Strong induction (3) 23/38 Let P be a unary predicate on N Strong - PowerPoint PPT Presentation

Strong induction (3) 23/38 Let P be a unary predicate on N Strong induction: Induction . . . Lecture 12 (Chapter 19) ( ℓ ) ∀ k [ k ∈ N : ∀ j [ j ∈ N ∧ j < k : P ( j )] ⇒ P ( k )] { Strong induction on ( ℓ ) : } ( m ) ∀ n [ n ∈ N : P ( n )] October 14, 2016 NB: Follows from ‘normal induction.’ What happened to the base case? / department of mathematics and computer science / department of mathematics and computer science Example Example 24/38 25/38 var k ; k ∈ N (IH) ∀ j [ j ∈ N ∧ j < k : P ( j )] Consider the sequence of numbers a 0 , a 1 , a 2 , . . . defined by case k = 0 : a 0 = 2 = 3 · 2 0 − 1 , so P (0) case k = 1 : a 1 = 5 = 3 · 2 1 − 1 , so P (1) a 0 := 2 Define a 0 , a 1 , a 2 , . . . by case k ≥ 2 : := 5 a 1 ∀ -elim with k − 1 : P ( k − 1) , so a k − 1 = 3 · 2 k − 1 − 1 := 2 a 0 a i +2 := 3 a i +1 − 2 a i ( for all i ∈ N ) . a 1 := 5 ∀ -elim with k − 2 : P ( k − 2) , so a k − 2 = 3 · 2 k − 2 − 1 a i +2 := 3 a i +1 − 2 a i ( i ∈ N ) . Then: Then: Define P on N by a k = 3 a k − 1 − 2 a k − 2 (IH) = 3(3 · 2 k − 1 − 1) − 2(3 · 2 k − 2 − 1) P ( n ) := [ a n = 3 · 2 n − 1] . → a 0 a 1 a 2 a 3 a 4 a 5 = 9 · 2 k − 1 − 3 − 3 · 2 k − 1 + 2 = 6 · 2 k − 1 − 1 2 5 11 23 47 95 → We prove that = 3 · 2 k − 1 , so P ( k ) val { Case distinction ( k = 0 ∨ k = 1 ∨ k ≥ 2 = = k ∈ N !): } ∀ n [ n ∈ N : P ( n )] Prove that ∀ n [ n ∈ N : a n = 3 · 2 n − 1] . P ( k ) ∀ j [ j ∈ N ∧ j < k : P ( j )] ⇒ P ( k ) ∀ k [ k ∈ N : ∀ j [ j ∈ N ∧ j < k : P ( j )] ⇒ P ( k )] { Strong induction: } ∀ n [ n ∈ N : P ( n )] / department of mathematics and computer science / department of mathematics and computer science

