1 Introduction
There have been many attempts to solve the versus problem. Since the CookLevin theorem [4], [7] in the early 70s, the problem has been the most important unsolved problem in computer science. Unfortunately most conventional methods of proof have failed. Relativizing proof [2], natural proof [9] or algebrization[1] could not solve the problem, and it was concluded that there is no simple proof for . However, this is a fallacy. There are many barriers to proofs of the P versus NP problem. But this does not mean that there is no simple proof.
In [10, page 24] is mentioned that the set
is complete.A similar problem is considered in this paper. Does a deterministic TM accept any word with length within steps if and are unary coded?
This set is complete. We analyze what happens when . With Rice’s theorem we show that one has to test every in the worst case. It is undecidable which are accepted by . So it is also undecidable which will be accepted first for .
Although we use diagonalization indirectly via the halting problem, the proof is not relativizing. The proof is consistent with . It does not violate the relativization barrier.
1.1 Procedure
In this paper, the problem of determining whether a tuple in the set is considered:
is contained. The problem is in .
Section 3 shows that for a given in worst case, there are steps to decide whether is accepted by .
Section 4 shows that in the worst case, all must be checked to decide whether . This means that the running time would be exponential.
2 Preliminaries
Let be a deterministic Turing machine with input alphabet . It is defined, that . In this paper, we investigate what happens when the running time of a TM is restricted. We additionally define:
Singletape TMs are used as Turing machines, as described in [3] with input alphabet .
3 Time Complexity of
In this section, the running time is considered to compute . It is concluded that to check whether a TM in steps accepts a word, steps must be calculated in the worst case.
Definition 1.
Let be a TM and be an input word. Then, is the running time of on input if terminates with it. is a partial function. is total iff terminates on every . And
Lemma 1.
For an arbitrary multitaped TM with input , it takes computation time in the worst case to find out whether is accepted by within steps, i.e., whether .
Proof.
Remark.
Lemma 1 applies to both singletape and multitape TMs.
4 Proof of
Lemma 2.
Let be a finite set of words and be a TM. To find out if every is not accepted by within steps, i.e. , one need a running time of in worst case.
Proof.
Let be an arbitrary deterministic TM and be a finite set of of input words. Let be a sequence with . If , then there exists a with . Thus:
Now let , where when . because is finite.
If , then there is a with . could be so large that it is not computable. For any , it is undecidable, according to the halting problem, whether . By Lemma 1, it takes steps to test whether in worst case. Let . As per Rice’s theorem, for any , it is undecidable, whether or any other subset of . Thus, in the worst case, every must be tested, especially, if no word from is accepted. This results in a runtime of . ∎
Remark.
That each must be tested independently of the other words from , can be seen especially if is a UTM and each is the code of a TM.
Consider the following set from Section 1.1:
As can be easily seen, the problem is whether an arbitrary tuple contained in is in . An NTM can be constructed, which writes symbols from to the input tape of , then executes and stops steps. Moreover, it holds:
By Lemma 2, the computational cost in the worst case is thus:
.
5 Acknowledgments
I would like to thank Günter Rote for helping me make the paper smoother. He gave me many helpful tips. Furthermore, I thank Paul Spirakis, Hermann Vosseler, Gabriel Wolf and Martin Dietzfelbinger for helping me improve some formulations. I also thank Michael Sipser, Richard Borcherds, Martin Ritzert, Steven Cook, Valentin Pickel, David Wellner and Thomas Karbe for their valuable comments, which helped me craft the formulations more vividly. I would also like to thank everyone who proofread the paper.
The paper was partly translated with www.DeepL.com/Translator (free version)
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