II. Recursive Function

II. Recursive Function
Yuxi Fu
BASICS, Shanghai Jiao Tong University
Hilbert’s Program
The epochal address of David Hilbert (23Jan.1862-14Feb.1934)
to the International Congress of Mathematicians (Paris, 1900):
1. Consistency of the Axioms of Arithmetic.
This is the second among Hilbert’s 23 problems published in 1900
(ten were announced in the Paris Congress).
2. Entscheidungsproblem (decision problem) of first order logic.
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Gödel’s Incompleteness Theorem
Hilbert retired in 1930, and gave a special speech to the annual
meeting of the Society of German Scientists and Physicians in his
birth place, Königsberg.
It was in this occasion he made his famous remark
”Wir müssen wissen. Wir werden wissen.”
Kurt Gödel (28April.1906-14Jan.1978), a young man from Vienna
and one year past his PhD, announced his Incompleteness
Theorem in a roundtable discussion at the Conference on
Epistemology held jointly with the Society meetings.
(one day before Hilbert’s speech!)
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History of Recursion Theory
Gödel’s remarkable idea is the arithmetization of syntax.
He used primitive recursive functions to do encoding.
In 1931 Gödel was aware of the Ackermann function.
By taking a suggestion from Herbrand, he developed in 1934 a
formal system of Herbrand-Gödel recursive functions.
Kleene (5Jan.1909-25Jan.1994) introduced in 1936 the
µ-recursive functions, based on the system of the primitive
recursive functions and an unbounded search operator.
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Synopsis
1. Primitive Recursive Function
2. Ackermann Function
3. Recursive Function
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1. Primitive Recursive Function
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Recursion Theory offers a mathematical model for the study of
effective calculability.
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I
All effective objects can be encoded by natural numbers.
I
All effective procedures can be modeled by functions from
numbers to numbers.
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Initial Function
def
1. The zero function 0 = λx1 . . . λxn .0.
def
2. The successor function s(x) = λx.x+1.
def
3. The projection function Uni (x1 , . . . , xn ) = λx1 . . . λxn .xi .
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Composition
Suppose f (y1 , . . . , yk ) is a k-ary function and g1 (e
x ), . . . , gk (e
x ) are
n-ary functions, where e
x abbreviates x1 , . . . , xn .
The composition function h(e
x ) is defined by
h(e
x ) = f (g1 (e
x ), . . . , gk (e
x )),
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Recursion
Suppose that f (e
x ) is an n-ary function and g (e
x , y , z) is an
(n+2)-ary function.
The recursion function h(e
x ) is defined by
h(e
x , 0) = f (e
x ),
h(e
x , y + 1) = g (e
x , y , h(e
x , y )).
(1)
(2)
Clearly there is a unique function that satisfies (1) and (2).
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Primitive Recursive Recursion
The set of primitive recursive function is the least set generated
from the initial functions, composition and recursion.
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Dummy Parameter
Proposition. Suppose that f (y1 , . . . , yk ) is a primitive recursive
and that xi1 , . . . , xik is a sequence of k variables from x1 , . . . , xn
(possibly with repetition). Then the function h given by
h(x1 , . . . , xn ) = f (xi1 , . . . , xik )
is primitive recursive.
Proof.
x ), . . . , Unik (e
x )).
h(e
x ) = f (Uni1 (e
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Basic Arithmetic Function
x + y:
x + 0 = x,
x + (y + 1) = (x + y ) + 1.
xy :
x0 = 0,
x(y + 1) = xy + x.
xy :
x 0 = 1,
x y +1 = x y x.
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Basic Arithmetic Function
x −̇1:
0−̇1 = 0,
(x + 1)−̇1 = x.
def
x −̇y =
x − y , if x ≥ y ,
:
0,
otherwise.
x −̇0 = x,
x −̇(y + 1) = (x −̇y )−̇1.
