A strategy for attacking Büchi’s problem
Pablo Sáez
Universidad del Bı́o-Bı́o, Chillán, Chile
Pucón, dec/14/10
Joint work with Xavier Vidaux
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem
◮
This work is concerned with Büchi’s system of n − 2 equations
x12 − 2x22 + x32 = 2
x22 − 2x32 + x42 = 2
...
2
xn−2
−
2
2xn−1
+
xn2
=2
over the integers, where n is given.
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem
◮
This work is concerned with Büchi’s system of n − 2 equations
x12 − 2x22 + x32 = 2
x22 − 2x32 + x42 = 2
...
2
xn−2
−
2
2xn−1
+
xn2
=2
over the integers, where n is given.
◮
Rewriting any equation as (xi2 − xi2+1 ) − (xi2+1 − xi2+2 ) = 2,
since
((n + 2)2 − (n + 1)2 ) − ((n + 1)2 − n2 ) = 2
always holds, a triplet of consecutive integers is always a
solution to an equation, namely a trivial solution.
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem (cont.)
◮
A solution to a Büchi system of equations for a given n is
called a Büchi sequence of length n.
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem (cont.)
◮
◮
A solution to a Büchi system of equations for a given n is
called a Büchi sequence of length n.
The fact is that
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem (cont.)
◮
◮
A solution to a Büchi system of equations for a given n is
called a Büchi sequence of length n.
The fact is that
◮
Clearly, trivial solutions can be extended up to an arbitrary n.
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem (cont.)
◮
◮
A solution to a Büchi system of equations for a given n is
called a Büchi sequence of length n.
The fact is that
◮
◮
Clearly, trivial solutions can be extended up to an arbitrary n.
Non-trivial solutions do exist. For instance
x1 = 0, x2 = 7, x3 = 10, for n = 3.
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem (cont.)
◮
◮
A solution to a Büchi system of equations for a given n is
called a Büchi sequence of length n.
The fact is that
◮
◮
◮
Clearly, trivial solutions can be extended up to an arbitrary n.
Non-trivial solutions do exist. For instance
x1 = 0, x2 = 7, x3 = 10, for n = 3.
Actually infinitely many non-trivial solutions are known for
n = 3 and for n = 4.
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem (cont.)
◮
◮
A solution to a Büchi system of equations for a given n is
called a Büchi sequence of length n.
The fact is that
◮
◮
◮
◮
Clearly, trivial solutions can be extended up to an arbitrary n.
Non-trivial solutions do exist. For instance
x1 = 0, x2 = 7, x3 = 10, for n = 3.
Actually infinitely many non-trivial solutions are known for
n = 3 and for n = 4.
No non-trivial Büchi sequence of length 5 is known.
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem (cont.)
◮
◮
A solution to a Büchi system of equations for a given n is
called a Büchi sequence of length n.
The fact is that
◮
◮
◮
◮
◮
Clearly, trivial solutions can be extended up to an arbitrary n.
Non-trivial solutions do exist. For instance
x1 = 0, x2 = 7, x3 = 10, for n = 3.
Actually infinitely many non-trivial solutions are known for
n = 3 and for n = 4.
No non-trivial Büchi sequence of length 5 is known.
The natural question is therefore (Büchi’s Problem): is it the
case that only trivial solutions exist for n = 5?
Pablo Sáez
A strategy for attacking Büchi’s problem
The problem (cont.)
◮
◮
A solution to a Büchi system of equations for a given n is
called a Büchi sequence of length n.
The fact is that
◮
◮
◮
◮
Clearly, trivial solutions can be extended up to an arbitrary n.
Non-trivial solutions do exist. For instance
x1 = 0, x2 = 7, x3 = 10, for n = 3.
Actually infinitely many non-trivial solutions are known for
n = 3 and for n = 4.
No non-trivial Büchi sequence of length 5 is known.
