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Review: Mathematical Induction Use induction to prove that the sum of the first n odd integers is n2. Prove a base case (n=1) Base case (n=1): the sum of the first 1 odd integer is 12. Yes, 1 = 12. Prove P(k)P(k+1) Assume P(k): the sum of the first k odd ints is k2. 1 + 3 + … + (2k - 1) = k2 Inductive hypothesis Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2 By inductive 1 + 3 + … + (2k-1) + (2k+1) = k2 + (2k + 1) hypothesis = (k+1)2 By arithmetic 1 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 1 Mathematical Induction - a cool example Deficient Tiling A 2n x 2n sized grid is deficient if all but one cell is tiled. 2n 2n 2 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 2 Mathematical Induction - a cool example • We want to show that all 2n x 2n sized deficient grids can be tiled with tiles, called triominoes, shaped like: 3 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 3 Mathematical Induction - a cool example Yes! • Is it true for all 21 x 21 grids? 4 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 4 Mathematical Induction - a cool example 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. 5 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 5 Mathematical Induction - a cool example 2k 2k 2k ? ? 2k+1 2k OK!! (by IH) ? 6 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 6 Mathematical Induction - a cool example 2k 2k OK!! OK!! 2k (by (by IH) IH) 2k+1 OK!! OK!! 2k (by (by IH) IH) 7 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 7 Mathematical Induction - a cool example 8 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 8 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. No. { x Z : x < 0 } has no Is the set of integers (Z) well least element. ordered? 9 Extensible Networking Platform - CSE 240 – Logic and Discrete Mathematics 9 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 10 - CSE 240 – Logic and Discrete Mathematics 10 Mathematical Induction - why does it work? Proof of Mathematical Induction: We prove that (P(0) (k P(k) P(k+1))) (n P(n)) Assume Proof by 1. P(0) contradiction. 2. k P(k) P(k+1) 3. n P(n) n P(n) Extensible Networking Platform 11 - CSE 240 – Logic and Discrete Mathematics 11 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. But by (2), P(k-1) P(k). What do we know? Contradicts P(k-1) true, P(k) -P(k) is false because it’s in S. false. -k 0 because P(0) is true. -P(k-1) is true because P(k) is the least element in S. Done. Extensible Networking Platform 12 - CSE 240 – Logic and Discrete Mathematics 12 Strong Mathematical Induction If P(0) and n0 (P(0) P(1) … P(n)) P(n+1) Then In our proofs, to show P(k+1), our n0 P(n) 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 13 - CSE 240 – Logic and Discrete Mathematics 13 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 14 - CSE 240 – Logic and Discrete Mathematics 14 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 15 - CSE 240 – Logic and Discrete Mathematics 15 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 16 - CSE 240 – Logic and Discrete Mathematics 16 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 17 - CSE 240 – Logic and Discrete Mathematics 17 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 18 - CSE 240 – Logic and Discrete Mathematics 18 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: Inductive (Recursive) Recursive Case Definition n (n 1)! if n 1 n! Base Case 1 if n 0 Extensible Networking Platform 19 - CSE 240 – Logic and Discrete Mathematics 19 Recursive Definitions Another VERY common example: Fibonacci Numbers 0 if n 0 Base Cases f (n) 1 if n 1 f (n 1) f (n 2) if n 1 Recursive Case éæ ön æ1 - 5 ön ù Is there a non-recursive 1 ê 1+ 5 definition for the Fibonacci f (n) = ç ÷ -ç ÷ú 5 êè 2 ø è 2 ø ú ë û Numbers? Extensible Networking Platform 20 - CSE 240 – Logic and Discrete Mathematics 20