La terminación unilateral por incumplimiento previsible como mecanismo para proteger

and NPSPACE

With all preliminary work done, we now can go on to the real problem, that is, to investigate the relationship between the classesNPf andNPSPACEf.

The main problem here, and the reason why we have difficulties even estab- lishing results that are trivial for regular, finite, Turing machines, is that there are machines for which we know that we havetime(x, T, p)≤αfor all halting computations p, which nonetheless can write down very complicated reals on the tape. An example of such a machine would be the machine, that, inωsteps, nondeterministically writes an 1 or a 0 on each of theω cells of the tape. After onlyωsteps, every real already occurs on one of the branches. So, whereas in the earlier cases the maxim ‘an algorithm never takes more space than it takes time’ always held, we have no certainty about that in this case, and even counterex- amples. As a result, to even prove something such asNPHK_⊆NPSPACEHK

we have to find, for every setAinNPHK_{,another machine, which also computes}

A, but which is guaranteed to be limited in space byNPSPACEHK. Whether

this is at all possible, remains an open question.

The situation, however, is not entirely hopeless: we may still try to ap- proach things, by showing that a complicated set, such the set of all reals cod- ing wellordered sets, occurs at a very low level in the NPSPACE-hierarchy. This turns out to be a feasible method indeed, and it appears that the set of non-wellordered sets appears in NPSPACEω+1. We first define the set WO:

Definition 7.17. The set WO is defined as {x∈R:xcodes a wellorder}. Proposition 7.18. WO∈NPSPACEω+1

Proof. We here use the fact that a realxcodes a wellorder if and only if it is a linear order that contains no infinitely descending chains. To see if a real xis

in WO, therefore, it will suffice to provide a nondeterministic Turing machine with an algorithm, that will first recognize any real that is not a linear order; and following that, if an infinitely descending chain is present, recognize it on at least one of the branches.

First of all, we can easily recognize if a realxcodes a linear orderR by first checking if it is total, then checking if it is antisymmetric, and then checking if it is transitive. These things can all be done by only writing finite sets or possibly Non the tape.

For the rest of the task, the task of recognizing an infinitely descending chain

s1, s2, . . . with siRsi+1 and si ∈ Field(R) fori ∈N, we should first note that

there must be another infinite sequencet1, t2, . . .withtiRti+1andti ∈Field(R)

fori∈N, such that for allti, the index oftiis lower than the index ofti+1. The

trick that we can use to keep the space complexity ‘low’, is to only keep track of the last element of the (possibly infinite) sequence, deleting all the earlier elements.

Now, we first need to find an initial element for the sequence, and then move along the field of the order from left to right. For each element we find, on one branch we pick it as the starting element of the sequence, and on another branch we do not. On the branch where we do, we put a flag on one of the scratch tapes at the index of the element, and put another flag on to indicate that we have found a first element. On the branch where we do not, we simply continue the search for a starting element.

When we have found at least one element for a sequence, we go on alongside the natural numbers, and for each number, we check if the element corresponding to it in the order is lower than our current lowest element. If it is, on one new branch we put a flag on this element as the new lowest element, deleting the previous lowest element, and on another branch we ignore it and keep the current lowest element.

When we reach a limit state, there are three possibilities:

1. The ‘found a first element’ flag is off, and (necessarily) the ‘lowest element’ tape is empty. We have not found an infinite descending sequence, and output 0.

2. The ‘found a first element’ flag is on, and there is exactly one 1 on the ‘lowest element’ tape. This means that this element has been chosen as ‘lowest element’ somewhere, but never been erased afterwards, so no lower elements have been chosen. Again, we have not found an infinite descending sequence, and output 0.

3. The ‘found a first element’ flag is on, and the ‘lowest element’ tape is empty. This means that every element that has been chosen as lowest element at one point, has been updated by a lower element afterwards. This however means that we have found an infinite descending sequence, and output 1.

This computation halts in all branches, and only writes very simple sets— finite ones containing one element, in fact—on the scratch tapes. It follows from this that the computation is inNPSPACEω+1.

We now refer to the following lemma about the analytical hierarchy (for background information on the analytical hierarchy, see e.g. [Ro]):

Lemma 7.19. For every Σ1

1 set A, there is a recursive function f, such that

x∈A ⇐⇒ f(x)∈WO

Proof. Corollary 20b in section 16.4 of [Ro] shows that, for every Π1 1 set B,

there is a recursive functionf, such thatx∈B ⇐⇒ f(x)∈WO. Now assume that A is Σ1

1. This implies that A is Π11, and there is a recursive function f,

such that x ∈ A ⇐⇒ f(x) ∈ WO. This same function f witnesses that

x∈A ⇐⇒ f(x)∈WO.

Because all Σ1

1 sets are reducible to the compliment of WO by a recursive

function, we can generalize Lemma 7.19 to the following result: Proposition 7.20. For anyΣ1

1 set A, we haveA∈NPSPACEω+2.

Proof. Because A is Σ1

1, we know that there must be some recursive function

f, such that x ∈ A ↔ f(x) ∈ WO. We can now proceed by first computing the function f on the input, which takes us ω time, and then performing the earlier algorithm to search an infinite descending sequence. Because, during the latter part of the computation, the content of the tape will only ever differ finitely many cells from the earlier resultf(x), we are guaranteed that the space complexity of any of the snapshots will never go beyond theω+ 2 level. It now follows that, indeed,A∈NPSPACEω+2.

Which gives us the following result:

Proposition 7.21. We haveNP+⊆NPSPACEω+2 and hence, for any or-

dinal αsuch thatω+ 1≤α≤ωCK

1 , also NPα⊆NPSPACEα. Proof. Theorem 5 from [DeHaSc] shows that NP0

+ = Σ11, and because the

functionf0(x) =ω1xsatisfies the conditions of Propositions 7.6 and 7.7,NP+=

Σ1

1 follows.

From this and Proposition 7.20, it is a direct consequence that NP+ ⊆

NPSPACEω+2, and the desired result follows. The second result then di-

rectly follows from the fact that NPα ⊆NP+ if α≤ωCK1 , and the fact that

NPSPACEω+1⊆NPSPACEα ifα≥ω+ 1.

Of course, this is not much: we have shown one instance of the problem NPα ⊆NPSPACEα, a problem which, in the case of finite Turing machine

computations and, as we will soon see, also in the case of nondeterministic Koepke machines, is trivially easy to solve.