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Basic Arithmetic Function
def
sg(x) =
0, if x = 0,
:
1, if x 6= 0.
sg(0) = 0,
sg(x + 1) = 1.
def
sg(x) =
1, if x = 0,
:
0, if x 6= 0.
sg(x) = 1−̇sg(x).
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Basic Arithmetic Function
|x − y | = (x −̇y ) + (y −̇x).
x!:
0! = 1,
(x + 1)! = x!(x + 1).
min(x, y ) = x −̇(x −̇y ).
max(x, y ) = x + (y −̇x).
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Basic Arithmetic Function
def
rm(x, y ) = the remainder when y is devided by x:
rm(x, y ) + 1 if rm(x, y ) + 1 < x,
def
rm(x, y + 1) =
0,
otherwise.
The recursive definition is given by
rm(x, 0) = 0,
rm(x, y + 1) = (rm(x, y ) + 1)sg(x −̇(rm(x, y ) + 1)).
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Basic Arithmetic Function
def
qt(x, y ) = the quotient when y is devided by x:
qt(x, y ) + 1, if rm(x, y ) + 1 = x,
def
qt(x, y + 1) =
qt(x, y ),
if rm(x, y ) + 1 6= x.
The recursive definition is given by
qt(x, 0) = 0,
qt(x, y + 1), = qt(x, y ) + sg(x − (rm(x, y ) + 1)).
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Basic Arithmetic Function
def
div(x, y ) =
1, if x divides y ,
:
0, otherwise.
div(x, y ) = sg(rm(x, y )).
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Bounded Sum and Bounded Product
Bounded sum:
X
f (e
x , y ) = 0,
y <0
X
y <z+1
Bounded product:
Y
f (e
x, y) =
X
f (e
x , y ) + f (e
x , z).
y <z
f (e
x , y ) = 1,
y <0
Y
y <z+1
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f (e
x, y) = (
Y
f (e
x , y )) · f (e
x , z).
y <z
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Bounded Sum and Bounded Product
By composition the following functions are also primitive recursive
e ) is primitive recursive:
if k(e
x, w
X
f (e
x , z)
e)
z<k(e
x ,w
and
Y
f (e
x , z).
e)
z<k(e
x ,w
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Bounded Minimization Operator
Bounded search:
µz<y (f (e
x , z) = 0)
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def
=
the least z < y , such that f (e
x , z) = 0;
y,
if there is no such z.
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Bounded Minimization Operator
Bounded search:
µz<y (f (e
x , z) = 0)
def
=
the least z < y , such that f (e
x , z) = 0;
y,
if there is no such z.
Proposition.
If f (e
x , z) is primitive recursive, then so is µz<y (f (e
x , z) = 0).
Proof.
µz<y (f (e
x , z) = 0) =
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P
Q
v <y (
x , u))).
u<v +1 sg(f (e
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Bounded Minimization Operator
e ) are primitive recursive functions, then so is
If f (e
x , z) and k(e
x, w
the function
e )(f (e
µz<k(e
x, w
x , z) = 0).
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Primitive Recursive Predicate
Suppose M(x1 , . . . , xn ) is an n-ary predicate of natural numbers.
The characteristic function cM (e
x ), where e
x = x1 , . . . , xn , is
1, if M(a1 , . . . , an ) holds,
cM (a1 , . . . , an ) =
0, if otherwise.
The predicate M(e
x ) is primitive recursive if cM is primitive
recursive.
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Closure Property
Proposition. The following statements are valid:
I
I
If R(e
x ) is a primitive recursive predicate, then so is ¬R(e
x ).
If R(e
x ), S(e
x ) are primitive recursive predicates, then the
following predicates are primitive recursive:
I
I
I
R(e
x ) ∧ S(e
x );
R(e
x ) ∨ S(e
x ).
If R(e
x , y ) is a primitive recursive predicate, then the following
predicates are primitive recursive:
I
I
∀z < y .R(e
x , z);
∃z < y .R(e
x , z).
Proof.
For example c∀z<y .R(ex ,z) (e
x, y) =
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Q
z<y
cR (e
x , z).
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Definition by Case
Suppose that f1 (e
x ), . . . , fk (e
x ) are primitive recursive functions, and
M1 (e
x ), . . . , Mk (e
x ) are primitive recursive predicates, such that for
every e
x exactly one of M1 (e
x ), . . . , Mk (e
x ) holds. Then the function
g (e
x ) given by

