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Unformatted text preview: 1 LIMITATIONS ON OUR UNDERSTANDING OF THE
BEHAVIOR OF SIMPLIFIED PHYSICAL SYSTEMS
by
Harvey M. Friedman
Distinguished University Professor of Mathematics,
Philosophy, and Computer Science
Ohio State University
August 18, 2007
Abstract. Results going back to Turing and Gödel provide us
with limitations on our ability to decide the truth or
falsity of mathematical assertions in a number of important
mathematical contexts.
There are two kinds of such limiting results that must be
carefully distinguished. Results of the first kind state
the nonexistence of any algorithm for determining whether
any statement among a given set of statements is true or
false.
Results of the second kind are much deeper and represent
much greater challenges. They point to a specific statement
A, among a given set of statements, where we can neither
prove nor refute A using accepted principles of
mathematical reasoning.
We give a brief survey of these limiting results. These
include limiting results of the first kind: from number
theory in mathematics, and from idealized computing devices
in theoretical computer science.
The highlight of the talk is a discussion of limiting
results of the first kind in the context of simplified
physical systems; and a discussion of limiting results of
the second kind. The simplified physical systems involve a
small number of bodies, operating in potentially infinite
one dimensional discrete space time. AGENDA
1. EXAMPLES OF ALGORITHMS. Arithmetic ops, gcd, primality,
factoring, solvability of equations.
2. NO ALGORITHMS. Robust model of computation. Behavior of
abstract machines, solvability of equations. 2
3. SIMPLIFIED PHYSICAL SYSTEMS. Linear Order Systems. No
algorithm for determining boundedness. Research plans.
4. SIMULATION OF TURING MACHINES.
5. AXIOMS FOR MATHEMATICS. The ZFC axioms.
6. INDIVIDUAL SENTENCES NOT DECIDED IN ZFC. Consequence of
no algorithms. Research plans. 1. EXAMPLES OF ALGORITHMS. Arithmetic ops, gcd,
primality, factoring, solvability of equations.
I want to provide some significant context by discussing
demonstrable successes before I discuss demonstrable
limitations.
Long before there were any appropriate models of
computation, there were actual interesting algorithms.
Schoolchildren are still taught the standard algorithms for
adding, subtracting, and multiplying two integers given in
base 10. They are also taught the standard division
algorithm with remainder.
Algorithms for these standard arithmetic operations have
been revisited in a very powerful way because of the
computer revolution. There is a real need for optimizing
speed as much as possible, for numerous applications.
New ideas were injected into this ancient business, taking
advantage of the nature of actual circuit designs,
including the capacity for parallelism.
Going beyond this universal staple, one of the most famous
of all algorithms is the Euclidean algorithm from Greek
times, which is believed to predate Euclid. The function of
this algorithm is this. When given two two positive
integers, n,m, in base 10, return the greatest common
divisor of n,m in base 10.
The most obvious algorithm, when presented with n,m,
maintains a number d which is the greatest integer found
thus far to divide both n,m. We start with 1, which divides
both n,m. Generally, we increment d by 1 and see if the
result, d+1, divides both n,m. If so, we replace d by d+1,
and pass on to d+1. If not, we leave d alone, and pass on
to d+2. 3
We continue in this way. But how do we know when to stop?
Conservatively, we can stop after we have dealt with
max(n,m).
Of course, this is about the "worst possible algorithm" in
terms of computing resources, imaginable. The Greeks did
far better, in the following clever way.
If n = m, then gcd(n,m) = n. So assume, say, n < m. Divide
n into m and get a remainder r1 < n. Divide r1 into n and
get a remainder r2 < r1. Divide r2 into r1 and get a
remainder r3 < r2. Keep doing this until the remainder is
0. The divisor r that generated this 0 remainder is
gcd(n,m). This is because every common divisor of n,m
divides r, and r is a common divisor of n,m.
The number of steps that the Euclidean algorithm takes is
at most roughly the total number of digits in the problem an enormous improvement over the "worst" algorithm.
There are also good candidates for the "worst" way to test
whether or not a positive integer is prime. Just divide n
by all of the numbers from 2 through n1 and see if you
find a factor.
For reference, there are approximately n/ln(n) primes below
n. This is called the Prime Number Theorem.
Primality testing has gone through a very interesting
evolution. The very best algorithms in practice has an
unusual component which is not normally present. Namely, a
probabilistic one.
In the Miller Rabin primality test, they cleverly
associate, to any odd positive integer n, a set S[n] Õ
[1,n1] with the following properties.
i. If n is prime then S[n] is empty.
ii. If n is composite then S[n] has at least (n1)/2
elements.
iii. It is highly efficient to test for membership in S[n].
We can now "test" whether n is prime, highly efficiently.
Generate a lot of numbers in [1,n1], "randomly". Say
a1,...,a200. Check for membership in S[n]. If at least one
of them lies in S[n] then we know that n is composite. If
none of them lie in S[n], then almost certainly n is prime. 4
Why? If it were composite then you would have a misleading
outcome 200 times in a row! Chance of that is 2200.
This algorithm raises a number of deep foundational,
philosophical, and metaphysical issues surrounding
randomness and truth  in a particularly rich and focused
way.
