Chương 5
Hàm đệ
đệ quy
Recursive Function Theory
Lý thuyế
ết tí
ính tố
án
thuy
ttính
to
thuyết
tốn
Gưdel's Incompleteness Theorem
Zero, Successor, Projector Functions
Functional Composition
PGS.TS. Phan Huy Khá
Khánh
Primitive Recursion
Proving Functions are Primitive Recursive
Ackermann's Function
(Theory
(Theory of
of Computation)
Computation)
Chương 5
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àm đệ
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Maths Functions
Computation
An example function:
Get N a set of Natural Numbers :
Range
Domain
N
N
f(n) = n2 + 1
3
N = { 0, 1, 2, … }
Building the functions on N
For examples :
x+y
x*y
xy
x2 + y2
10
f(3) = 10
We need a way to define functions
are computable functions
We need a set of basic functions
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Complicated Functions
Function is computable
Factorial function:
x ( y + z)
n! = n (n-1) (n-2) … 2 1
is complicated functions from the addition and
multiplication function
is computable :
is computable:
there is a sequence of multiplication operations
there is a sequence of operations
of the addition and the multiplication
The factorial function is not alone the composition of the
addition and multiplication operations
The number of multiplication oprations depends on
Attention :
n
There are also many functions that are not composed
from the basis functions
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Recursivity
Function is computable
Why is computable?
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)
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
Gödel's Incompleteness Theorem
“Any interesting consistent system must be incomplete;
that is, it must contain some unprovable propositions”
propositions”
Basic set of Recursive primitive functions
Primitive Recursive Functions :
Mechanism for composition of functions
by combining previously-defined functions
composition
Clearly
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Hierarchy of Functions
1. PrimitivePrimitive-Recursive Functions
2. Recursive (
(-recursive) Functions
and/or recursive definitions
they are infinite in number
3. Interesting wellwell-defined Functions but "unprovable
"unprovable""
BB Function
Some can have any arity (unary, binary, …)
f(n1, n2, …, nm), m 1
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Primitive Recursive Functions
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Zero, Successor, Projector Functions
Defined over the domain I = set of all nonnon-negative
integers
or domain I×I
or domain I×I×I, etc.
etc.
Zero function:
z(x)
(x) = 0, for all x I
Successor function:
s(x)
(x) = x+1
Definition:
Functions are said to be Primitive Recursive
if they can be built
Projector functions:
p1(x1, x2) = x1
from the basic functions (zero, successor, and projection)
using functional composition and/or primitive recursion
p2(x1, x2) = x2
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Example of Primitive Recursives
Subtraction
pred(0) = 0
pred(x+1) = x
Constants are Primitive Recursive:
2 = s(s(z(x)))
s(s(z(x)))
3 = s(s(s(z(x))))
s(s(s(z(x))))
5 = s(s(s(s(s(z(x))))))
monus(x, 0)
0) = x // called subtr in text
monus(x, y+1)
y+1) = pred(monus(x, y))
y))
absdiff(x, y)
y) = monus(x, y)
y) + monus(y, x)
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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Operators
Other Primitive Recursive Functions
Relational Operators
equal(x, y)
y) = test(absdiff(x, y))
y))
geq(x, y)
y) = test(monus(y, x))
x))
leq(x, y)
y) = test(monus(x, y))
y))
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)
test(0) = 1
test(x+1) = 0
gt(x, y)
y) = test(leq(x, y))
y))
lt(x, y)
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))
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Test for Prime
numdiv(x)
numdiv(x) = divisors_leq(x, x)
x)
divisors_leq(x, 0)
0) = 0
divisors_leq(x, y+1)
= divisors_leq(x, y)
y) + test(remaind(x, y+1))
y+1))
Square Root
sqrt(0) = 0
sqrt(x+1) = sqrt(x) +
equal(x+1, (s(sqrt(x))*s(sqrt(x))))
is_prime(x)
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
Functional Composition
(can’
(can’t use Euclidean Algorithm—
Algorithm—not P.R.)
