Recursive Definitions and Structural Induction

1. Recursive Definitions and Structural Induction CS/APMA 202 Rosen section 3.4 Aaron Bloomfield

2. Recursion • Recursion means defining something, such as a function, in terms of itself • For example, let f(x) = x! • We can define f(x) as f(x) = x * f(x-1)

3. Recursion example • Rosen, section 3.4, question 1 • Find f(1), f(2), f(3), and f(4), where f(0) = 1 • Let f(n+1) = f(n) + 2 • f(1) = f(0) + 2 = 1 + 2 = 3 • f(2) = f(1) + 2 = 3 + 2 = 5 • f(3) = f(2) + 2 = 5 + 2 = 7 • f(4) = f(3) + 2 = 7 + 2 = 9 • Let f(n+1) = 3f(n) • f(1) = 3 * f(0) = 3*1 = 3 • f(2) = 3 * f(1) = 3*3 = 9 • f(3) = 3 * f(2) = 3*9 = 27 • f(4) = 3 * f(3) = 3*27 = 81

4. Recursion example • Rosen, section 3.4, question 1 • Find f(1), f(2), f(3), and f(4), where f(0) = 1 • Let f(n+1) = 2f(n) • f(1) = 2f(0) = 21 = 2 • f(2) = 2f(1) = 22 = 4 • f(3) = 2f(2) = 24 = 16 • f(4) = 2f(3) = 216 = 65536 • Let f(n+1) = f(n)2 + f(n) + 1 • f(1) = f(0)2 + f(0) + 1 = 12 + 1 + 1 = 3 • f(2) = f(1)2 + f(0) + 1 = 32 + 3 + 1 = 13 • f(3) = f(2)2 + f(0) + 1 = 132 + 13 + 1 = 183 • f(4) = f(3)2 + f(0) + 1 = 1832 + 183 + 1 = 33673

5. Fractals • A fractal is a pattern that uses recursion • The pattern itself repeats indefinitely

6. Fractals

7. Fibonacci sequence • Definition of the Fibonacci sequence • Non-recursive: • Recursive: F(n) = F(n-1) + F(n-2) or: F(n+1) = F(n) + F(n-1) • Must always specify base case(s)! • F(1) = 1, F(2) = 1 • Note that some will use F(0) = 1, F(1) = 1

8. Fibonacci sequence in Java long Fibonacci (int n) { if ( (n == 1) || (n == 2) ) return 1; else return Fibonacci (n-1) + Fibonacci (n-2); } long Fibonacci2 (int n) { return (long) ((Math.pow((1.0+Math.sqrt(5.0)),n)- Math.pow((1.0-Math.sqrt(5.0)),n)) / (Math.sqrt(5) * Math.pow(2,n))); }

9. Recursion definition • From “The Hacker’s Dictionary”: • recursion n. See recursion. See also tail recursion.

10. Bad recursive definitions • Consider: • f(0) = 1 • f(n) = 1 + f(n-2) • What is f(1)? • Consider: • f(0) = 1 • f(n) = 1+f(-n) • What is f(1)?

11. Defining sets via recursion • Same two parts: • Base case (or basis step) • Recursive step • Example: the set of positive integers • Basis step: 1  S • Recursive step: if x  S, then x+1  S

12. Defining sets via recursion • Rosen, section 3.4, question 24: give recursive definitions for: • The set of odd positive integers • 1  S • If x  S, then x+2  S • The set of positive integer powers of 3 • 3  S • If x  S, then 3*x  S • The set of polynomials with integer coefficients • 0  S • If p(x)  S, then p(x) + cxn  S • c  Z, n  Z and n≥ 0

13. Defining strings via recursion • Terminology •  is the empty string: “” •  is the set of all letters: { a, b, c, …, z } • The set of letters can change depending on the problem • We can define a set of strings * as follows • Base step:   * • If w  * and x  , then wx  * • Thus, * s the set of all the possible strings that can be generated with the alphabet • Is this countably infinite or uncountably infinite?

