1 Chapter 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, …} = {a n c n | n ∈ N}. Solution: Basis: ٨ ∈ S. Induction: If x ∈ S then axc ∈ S.

7 Example 5 Find an inductive definition for S = {a n+1 bc n | 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 = {ƒ n (a) | n ∈ N} ⋃ {ƒ n (b) | n ∈ N} = {ƒ n (x) | x ∈ {a, b} and n ∈ N}.

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

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 = {<>,,, …}. 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) = … + 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) = ( … + 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) = ٨ t = 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( ) = x 1 + … + x n. 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( ) = x 1 + … + x n = x 1 + (x 2 + … + x n ) = x 1 + f( ) = head( ) + f(tail( ). 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, ) =. 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( ) = x 1 x 2 + x 2 x 3 + … + x n-1 x n. Solution: Let f(<>) = 0 and f( ) = 0. Then for n ≥ 2 we can write f( ) = x 1 x 2 + (x 2 x 3 + … + x n-1 x n ) = x 1 x 2 + f( ). So we have the following recursive definition. –f(<>) = 0 –f( ) = 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 b c de

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), )) 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) =. Solution: Use the definition to discover a solution as follows: ƒ(x) = = x :: = x :: ƒ(x 2 ). So define ƒ(x) = x :: ƒ(x 2 ).

34 More examples Example 13. What sequence is defined by g(x, k) = x k :: g(x, k + 1)? Answer: g(x, k) = x k :: g(x, k + 1) = x k :: x k+1 :: g(x, k + 2) =… =. Example 14. How do we obtain the sequence ? A Solution. Define ƒ(x) = h(x, 1), where h(x, k) = x k :: h(x, k + 2). Example 15. How do we obtain the sequence ? A Solution: Use h(x, 0) from Example 14.

35 The End of Chapter 3 - 1