We now come to factoring. It is widely believed that
factoring a positive integer is "hard" in the sense that it
is going to take an unfeasible amount of time, generally,
to factor positive integers randomly chosen with, say,
hundreds of digits.
There has been a huge success in building up a theory and
practice of cryptographic schemes based on the assumption
that factoring positive integers chosen randomly is too
difficult for anyone.
Nevertheless, there are methods that are considerably
better than the "worst", which is to try everything. There
is a family of methods called the sieve method, which
originated in ancient times.
We now come to the solvability of equations. Let me
concentrate on a single polynomial equation with integer
coefficients. There are three key parameters: the number of
variables, the degree, and the size of the coefficients.
1. There is an algorithm for testing solvability with real
or complex number unknowns.
2. There is an algorithm for testing solvability for
degree 2, with integer or rational or real or complex
number unknowns.
3. There is no algorithm for testing solvability for 9
positive integer unknowns.
4. There is no algorithm for testing solvability for
degree 4, with integer unknowns.
5. Nobody knows if there is an algorithm for testing
solvability with rational unknowns. 5
6. Nobody knows if there is an algorithm for testing
solvability for degree 3, with 2 rational unknowns. There
is an algorithm for degree 3, with 2 integer unknowns.
There is, and has been, a huge amount of work trying to
deal with the theoretical and practical issues surrounding
the implementation of 1 and fragments, (first due to Alfred
Tarski). 2. NO ALGORITHMS. Robust model of computation.
Behavior of abstract machines, solvability of
equations.
In order to establish that there is no algorithm for
determining whether a property holds of inputs from a
class, or no algorithm for returning information related to
inputs from a class, one relies on a standard robust model
of computation.
The first presentation of such a model was by Alan Turing.
It was rather specialized, and considerable effort went
into establishing its robustness.
Turing's formal computing devices were, on the surface,
rather limited and restricted. So it was not entirely clear
what would happen if one enlarged them without destroying
the general features that they possessed that made them
obviously algorithmic.
After much work along these lines, it became generally
accepted that an extremely robust analysis had been given.
In particular, the notions of partial recursive and
recursive functions on nonnegative integers and on finite
strings from a finite alphabet, as well as of recursively
enumerable and recursive sets of nonnegative integers and
finite strings from a finite alphabet, were fundamental,
and "correctly analyze" clear informal notions.
In fact, this feeling was codified by Alonzo Church into
what is called "Church's Thesis".
I am one of the very few people who has some optimism
there is some strikingly simple condition that can be
placed on "computable" or "algorithm", which, in some
"obviously" encompasses all "imaginable" computations
algorithms, and which is then proved to be equivalent that
way,
and
to 6
the Turing model. Then we have an indisputable "proof" of
Church's Thesis.
The first examples of algorithmically unsolvable problems
were obtained by turning the models of computation on their
heads.
Suppose there is an algorithm for testing whether a given
Turing machine halts at a given input.
I.e., let TM do the following. Given any TM' and string x
as input to TM, TM answers yes/no whether TM' halts with
input x.
It is then easy to adjust TM to TM* that does the
following. Given any TM' and string x as input to TM', TM*
does not halt if TM' halts with input x, and TM* halts if
TM' does not halt with input x.
By setting TM' = x = TM*, we get
With TM* and TM* as input to TM*, TM* does not halt if TM*
halts with input TM*, and TM* halts if TM* does not halt
with input TM*.
This is clearly impossible.
However theoretically satisfying this kind of thing might
be to some, the down to earth mathematician is far more
moved by their being a no algorithm result that is
considerably closer to their mathematical interests.
There is a shortage of such negative results along these
lines. I already mentioned the no algorithm results
concerning solvability of polynomial equations in integer
unknowns.
Another important one is that there is no algorithm for
testing whether a given finitely generated group is
trivial.
A related one in topology is: there is no algorithm for
testing whether two closed 4manifolds are homeomorphic. 7 3. SIMPLIFIED PHYSICAL SYSTEMS. Linear Order
Systems. No algorithm for determining Boundedness.
Research plans.
We now describe a very simplified kind of discrete physical
system, based on finitely many bodies. We call these Linear
Order Systems.
We prove that there is no way to analyze the behavior of
such systems over eternity, in the following sense. There
is no algorithm that determines, for any given initial
configuration of the bodies, whether the evolution is
contained in a finite region of space.
Moreover, the result is proved in the following strong
form:
There is a specific Linear Order System with 12 bodies such
that there is no algorithm that determines, for any given
initial configuration of the 12 bodies, whether the
evolution is contained in a finite region of space.
There has been considerable work on no algorithm results
for simple machines. Below, we indicate how this work
builds on existing work, and takes us into exciting new
directions.
We are anxious to begin research in a number of directions.
There are obvious extensions of Linear Order Systems to
Planar Systems and Space Systems. We expect that the number
of bodies required for the analogous results will be
considerably smaller than 12.
Linear Order Systems involve quite general laws of motion,
with few bodies. It will be important to extend the work to
cover much more restricted laws of motion, perhaps with
many more, but not too many more, bodies.
