The Complexity of
Mathematical Problems
C. S. Calude (UoA) and E. Calude (Massey U)
In Honour of Eric Goles 60th Birhday
Valparaiso, November 2011
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Do the following statements
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Do the following statements
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the four colour theorem,
Do the following statements
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the four colour theorem,
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Fermat’s great theorem,
Do the following statements
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the four colour theorem,
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Fermat’s great theorem,
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the Riemann hypothesis,
Do the following statements
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the four colour theorem,
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Fermat’s great theorem,
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the Riemann hypothesis,
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the Collatz’s conjecture?
Do the following statements
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the four colour theorem,
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Fermat’s great theorem,
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the Riemann hypothesis,
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the Collatz’s conjecture?
share a common mathematical property?
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Do the following statements
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the four colour theorem,
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Fermat’s great theorem,
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the Riemann hypothesis,
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the Collatz’s conjecture?
share a common mathematical property?
And, if there is such a property, how can we use it for a better
understanding of these statements?
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Computability and Complexity
1
Universality theorem. There exists (and can be constructed) a
(Turing) machine U—called universal—such that for every machine
V there exists a constant c = cU,V such that for every program σ
there exists a σ 0 for which the following two conditions hold:
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U(σ 0 ) = V (σ),
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|σ 0 | ≤ |σ| + c.
Computability and Complexity
The halting problem for a machine V is the function ΛV defined
by
1, if V (σ) = ∞,
ΛV (σ) =
0, otherwise .
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2
Computability and Complexity
The halting problem for a machine V is the function ΛV defined
by
1, if V (σ) = ∞,
ΛV (σ) =
0, otherwise .
Undecidability theorem. If U is universal, then ΛU is
incomputable, i.e. the halting problem for a universal machine is
undecidable.
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2
Π1 –problems
A problem π of the form
∀σP(σ),
where P is a computable predicate is called a Π1 –problem.
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Π1 –problems
A problem π of the form
∀σP(σ),
where P is a computable predicate is called a Π1 –problem.
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Any Π1 –problem is finitely refutable.
Π1 –problems
A problem π of the form
∀σP(σ),
where P is a computable predicate is called a Π1 –problem.
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Any Π1 –problem is finitely refutable.
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For every Π1 –problem π = ∀σP(σ) we associate the program
σπ = inf{n : P(n) = false}
which satisfies:
π is true iff U(σπ ) = ∞.
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Π1 –problems
A problem π of the form
∀σP(σ),
where P is a computable predicate is called a Π1 –problem.
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Any Π1 –problem is finitely refutable.
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For every Π1 –problem π = ∀σP(σ) we associate the program
σπ = inf{n : P(n) = false}
which satisfies:
π is true iff U(σπ ) = ∞.
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Solving the halting problem for U solves all Π1 –problems.
Examples
The problems
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the four colour theorem,
Examples
The problems
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the four colour theorem,
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Fermat’s great theorem,
Examples
The problems
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the four colour theorem,
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Fermat’s great theorem,
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the Riemann hypothesis,
Examples
The problems
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the four colour theorem,
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Fermat’s great theorem,
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the Riemann hypothesis,
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the Collatz’s conjecture
Examples
The problems
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the four colour theorem,
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Fermat’s great theorem,
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the Riemann hypothesis,
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the Collatz’s conjecture
are all Π1 –problems.
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Examples
The problems
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the four colour theorem,
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Fermat’s great theorem,
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the Riemann hypothesis,
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the Collatz’s conjecture
are all Π1 –problems.
Of course, not all problems are Π1 –problems. For example, the
twin prime conjecture.
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Complexity
Complexity
CU (π) = min{|ΠP | : π = ∀nP(n)}.
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Complexity
Complexity
CU (π) = min{|ΠP | : π = ∀nP(n)}.
Invariance theorem. If U, U 0 are universal, then there exists a
constant c = cU,U 0 such that for all π = ∀nP(n), P computable:
|CU (π) − CU 0 (π)| ≤ c.
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Complexity
Complexity
CU (π) = min{|ΠP | : π = ∀nP(n)}.
Invariance theorem. If U, U 0 are universal, then there exists a
constant c = cU,U 0 such that for all π = ∀nP(n), P computable:
|CU (π) − CU 0 (π)| ≤ c.
Incomputability theorem. If U is universal, then CU is
incomputable.
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Complexity Classes
Because of the incomputability theorem, we work with upper
bounds for CU . As the exact value of CU is not important, we
classify Π1 –problems into the following classes:
CU,n = {π : π is a Π1 –problem, CU (π) ≤ n kbit}.
