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Chapter 3 - 1

Construction Techniques

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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.

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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.

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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).

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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, … }.

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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.

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Example 5

• Find an inductive definition for S = {an+1bcn | n ∈ N}.

• Solution: Basis: ab ∈ S.

• Induction: If x ∈ S then axc ∈ S.

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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}.

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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 >, …}.

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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.

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Example 9

• Find an inductive definition for S = {<>, <<>>, <<<>>>, …}.

• Solution: Basis: <> ∈ S.

Induction: x ∈ S implies x :: <> ∈ S.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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).

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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.

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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).

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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)).

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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⌋.

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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)).

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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, <>) = falseisin(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)).

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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.

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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)).

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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.

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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

d e

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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).

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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).

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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).

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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.

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The End of Chapter 3 - 1