We now present the Linear Order Systems.
Space is identified with the integer number line
...2,1,0,1,2,...
We will not assume any reference point (e.g., 0).
Time is identified with the nonnegative integer number line 8 0,1,2,...
REMARK. An interesting direction of research is to allow
arbitrary integers as times. We suspect that there are very
interesting no algorithm results concerning what
configurations can occur in a given Linear Order System.
In an n body Linear Order System, we have n bodies,
B1,...,Bn, where n ≥ 1. We write
Bi[t], 1 £ i £ n,
for the position of body Bi at time t. Each Bi[t] is an
integer.
Thus the initial configuration of the n bodies is given by
B1[0],...,Bn[0].
Motion is deterministic, and very restricted. At each time
> 0, each body will move to the left one unit, to the right
one unit, or stay at the same position. I.e., we have
Bi[t+1]  Bi[t] Œ {1,0,1}.
In a Linear Order System, motion is completely specified in
terms of the relative order of the n bodies.
The relative order of the n bodies B1,...,Bn at time t is
just the relative order of the n integers B1[t],...,Bn[t].
This is merely a tabulation of which of these integers is
less than which others. Obviously, this information also
tells us which of these integers is equaled to which
others, and which of these integers is greater than which
others.
There is a standard, compact way of specifying the relative
order of B1[t],...,Bn[t]. It is by a listing
C1[t] ? C2[t] ? ... ? Cn[t]
where each ? is either < or =, and C1,...,Cn is a listing
of the bodies B1,...,Bn. A useful convention is that if we
see
Ci[t] = Cj[t] 9 adjacently in the listing, then we require i < j.
How can we justify this setup physically?
Think of each body as, e.g., constantly transmitting
radiation with its unique signature. So each body is
constantly aware of the presence of the other n1 bodies,
and knows which is which. In particular, each body is
constantly aware of the presence of the other bodies to the
limited extent of: is it to my left, to my right, or right
here (at the same position)?
Furthermore, assume that each body can sense, given what it
receives, the relative distance of each of the bodies from
itself  at least to the extent of knowing which ones are
closer or further away than others. E.g., by the strength
of the received signals.
From this information, each body can obviously infer the
relative order of ALL of the n bodies.
Each body then moves, by 0 or 1 units in each direction,
according to this information only.
We are considering arbitrary rules of this kind, so that
there is no assumption that different bodies may react in
the same way to the same information.
Here is just one example of further research: suppose each
body only knows which bodies are to the left of it, to the
right of it, and at the same position as it. This is a very
restricted class of Order Systems. Then do we have a no
algorithm theorem, and how many bodies do we need for such
a no algorithm theorem?
But let us stick to what we know now at this first stage of
development.
ONE BODY LINEAR ORDER SYSTEMS. There are exactly three one
body Linear Order Systems. Always moving to the right,
always moving to the left, always not moving. For any
initial configuration, the first two are unbounded, and the
third is bounded.
TWO BODY LINEAR ORDER SYSTEMS. There are three relative
orders. 10 B1 = B2.
B1 < B2.
B2 < B1.
B1,B2 have to be assigned a number from 1,0,1 under each
of the relative orders. There are 9 possibilities for this,
and so we obtain 9^3 = 729 two body Linear Order Systems.
It should be quite manageable  and interesting  to give a
complete analysis of which of these laws lead to
boundedness under which initial configurations, especially
with the help of a computer to dispense with trivial cases,
and to indicate what is true (that still needs to be
proved), and cataloging everything.
THREE BODY LINEAR ORDER SYSTEMS. There are 13 relative
orders.
B1
B1
B2
B2
B3
B3
B1
B1
B2
B1
B2
B3
B1 <
<
<
<
<
<
=
=
=
<
<
<
= B2
B3
B1
B3
B1
B2
B2
B3
B3
B2
B1
B1
B2 <
<
<
<
<
<
<
<
<
=
=
=
= B3.
B2.
B3.
B1.
B2.
B1
B3.
B2.
B1.
B3.
B3.
B2.
B3. B1,B2,B3 have to be assigned a number from 1,0,1 under
each of the relative orders. This consists of 39 numbers
drawn from {1,0,1}. Thus there are 27^13 = 3^39 three body
Linear Order Systems. The challenge is to construct an
efficient algorithm that determines boundedness for any
three body Linear Order System under any given initial
configuration of the three bodies. We conjecture that this
can be done.
...
TWELVE BODY LINEAR ORDER SYSTEMS. Our theorem asserts that
a boundedness test is impossible for at least one nine body
Linear Order System. 11
Officially, any 12 body Linear Order System involves 12
numbers from {1,0,1} assigned to each relative order for
B1,....,B12. The number of such relative orders is rather
large. E.g., considerably more than 12!. So each law of
motion for 12 bodies officially requires a table of
considerably more than 12(12!) numbers from 1,0,1.
However, there are many opportunities for providing far
shorter descriptions of Linear Order Systems.
A partial relative order for n bodies, is a set of
conditions of the form
Bi
Bi
Bi
Bi =
<
£
≠ Bj
Bj
Bj
Bj where 1 £ i,j £ n. We can now write laws of motion in the
following form.
