Recursive Function Theory
Gödel's Incompleteness Theorem
Zero, Successor, Projector Functions
Functional Composition
Primitive Recursion
Proving Functions are Primitive Recursive
Ackermann's Function
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1Lý thuyết tính toán
(Theory of Computation) i
PGS.TS. Phan Huy Khánh
khanhph@vnn.vn
Chương 5
Hàm đệ quy
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Chương 5 Hàm đệ quy
Recursive Function Theory
Gödel's Incompleteness Theorem
Zero, Successor, Projector Functions
Functional Composition
Primitive Recursion
Proving Functions are Primitive Recursive
Ackermann's Function
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Maths Functions
An example function:
Domain Range
We need a way to define functions
We need a set of basic functions
N
3
N
10
f(n) = n2 + 1
f(3) = 10
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Computation
Get N a set of Natural Numbers :
N = { 0, 1, 2, }
Building the functions on N
For examples :
x + y
x * y
xy
x2 + y2
are computable functions
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Complicated Functions
x ( y + z)
is complicated functions from the addition and
multiplication function
is computable:
there is a sequence of operations
of the addition and the multiplication
Attention :
There are also many functions that are not composed
from the basis functions
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Factorial function:
n! = n (n-1) (n-2) 2 1
is computable :
there is a sequence of multiplication operations
The factorial function is not alone the composition of the
addition and multiplication operations
The number of multiplication oprations depends on n
Function is computable
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Factorial function is a recursive definition:
0! = 1
(n + 1) ! = (n + 1) n !
Uses the recursivity to define some functions
f(n + 1)
is defined from:
f(n)
Start at:
f(0)
Recursivity
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Function is computable
Why is computable?
Basic primitive recursive functions:
Computation on the natural number N
Primitive Recursive Function:
Any function built from the basic primitive recursive
functions
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Computable functions
Basic set of Recursive primitive functions
Primitive Recursive Functions :
Mechanism for composition of functions
by combining previously-defined functions
composition and/or recursive definitions
Clearly they are infinite in number
Some can have any arity (unary, binary, )
f(n1, n2, , nm), m 1
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Gödel's Incompleteness Theorem
“Any interesting consistent system must be incomplete;
that is, it must contain some unprovable propositions”
Hierarchy of Functions
1. Primitive-Recursive Functions
2. Recursive (-recursive) Functions
3. Interesting well-defined Functions but "unprovable"
BB Function
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Primitive Recursive Functions
Defined over the domain I = set of all non-negative
integers
or domain I×I
or domain I×I×I, etc.
Definition:
Functions are said to be Primitive Recursive
if they can be built
from the basic functions (zero, successor, and projection)
using functional composition and/or primitive recursion
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Zero, Successor, Projector Functions
Zero function:
z(x) = 0, for all x I
Successor function:
s(x) = x+1
Projector functions:
p1(x1, x2) = x1
p2(x1, x2) = x2
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Example of Primitive Recursives
Constants are Primitive Recursive:
2 = s(s(z(x)))
3 = s(s(s(z(x))))
5 = s(s(s(s(s(z(x))))))
Addition & Multiplication
add(x, 0) = x
add(x, y+1) = s(add(x, y))
mult(x, 0) = 0
mult(x, y+1) = add(x, mult(x, y))
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Subtraction
pred(0) = 0
pred(x+1) = x
monus(x, 0) = x // called subtr in text
monus(x, y+1) = pred(monus(x, y))
absdiff(x, y) = monus(x, y) + monus(y, x)
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Other Primitive Recursive Functions
Factorial & Exponentiation
fact(0) = 1
fact(n+1) = mult(s(n), fact(n))
exp(x, 0) = 1
exp(x, n+1) = mult(x, exp(x, n))
Test for Zero (Logical Complement)
test(0) = 1
test(x+1) = 0
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Operators
Relational Operators
equal(x, y) = test(absdiff(x, y))
geq(x, y) = test(monus(y, x))
leq(x, y) = test(monus(x, y))
gt(x, y) = test(leq(x, y))
lt(x, y) = test(geq(x, y))
Minimum & Maximum
min(x, y) = lt(x, y)*x + geq(x, y)*y
max(x, y) = geq(x, y)*x + lt(x, y)*y
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Division
remaind(numerator, denominator) = rem(denominator, numerator)
rem(x, 0) = 0
rem(x, y+1) = s(rem(x, y))*test(equal(x, s(rem(x, y))))
div(numerator, denominator) = dv(denominator, numerator)
dv(x, 0) = 0
dv(x, y+1) = dv(x, y) + test(remaind(y+1, x))
Square Root
sqrt(0) = 0
sqrt(x+1) = sqrt(x) +
equal(x+1, (s(sqrt(x))*s(sqrt(x))))
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Test for Prime
numdiv(x) = divisors_leq(x, x)
divisors_leq(x, 0) = 0
divisors_leq(x, y+1)
= divisors_leq(x, y) + test(remaind(x, y+1))
is_prime(x) = equal(numdiv(x), 2)
{ a b mod c }
congruent(a, b, c)
= equal(remaind(a, c), remaind(b, c))
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Greatest Common Divisor
(can’t use Euclidean Algorithm—not P.R.)
