sized grid is deficient if all but one cell is tiled. 2n2n 2n2n

Extensible Networking Platform 3 3 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - a cool example We want to show that all 2 n x 2 n sized deficient grids can be tiled with tiles, called triominoes, shaped like:

Extensible Networking Platform 4 4 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - a cool example Is it true for all 2 1 x 2 1 grids? Yes!

Extensible Networking Platform 5 5 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - a cool example Inductive Hypothesis: We can tile any 2 k x 2 k deficient board using our fancy designer tiles. Use this to prove: We can tile any 2 k+1 x 2 k+1 deficient board using our fancy designer tiles.

Extensible Networking Platform 6 6 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - a cool example 2k2k 2k2k 2k2k 2k2k 2 k+1 OK!! (by IH) ? ? ?

Extensible Networking Platform 7 7 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - a cool example 2k2k 2k2k 2k2k 2k2k 2 k+1 OK!! (by IH)

Extensible Networking Platform 8 8 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - a cool example

Extensible Networking Platform 9 9 - CSE 240 – Logic and Discrete Mathematics 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.

Extensible Networking Platform 10 10 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - why does it work? Is the set of non-negative reals (R) well ordered? No. { x R : x > 1 } has no least element.

Extensible Networking Platform 11 11 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - why does it work? Proof of Mathematical Induction: We prove that (P(0) ( k P(k) P(k+1))) ( n P(n)) Proof by contradiction. Assume 1.P(0) 2. k P(k) P(k+1) 3. n P(n) n P(n)

Extensible Networking Platform 12 12 - CSE 240 – Logic and Discrete Mathematics Mathematical Induction - why does it work? Assume 1.P(0) 2. n P(n) P(n+1) 3. n P(n) n P(n) Let S = { n : P(n) } Since N is well ordered, S has a least element. Call it k. What do we know? -P(k) is false because its in S. -k 0 because P(0) is true. -P(k-1) is true because P(k) is the least element in S. But by (2), P(k-1) P(k). Contradicts P(k-1) true, P(k) false. Done.

Extensible Networking Platform 13 13 - CSE 240 – Logic and Discrete Mathematics Strong Mathematical Induction If P(0) and n 0 (P(0) P(1) … P(n)) P(n+1) Then n 0 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.

Extensible Networking Platform 14 14 - CSE 240 – Logic and Discrete Mathematics 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

Extensible Networking Platform 15 15 - CSE 240 – Logic and Discrete Mathematics 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

Extensible Networking Platform 16 16 - CSE 240 – Logic and Discrete Mathematics 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. How is this different than regular induction?

Extensible Networking Platform 17 17 - CSE 240 – Logic and Discrete Mathematics 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

Extensible Networking Platform 18 18 - CSE 240 – Logic and Discrete Mathematics 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 Why does this work?

Extensible Networking Platform 19 19 - CSE 240 – Logic and Discrete Mathematics Recursive Definitions We completely understand the function f(n) = n!, right? As a reminder, heres the definition: n! = 1 · 2 · 3 · … · (n-1) · n, n 1 Inductive (Recursive) Definition But equivalently, we could define it like this: Recursive CaseBase Case

Extensible Networking Platform 20 20 - CSE 240 – Logic and Discrete Mathematics Recursive Definitions Another VERY common example: Fibonacci Numbers Recursive CaseBase Cases Is there a non-recursive definition for the Fibonacci Numbers?

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