Partial relative order #1. B1 moves ?,...,Bn moves ?
Partial relative order #2. B1 moves ?,...,Bn moves ?
...
Partial relative order #m. B1 moves ?,...,Bn moves ?
Otherwise. B1 moves ?,...,Bn moves ?.
We demand that these partial relative orders are mutually
exclusive.
Let us call the total number of partial relative orders
used, the presentation complexity.
Our intractable 12 body law of motion has presentation
complexity far lower than the theoretical limit, which is
greater than 12!, and probably can be massaged to have
still lower presentation complexity – perhaps down to a few
dozen.
QUESTION. What can we say about the triples (n,r) for which
there is an n body Order System of presentation complexity
at most r, such that boundedness for arbitrary initial
configurations cannot be determined algorithmically?
Before getting into the details, we make some basic remarks
on the relation between this work and previous work on
abstract machines. 12 Interpretations of the no algorithms results on Turing
machines, in terms of the motion of bodies, involves having
arbitrarily large numbers of bodies, with no a priori
bound. This is because they involve arbitrarily long
strings of bits, or symbols from a finite alphabet.
Linear Order Systems are more closely related to register
machines. But in Linear Order Systems, all abstract state
and all flow of control must emanate from the configuration
of the bodies alone  not from an ordered list of
instructions and list of abstract states.
Of course, we rely heavily on previous work on Turing
machines and (ideas from) register machines.
We anticipate a steady stream of restrictions on Linear
Order Systems (such as the one mentioned earlier) and also
the introduction of other models which take more and more
aspects of actual physical systems into account.
The mathematics will then get increasingly deeper and more
involved, getting further away from the earlier no
algorithms work, which was not primarily motivated by this
kind of direct consideration of physical systems. 4. SIMULATION OF TURING MACHINES.
We now show how to simulate the action of any Turing
machine TM with two symbols and n states, by a Linear Order
System, in the present sense, with considerably fewer than
n bodies.
Because we are focused only on the determination of
boundedness, we will make the assumption that the TM has no
halting instructions. I.e., the TM is halting free. We will
see that this requirement may cost at most one state.
Let the TM have symbols 0,1, and n states q1,...,qn. We use
the standard quintuple formulation of TMs, with the two way
infinite tape. We will use q1 as the initial state.
Inputs consist of two strings x,y Œ {0,1}*, of length ≥ 0,
and a state qi. Initialization is by
...0001x qi y1000... 13
with the reading head on the first bit of y1, and the
device in state q1.
TM operates by a complete set of 2n quintuples of the form
qi b c L qj
qi b c R qj
qi b   where 1 £ i,j £ n, L = left, R = right, and b,c Œ {0,1}.
Here qi b    indicates halting when in state qi and
reading bit b.
At any time, the TM is in the form
000...x q y000...
where x,y Œ {0,1}* and q Œ {q1,...,qn}. Here y is nonempty.
The reading head is reading the first bit of y1.
In the simulation, we will use five bodies, B1,B2,B3 which
are, however, augmented with some additional apparatus that
is strictly forbidden in Linear Order Systems:
i. 0 as a reference point.
ii. Relative orders will use 0. I.e., they take the form
0 ? C1 ? C2 ? C3, where {C1,C2,C3} = {B1,B2,B3}.
iii. The n abstract states q1,...,qn.
iv. Two special states S1,S2.
v. Thus the "global states" will be the relative order (in
the sense of ii) together with the q state (one of the n
abstract states q1,...,qn), and the S1,S2 states.
vi. We assign 1,0,1 for movement, as usual.
We will then eliminate the use of the q,S1, and S2 states,
thereby conforming to the Linear Order System model  with
the exception that we will still be using the reference
point 0. To accomplish this, we need to introduce ≥
log3(14n) bodies.
Finally, we can treat the reference point 0 as an
additional body that does not move, resulting in a total of
4 + Èlog3(14n)˘ bodies.
For each state qi of TM, we define a law of motion M(qi),
freely using partial global states. We will then put the
M(qi) together in the obvious way. 14 M(qi) will depend on the unique quintuples in TM that start
with qi. Let these be
qi 0 a a qj
qi 1 b b qk
It is convenient to split this into 16 cases, according to
the values of a,b Œ {0,1} and a,b Œ {L,R}.
We use B↑ for adding 1, and BØ for subtracting 1.
Initialization of M(qi) is with
B1 = the integer given by x.
B2 = the integer given by rev(y).
B3 = 0.
S1 = 0.
S2 = 0.
q = qi.
Here "rev" means reverse. x div 2 is the floor of x/2. x
mod 2 is the parity of x (0 or 1).
Note that we have used S1,S2 instead of B4,B5. This is
convenient, since we use S1,S2 like "states", where S1
always stays within [0,6], and S2 always stays within
{0,1].
CASE 1. a = b = L. b = c = 0. We must transition to
B1 = x div 2.
B2 = 2y + (x mod 2) if y even; 2(y1) + (x mod 2) if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qi if y even; qk if y odd.
1.1. S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
1.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
1.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2.