14. Defining strings via recursion • Let  = { 0, 1 } • Thus, * is the set of all binary numbers • Or all binary strings • Or all possible computer files

15. String length via recursion • How to define string length recursively? • Basis step: l() = 0 • Recursive step: l(wx) = l(w) + 1 if w  * and x   • Example: l(“aaa”) • l(“aaa”) = l(“aa”) + 1 • l(“aa”) = l(“a”) + 1 • l(“a”) = l(“”) + 1 • l(“”) = 0 • Result: 3

16. Today’s demotivators

17. Strings via recursion example • Rosen, section 3.4, question 38: Give a recursive definition for the set of string that are palindromes • We will define set P, which is the set of all palindromes • Basis step:  P • Second basis step: x P when x   • Recursive step: xpx P if x   and p  P

18. Strings and induction example • This requires structural induction, which will be covered later in this slide set

19. Recursion pros • Easy to program • Easy to understand

20. Recursion cons • Consider the recursive Fibonacci generator • How many recursive calls does it make? • F(1): 1 • F(2): 1 • F(3): 3 • F(4): 5 • F(5): 9 • F(10): 109 • F(20): 13,529 • F(30): 1,664,079 • F(40): 204,668,309 • F(50): 25,172,538,049 • F(100): 708,449,696,358,523,830,149  7 * 1020 • At 1 billion recursive calls per second (generous), this would take over 22,000 years • But that would also take well over 1012 Gb of memory!

21. Trees • Rooted trees: • A graph containing nodes and edges • Cannot contain a cycle! Cycle not allowed in a tree

22. Rooted trees • Recursive definition: • Basis step: A single vertex r is a rooted tree • Recursive step: • Let T1, T2, …, Tn be rooted trees • Form a new tree with a new root r that contains an edge to the root of each of the trees T1, T2, …, Tn

23. (Extended) Binary trees • Recursive definition • Basis step: The empty set is an extended binary tree • Recursive step: • Let T1, and T2 be extended binary trees • Form a new tree with a new root r • Form a new tree such that T1 is the left subtree, and T2 is the right subtree

24. Full binary trees • Recursive definition • Basis step: A full binary tree consisting only of the vertex r • Recursive step: • Let T1, and T2 be extended binary trees • Form a new tree with a new root r • Form a new tree T such that T1 is the left subtree, and T2 is the right subtree • This is denoted by T = T1∙T2 • Note the only difference between a regular binary tree and a full one is the basis step

25. Binary tree height • h(T) denotes the height of tree T • Recursive definition: • Basis step: The height of a tree with only one node r is 0 • Recursive step: • Let T1 and T2 be binary trees • The binary tree T = T1∙T2 has heighth(T) = 1 + max ( h(T1), h(T2) ) • This definition can be generalized to non-binary trees

26. Binary tree size • n(T) denotes the number of vertices in tree T • Recursive definition: • Basis step: The number of vertices of an empty tree is 0 • Basis step: The number of vertices of a tree with only one node r is 1 • Recursive step: • Let T1 and T2 be binary trees • The number of vertices in binary tree T = T1∙T2 is:n(T) = 1 + n(T1) + n(T2) • This definition can be generalized to non-binary trees

27. A bit of humor: Computer terminology

28. Recursion vs. induction • Consider the recursive definition for factorial: • f(0) = 1 • f(n) = n * f(n-1) • Sort of like induction Base case The “step”

29. Recursion vs. induction • Rosen, section 3.4, example 7 (page 262) • Consider the set of all integers that are multiples of 3 • { 3, 6, 9, 12, 15, … } • { x | x = 3k and k Z+ } • Recursive definition: • Basis step: 3  S • Recursive step: If x S and y S, then x+y S

30. Recursion vs. induction • Proof via induction: prove that S contains all the integers that are divisible by 3 • Let A be the set of all ints divisible by 3 • Show that S = A • Two parts: • Show that S A • Let P(n) = 3n  S • Base case: P(1) = 3*1  S • By the basis step of the recursive definition • Inductive hypothesis: assume P(k) = 3*k  S is true • Inductive step: show that P(k+1) = 3*(k+1) is true • 3*(k+1) = 3k+3 • 3k  S by the inductive hypothesis • 3  S by the base case • Thus, 3k+3  S by the recursive definition • Show that A S • Done in the text, page 267 (not reproduced here)