gcd(a, 0)
0) = a
gcd(a, b+1)
b+1) = find_gcd(a, b+1, b+1)
b+1)
f(x, y)
y) = h(g1 (x, y), g2(x, y))
y))
find_gcd(a, b, 0)
0) = 1
find_gcd(a, b, c+1)
c+1) =
(c+1)*test_rem(a,
(c+1)*test_rem(a, b, c+1)
c+1) +
find_gcd(a, b, c)*test(test_rem(a, b, c+1))
c+1))
from previously defined functions g1, g2, and h
e.g.:
test_rem(a, b, c)
c) =
test(remaind(a, c))*test(remaind(b, c))
c))
min(x, y)
y) = lt(x, y)*x
y)*x + geq(x, y)*y
y)*y
h(x, y)
y) = add(x, y)
y)
g1(x, y)
y) = mult(lt(x, y), p1(x, y))
y))
g2(x, y)
y) = mult(geq(x, y), p2(x, y))
y))
h = mult(), g 1=lt(), g2=p1()
h = mult(), g 1=geq(), g2=p2()
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Ackermann's Function
Primitive Recursion
We can actually give an example of a total Turing-computable function
that is not primitive recursive, namely Ackermann’s function:
Composition:
f(x, 0)
0) = g1(x)
f(x, y+1)
y+1) = h(g2(x, y), f(x, y)
y))
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.
Note: Last argument defined at zero and y+1 only
e.g.:
exp(x, 0)
0) = 1
exp(x, n+1)
n+1) = x * exp(x, n)
g1(x) = s(z(x))
h(x, y)
y) = mult(x, y)
y)
g2(x, y)
y) = p1(x, y)
y)
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Ackermann's Function
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
Ackermann's Function is NOT Primitive Recursive
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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…
course…
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"Meaning" of Ackermann's Function
Rates of growth
(addition, multiplication, exponentiation, tetration)
tetration)
Growth of Ackerman’
Ackerman’s function:
A(0, n) = n+1 ;
A(3, n) = 2n + 3 – 3
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(1,
A(1, n) = n+2 ;
A(4,
A(4, n) = 2 - 3
A(2,
A(2, n) = 2n+3
with n powers of 2
Ackerman’
Ackerman’s function and friends
• A(m.n)
n3
A(3,0) 5; A(3,1) 13; A(3,2) 29; A(3,3) 61; A(3, 4) 125; A(3,n) 2
3
Iterated exponentials
2{n 3 times}
65534
3; A(4, n) 2
2
nn
n
Exponential functions
... 2
2
A(4, 0) 13; A(4,1) 65531; A(4, 2) 2
•
• 3n
3
• n!
• nn
Polynomial functions
• 2n+5
• n3+3n2+2n+1
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Countable Sets
Recursively Enumerable Languages
Countable if it can be put into a 11-toto-1 correspondence
with the positive integers
A language is said to be recursively enumerable
if there exists a Turing machine that accepts it.
You should already be familiar with the enumeration
procedure for the set of RATIONAL numbers
[diagonalization, page 278]
Quick review…
review…
This says nothing about what the machine will do
if it is presented with a word that is not in the language
You should already be familiar with the fact (and proof)
that the REAL numbers are NOT countable
That is, if the accepting machine is started on a word
in the language, it will halt in qf
(i.e. whether it halts in a nonnon-final state or loops)
Quick review…
review…
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Existence of Languages that are not
Recursively Enumerable
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
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.
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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
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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’
TM’s from last period
Once a string of 0’
0’s and 1’
1’s was verified as a valid TM,
we would simply run it (while nonnon-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
(and always halt)
halt)
whether or not a given TM will halt on a blank tape
Stay tuned for the unsolvability of the Halting Problem...
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
•?
• n
The computable total functions
• A(m,n)
The primitive recursive functions
• add(m,n)
add(m,n)
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