gcd(a, 0) = a
gcd(a, b+1) = find_gcd(a, b+1, b+1)
find_gcd(a, b, 0) = 1
find_gcd(a, b, c+1) =
(c+1)*test_rem(a, b, c+1) +
find_gcd(a, b, c)*test(test_rem(a, b, c+1))
test_rem(a, b, c) =
test(remaind(a, c))*test(remaind(b, c))
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Functional Composition
f(x, y) = h(g1(x, y), g2(x, y))
from previously defined functions g1, g2, and h
e.g.:
min(x, y) = lt(x, y)*x + geq(x, y)*y
h(x, y) = add(x, y)
g1(x, y) = mult(lt(x, y), p1(x, y))
h = mult(), g1=lt(), g2=p1()
g2(x, y) = mult(geq(x, y), p2(x, y))
h = mult(), g1=geq(), g2=p2()
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Primitive Recursion
Composition:
f(x, 0) = g1(x)
f(x, y+1) = h(g2(x, y), f(x, y))
Note: Last argument defined at zero and y+1 only
e.g.:
exp(x, 0) = 1
exp(x, n+1) = x * exp(x, n)
g1(x) = s(z(x))
h(x, y) = mult(x, y)
g2(x, y) = p1(x, y)
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Ackermann's Function
We can actually give an example of a total Turing-computable function
that is not primitive recursive, namely Ackermann’s function:
A(0, n) = n+1
A(m+1, 0) = A(m, 1)
A(m+1, n+1) = A(m, A(m+1, n))
For example,
A(0, 0) = 1
A(0, 1) = 2
A(1, 1) = A(0, A(1, 0)) = A(0, A(0, 1))
= A(0, 1) + 1 = 3.
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Ackermann's Function
Theorem
For every unary primitive recursive function f,
there is some m such that f(m) < A(m, m)
So A cannot be primitive recursive itself
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Ackermann's Function
Ackermann's Function is NOT Primitive Recursive
Just because it is not defined using the "official" rules of
primitive recursion is not a proof that it IS NOT primitive
recursive
Perhaps there is another definition that uses primitive
recursion
(NOT!) Proof is beyond the scope of this course
525/32
"Meaning" of Ackermann's Function
(addition, multiplication, exponentiation, tetration)
A(1,0) 2; A(1,1) 3; A(1,2 ) 4; A(1,n) 2 (n 3) 3
A(2,0) 3; A(2,1) 5; A(2,2) 7; A(2,n) 2 *(n 3) 3
A(3,0) 5; A(3,1) 13; A(3,2) 29; A(3,3) 61; A(3, 4) 125; A(3,n) 2
n3
3
A(4,0) 13; A(4,1) 65531; A(4,2) 2
65534
3; A(4,n) 2
2
2
... 2
2{n 3 times}
3
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Rates of growth
Ackerman’s function and friends
Iterated exponentials
Exponential functions
• A(m.n)
• 3n • n!
Polynomial functions
• n3+3n2+2n+1• 2n+5
• nn
• n n
n
Growth of Ackerman’s function:
A(0, n) = n+1 ; A(1, n) = n+2 ; A(2, n) = 2n+3
A(3, n) = 2n + 3 – 3 A(4, n) = 2 - 3 with n powers of 2
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Countable Sets
Countable if it can be put into a 1-to-1 correspondence
with the positive integers
You should already be familiar with the enumeration
procedure for the set of RATIONAL numbers
[diagonalization, page 278]
Quick review
You should already be familiar with the fact (and proof)
that the REAL numbers are NOT countable
Quick review
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Recursively Enumerable Languages
A language is said to be recursively enumerable
if there exists a Turing machine that accepts it.
That is, if the accepting machine is started on a word
in the language, it will halt in qf
This says nothing about what the machine will do
if it is presented with a word that is not in the language
(i.e. whether it halts in a non-final state or loops)
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Recursive Languages
A language, L, is recursive if there exists a Turing machine
that accepts L and halts on every w in +
That is, there exists a membership decision procedure for L
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Existence of Languages that are not
Recursively Enumerable
Let S be an infinite countable set
Then its powerset 2S is not countable
Proof by diagonalization
Recall the fact that the REAL numbers are not countable
For any nonempty , there exist languages that are not
recursively enumerable.
Every subset of * is a language
Therefore there are exactly 2* languages
However, there are only a countable number of Turing
machines
Therefore there exist more languages than Turing machines
to accept them
631/32
Recursively Enumerable but not Recursive
We can list all Turing machines that eventually halted
on a given input tape (say blank)
Recall the enumeration procedure for TM’s from last period
Once a string of 0’s and 1’s was verified as a valid TM,
we would simply run it (while non-deterministically continuing
to list other machines). [Note how long this would take!]
A halt on the part of the simulation (recall the Universal
Turing Machine) would trigger adding the TM in question to
the list of those that halted. (copying it to another tape?)
However, we cannot determine (and always halt)
whether or not a given TM will halt on a blank tape
Stay tuned for the unsolvability of the Halting Problem...
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The hierarchy of functions
Recall that a function f :
Nk N is total if f is defined on every input from Nk
and is partial if we don’t insist that it has to be total.
All partial functions from Nk to N
The computable partial functions
The computable total functions
The primitive recursive functions
• ?
• n
• A(m,n)
• add(m,n)
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