◮
The natural question is therefore (Büchi’s Problem): is it the
case that only trivial solutions exist for n = 5?
◮
Or, in general, is there an N (constant, fixed) such that only
trivial solutions exist for n = N?
Pablo Sáez
A strategy for attacking Büchi’s problem
The importance of Büchi’s Problem
◮
Consider the language L2 = {0, 1, +, P2 } where P2 (x) means
’x is a square’
Pablo Sáez
A strategy for attacking Büchi’s problem
The importance of Büchi’s Problem
◮
Consider the language L2 = {0, 1, +, P2 } where P2 (x) means
’x is a square’
◮
Assume that Büchi’s problem has a positive answer, say, there
is no non-trivial Büchi-sequence of length 5.
Pablo Sáez
A strategy for attacking Büchi’s problem
The importance of Büchi’s Problem
◮
Consider the language L2 = {0, 1, +, P2 } where P2 (x) means
’x is a square’
◮
Assume that Büchi’s problem has a positive answer, say, there
is no non-trivial Büchi-sequence of length 5.
◮
The p.e. L2 -formula
φ(u1 , . . . , u5 ) :
5
^
!
P2 (ui )
i =1
∧
3
^
ui + ui +2 = 2 + 2ui +1
i =1
!
would define ’u1 , . . . , u5 are squares of consecutive integers’
Pablo Sáez
A strategy for attacking Büchi’s problem
The importance of Büchi’s Problem
◮
Consider the language L2 = {0, 1, +, P2 } where P2 (x) means
’x is a square’
◮
Assume that Büchi’s problem has a positive answer, say, there
is no non-trivial Büchi-sequence of length 5.
◮
The p.e. L2 -formula
φ(u1 , . . . , u5 ) :
5
^
!
P2 (ui )
i =1
∧
3
^
ui + ui +2 = 2 + 2ui +1
i =1
!
would define ’u1 , . . . , u5 are squares of consecutive integers’
◮
Hence, the relation ’y = x 2 ’ would be defined by the p.e.
L2 -formula
ψ(x, y ) : ∃u1 . . . ∃u5 φ(u1 , . . . , u5 ) ∧ y = u1 ∧ u2 = y + 2x + 1
Pablo Sáez
A strategy for attacking Büchi’s problem
Büchi + H10 ⇒ diagonal quadratic systems are
undecidable
◮
A positive answer to BP(Z) implies that multiplication is p.e.
L2 -definable over Z:
µ(x, y , z) : ∃u∃v ψ(x + y , u + 2z + v ) ∧ ψ(x, u) ∧ ψ(y , v )
Pablo Sáez
A strategy for attacking Büchi’s problem
Büchi + H10 ⇒ diagonal quadratic systems are
undecidable
◮
A positive answer to BP(Z) implies that multiplication is p.e.
L2 -definable over Z:
µ(x, y , z) : ∃u∃v ψ(x + y , u + 2z + v ) ∧ ψ(x, u) ∧ ψ(y , v )
◮
Hence, the positive existential theory of Z over L2 would be
undecidable.
Pablo Sáez
A strategy for attacking Büchi’s problem
Büchi + H10 ⇒ diagonal quadratic systems are
undecidable
◮
A positive answer to BP(Z) implies that multiplication is p.e.
L2 -definable over Z:
µ(x, y , z) : ∃u∃v ψ(x + y , u + 2z + v ) ∧ ψ(x, u) ∧ ψ(y , v )
◮
Hence, the positive existential theory of Z over L2 would be
undecidable.
◮
This would imply undecidability for systems of the form
X
X
aij xi2 +
bij yi = cj j = 1, . . . , n
Pablo Sáez
A strategy for attacking Büchi’s problem
Some observations
◮
Let us call Γ2 the set of integer Büchi sequences of length 3.