31. What did we just do? • Notice what we did: • Showed the base case • Assumed the inductive hypothesis • For the inductive step, we: • Showed that each of the “parts” were in S • The parts being 3k and 3 • Showed that since both parts were in S, by the recursive definition, the combination of those parts is in S • i.e., 3k+3  S • This is called structural induction

32. Structural induction • A more convenient form of induction for recursively defined “things“ • Used in conjunction with the recursive definition • Three parts: • Basis step: Show the result holds for the elements in the basis step of the recursive definition • Inductive hypothesis: Assume that the statement is true for some existing elements • Usually, this just means assuming the statement is true • Recursive step: Show that the recursive definition allows the creation of a new element using the existing elements

33. Tree structural induction example • Rosen, section 3.4, question 43 • Show that n(T) ≥ 2h(T) + 1 • Basis step: Let T be the full binary tree of just one node r • h(T) = 0 • n(T) = 1 • n(T) ≥ 2h(T) + 1 • 1 ≥ 2*0 + 1 • 1 ≥ 1

34. Tree structural induction example • Show that n(T) ≥ 2h(T) + 1 • Inductive hypothesis: • Let T1 and T2 be full binary trees • Assume that n(T1) ≥ 2h(T1) + 1 for some tree T1 • Assume that n(T2) ≥ 2h(T2) + 1 for some tree T2 • Recursive step: • Let T = T1 ∙ T2 • Here the ∙ operator means creating a new tree with a root note r and subtrees T1 and T2 • New element is T • By the definition of height and size, we know: • n(T) = 1 + n(T1) + n(T2) • h(T) = 1 + max ( h(T1), h(T2) ) • Therefore: • n(T) = 1 + n(T1) + n(T2) • ≥ 1 + 2h(T1) + 1 + 2h(T2) + 1 • ≥ 1 + 2*max ( h(T1), h(T2) ) the sum of two non-neg #’s is at least as large as the larger of the two • = 1 + 2*h(T) • Thus, n(T) ≥ 2h(T) + 1

35. String structural induction example • Rosen, section 3.4, question 32 • Part (a): Give the definition for ones(s), which counts the number of ones in a bit string s • Let  = { 0, 1 } • Basis step: ones() = 0 • Recursive step: ones(wx) = ones(w) + x • Where x  and w  * • Note that x is a bit: either 0 or 1

36. String structural induction example • Part (b): Use structural induction to prove that ones(st) = ones(s) + ones(t) • Basis step: t =  • ones (s∙) = ones(s) = ones(s)+0 = ones(s) + ones() • Inductive hypothesis: Assume ones(s∙t) = ones(s) + ones(t) • Recursive step: Want to show that ones(s∙t∙x) = ones(s) + ones(t∙x) • Where s, t  * and x   • New element is ones(s∙t∙x) • ones (s∙t∙x) = ones ((s∙t)∙x)) by associativity of concatenation • = x+ones(s∙t) by recursive definition • = x + ones(s) + ones(t) by inductive hypothesis • = ones(s) + (x + ones(t)) by commutativity and assoc. of + • = ones(s) + ones(t∙x) by recursive definition • Proven!

37. Quick survey • I feel I understand structural induction… • Very well • With some review, I’ll be good • Not really • Not at all

38. Human stupidity

39. Induction methods compared

40. Induction types compared • Show that F(n) < 2n • Where F(n) is the nth Fibonacci number • Actually F(n) < 20.7*n, but we won’t prove that here • Fibonacci definition: • Basis step: F(1) = 1 and F(2) = 1 • Recursive step: F(n) = F(n-1) + F(n-2) • Base case (or basis step): Show true for F(1) and F(2) • F(1) = 1 < 21 = 2 • F(2) = 1 < 22 = 4

41. Via weak mathematical induction • Inductive hypothesis: Assume F(k) < 2k • Inductive step: Prove F(k+1) < 2k+1 • F(k+1) = F(k) + F(k-1) • We know F(k) < 2k by the inductive hypothesis • Each term is less than the next, therefore F(k) > F(k-1) • Thus, F(k-1) < F(k) < 2k • Therefore, F(k+1) = F(k) + F(k-1) < 2k + 2k = 2k+1 • Proven!

