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Chapter 3 - 1

Chapter 3 - 1. Construction Techniques. Section 3.1 Inductively Defined Sets. To define a set S inductively is to do three things: Basis: Specify one or more elements of S. Induction: Specify one or more rules to construct elements of S from existing elements of S.

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Chapter 3 - 1

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  1. Chapter 3 - 1 Construction Techniques

  2. Section 3.1 Inductively Defined Sets • To define a set S inductively is to do three things: • Basis: Specify one or more elements of S. • Induction: Specify one or more rules to construct elements of S from existing elements of S. • Closure: Specify that no other elements are in S (always assumed). • Note: The basis elements and the induction rules are called constructors.

  3. Example 1 • Find an inductive definition for S = {3, 16, 29, 42, …}. • Solution: Basis: 3 ∈ S. Induction: If x ∈ S then x + 13 ∈ S. • The constructors are 3 and the operation of adding 13. • Also, without closure, many sets would satisfy the basis and induction rule. e.g., 3 ∈ Z and x ∈ Z implies x + 13 ∈ Z.

  4. Example 2 • Find an inductive definition for S = {3, 4, 5, 8, 9, 12, 16, 17, 20, 24, 33,…}. • Solution: To simplify things we might try to “divide and conquer” by writing S as the union of more familiar sets as follows: S = {3, 5, 9, 17, 33, …} ⋃ {4, 8, 12, 16, 20, 24, …}. • Basis: 3, 4 ∈ S. • Induction: If x ∈ S then (if x is odd then 2x – 1 ∈ S else x + 4 ∈ S).

  5. Example 3 • Describe the set S defined inductively as follows: • Basis: 2 ∈ S; • Induction: x ∈ S implies x ± 3 ∈ S. • Solution: S = {2, 5, 8, 11, … } ⋃ {–1, –4, –7, –10, … }.

  6. Example 4 • Find an inductive definition for S = {٨, ac, aacc, aaaccc, …} = {ancn | n ∈ N}. • Solution: Basis: ٨∈ S. Induction: If x ∈ S then axc ∈ S.

  7. Example 5 • Find an inductive definition for S = {an+1bcn | n ∈ N}. • Solution: Basis: ab ∈ S. • Induction: If x ∈ S then axc ∈ S.

  8. Example 6 • Describe the set S defined by: • Basis: a, b ∈ S • Induction: x ∈ S implies ƒ(x) ∈ S. • Solution: S = {a, ƒ(a), ƒ(ƒ(a)), …} ⋃ {b, ƒ(b), ƒ(ƒ(b)), …}, which could also be written as S = {ƒn(a) | n ∈ N} ⋃ {ƒn(b) | n ∈ N} = {ƒn(x) | x ∈ {a, b} and n ∈ N}.

  9. Example 7 • Describe the set S defined by: • Basis: < 0 > ∈ S • Induction: x ∈ S implies cons(1, x) ∈ S. • Solution: S = {< 0 >, < 1, 0 >, <1, 1, 0 >, …}.

  10. Infix notation • cons(h, t) = h :: t. Associate to the right. e.g., x :: y :: z = x :: (y :: z). • Example 8. Find an inductive definition for S = {<>, <a, b >, <a, b, a, b >, …}. • Solution: • Basis: <> ∈ S. • Induction: x ∈ S implies a :: b :: x ∈ S.

  11. Example 9 • Find an inductive definition for S = {<>, <<>>, <<<>>>, …}. • Solution: Basis: <> ∈ S. Induction: x ∈ S implies x :: <> ∈ S.

  12. Notation for Binary Trees • Let t(L, x, R) denote the tree with root x, left subtree L, and right subtree R. Let <> denote the empty binary tree. If T = t(L, x, R), then root(T) = x, left(T) = L, and right(T) = R.

  13. Example 10 • Describe the set S defined inductively as follows: • Basis: t(<>, •, <>) ∈ S. • Induction: T ∈ S implies t(T, •, t(<>, •, <>)) ∈ S. • Solution (picture): The first few trees constructed from the definition are pictured as follows: and so on.

  14. Example 11 • Find an inductive definition for the set S of binary trees indicated by the following picture. and so on. • Solution: Basis: t(<>, •, <>) ∈ S. Induction: T ∈ S implies t(t(left(T), •, <>), •, t(<>, •, right(T))) ∈ S.

