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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  MIU
 MI  MII  MIIII  MUI
 MI  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.
Arithmetic Expressions

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.
Arithmetic Expressions

Derive the arithmetic expression
(2 + 3) * (10 / 2 – 4)
Pointer Arithmetic
Pointer- and integer-valued expressions
may be defined recursively.
 Base

Any 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 expression
 p – 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 pointervalued?







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.

Recursively Defined
Functions

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.
Recursively Defined
Functions

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+
by
A(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.

The Ackermann Function
A(0, 0) = 1.
 A(0, 1) = 2.
 A(0, 2) = 3.
 A(0, 3) = 4.
 In general, A(0, n) = n + 1.

The Ackermann Function
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.

The Ackermann Function
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.

The Ackermann Function
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.

The Ackermann Function
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) = ?
