Computer programs and natural number
arithmetic
Wong Tin Lok
27 July, 2011
Plan
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What is a computer program?
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What can(not) be achieved by a computer program?
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Can a computer solve Diophantine equations?
Characteristics of a program
A (computer) program. . .
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is a finite sequence of instructions that can be run by a
machine;
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can take a finite piece of data as input;
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can return a finite piece of information as output after a finite
amount of time;
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may perform an infinite loop without giving an output
(depending on the input).
Deciding a set of natural numbers
Think of k = 1 first.
Convention
N = {0, 1, 2, . . .} and k ∈ N.
Definition
A set A ⊆ Nk is decidable if there is a program P which when
given any input x̄ ∈ Nk ,
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outputs true if x̄ ∈ A;
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outputs false if x̄ 6∈ A.
Global assumptions
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All our programs take a fixed finite number of natural
numbers as inputs.
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Outputs of our programs can only be true or false, if any.
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There is no time or memory restriction on our computers.
We are only interested in the relationship between
the inputs and the outputs of our programs.
What is a program?
Programs can use
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variables v0 , v1 , . . .;
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assignments ←;
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conditionals and Boolean operations if-then, &, , not, =;
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arithmetical operations and relations 0, 1, +, ×, 6;
for loops; and
restricted versions
repeat-until loops.
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Church–Turing Thesis (Definition of a program)
Every program is equivalent to one that uses only the above.
Example: prime numbers
Consider A = {x ∈ N : x is a prime number}
(
)
∀u,v 6x x = uv → u = 1 ∨ v = 1
= x ∈N
& x 6= 1
·
input x ∈ N.
answer ← true.
for u ← 0, . . . , x do:
for v ← 0, . . . , x do:
if x = uv & u 6= 1 & v 6= 1, then answer ← false.
if x = 1, then answer ← false.
output answer .
No repeat-until is
involved.
Observation
This program stops on all inputs.
Exercise: coprime numbers
Write programs that decide
(a) B = {(x, y ) ∈ N2 : hcf(x, y ) = 1};
or
(b) B 0 = {(x, y ) ∈ N2 : hcf(x, y ) 6= 1}.
Question
How to perform negation, conjunction, and disjunction on
programs in general?
not
and
Example: Fermat Last Theorem
Consider C17 = x ∈ N : ∃z ∃y 6z (y 6= 0 & x 17 + y 17 = z 17 ) .
input x ∈ N.
repeat, starting with z ← 0:
for y ← 0, . . . , z do:
if y 6= 0 & x 17 + y 17 = z 17 , then answer ← true.
next z until answer = true.
output answer .
Observation
Wiles and Taylor (1995) proved that C17 = {0}. However, the
program above does not stop on input 1, for example!
Semidecidability
(Semi- means half-.)
Definition
A set A ⊆ Nk is semidecidable if there is a program P such that
A = {x̄ ∈ Nk : P outputs true on input x̄}.
Example
C17 is semidecidable.
Remark
All decidable sets are semidecidable.
Programs and arithmetic
Programs
if-then, &, , not
0, 1, +, ×, 6
for loops
repeat-until loops
Formulas in arithmetic
→, &, ∨, ¬
0, 1, +, ×, 6
∀v 6t and ∃v 6t
∃v
Correspondence Principle
The above table gives a correspondence between programs P and
formulas ϕ(x̄) in arithmetic such that
{x̄ ∈ N : P outputs true on input x̄} = {x̄ ∈ N : ϕ(x̄)}.
Decidability is arithmetical
Definition
Formulas that correspond to a program are called Σ1 . They are of
the form
∃v̄ χ(v̄ , x̄),
where χ(v̄ , x̄) is made up of the Boolean operations, the
arithmetical operations and relations, and bounded quantifiers
(i.e., ∀v 6t and ∃v 6t).
Correspondence Principle (restated)
The semidecidable sets are exactly those of the form
{x̄ ∈ Nk : ϕ(x̄)},
where ϕ(x̄) is Σ1 .
Example: an infinite loop
input x ∈ N.
answer ← false.
repeat, starting with z ← 0:
if z 6= z, then answer ← true.
next z until answer = true.
output answer .
This corresponds to the formula
∃z (z 6= z).
Undecidable sets
Question
Are there undecidable sets?
Answer
Yes, by counting.
Question
Are all semidecidable sets decidable?
Remark
Our previous example C17 of a semidecidable set is decidable.
The Halting Set
(Halt means stop.)
Definition
H = {(pPq, x) ∈ N2 : the program P stops on input x}.
Observation
Every program P is a natural number, which we denote by pPq.
Proposition
H is semidecidable.
Proof.
Given an input (pPq, x) ∈ N2 , run the program P on input x.
Theorem
H is not decidable.