42. Via strong mathematical induction • Inductive hypothesis: Assume F(1) < 21, F(2) < 22, …, F(k-1) < 2k-1, F(k) < 2k • Inductive step: Prove F(k+1) < 2k+1 • F(k+1) = F(k) + F(k-1) • We know F(k) < 2k by the inductive hypothesis • We know F(k-1) < 2k-1 by the inductive hypothesis • Therefore, F(k) + F(k-1) < 2k + 2k-1 < 2k+1 • Proven!

43. Via structural induction • Inductive hypothesis: Assume F(n) < 2n • Recursive step: • Show true for “new element”: F(n+1) • We know F(n) < 2n by the inductive hypothesis • Each term is less than the next, therefore F(n) > F(n-1) • Thus, F(n-1) < F(n) < 2n • Therefore, F(n) + F(n-1) < 2n + 2n = 2n+1 • Proven!

44. Another way via structural induction • Inductive hypothesis: Assume F(n) < 2n and F(n-1) < 2n-1 • The difference here is we are using two “old” elements versus one, as in the last slide • Recursive step: • Show true for “new element”: F(n+1) • F(n+1) = F(n) + F(n-1) • We know F(n) < 2n by the inductive hypothesis • We know F(n-1) < 2n-1 by the inductive hypothesis • Therefore, F(n) + F(n-1) < 2k + 2k-1 < 2k+1 • Proven!

45. But wait! • In this example, the structural induction proof was essentially the same as the weak or strong mathematical induction proof • It’s hard to find an example that works well for all of the induction types • Structural induction will work on some recursive problems which weak or strong mathematical induction will not • Trees, strings, etc.

46. A bit of humor…

47. Section 3.4, question 8 • Give the recursive definition of the following sequences • Note that many answers are possible! • an = 4n – 2 • Terms: 2, 6, 10, 14, 16, etc. • a1 = 2 • an = an-1 + 4 • an = 1 + (-1)n • Terms: 0, 2, 0, 2, 0, 2, etc. • a1 = 0, a2 = 2 • an = an-2 • an = n(n+1) • Terms: 2, 6, 12, 20, 30, 42, etc. • a1 = 2 • an = an-1 + 2*n • an = n2 • Terms: 1, 4, 9, 16, 25, 36, 49, etc. • a1 = 1 • an = an-1 + 2n - 1

48. Section 3.4, question 12 • Show that f12 + f22 + f32 + … + fn2 = fnfn+1 • Base case: n = 1 • f12 = f1f2 • 12 = 1*1 • Inductive hypothesis: Assume • f12 + f22 + f32 + … + fk2 = fkfk+1 • Inductive step: Prove • f12 + f22 + f32 + … + fk2 + fk+12 = fk+1fk+2

49. Section 3.4, question 12 • Inductive hypothesis: Assume • f12 + f22 + f32 + … + fk2 = fkfk+1 • Inductive step: Prove • f12 + f22 + f32 + … + fk2 + fk+12 = fk+1fk+2 • fkfk+1 + fk+12 = fk+1fk+2 • fkfk+1 + fk+12 = fk+1 (fk + fk+1) • fkfk+1 + fk+12 = fkfk+1 + fk+12

50. Section 3.4, question 13 • Show that f1 + f2 + f3 + … + f2n-1 = f2n • Base case: n = 1 • f1 = f2*1 • 1 = 1 • Inductive hypothesis: Assume • f1 + f2 + f3 + … + f2k-1 = f2k • Inductive step: Prove • f1 + f2 + f3 + … + f2k-1 + f2(k+1)-1 = f2(k+1) • f1 + f2 + f3 + … + f2k-1 + f2k+1 = f2k+2