General Recursive Definitions

General Recursive Definitions

Lecture 37

Section 8.4

Fri, Mar 25, 2005

Recursively Defined Sets

The definition of a recursively defined set consists of three parts.Base – Specific objects are in the set.Recursion – Rules describing how to form

new members from other members.Restriction – No other elements are in the

set.

Boolean Expressions

Define Boolean expressions as follows.Base – Each lowercase letter of the

alphabet represents a Boolean expression and T and F are Boolean expressions.

Recursion – If p and q are Boolean expressions, then so are p q, p q, p and (p).

Restriction – There are no other Boolean expressions.

Strings Not Containing the Substring 11

Define a set of strings as follows.Base – , 0, and 1 are in the set.Recursion – If s is in the set, then so is 0s

and 10s.Restriction – No other strings are in the set.

The set consists of all strings that do not contain the substring 11.

The MIU System

Define the strings in the MIU system as follows.Base – MI is in the MIU system.

The MIU System

Recursion• If xI is in the MIU system, then so is xIU.• If Mx is in the MIU system, then so is Mxx.• If xIIIy is in the MIU system, then so is xUy.• If xUUy is in the MIU system, then so is xUy.

Restriction – There are no other strings in the MIU system.

The MIU System

Derive a few strings in the MIU system.MI MIUMI MII MIIII MUIMI MII MIIII MIIIIIIII MUIIIII

MUUII MUII

The MU Puzzle

Is MU in the MIU system?

The MU Puzzle

Let n be the number of I’s in a string in the MIU system.

If we apply Rule 1, then the resulting string still contains n I’s.

If we apply Rule 2, then the resulting string contains 2n I’s.

If we apply Rule 3, then the resulting string contains n – 3 I’s.

If we apply Rule 4, then the resulting string still contains n I’s.

The MU Puzzle

Initially, n is 1. Only Rules 2 and 3 change the number of

I’s. Thus, the following changes are possible.

n 2n.n n – 3.

The MU Puzzle

The first change will map A multiple of 3 to a multiple of 3, and A non-multiple of 3 to a non-multiple of 3.

The second change will also map A multiple of 3 to a multiple of 3, and A non-multiple of 3 to a non-multiple of 3.

Given that n = 1 in the initial string and 1 is not a multiple of 3, it is impossible to eliminate I from the strings.

Arithmetic Expressions

Define a set of arithmetic expressions as follows.Base – Any number is an arithmetic

expression.


Recursion – If x and y are arithmetic expressions, then so are

• x + y• x – y• x * y• x / y• (x)

Restriction – There are no other arithmetic expressions.


Derive the arithmetic expression

(2 + 3) * (10 / 2 – 4)

Pointer Arithmetic

Pointer- and integer-valued expressions may be defined recursively.

BaseAny letter may represent a pointer-valued

expression.Any letter may represent an integer-valued

expression.

Pointer Arithmetic

Recursion – Let p and q be pointer-valued expressions and i and j be integer-valued expressions.i + j is an integer-valued expression.i – j is an integer-valued expressionp – q is an integer-valued expression.(i) is an integer-valued expression.

Pointer Arithmetic

p + i is a pointer-valued expression.p – i is a pointer-valued expression.(p) is a pointer-valued expression.

Restriction – No other expressions are in the set.

Pointer Arithmetic

Which expressions are legal expressions? Which are integer-valued and which are pointer-

valued? p + (q + i) (p + q) + i (p – q) + i (p – q) + (r – s) (p + i) – (q – j) (p – (q – j)) + i (p – q) + (j + i)

Recursively Defined Functions

Mathematical functions may be defined recursively.

The classic example is the factorial function:0! = 1,n! = n (n – 1)! for all n 1.


Given the “successor” function

succ(n) = n + 1,

addition of nonnegative integers may be defined recursively.

• sum(0, 0) = 0,• sum(m, 0) = succ(sum(m – 1, 0)) for all m 1,• sum(m, n) = succ(sum(m, n – 1)) for all n 1.


Given the sum() function just defined, how could we define multiplication of nonnegative integers?prod(m, n) = ?

The Ackermann Function

Define the Ackermann function

A : Znonneg Znonneg Z+

byA(0, n) = n + 1, n 0.A(m, 0) = A(m – 1, 1), m > 0.A(m, n) = A(m – 1, A(m, n – 1)), m > 0, n >

0.


A(0, 0) = 1. A(0, 1) = 2. A(0, 2) = 3. A(0, 3) = 4. In general, A(0, n) = n + 1.


A(1, 0) = A(0, 1) = 2. A(1, 1) = A(0, A(1, 0)) = A(0, 2) = 3. A(1, 2) = A(0, A(1, 1)) = A(0, 3) = 4. A(1, 3) = A(0, A(1, 2)) = A(0, 4) = 5. In general, A(1, n) = n + 2.


A(2, 0) = A(1, 1) = 3. A(2, 1) = A(1, A(2, 0)) = A(1, 3) = 5. A(2, 2) = A(1, A(2, 1)) = A(1, 5) = 7. A(2, 3) = A(1, A(2, 2)) = A(1, 7) = 9. In general, A(2, n) = 2n + 3.


A(3, 0) = A(2, 1) = 5. A(3, 1) = A(2, A(3, 0)) = A(2, 5) = 13. A(3, 2) = A(2, A(3, 1)) = A(2, 13) = 29. A(3, 3) = A(2, A(3, 2)) = A(2, 29) = 61. In general, A(3, n) = 2n + 3 – 3.


A(4, 0) = A(3, 1) = 13. A(4, 1) = A(3, A(4, 0)) = A(3, 13) = 216 – 3

= 65533. A(4, 2) = A(3, 65533) = 265533 – 3. A(4, 3) = ? A(4, 4) = ? A(5, 5) = ?

General Recursive Definitions

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