Padding argument

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Show that, for any real numbers r≥1r \geq 1 and 0<ϵ<10<\epsilon <1,

NSPACE⁡(nr)≠⊂NSPACE⁡(nr+ϵ). \operatorname {NSPACE}\left(n^{r}\right) \stackrel{\subset }{\neq } \operatorname {NSPACE}\left(n^{r+\epsilon }\right).
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(a)

Show that Lemma 6.31 still holds if we replace the conditions s2(n)≥ns_{2}(n) \geq n and f(n)≥nf(n) \geq n by s2(n)≥log⁡ns_{2}(n) \geq \log n and log⁡(f(n))=O(s2(n))\log (f(n))= O\left(s_{2}(n)\right). [Hint: Note that NTM M3M_{3} can simulate M2M_{2} on input y=xSf(∣x∣)−∣x∣y=x S^{f(\left|x\right|)-\left|x\right|} without writing down the string yy on the second tape. Instead, it may simply write down, in the second tape, the position kk of the input tape head of M2M_{2} and use kk to determine what input symbol M2M_{2} 's tape head is scanning.]

(b)

Show that, for any real numbers r>0r>0 and 0<ϵ<10<\epsilon <1,

NSPACE⁡(nr)≠⊂NSPACE⁡(nr+ϵ). \operatorname {NSPACE}\left(n^{r}\right) \stackrel{\subset }{\neq } \operatorname {NSPACE}\left(n^{r+\epsilon }\right).

Show that if t1(n),t2(n)t_{1}(n), t_{2}(n), and f(n)f(n) are fully time-constructible functions with t2(n)≥nt_{2}(n) \geq n and f(n)≥nf(n) \geq n, then NTIME⁡(t1(n))⊆NTIME⁡(t2(n))\operatorname {NTIME}\left(t_{1}(n)\right) \subseteq \operatorname {NTIME}\left(t_{2}(n)\right) implies NTIME⁡(t1(f(n)))⊆NTIME⁡(t2(f(n)))\operatorname {NTIME}\left(t_{1}(f(n))\right) \subseteq \operatorname {NTIME}\left(t_{2}(f(n))\right).

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(a)

Show that EXP≠NP\mathrm{EXP} \neq \mathrm{NP}.

(b)

Show that EXP ≠⋃c>0DTIME⁡(2nc)\neq \bigcup_{c>0} \operatorname {DTIME}\left(2^{n^{c}}\right).

Show that if P=NPP=\mathrm{NP}, then

⋃c>0DTIME⁡(2nc)=⋃c>0NTIME⁡(2nc). \bigcup _{c>0} \operatorname {DTIME}\left(2^{n^{c}}\right)=\bigcup _{c>0} \operatorname {NTIME}\left(2^{n^{c}}\right).
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Consider the function pad : Σ∗×N⟶Σ∗#∗\Sigma^{*} \times \mathcal{N} \longrightarrow \Sigma^{*} \#^{*} that is defined as follows. Let pad⁡(s,l)=s#j\operatorname {pad}(s, l)=s \#^{j}, where j=max⁡(0,l−m)j=\max (0, l-m) and mm is the length of ss. Thus, pad⁡(s,l)\operatorname {pad}(s, l) simply adds enough copies of the new symbol to the end of ss so that the length of the result is at least ll. For any language AA and function f:N⟶Nf: \mathcal{N} \longrightarrow \mathcal{N}, define the language pad⁡(A,f)\operatorname {pad}(A, f) as

pad⁡(A,f)={pad⁡(s,f(m))∣ where s∈A and m is the length of s}. \operatorname {pad}(A, f)=\left\{ \operatorname {pad}(s, f(m)) \mid \text{ where } s \in A \text{ and } m \text{ is the length of } s\right\} .

Prove that if A∈TIME⁡(n6)A \in \operatorname {TIME}\left(n^{6}\right), then pad⁡(A,n2)∈TIME⁡(n3)\operatorname {pad}\left(A, n^{2}\right) \in \operatorname {TIME}\left(n^{3}\right).

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Prove that if NEXPTIME ≠\neq EXPTIME, then P≠NP\mathrm{P} \neq \mathrm{NP}. You may find the function pad, defined in Problem 9.13, to be helpful.

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ᴬ Define pad as in Problem 9.13.

Problem 9.13: Consider the function pad : Σ∗×N⟶Σ∗#∗\Sigma^{*} \times \mathcal{N} \longrightarrow \Sigma^{*} \#^{*} that is defined as follows. Let pad⁡(s,l)=s#j\operatorname {pad}(s, l)=s \#^{j}, where j=max⁡(0,l−m)j=\max (0, l-m) and mm is the length of ss. Thus, pad⁡(s,l)\operatorname {pad}(s, l) simply adds enough copies of the new symbol to the end of ss so that the length of the result is at least ll. For any language AA and function f:N⟶Nf: \mathcal{N} \longrightarrow \mathcal{N}, define the language pad⁡(A,f)\operatorname {pad}(A, f) as

pad⁡(A,f)={pad⁡(s,f(m))∣ where s∈A and m is the length of s}. \operatorname {pad}(A, f)=\left\{ \operatorname {pad}(s, f(m)) \mid \text{ where } s \in A \text{ and } m \text{ is the length of } s\right\} .
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(a)

Prove that for every AA and natural number k,A∈Pk, A \in \mathrm{P} iff pad⁡(A,nk)∈P\operatorname {pad}\left(A, n^{k}\right) \in \mathrm{P}.

(b)

Prove that P≠SPACE⁡(n)\mathrm{P} \neq \operatorname {SPACE}(n).

References

[DuKo01] Du D., Ko K. Problem Solving in Automata, Languages, and Complexity. John Wiley & Sons, 2001.
[Sip13] Sipser M. Introduction to the Theory of Computation. 3rd ed., Cengage Learning, 2013.