Notice that if (x, y , z) ∈ Γ2 then (3x + 4y , 2x + 3y , z) is also
in Γ2 . Indeed:
(3x + 4y )2 − 2(2x + 3y )2 =
9x 2 + 24xy + 16y 2 − 2(4x 2 + 12xy + 9y )2 =
x 2 − 2y 2
Pablo Sáez
A strategy for attacking Büchi’s problem
Some observations
◮
Let us call Γ2 the set of integer Büchi sequences of length 3.
Notice that if (x, y , z) ∈ Γ2 then (3x + 4y , 2x + 3y , z) is also
in Γ2 . Indeed:
(3x + 4y )2 − 2(2x + 3y )2 =
9x 2 + 24xy + 16y 2 − 2(4x 2 + 12xy + 9y )2 =
x 2 − 2y 2
◮
Also, obviously if (x, y , z) ∈ Γ2 then (z, y , x) ∈ Γ2 , since
Büchi’s equation is symmetric in x and z.
Pablo Sáez
A strategy for attacking Büchi’s problem
An action on Γ2
◮
Moreover matrixes




0 0 1
3 4 0
B =  2 3 0 ,J =  0 1 0 
1 0 0
0 0 1
have determinant 1 and −1 respectively, so they are in
GL3 (Z).
Pablo Sáez
A strategy for attacking Büchi’s problem
An action on Γ2
◮
Moreover matrixes




0 0 1
3 4 0
B =  2 3 0 ,J =  0 1 0 
1 0 0
0 0 1
have determinant 1 and −1 respectively, so they are in
GL3 (Z).
◮
So H =< B, J > actually acts on Γ2 .
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits
◮
Actually, there are five orbits: those of elements in the set
∆2 = {(1, 0, 1), (−1, 0, −1), (1, 0, −1), (2, 1, 0), (−2, 1, 0)}.
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits
◮
Actually, there are five orbits: those of elements in the set
∆2 = {(1, 0, 1), (−1, 0, −1), (1, 0, −1), (2, 1, 0), (−2, 1, 0)}.
◮
To see this, observe that, except for (±1, 0, ±1) all Büchi
sequences of length 3 are either strictly increasing or strictly
decreasing, in absolute value (write Büchi’s equation as
x 2 − y 2 = y 2 − z 2 + 2).
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits
◮
Actually, there are five orbits: those of elements in the set
∆2 = {(1, 0, 1), (−1, 0, −1), (1, 0, −1), (2, 1, 0), (−2, 1, 0)}.
◮
To see this, observe that, except for (±1, 0, ±1) all Büchi
sequences of length 3 are either strictly increasing or strictly
decreasing, in absolute value (write Büchi’s equation as
x 2 − y 2 = y 2 − z 2 + 2).
◮
Now, if (x, y , z) is strictly decreasing (assume x and y are
positive) multiplying by B −1 we get (3x − 4y , −2x + 3y , z).
So the new value for y is y − 2(x − y ). It decreases.
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits
◮
Actually, there are five orbits: those of elements in the set
∆2 = {(1, 0, 1), (−1, 0, −1), (1, 0, −1), (2, 1, 0), (−2, 1, 0)}.
◮
To see this, observe that, except for (±1, 0, ±1) all Büchi
sequences of length 3 are either strictly increasing or strictly
decreasing, in absolute value (write Büchi’s equation as
x 2 − y 2 = y 2 − z 2 + 2).
◮
Now, if (x, y , z) is strictly decreasing (assume x and y are
positive) multiplying by B −1 we get (3x − 4y , −2x + 3y , z).
So the new value for y is y − 2(x − y ). It decreases.
◮
But we have −2x + 3y > −y whenever 2y > x. So y
decreases in absolute value whenever
(x, y , z) 6∈ {(x, y , z) ∈ Γ2 /2y ≤ x}(= Ω2 ).
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits (cont.)
◮
But Ω2 = {(±2, ±1, 0), (±1, 0, ±1), (0, ±1, ±2)}.