  15. Example 12 • Find an inductive definition for the set S = {a}* × N. • Solution: Basis: (٨, 0) ∈ S. Induction: (s, n) ∈ S implies (as, n), (s, n + 1) ∈ S.

  16. Example 13 • Find an inductive definition for the set S = {(x, –y) | x, y ∈ N and x ≥ y}. • Solution: To get an idea about S we can write out a few tuples: (0, 0), (1, 0), (1, –1), (2, 0), (2, –1), (2, –2), and so on. We can also get an idea about S by graphing a few points, as indicated in the picture. One solution can be written as follows: • Basis: (0, 0) ∈ S. • Induction: (x, y) ∈ S implies (x + 1, y), (x + 1, y – 1) ∈ S. Notice that this definition constructs some repeated points. For example, (2, –1) is constructed twice.

  17. Quiz (2 minutes) • Try to find a solution that does not construct repeated elements. • Solution: We might use two separate rules. One rule to construct the diagonal points and one rule to construct horizontal lines that start at the diagonal points. • Basis: (0, 0) ∈ S. • Induction: 1. (x, y) ∈ S implies (x + 1, y) ∈ S. 2. (x, –x) ∈ S implies (x + 1, – (x + 1)) ∈ S.

  18. Section 3.2 Recursively Defined Functions and Procedures • A function f is recursively defined if at least one value f(x) is defined in terms of another value f(y), where x ≠ y. Similarly, a procedure P is recursively defined if the action of P(x) is defined in terms of another action P(y), where x ≠ y.

  19. Technique for recursive definitions • Technique for recursive definitions when the argument domain is inductively defined. 1. Specify a value f(x), or action P(x), for each basis element x of S. 2. Specify rules that, for each inductively defined element x in S, define the value f(x), or action P(x), in terms of previously defined values of f or actions of P.

  20. Example 1 • Find a recursive definition for the function f: N → N defined by f(n) = 0 + 3 + 6 + … + 3n. • Solution: Notice that N is an inductively defined set: 0 ∈ N; n ∈ N implies n + 1 ∈ N. So we need to give f(0) a value in N and we need to define f(n + 1) in terms of f(n). The given definition of f tells us to set f(0) = 0. To discover a definition for f(n + 1) we can write f(n + 1) = (0 + 3 + 6 + … + 3n) + 3(n + 1) = f(n) + 3(n + 1). So we have a recursive definition for f • f(0) = 0 • f(n + 1) = f(n) + 3(n + 1).

  21. Two alternative definitions • f(0) = 0 f(n) = f(n - 1) + 3n (n > 0). • (if-then-else form): f(n) = if n = 0 then 0 else f(n - 1) + 3n.

  22. Example 2 • Find a recursive definition for cat : A* × A* → A* defined by cat(s, t) = st. • Solution: Notice that A* is inductively defined: ٨∈ A*; a ∈ A and x ∈ A* imply ax ∈ A*, where ax denotes the string version of cons. We can define cat recursively using the first argument. The definition of cat gives cat(٨, t) = ٨t = t. For the recursive part we can write cat(ax, t) = axt = a(xt) = acat(x, t). So we have a definition: • cat(٨, t) = t • cat(ax, t) = acat(x, t). • If-then-else form using head and tail for strings: cat(s, t) = if s = ٨ then t else head(s)cat(tail(s), t).

  23. Example 3 • Find a definition for f: lists(Q) → Q defined by f(<x1, …, xn>) = x1 + … + xn. • Solution: The set lists(Q) is inductively defined: <> ∈ lists(Q); h ∈ Q and t ∈ lists(Q) imply h :: t ∈ lists(Q). To discover a recursive definition, we can use the definition of f as follows: f(<x1, …, xn>) = x1 + … + xn = x1 + (x2 + … + xn) = x1 + f(<x2, … , xn>) = head(<x1, …, xn>) + f(tail(<x1, …, xn>). So if we let f(<>) = 0, we have a recursive definition: • f(<>) = 0 • f(h :: t) = h + f(t). • If-then-else form: f(L) = if L = <> then 0 else head(L) + f(tail(L)).

  24. Example 4 • Given f: N → N defined recursively by • f(0) = 0 • f(1) = 0 • f(x + 2) = 1 + f(x). The if-then-else form for f can be written as follows: f(x) = if x = 0 or x = 1 then 0 else 1 + f(x - 2). What does f do? • Answer: List a few values to get the idea. For example, map(f, <0, 1, 2, 3, 4, 5, 6, 7, 8, 9>) = <0, 0, 1, 1, 2, 2, 3, 3, 4, 4>. So f(x) returns the floor of x/2. i.e., f(x) = ⌊x/2⌋.