1.4. S1 = 1, B3 > 0 ﬁ B2↑, B3Ø.
1.5. S1 = 1, S2 = 0, B3 = 0 ﬁ S1 = 2.
1.6. S1 = 1, S2 = 1, B3 = 0 ﬁ B2Ø, S1 = 2.
We now have B2 = y if y even; y1 if y odd.
1.7. S1 = 2, B3 < B2 ﬁ B3↑.
1.8. S1 = 2, B2 = B3 ﬁ S1 = 3. 15
1.9. S1 = 3, B3 > 0 ﬁ B2↑.
1.10. S1 = 3, B3 = 0 ﬁ S1 = 4.
We now have B2 = 2y if y even; 2(y1) if y odd.
1.11. S1 = 4, B3 < B1 ﬁ B1↑, B3Ø.
1.12. S1 = 4, B1 = B3 ﬁ S1 = 5.
1.13. S1 = 4, B1 < B3 ﬁ B2↑, S1 = 5.
We now have B1 = x div 2, B2 = previous B2 if x is even;
previous B2 plus 1 if x is odd.
1.14. S1 = 5, B3 > 0 ﬁ B3Ø, S1 = 6.
1.15. S1 = 6, S2 = 0, B3 = 0 ﬁ S1 = 0, q = qi.
1.16. S1 = 6, S2 = 1, B3 = 0 ﬁ S1 = 0, S2 = 0, q = qk.
CASE 2. a = b = L. b = c = 1. We transition to
B1 = x div 2.
B2 = 2(y+1) + (x mod 2) if y even; 2y + (x mod 2) if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y mod 2 = 0; qk otherwise.
We have only
if y even; y
2.4. S1 = 1,
2.5. S1 = 1,
2.6. S1 = 1, to adjust 1.4  1.6 so that we have B2 = y+1
if y odd.
0 < B3 ﬁ B2↑, B3Ø.
S2 = 0, B3 = 0 ﬁ B2↑, S1 = 2.
S2 = 1, B3 = 0 ﬁ S1 = 2. CASE 3. a = b = L. b = 0, c = 1. We transition to
B1 = x div 2.
B2 = 2y + (x mod 2).
B3 = 0.
S1 = 0.
q = qj if y mod 2 = 0; qk otherwise.
We have only to adjust 1.4  1.6 so that we have B2 = y.
3.4. S1 = 1, B3 > 0 ﬁ B2↑, B3Ø.
CASE 4. a = b = L. b = 1, c = 0. We transition to
B1 = x div 2.
B2 = 2(y+1) + (x mod 2) if y even; 2(y1) + (x mod 2) if y
odd.
B3 = 0.
S1 = 0.
q = qj if y mod 2 = 0; qk otherwise.
We have only to adjust 1.4  1.6 so that we have B2 = y+1
if y even; y1 if y odd.
4.4. S1 = 1, 0 < B3 ﬁ B2↑, B3Ø. 16
4.5. S1 = 1, S2 = 0, B3 = 0 ﬁ B2↑, S1 = 2.
4.6. S1 = 1, S2 = 1, B3 = 0 ﬁ B2Ø, S1 = 2.
CASE 5. a = b = R. b = c = 0. We transition to
B1 = 2x.
B2 = y div 2.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
5.1. S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
5.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
5.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have B2 = y div 2, S2 = y mod 2.
5.4. S1 = 1, 0 < B3 ﬁ B3Ø.
5.5. S1 = 1, B3 = 0 ﬁ S1 = 2.
5.6. S1 = 2, B3 < B1 ﬁ B3↑.
5.7. S1 = 2, B1 = B3 ﬁ S1 = 3.
5.8. S1 = 3, B3 > 0 ﬁ B1↑, B3Ø.
We now have B1 = 2x.
5.9. S1 = 3, S2 = 0, B3 = 0 ﬁ S1 = 0, q = qj.
5.10. S1 = 3, S2 = 1, B3 = 0 ﬁ S1 = 0, S2 = 0, q = qk.
CASE 6. a = b = R. b = c = 1. We transition to
B1 = 2x+1.
B2 = y div 2.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y mod 2 = 0; qk otherwise.
We have only to adjust 5.9, 5.10 so that we have B1 = 2x+1.
6.9. S1 = 3, S2 = 0, B3 = 0 ﬁ B1↑, S1 = 0, q = qj.
6.10. S1 = 3, S2 = 1, B3 = 0 ﬁ B1↑, S1 = 0, S2 = 0, q = qk.
CASE 7. a = b = R. b = 0, c = 1. We transition to
B1 = 2x if y even; 2x+1 if y odd.
B2 = y div 2.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y mod 2 = 0; qk otherwise.
We have only to adjust 5.10 so that we have B1 = 2x if y
even; 2x+1 if y odd. 17 7.10. S1 = 3, B2 = 1, B3 = 0 ﬁ B1↑, S1 = 0, S2 = 0, q = qk.
CASE 8. a = b = R. b = 1, c = 0. We transition to
B1 = 2x+1 if y even; 2x if y odd.
B2 = y div 2.
B3 = 0.
S1 = 0.
S2 = 0.
q = qi if y mod 2 = 0; qk otherwise.
We have only to adjust 5.9 so that we have B1 = 2x+1 if y
even; 2x if y odd.