The Halting Set H is not decidable
Proof
Suppose there is a program K that decides H. Let L be the
following program.
input pPq ∈ N.
if P stops on input pPq,
then perform an infinite loop,
else answer ← true.
output answer .
L stops on input pPq
Then
⇔ P does not stop on input pPq.
Question
What happens when we run L on input pLq?
Russell’s Paradox!
Running L on input pLq
Proof (continued).
Recall that
L stops on input pPq
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⇐
⇒ P does not stop on input pPq.
If L stops on input pLq, then L does not stop on input pLq.
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If L does not stop on input pLq, then L stops on input pLq.
This is a contradiction.
No program can check whether
a program stops on an input.
How mathematical is H?
The halting set is not quite a ‘mathematically interesting’ object.
Open problem
Find a ‘mathematically interesting’ semidecidable set that is not
decidable.
Gödel’s Incompleteness Theorem
Gödel Incompleteness Theorem (1931, rudimentary form)
No program can tell us the truth of any given statement about N.
Proof.
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By the Correspondence Principle, there is a formula ϕ(x, y ) in
arithmetic that corresponds to the Halting Set H.
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Since H is not decidable, there is no program that tells us the
truth of ϕ(x, y ) correctly for all x, y ∈ N.
Diophantine equations
Diophantine equations are polynomial equations with integer
coefficients to be solved in the integers (or in the natural numbers).
Examples
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{(x, y , z) ∈ N3 : x 2 + y 2 = z 2 }.
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{(x, y , z, w ) ∈ N4 : x 4 + y 4 + z 4 + w 4 = (x + y + z + w )4 }.
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{(a, b, c) ∈ N3 : ∃x (ax 2 + bx + c = 0)}.
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{(x, y , z) ∈ N3 : ∃w (x 3 + y 3 + z 3 = w 3 )}.
Diophantine sets
Definition
A set A ⊆ Nk is Diophantine if
A = {x̄ ∈ Nk : ∃v̄ p(x̄, v̄ ) = q(x̄, v̄ )}
for some polynomials p, q with coefficients in N.
Exercise
Convince yourself that every Diophantine set is semidecidable.
Hilbert’s Tenth Problem
Hilbert’s Tenth Problem (1900)
Find a general algorithm to solve Diophantine equations.
Hilbert’s Tenth Problem (simplified form)
Show that every Diophantine set is decidable.
Solution (MRDP 1970)
This is not possible!
The solution
MRDP Theorem (Matiyasevich–Robinson–Davis–Putnam)
Every semidecidable set is Diophantine.
Corollary
There is an undecidable Diophantine set.
Proof.
By the MRDP Theorem, the Halting Set H is a Diophantine set
that is not decidable.
How mathematical is the Halting Set?
Theorem (Jones 1980)
H = {(ρ, x) ∈ N2 : ψ(ρ, x)}, where ψ(ρ, x) is
∃a∃b∃c∃d∃e∃f ∃g ∃h∃i∃j∃k∃l∃m∃n∃o∃p∃q∃r ∃s∃t∃u∃w ∃y ∃z∃α∃γ∃η∃θ∃λ∃τ ∃ϕ
60 elg 2 + α − (b − xy )q 2 q − b 5
λ + q 4 − 1 − λb 5 k − r − 1 − hp − h
l − u − tθ e − y − mθ n − q 16 p − 2ws 2 r 2 n2 p 2 k 2 − k 2 + 1 − τ 2
2
c − 2r − 1 − ϕ θ + 2z − b 5
(zuy )2 + u + y − ρ a − (wn2 + 1)rsn2
4(c − ksn2 )2 + η − k 2 d − bw − ca + 2c − 4αγ + 5γ d 2 − (a2 − 1)c 2 − 1
f 2 − (a2 − 1)i 2 c 4 − 1 (d + of )2 − (a + f 2 (d 2 − a))2 − 1 (2r + 1 + jc)2 − 1
!
g + eq 3 + lq 5 + 2(e − zλ)(1 + xb 5 + g )4 + λb 5 + λb 5 q 4 q 4 (n2 − n)
= 0.
+ q 3 − bl + l + θλq 3 + (b 5 − 2)q 5 (n2 − 1) − r
Connections with complexity theory
MRDP Theorem
Every semidecidable set is Diophantine.
Definition
A formula is Σ0 if it is made up of the Boolean operations, the
arithmetical operations and relations, and bounded quantifiers.
Open question (Paris)
Is the MRDP Theorem provable using only Σ0 -induction?
Observation (Wilkie)
If the answer is yes, then NP = co-NP.
Conclusion
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There is a correspondence between programs and arithmetic.
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Decidability is an arithmetical property.
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The Halting Set is an important undecidable set.
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An analysis of undecidable sets results in number-theoretic
problems that cannot be solved by a program.
Webpage: http://mat140.bham.ac.uk/∼wongtl/
E-mail: wongtl03c62@gmail.com