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Some Results
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CU,1 : Legendre’s conjecture (there is a prime number between
n2 and (n + 1)2 , for every positive integer n), Fermat’s last
theorem (there are no positive integers x, y , z satisfying the
equation x n + y n = z n , for any integer value n > 2) and
Goldbach’s conjecture (every even integer greater than 2 can
be expressed as the sum of two primes)
Some Results
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CU,1 : Legendre’s conjecture (there is a prime number between
n2 and (n + 1)2 , for every positive integer n), Fermat’s last
theorem (there are no positive integers x, y , z satisfying the
equation x n + y n = z n , for any integer value n > 2) and
Goldbach’s conjecture (every even integer greater than 2 can
be expressed as the sum of two primes)
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CU,2 : Dyson’s conjecture (the reverse of a power of two is
never a power of five)
Some Results
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CU,1 : Legendre’s conjecture (there is a prime number between
n2 and (n + 1)2 , for every positive integer n), Fermat’s last
theorem (there are no positive integers x, y , z satisfying the
equation x n + y n = z n , for any integer value n > 2) and
Goldbach’s conjecture (every even integer greater than 2 can
be expressed as the sum of two primes)
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CU,2 : Dyson’s conjecture (the reverse of a power of two is
never a power of five)
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CU,3 : the Riemann hypothesis (all non-trivial zeros of the
Riemann zeta function have real part 1/2)
Some Results
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CU,1 : Legendre’s conjecture (there is a prime number between
n2 and (n + 1)2 , for every positive integer n), Fermat’s last
theorem (there are no positive integers x, y , z satisfying the
equation x n + y n = z n , for any integer value n > 2) and
Goldbach’s conjecture (every even integer greater than 2 can
be expressed as the sum of two primes)
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CU,2 : Dyson’s conjecture (the reverse of a power of two is
never a power of five)
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CU,3 : the Riemann hypothesis (all non-trivial zeros of the
Riemann zeta function have real part 1/2)
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CU,4 the four colour theorem (the vertices of every planar
graph can be coloured with at most four colours so that no
two adjacent vertices receive the same colour)
More Results and Open Questions
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CU,5 : ?
More Results and Open Questions
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CU,5 : ?
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CU,6 : ?
More Results and Open Questions
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CU,5 : ?
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CU,6 : ?
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CU,7 : Euler’s integer partition theorem (the number of
partitions of an integer into odd integers is equal to the
number of partitions into distinct integers).
More Results and Open Questions
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CU,5 : ?
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CU,6 : ?
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CU,7 : Euler’s integer partition theorem (the number of
partitions of an integer into odd integers is equal to the
number of partitions into distinct integers).
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In which class is the Collatz conjecture? ( given any positive
integer a1 there exists a natural N such that aN = 1, where
an /2,
if an is even,
an+1 =
3an + 1, otherwise .
Inductive Complexity and Complexity Classes of First Order
By transforming each program ΠP for U into a program Πind,1
for
P
U ind (U working in “inductive mode”) we can define the inductive
complexity of first order by
CUind,1 (π) = min{|Πind,1
| : π = ∀nP(n)},
P
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Inductive Complexity and Complexity Classes of First Order
By transforming each program ΠP for U into a program Πind,1
for
P
U ind (U working in “inductive mode”) we can define the inductive
complexity of first order by
CUind,1 (π) = min{|Πind,1
| : π = ∀nP(n)},
P
the inductive complexity classes of order one by
ind,1
Cind,1
(π) ≤ n kbit},
U,n = {π : π is a Π1 –statement, CU
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Inductive Complexity and Complexity Classes of First Order
By transforming each program ΠP for U into a program Πind,1
for
P
U ind (U working in “inductive mode”) we can define the inductive
complexity of first order by
CUind,1 (π) = min{|Πind,1
| : π = ∀nP(n)},
P
the inductive complexity classes of order one by
ind,1
Cind,1
(π) ≤ n kbit},
U,n = {π : π is a Π1 –statement, CU
and prove that
CU,n = Cind,1
U,n .
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Inductive Complexity and Complexity Classes of Higher Orders
By allowing inductive programs of order 1 as routines we get
inductive programs of order 2, so we can define the inductive
complexity of second order (for more complex problems)
CUind,2 (ρ) = min{|MRind,2 | : ρ = ∀n∃iR(n, i)},
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Inductive Complexity and Complexity Classes of Higher Orders
By allowing inductive programs of order 1 as routines we get
inductive programs of order 2, so we can define the inductive
complexity of second order (for more complex problems)
CUind,2 (ρ) = min{|MRind,2 | : ρ = ∀n∃iR(n, i)},
and the inductive complexity class of second order:
ind,2
Cind,2
(ρ) ≤ n kbit}.
U,n = {ρ : ρ = ∀n∃iR(n, i), CU
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Inductive Complexity and Complexity Classes of Higher Orders
By allowing inductive programs of order 1 as routines we get
inductive programs of order 2, so we can define the inductive
complexity of second order (for more complex problems)
CUind,2 (ρ) = min{|MRind,2 | : ρ = ∀n∃iR(n, i)},
and the inductive complexity class of second order:
ind,2
Cind,2
(ρ) ≤ n kbit}.
U,n = {ρ : ρ = ∀n∃iR(n, i), CU
The Collatz conjecture is in the class Cind,2
U,3 .
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Two open problems
What is the complexity of
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Two open problems
What is the complexity of
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P vs NP problem?
Two open problems
What is the complexity of
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P vs NP problem?
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Poincaré’s conjecture?
Thank you
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Thank you
VIVE ERIC!
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