8.9. S1 = 3, S2 = 0, B3 = 0 ﬁ B1↑, S1 = 0, q = qj.
CASE 9. a = L, b = R, b = c = 0. We transition to
B1 = x div 2 if y even; 2x if y odd.
B2 = 2y + (x mod 2) if y even; y div 2 if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
9.1 S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
9.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
9.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2, B2 = y div 2.
We now list clauses pertaining to y even; i.e., S2 = 0.
Note that we can repeat Case 1 with S2 = 0, after 1.3.
9.4. S1 = 1, S2 = 0, B3 > 0 ﬁ B2↑, B3Ø.
9.5. S1 = 1, S2 = 0, B3 = 0 ﬁ S1 = 2.
We now have B2 = y.
9.6. S1 = 2, S2 = 0, B3 < B2 ﬁ B3↑.
9.7. S1 = 2, S2 = 0, B2 = B3 ﬁ S1 = 3.
9.8. S1 = 3, S2 = 0, B3 > 0 ﬁ B2↑.
9.9. S1 = 3, S2 = 0, B3 = 0 ﬁ S1 = 4.
We now have B2 = 2y.
9.10. S1 = 4, S2 = 0, B3 < B1 ﬁ B1↑, B3Ø.
9.11. S1 = 4, S2 = 0, B1 = B3 ﬁ S1 = 5.
9.12. S1 = 4, S2 = 0, B1 < B3 ﬁ B2↑, S1 = 5.
We now have B1 = x div 2, B2 = 2y + x mod 2.
9.13. S1 = 5, S2 = 0, B3 > 0 ﬁ B3Ø, S1 = 6.
9.14. S1 = 6, S2 = 0, B3 = 0 ﬁ S1 = 0, q = qj.
This is final for y even. 18 We now list clauses pertaining y odd; i.e., S2 = 1. Note
that we can repeat Case 5 with S1 = 1, after 5.3.
9.15. S1 = 1, S2 = 1, 0 < B3 ﬁ B3Ø.
9.16. S1 = 1, S2 = 1, B3 = 0 ﬁ S1 = 2.
9.17. S1 = 2, S2 = 1, B3 < B1 ﬁ B3↑.
9.18. S1 = 2, S2 = 1, B1 = B3 ﬁ S1 = 3.
9.19. S1 = 3, S2 = 1, B3 > 0 ﬁ B1↑, B3Ø.
We now have B1 = 2x.
9.20. S1 = 3, S2 = 1, B3 = 0 ﬁ S1 = 0, S2 = 0, q = qk.
This is final for y odd.
CASE 10. a = L, b = R, b = c = 1. We transition to
B1 = x div 2 if y even; 2x+1 if y odd.
B2 = 2(y+1) + (x mod 2) if y even; y div 2 if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
10.1 S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
10.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
10.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2, B2 = y div 2.
We have only to adjust 9.20 so that B1 = 2x+1, and 9.5 so
that B2 = y+1.
10.5. S1 = 1, S2 = 0, B3 = 0 ﬁ B1↑, S1 = 2.
10.20. S1 = 3, S2 = 1, B3 = 0 ﬁ B1↑, S1 = 0, S2 = 0, q =
qk.
CASE 11. a = L, b = R, b = 0, c = 1. We transition to
B1 = x div 2 if y even; 2x+1 if y odd.
B2 = 2y + (x mod 2) if y even; y div 2 if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
11.1 S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
11.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
11.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2, B2 = y div 2.
We have only to adjust 9.20 so that B1 = 2x+1. 19 11.20. S1 = 3, S2 = 1, B3 = 0 ﬁ B1↑, S1 = 0, S2 = 0, q =
qk.
CASE 12. a = L, b = R, b = 1, c = 0. We transition to
B1 = x div 2 if y even; 2x if y odd.
B2 = 2(y+1) + (x mod 2) if y even; y div 2 if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
12.1 S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
12.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
12.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2, B2 = y div 2.
We have only to adjust 9.5 so that B2 = y+1.
12.5. S1 = 1, S2 = 0, B3 = 0 ﬁ B2↑, S1 = 2.
CASE 13. a = R, b = L. b = c = 0. We transition to
B1 = 2x if y even; x div 2 if y odd.
B2 = y div 2 if y even; 2(y1) + (x mod 2) if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
13.1 S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
13.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
13.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2, B2 = y div 2.
We now list clauses pertaining to y even; i.e., S2 = 0.
Note that we can repeat Case 5 with S2 = 0, after 5.3.
13.4. S1 = 1, S2 = 0, 0 < B3 ﬁ B3Ø.
13.5. S1 = 1, S2 = 0, B3 = 0 ﬁ S1 = 2.
13.6. S1 = 2, S2 = 0, B3 < B1 ﬁ B3↑.
13.7. S1 = 2, S2 = 0, B1 = B3 ﬁ S1 = 3.
13.8. S1 = 3, S2 = 0, B3 > 0 ﬁ B1↑, B3Ø.
We now have B1 = 2x.
13.9. S1 = 3, S2 = 0, B3 = 0 ﬁ S1 = 0, q = qj.
This is final for y even. 20
We now list clauses pertaining to y odd; i.e.., S2 = 1.