f1 (e
x ), if M1 (e
x ) holds,



 f2 (e
x ), if M2 (e
x ) holds,
g (e
x) =
..

.



fk (e
x ), if Mk (e
x ) holds.
is primitive recursive.
Proof.
g (e
x ) = cM1 (e
x )f1 (e
x ) + . . . + cMk (e
x )fk (e
x ).
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More Arithmetic Function
Proposition. The following functions are primitive recursive.
(i) D(x) = the number of divisors of x;
1, if x is prime,
(ii) Pr (x) =
;
0, if x is not prime.
(iii) px = the x-th

 k,
(iv) (x)y =

0,
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prime number;
k is the exponent of py in the prime
factorisation of x, for x, y > 0,
.
if x = 0 or y = 0.
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More Arithmetic Function
Proof.
(i) D(x) =
P
y <x+1 div(y , x).
(ii) Pr (x) = sg(|D(x) − 2|).
(iii) px can be recursively defined as follows:
p0 = 0,
px+1 = µz < (2 + px !) 1−̇(z −̇px )Pr (z) = 0 .
(iv) (x)y = µz<x(div(pz+1
y , x) = 0).
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Encoding a Finite Sequence
Suppose s = (a1 , a2 , . . . , an ) is a finite sequence of numbers.
It can be coded by the following number
b = pa11 +1 pa22 +1 . . . pann +1 .
Then the length of s can be recovered from
µz<b((b)z+1 = 0),
and the i-th component can be recovered from
(b)i −̇1.
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Not all Computable Functions are Primitive Recursive
Using the fact that all primitive recursive functions are total, a
diagonalisation argument shows that non-primitive recursive
computable functions must exist.
The same diagonalisation argument applies to all finite
axiomatizations of computable total function.
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2. Ackermann Function
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Ackermann Function
The Ackermann function [1928] is defined as follows:
ψ(0, y ) ' y + 1,
ψ(x + 1, 0) ' ψ(x, 1),
ψ(x + 1, y + 1) ' ψ(x, ψ(x + 1, y )).
The equations clearly define a total function.
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Ackermann Function is Monotone
The Ackermann function is monotone:
ψ(n, m) < ψ(n, m + 1),
ψ(n, m) < ψ(n + 1, m).
The Ackermann function grows faster on the first parameter:
ψ(n, m + 1) ≤ ψ(n + 1, m)
ψ(n, m) + C is dominated by ψ(J, m) for some large enough J:
ψ(n, m) + ψ(n0 , m) < ψ(max(n, n0 ) + 4, m),
ψ(n, m) + m < ψ(n + 4, m).
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Ackermann Function Grows Fast
Lemma. Let f (e
x ) be a k-ary primitive recursive function. Then
there exists some J such that for all n1 , . . . , nk we have that
f (n1 , . . . , nk ) < ψ(J,
k
X
nk ).
i=1
Proof.
The proof is by structural induction.
(i) f is one of the initial functions. In this case take J to be 1.
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Ackermann Function Grows Fast
(ii) f is the composition function h(g1 (e
x ), . . . , gm (e
x )). Then
f (e
n) = h(g1 (e
n), . . . , gm (e
n))
< ψ(J0 ,
m
X
gi (e
n)) < ψ(J0 ,
i=1
m
X
ψ(Ji ,
i=1
< ψ(J0 , ψ(J ∗ ,
k
X
k
X
nj )) < ψ(J ∗ , ψ(J ∗ + 1,
j=1
= ψ(J ∗ + 1,
k
X
nj ))
j=1
nj + 1) ≤ ψ(J ∗ + 2,
j=1
k
X
nj ))
j=1
k
X
nj ).
j=1
Now set J = J ∗ + 2.
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Ackermann Function Grows Fast
Suppose f is defined by the recursion:
f (e
x , 0) ' h(e
x ),
f (e
x , y + 1) ' g (e
x , y , f (e
x , y )).
Then h(e
n) < ψ(Jh ,
P
e
n) and g (e
n, m, p) < ψ(Jg ,
P
e
n + m + p).
It is easy to prove
f (n1 , . . . , nk , m) < ψ(J,
k
X
nk + m)
i=1
by induction on m.
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Faster than any Primitive Recursive Function
Now suppose ψ(x, y ) was primitive recursive.
By composition ψ(x, x) would be primitive recursive.
According to the Lemma
ψ(n, n) < ψ(J, n)
for some J and all n, which would lead to the contradiction
ψ(J, J) < ψ(J, J).
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Faster than any Primitive Recursive Function
The Ackermann function grows faster than every primitive
recursive function.
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3. Recursive Function
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Minimization Operator, or Search Operator
Minimization function, or µ-function, or search function:


 the least y such that

f (e
x , y ) is defined for all z ≤ y , and
µy (f (e
x , y ) = 0) '
f
(e
x
, y ) = 0,



undefined if otherwise.
Here ' is the computational equality.
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Minimization Operator, or Search Operator
Minimization function, or µ-function, or search function:


 the least y such that

f (e
x , y ) is defined for all z ≤ y , and
µy (f (e
x , y ) = 0) '
f
(e
x
, y ) = 0,



undefined if otherwise.
Here ' is the computational equality.
The recursion operation is a well-founded going-down procedure.
The search operation is a possibly divergent going-up procedure.
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Recursive Function
The set of recursive functions is the least set generated from the
initial functions, composition, recursion and minimization.
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Decidable Predicate
A predicate R(e
x ) is decidable if its characteristic function
1, if R(e
x ) is true,
def
cR (e
x) =
0, otherwise.
is a recursive function.
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Decidable Predicate
A predicate R(e
x ) is decidable if its characteristic function
1, if R(e
x ) is true,
def
cR (e
x) =
0, otherwise.
is a recursive function. The predicate R(e
x ) is partially decidable if
its partial characteristic function
1, if R(e
x ) is true,
def
χR (e
x) =
↑, otherwise.
is a recursive function.
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Closure Property
Proposition. The following statements are valid:
I
I
If R(e
x ) is decidable, then so is ¬R(e
x ).
If R(e
x ), S(e
x ) are (partially) decidable, then the following
predicates are (partially) decidable:
I
I
I
If R(e
x , y ) is (partially) decidable, then the following predicates
are (partially) decidable:
I
I
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R(e
x ) ∧ S(e
x );
R(e
x ) ∨ S(e
x ).
∀z < y .R(e
x , y );
∃z < y .R(e
x , y ).
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Definition by Cases
Suppose f1 (e
x ), . . . , fk (e
x ) are recursive and M1 (e
x ), . . . , Mk (e
x ) are
partially decidable. For every e
x at most one of M1 (e
x ), . . . , Mk (e
x)
holds. Then the function g (e
x ) given by

x ), if M1 (e
x ) holds,

 f1 (e

 f2 (e
x ), if M2 (e
x ) holds,
g (e
x) '
..

.



fk (e
x ), if Mk (e
x ) holds.
is recursive.
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Minimization via Decidable Predicate
Suppose R(x, y ) is a partially decidable predicate. The function
g (x) = µyR(e
x, y)
the least y such that R(e
x , y ) holds, if there is such a y ,
=
undefined,
otherwise.
is recursive.
Proof.
g (e
x ) = µy (sg(χR (e
x , y )) = 0).
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Comment
The µ-operator allows one to define partial functions.
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Comment
The µ-operator allows one to define partial functions.
The diagonalisation argument does not apply to the set F of
recursive functions.
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Comment
The µ-operator allows one to define partial functions.
The diagonalisation argument does not apply to the set F of
recursive functions.
Using the µ-operator, one may define total functions that are not
primitive recursive.
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Minimization Operator is a Search Operator
It is clear from the above proof why the minimization operator is
sometimes called a search operator.
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Definable Function
A function is definable if there is a recursive function calculating it.
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Ackermann Function is Recursive
Proposition. The Ackermann function is recursive.
Proof.
A finite set S of triples is said to be suitable if the followings hold:
(i) if (0, y , z) ∈ S then z = y + 1;
(ii) if (x + 1, 0, z) ∈ S then (x, 1, z) ∈ S;
(iii) if (x + 1, y + 1, z) ∈ S then ∃u.((x + 1, y , u)∈S ∧ (x, u, z)∈S).
A triple (x, y , z) can be coded up by 2x 3y 5z .
A set {u1 , . . . , uk } can be coded up by pu1 · · · puk .
Let R(x, y , v ) be “v is a legal code and ∃z < v .(x, y , z)∈Sv ”.
The Ackermann function ψ(x, y ) ' µz((x, y , z)∈SµvR(x,y ,v ) ).
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Do recursive functions capture all computable functions?
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Do recursive functions capture all computable functions?
Gödel himself wasn’t very sure about it in 1934.
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