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits (cont.)
◮
But Ω2 = {(±2, ±1, 0), (±1, 0, ±1), (0, ±1, ±2)}.
◮
A similar analysis can be made when x and y are both
negative. If they are of different signs, multiply by B to get
the same fact (y decreases in absolute value).
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits (cont.)
◮
But Ω2 = {(±2, ±1, 0), (±1, 0, ±1), (0, ±1, ±2)}.
◮
A similar analysis can be made when x and y are both
negative. If they are of different signs, multiply by B to get
the same fact (y decreases in absolute value).
◮
If the sequence is increasing in absolute value, multiply by J.
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits (cont.)
◮
But Ω2 = {(±2, ±1, 0), (±1, 0, ±1), (0, ±1, ±2)}.
◮
A similar analysis can be made when x and y are both
negative. If they are of different signs, multiply by B to get
the same fact (y decreases in absolute value).
◮
If the sequence is increasing in absolute value, multiply by J.
◮
Therefore, appropriately multiplying by B, J or B −1 a finite
number of times, one gets to Ω2 .
Pablo Sáez
A strategy for attacking Büchi’s problem
The orbits (cont.)
◮
But Ω2 = {(±2, ±1, 0), (±1, 0, ±1), (0, ±1, ±2)}.
◮
A similar analysis can be made when x and y are both
negative. If they are of different signs, multiply by B to get
the same fact (y decreases in absolute value).
◮
If the sequence is increasing in absolute value, multiply by J.
◮
Therefore, appropriately multiplying by B, J or B −1 a finite
number of times, one gets to Ω2 .
◮
So we get the five orbits after analyzing the relationships
between the elements in Ω2 .
Pablo Sáez
A strategy for attacking Büchi’s problem
The group H
◮
Since H is generated by B and J, and since J −1 = J one can
write every element M ∈ H as an expression of the form
J b B n1 JB n2 J...JB nk J c
where b and c are 0 or 1 and the ni are non-zero integers.
Pablo Sáez
A strategy for attacking Büchi’s problem
The group H
◮
Since H is generated by B and J, and since J −1 = J one can
write every element M ∈ H as an expression of the form
J b B n1 JB n2 J...JB nk J c
where b and c are 0 or 1 and the ni are non-zero integers.
◮
Observe that H acts on tuples (x, y , z) such that
x 2 − 2y 2 + z 2 = a independently of the value of a.
Pablo Sáez
A strategy for attacking Büchi’s problem
The group H
◮
Since H is generated by B and J, and since J −1 = J one can
write every element M ∈ H as an expression of the form
J b B n1 JB n2 J...JB nk J c
where b and c are 0 or 1 and the ni are non-zero integers.
◮
Observe that H acts on tuples (x, y , z) such that
x 2 − 2y 2 + z 2 = a independently of the value of a.
◮
In particular, it acts on the columns of the identity matrix, for
which the values of a are 1,-2,1.
Pablo Sáez
A strategy for attacking Büchi’s problem
The group H
◮
Since H is generated by B and J, and since J −1 = J one can
write every element M ∈ H as an expression of the form
J b B n1 JB n2 J...JB nk J c
where b and c are 0 or 1 and the ni are non-zero integers.
◮
Observe that H acts on tuples (x, y , z) such that
x 2 − 2y 2 + z 2 = a independently of the value of a.
◮
In particular, it acts on the columns of the identity matrix, for
which the values of a are 1,-2,1.
◮
So, for every M ∈ H, its first column is in Γ1 , its second
column is in Γ−2 and its third column is in Γ1 .
Pablo Sáez
A strategy for attacking Büchi’s problem
Presentation of the group H
◮
Studying Γ1 , one can prove that, unless k = 0, 1 matrix M,
when multiplied to the right by a vector from the identity
matrix with a = 1, produces a vector whose second
component always increases when Σki=1 |ni | increases (the
details are quite tedious).