Note that we can repeat Case 1 for S2 = 1, after 1.3.
13.10. S1 = 1, S2 = 1, B3 > 0 ﬁ B2↑, B3Ø.
13.11. S1 = 1, S2 = 1, B3 = 0 ﬁ B2Ø, S1 = 2.
We now have B2 = y1.
13.12. S1 = 2, S2 = 1, B3 < B2 ﬁ B3↑.
13.13. S1 = 2, S2 = 1, B2 = B3 ﬁ S1 = 3.
13.14. S1 = 3, S2 = 1, B3 > 0 ﬁ B2↑.
13.15. S1 = 3, S2 = 1, B3 = 0 ﬁ S1 = 4.
We now have B2 = 2(y1). Also B3 = 0.
13.16. S1 = 4, S2 = 1, B3 < B1 ﬁ B1↑, B3Ø.
13.17. S1 = 4, S2 = 1, B1 = B3 ﬁ S1 = 5.
13.18. S1 = 4, S2 = 1, B1 < B3 ﬁ B2↑, S1 = 5.
We now have B1 = x div 2, B2 = previous B2 plus 1.
13.19. S1 = 5, S2 = 1, B3 > 0 ﬁ B3Ø, S1 = 6.
13.20. S1 = 6, S2 = 1, B3 = 0 ﬁ S1 = 0, S2 = 0, q = qk.
This is final for y odd.
CASE 14. a = R, b = L. b = c = 1. We transition to
B1 = 2x+1 if y even; x div 2 if y odd.
B2 = y div 2 if y even; 2y + (x mod 2) if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
14.1 S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
14.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
14.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2, B2 = y div 2.
We have only to adjust clause 13.9 so that B1 = 2x+1, and
clause 13.11 so that B2 = y.
14.9. S1 = 3, S2 = 0, B3 = 0 ﬁ B1↑, S1 = 0, q = qj.
14.11. S1 = 1, S2 = 1, B3 = 0 ﬁ S1 = 2.
CASE 15. a = R, b = L. b = 0, c = 1. We transition to
B1 = 2x if y even; x div 2 if y odd.
B2 = y div 2 if y even; 2y + (x mod 2) if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
15.1 S1 = 0, B3 < B2 ﬁ B2Ø, B3↑. 21
15.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
15.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2, B2 = y div 2.
We have only to adjust clause 13.11 so that B2 = y.
15.11. S1 = 1, S2 = 1, B3 = 0 ﬁ S1 = 2.
CASE 16. a = R, b = L. b = 1, c = 0. We transition to
B1 = 2x+1 if y even; x div 2 if y odd.
B2 = y div 2 if y even; 2(y1) + (x mod 2) if y odd.
B3 = 0.
S1 = 0.
S2 = 0.
q = qj if y even; qk if y odd.
16.1 S1 = 0, B3 < B2 ﬁ B2Ø, B3↑.
16.2. S1 = 0, B2 = B3 ﬁ S1 = 1.
16.3. S1 = 0, B2 < B3 ﬁ S1 = 1, S2 = 1.
We now have S2 = y mod 2, B2 = y div 2.
We have only to adjust clause 13.9 so that B1 = 2x+1.
16.9. S1 = 3, S2 = 0, B3 = 0 ﬁ B1↑, S1 = 0, q = qj.
This completes the description of M(qi).
Note that in each of the 16 cases, S1 stays in [0,6] and S2
stays in [0,1]. Hence there is a total of most 14 states of
S1,S2 in each M(qi). So there is a total of at most 14n
auxiliary states.
A more careful analysis shows that in each M(qi), there are
at most 11 states of S1,S2. Hence there is a total of at
most 11n auxiliary states.
Suppose log2(m) ≥ 11n. Then we can replace the auxiliary
states with m new bodies which occupy positions 0,1 only.
We can transition in any way from one of these to any
other.
Hence we have accomplished the simulation of TM with 3 +
log2(m) bodies, using 0 as a reference point. We can view
the reference point as another body that does not move.
Hence we have accomplished the simulation with a 4 +
log2(m) body Linear Order System. 22
We can now prove the following.
THEOREM 4.1. Suppose there is a TM with 2 symbols and n
states, for which the halting problem is algorithmically
unsolvable. Suppose m ≥ log(11(n+1)). Then there is a 4+m
body Linear Order System LOS whose boundedness problem is
algorithmically unsolvable. If the former is complete r.e.
then the latter is complete r.e.
Proof: Let TM be as given. Let TM' be obtained from TM by
replacing each halting quintuple qi b HALT in TM by the
quintuples
qi b b L qn+1
qn+1 0 0 R qi
qn+1 1 1 R qi
Obviously, any run of TM that doesn't halt is a run of TM'.
Now consider any run of TM that does halt, say with qi b
HALT.
Then TM' will also arrive at qi b, and execute the
following instructions indefinitely in a loop:
qi b b L qn+1
qn+1 c c R qi
qi b b L qn+1
Hence the corresponding run of TM' is bounded.
Suppose that boundedness for TM' is algorithmically
decidable. We now algorithmically solve the halting problem
for TM.