  25. Example 5 • Find a recursive definition for f: lists(Q) → Q defined by f(<x1, …, xn>) = x1x2 + x2x3 + … + xn-1xn. • Solution: Let f(<>) = 0 and f(<x>) = 0. Then for n ≥ 2 we can write f(<x1, …, xn>) = x1x2 + (x2x3 + … + xn-1xn) = x1x2 + f(<x2, …, xn>). So we have the following recursive definition. • f(<>) = 0 • f(<x>) = 0 • f(h :: t) = h • head(t) + f(t). • If-then-else form: f(L) = if L = <> or tail(L) = <> then 0 else head(L) • head(tail(L)) + f(tail(L)).

  26. Example 6 • Find a recursive definition for isin : A × lists(A) → {true, false} where isin(x, L) means that x is in the list L. • Solution: isin(x, <>) = false isin(x, x :: t) = true isin(x, h :: t) = isin(x, t). • If-then-else form: isin(x, L) = if L = <> then false else if x = head(L) then true else isin(x, tail(L)).

  27. Example 7 • Find a recursive definition for sub : lists(A) × lists(A) → {true, false} where sub(L, M) means the elements of L are elements of M. • Solution: sub(<>, M) = true sub(h :: t, M) = if isin(h, M) then sub(t, M) else false. • If-then-else form: sub(L, M) = if L = <> then true else if isin(head(L), M) then sub(tail(L), M) else false.

  28. Example 8 • Find a recursive definition for intree : Q × binSearchTrees(Q) → {true, false} where intree(x, T) means x is in the binary search tree T. • Solution: intree(x, <>) = false intree(x, tree(L, x, R)) = true intree(x, tree(L, y, R)) = if x < y then intree(x, L) else intree(x, R). • If-then-else form: intree(x, T) = if T = <> then false else if x = root(T) then true else if x < root(T) then intree(x, left(T)) else intree(x, right(T)).

  29. Traversing Binary Trees • The three standard procedures to traverse a binary tree are defined recursively as follows: • preorder(T): if T ≠ <> then visit root(T); preorder(left(T)); preorder(right(T)) fi. • inorder(T): if T ≠ <> then inorder(left(T)); visit root(T); inorder(right(T)) fi • postorder(T): if T ≠ <> then postorder(left(T)); postorder(right(T)); visit root(T) fi.

  30. Example 9 • Traverse the following tree in each of the three orders. • Solution: • Preorder: a b c d e • Inorder: b a d c e • Postorder: b d e c a a c b d e

  31. Example 10 • Find a recursive definition for post : binaryTrees(A) → lists(A) where post(T) is the list of nodes from a postorder traversal of T. • Solution: post(<>) = <> post(tree(L, x, R)) = cat(post(L), cat(post(R), <x >)) where cat concatenates two lists and can be defined by, • cat(<>, L) = L • cat(h :: t, L) = h :: cat(t, L).

  32. Example 11 • Find a recursive definition for ƒ : binaryTrees(Q) → Q where ƒ(T) is the sum of the nodes in T. • Solution: ƒ( <>) = 0 ƒ(tree(L, x, R)) = x + ƒ(L) + ƒ(R).

  33. Infinite Sequences • We can construct recursive definitions for infinite sequences by defining a value ƒ(x) in terms of x and ƒ(y) for some value y in the sequence. • Example 12. Suppose we want to represent the infinite sequence ƒ(x) = <x, x2, x4, x8, … >. • Solution: Use the definition to discover a solution as follows: ƒ(x) = <x, x2, x4, x8, …> = x :: <x2, x4, x8, …> = x :: ƒ(x2). So define ƒ(x) = x :: ƒ(x2).

  34. More examples • Example 13. What sequence is defined by g(x, k) = xk :: g(x, k + 1)? • Answer: g(x, k) = xk :: g(x, k + 1) = xk :: xk+1 :: g(x, k + 2) =… = < xk, xk+1, xk+2, … >. • Example 14. How do we obtain the sequence <x, x3, x5, x7, … >? • A Solution. Define ƒ(x) = h(x, 1), where h(x, k) = xk :: h(x, k + 2). • Example 15. How do we obtain the sequence <1, x2, x4, x6, x8, … >? • A Solution: Use h(x, 0) from Example 14.

  35. The End of Chapter 3 - 1

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