Pablo Sáez
A strategy for attacking Büchi’s problem
Presentation of the group H
◮
Studying Γ1 , one can prove that, unless k = 0, 1 matrix M,
when multiplied to the right by a vector from the identity
matrix with a = 1, produces a vector whose second
component always increases when Σki=1 |ni | increases (the
details are quite tedious).
◮
So the above matrix cannot be I , unless k = b = c = 0.
Pablo Sáez
A strategy for attacking Büchi’s problem
Presentation of the group H
◮
Studying Γ1 , one can prove that, unless k = 0, 1 matrix M,
when multiplied to the right by a vector from the identity
matrix with a = 1, produces a vector whose second
component always increases when Σki=1 |ni | increases (the
details are quite tedious).
◮
So the above matrix cannot be I , unless k = b = c = 0.
◮
So the presentation of group H is: J 2 (only), i. e. H is
isomorphic to the free product Z2 × Z.
Pablo Sáez
A strategy for attacking Büchi’s problem
Presentation of the group H
◮
Studying Γ1 , one can prove that, unless k = 0, 1 matrix M,
when multiplied to the right by a vector from the identity
matrix with a = 1, produces a vector whose second
component always increases when Σki=1 |ni | increases (the
details are quite tedious).
◮
So the above matrix cannot be I , unless k = b = c = 0.
◮
◮
So the presentation of group H is: J 2 (only), i. e. H is
isomorphic to the free product Z2 × Z.
We have got, so to speak, the search space for Büchi
sequences of length 5, considered as three sub-sequences of
length 3.
Pablo Sáez
A strategy for attacking Büchi’s problem
Further observations
◮
Taking Büchi’s equation x 2 − 2y 2 + z 2 = 2 modulo 2 one sees
that x and z have the same parity (both even, or both odd).
Pablo Sáez
A strategy for attacking Büchi’s problem
Further observations
◮
◮
Taking Büchi’s equation x 2 − 2y 2 + z 2 = 2 modulo 2 one sees
that x and z have the same parity (both even, or both odd).
And considering it modulo 4, one sees that x and y have
different parity.
Pablo Sáez
A strategy for attacking Büchi’s problem
Further observations
◮
Taking Büchi’s equation x 2 − 2y 2 + z 2 = 2 modulo 2 one sees
that x and z have the same parity (both even, or both odd).
◮
And considering it modulo 4, one sees that x and y have
different parity.
◮
So Büchi sequences are either of the form even, odd, even
(even sequence); or odd, even, odd (odd sequence).
Pablo Sáez
A strategy for attacking Büchi’s problem
Further observations
◮
Taking Büchi’s equation x 2 − 2y 2 + z 2 = 2 modulo 2 one sees
that x and z have the same parity (both even, or both odd).
◮
And considering it modulo 4, one sees that x and y have
different parity.
◮
So Büchi sequences are either of the form even, odd, even
(even sequence); or odd, even, odd (odd sequence).
◮
And, since B is I (modulo 2), one concludes that orbits
consist of sequences of the same parity.
Pablo Sáez
A strategy for attacking Büchi’s problem
Further observations
◮
Taking Büchi’s equation x 2 − 2y 2 + z 2 = 2 modulo 2 one sees
that x and z have the same parity (both even, or both odd).
◮
And considering it modulo 4, one sees that x and y have
different parity.
◮
So Büchi sequences are either of the form even, odd, even
(even sequence); or odd, even, odd (odd sequence).
◮
And, since B is I (modulo 2), one concludes that orbits
consist of sequences of the same parity.
◮
So in a Büchi sequence of length 5 subsequences (x1 , x2 , x3 )
and (x3 , x4 , x5 ) may lie in the same orbit. And subsequence
(x2 , x3 , x4 ) necessarily lies in a different orbit.
Pablo Sáez
A strategy for attacking Büchi’s problem
Two lemmata before the end...