Let X be the run of TM at input a. Let X* be the run of TM'
at input a.
case 1. X* is unbounded. Then X cannot halt, for then X*
would be bounded. We have solved the halting problem for TM
in this case.
case 2. X* is bounded. Then X* goes into a loop, and we can
examine the quintuples executed in X* to see if any of them
are qi b, where qi b HALT is in TM. It is obvious that X
halts if and only if we find such. 23
This solves the halting problem for TM, contrary to the
hypothesis. Therefore boundedness for TM' is
algorithmically undecidable.
Moreover, if the halting problem for TM is complete r.e.,
then boundedness for TM' is complete r.e., by the same
argument.
We now let m be as given. I.e., the number of order types
of m tuples is at most 14(n+1). We let LOS be the 4+m body
Linear Order System that simulates TM' at any input, as
given above. Then boundedness of TM' is equivalent to
boundedness of LOS. QED
LEMMA 4.2. There is a TM with 2 symbols and 19 states whose
halting problem is complete r.e.
Proof: See
Baiocchi: Three Small Universal Turing Machines. In:
Springer
LNCS 2055, Machines, Computation, and Universality, 2001,
pp. 110.
QED
THEOREM 4.3. There is a 12 body Linear Order System such
that there is no algorithm for determining whether an
arbitrary initial configuration has bounded evolution. In
fact, the boundedness problem is complete r.e.
Proof: If we use Theorem 4.1 with 11(n+1), then we need 2^8
= 128 ≥ 11(20) = 220. QED 5. AXIOMS FOR MATHEMATICS. The ZFC axioms.
For the next section, we need to be aware that already by
the early part of the 20th century, the standard axiomatic
basis for mathematics had been carefully formulated and
reasonably well accepted.
This is the so called ZFC axioms, or Zermelo Frankel set
theory with the axiom of choice.
The vast preponderance of mathematics can, without
difficulty, be proved within ZFC. 24 Notable exceptions include various unusually and heavily
set theoretic problems, plus some examples of ours that are
much more down to earth, and will be discussed in the soon
forthcoming book
Boolean Relation Theory and Incompleteness Phenomena. 6. INDIVIDUAL SENTENCES NOT DECIDED IN ZFC.
Consequence of no algorithms. Research plans.
There is a generally much more delicate, deeper, and more
difficult kind of "undecidability", that is often confused
with algorithmic undecidability.
This concerns a SINGLE STATEMENT. Not an infinite family of
statements.
There is no clear way to get at the hardness of a SINGLE
QUESTION using algorithms. Obviously, there is an algorithm
for deciding whether a single given statement S is true or
not.
case 1. S is true. Let ALG be the algorithm that always
says true. Then ALG correctly decides the truth value of S.
case 2. S is false. Let ALG be the algorithm that always
says false. Then ALG correctly decides the truth value of
S.
So how do we talk about "undecidability" in the context of
a single statement S?
One simple way is to assert that it is not provable or
refutable in ZFC.
From the early by now standard work in mathematical logic,
we have the following.
THEOREM. Let S be a set of positive integers defined in
ZFC. There exists a positive integer n such that the
statement “n Œ S” is neither provable nor refutable in ZFC.
With a little bit of fiddling, this tells us that there is
a 12 body Order System with initial conditions, for which 25
the statement “is it bounded” is neither provable nor
refutable in ZFC.
But this misses the essential point. If we pass through
this classic Theorem, the number of digits needed to
present the initial condition is grotesque.
On the other hand, if we require that the bodies start out
at the same place, then the number of bodies needed becomes
grotesque. CONJECTURE
There is a 20 body Linear Order System such that ZFC
neither proves nor refutes boundedness with the initial
configuration having all bodies at the same point.
As remarked before, an arbitrary 20 body Linear Order
System may have so little in the way of symmetry that it
may be unfeasible to describe it.
Here is a much strong conjecture.
There is a 20 body Linear Order System of presentation
complexity at most 50 such that ZFC neither proves nor
refutes boundedness with the initial configuration having
all bodies at the same point.
In a discussion yesterday with Professor Inwagen, he asked
me what I could hope to say about the situation with
exactly two bodies.
I saw right away that this might well turn out to be an
extremely fruitful path to follow.
We can very naturally allow the motion of each body to
depend on the exact distance (positive or negative or 0) to
the other, up to magnitude, say, 100. For signed distances
of magnitude greater than 100, we would require a uniform
choice of motion.
If there proves to be insufficient “juice” to get these
kinds of negative results, then we can allow the motion of
each body to be given by any number from 100 to 100.
The conjecture would be that we get algorithmic
undecidability with general initial position, and we get 26
ZFC undecidability with initial position having both
particles at the same position.
At the other extreme, physicists may be more attracted to
formulations where the number of particles is fixed but
quite large, spread out with a realistic initial
configuration, but where the “laws of motion” are extremely
simple – and the same for each body. Each body has a
position and also one of a very few states, which determine
how it interacts with other bodies in its vicinity.
It is very likely that this modest initial work can be
crafted to apply well to such systems. ...
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 Fall '08
 JOSHUA
 Math

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