Cutting the cake (1) Cutting the cake (2) 26/38 27/38 1 6 5 2 3 4 10 13 12 11 9 14 8 7 18 19 20 15 16 17 1 = 2 1 − 1 4 = 2 3 − 1 2 = 2 2 − 1 8 = 2 4 − 1 23 21 22 24 pieces pieces pieces pieces 25 27 28 29 30 26 31 Conjecture The number of pieces obtained by cutting the cake using n points on The maximum number of portions we can get using 6 points on the the edge of the cake is 2 n − 1 . edge of the cake is 31 � = 32 = 2 6 − 1 . Our conjecture fails for n = 6 ! / department of mathematics and computer science / department of mathematics and computer science Example Example (cont.) 28/38 29/38 (1) var p ; p ∈ N ∧ p > 7 Theorem: (2) ∀ j [ j ∈ N ∧ 7 < j < p : P ( j )] Every postage greater than 7 cent can be formed with only 3 -cent and (3) Case p = 8 : p = 1 · 3 + 1 · 5 , so ( ∃ ∗ -intro) P ( p ) 5 -cent stamps. (4) Case p = 9 : p = 3 · 3 + 0 · 5 , so ( ∃ ∗ -intro) P ( p ) (5) Case p = 10 : p = 0 · 3 + 2 · 5 , so ( ∃ ∗ -intro) P ( p ) Proof: (6) Case p ≥ 11 : 7 < p − 3 < p , so ( ∀ -elim) P ( p − 3) . Define the unary predicate P on N by Pick k, ℓ with k, ℓ ∈ N and p − 3 = k · 3 + ℓ · 5 (7) Then p = ( p − 3) + 3 = ( k · 3 + ℓ · 5) + 3 = ( k + 1) · 3 + ℓ · 5 So ( ∃ ∗ -intro) P ( p ) P ( p ) := ∃ k,ℓ [ k, ℓ ∈ N : p = k · 3 + ℓ · 5] . val { Case distinction: p ∈ N ∧ p > 7 = = p = 8 ∨ p = 9 ∨ p = 10 ∨ p ≥ 11 } (8) (9) P ( p ) To prove: ∀ p [ p ∈ N ∧ p > 7 : P ( p )] . (10) ∀ j [ j ∈ N ∧ 7 < j < p : P ( j )] ⇒ P ( p ) [Proof on the next slide] (11) ∀ p [ p ∈ N ∧ p > 7 : ∀ j [ j ∈ N ∧ 7 < j < p : P ( j )] ⇒ P ( p )] (12) { Strong induction: } (13) ∀ p [ p ∈ N ∧ p > 7 : P ( p )] / department of mathematics and computer science / department of mathematics and computer science

Example (proof in textual form) Induction checklist 32/38 34/38 Proof: We prove, with strong induction on p , that every postage p greater than 7 can be formed using only 3 -cent and 5 cent stamps. Let p be an arbitrary postage > 7 . Suppose: every postage p ′ with 7 < p ′ < p can be formed using 3 -cent and An inductive proof always includes the following ingredients: 5 -cent stamps (IH). 1. one or more basis cases; We now distinguish four cases: 2. one or more step cases; ◮ If p = 8 , then p can be formed with one 3 -cent stamp, and one 5 -cent 3. a clearly and explicitly stated induction hypothesis; and stamp. 4. one or more applications of the induction hypothesis. ◮ If p = 9 , then p can be formed with three 3 -cent stamps. ◮ If p = 10 , then p can be formed with two 5 -cent stamps. ◮ Suppose: p ≥ 11 . Then 7 < p − 3 < p , so by (IH) p − 3 can be formed using k 3 -cent stamps and ℓ 5 -cent stamps ( k, ℓ ∈ N ). Hence, p can be formed with k + 1 3 -cent stamps and ℓ 5 -cent stamps. Thereby, the result is proved. / department of mathematics and computer science / department of mathematics and computer science Example: tromino puzzles Example: tromino puzzles 35/38 36/38 Question: is it possible to tile every n × n -board with a single ‘unusable’ field? A tromino is a tile of the shape Answer: No! Obviously, tiling is, e.g., not possible if n 2 − 1 is not divisible by 3 . Theorem: An 8 × 8 board with an arbitrary ‘unusable’ field (coloured black) can Every 2 n × 2 n board with a single unusable field can be tiled with be tiled with trominoes: trominoes. Proof: The proof is by induction on n . ◮ (B ASIS ) If n = 0 , then the board entirely consists of the unusable field, and the rest of the board can be (trivially) tiled with zero trominoes. ◮ (S TEP ) [see next slide] / department of mathematics and computer science / department of mathematics and computer science

Example: tromino puzzles 38/38 Proof (cont.): ◮ (S TEP ) Suppose: every 2 n × 2 n board with single unusable field can be tiled (induction hypothesis). Consider arbitrary 2 n +1 × 2 n +1 board with 2 n 2 n single unusable field. 2 n It consists of four ‘sub-boards’ of 2 n × 2 n ; 2 n +1 unusable field is in one of the sub-boards. 2 n Place tromino on inner corners of the other three sub-boards. 2 n +1 By the induction hypothesis, each of the four 2 n × 2 n sub-boards can be tiled, and hence the 2 n +1 × 2 n +1 board can be tiled. / department of mathematics and computer science