We give two lemmata. In both cases the proof consists only in
verifying all the cases, so we skip them.
◮
Modulo 8, all Büchi sequences of length 3 (hence, also the
longer ones) are trivial.
Pablo Sáez
A strategy for attacking Büchi’s problem
Two lemmata before the end...
We give two lemmata. In both cases the proof consists only in
verifying all the cases, so we skip them.
◮
Modulo 8, all Büchi sequences of length 3 (hence, also the
longer ones) are trivial.
◮
All sequences (x, y , z) in the orbit of (2, 1, 0) (resp. (−2, 1, 0))
verify: either x or y is congruent to 2 (resp −2) modulo 8.
Pablo Sáez
A strategy for attacking Büchi’s problem
Two lemmata before the end...
We give two lemmata. In both cases the proof consists only in
verifying all the cases, so we skip them.
◮
Modulo 8, all Büchi sequences of length 3 (hence, also the
longer ones) are trivial.
◮
All sequences (x, y , z) in the orbit of (2, 1, 0) (resp. (−2, 1, 0))
verify: either x or y is congruent to 2 (resp −2) modulo 8.
◮
So, in an even Büchi sequence of length 8, there is necessarily
a subsequence of length 5 whose central element is either 2 or
−2.
Pablo Sáez
A strategy for attacking Büchi’s problem
Two lemmata before the end...
We give two lemmata. In both cases the proof consists only in
verifying all the cases, so we skip them.
◮
Modulo 8, all Büchi sequences of length 3 (hence, also the
longer ones) are trivial.
◮
All sequences (x, y , z) in the orbit of (2, 1, 0) (resp. (−2, 1, 0))
verify: either x or y is congruent to 2 (resp −2) modulo 8.
◮
So, in an even Büchi sequence of length 8, there is necessarily
a subsequence of length 5 whose central element is either 2 or
−2.
◮
That is, we would have necessarily two subsequences of length
3, α and β, that would necessarily lie on the same orbit. So
Mα = β needs to hold, for some M ∈ H.
Pablo Sáez
A strategy for attacking Büchi’s problem
And the end...
◮
So one can pose a system of equations over the group H that
holds if and only if a Büchi sequence of length 5 has its two
extreme subsequences of length 3 in the same orbit. And there
is always such a subsequence in a Büchi sequence of length 8.
Pablo Sáez
A strategy for attacking Büchi’s problem
And the end...
◮
So one can pose a system of equations over the group H that
holds if and only if a Büchi sequence of length 5 has its two
extreme subsequences of length 3 in the same orbit. And there
is always such a subsequence in a Büchi sequence of length 8.
◮
So Büchi’s problem would be solved, for N = 8, if the system
has only trivial solutions.
Pablo Sáez
A strategy for attacking Büchi’s problem
And the end...
◮
So one can pose a system of equations over the group H that
holds if and only if a Büchi sequence of length 5 has its two
extreme subsequences of length 3 in the same orbit. And there
is always such a subsequence in a Büchi sequence of length 8.
◮
So Büchi’s problem would be solved, for N = 8, if the system
has only trivial solutions.
◮
A possibility is to do induction on the length of M as a word.
Pablo Sáez
A strategy for attacking Büchi’s problem
And the end...
◮
So one can pose a system of equations over the group H that
holds if and only if a Büchi sequence of length 5 has its two
extreme subsequences of length 3 in the same orbit. And there
is always such a subsequence in a Büchi sequence of length 8.
◮
So Büchi’s problem would be solved, for N = 8, if the system
has only trivial solutions.
◮
A possibility is to do induction on the length of M as a word.
◮
We have some more pieces of information also. For instance,
the first step of the induction. Also, if α is an even Büchi
sequence then there are unique δ ∈ ∆2 , M ∈ H such that
α = Mδ.
Pablo Sáez
A strategy for attacking Büchi’s problem