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### 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.
- Closure: Specify that no other elements are in S (always assumed).
- Note: The basis elements and the induction rules are called constructors.

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.

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).

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, … }.

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.

Example 5

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

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}.

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 >, …}.

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.

Example 9

- Find an inductive definition for S = {<>, <<>>, <<<>>>, …}.
- Solution: Basis: <> ∈ S.

Induction: x ∈ S implies x :: <> ∈ S.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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.

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).

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)).

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⌋.

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)).

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)).

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.

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)).

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.

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

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).

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).

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).

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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