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Problem 16

Prove that for any integer \(p>1\), if \(p\) isn't pseudoprime, then \(p\) fails the Fermat test for at least half of all numbers in \(\mathcal{Z}_{p}^{+}\).

Expert verified

If 'p>1' is not a pseudoprime, then it fails the Fermat test for at least half of all integers in \(Z_p^+\). The key idea to prove this result is showing that at least half of the integers in this multiplicative group have at least one factor that is not coprime with 'p'. Through the relations between the Fermat test, Euler's totient function, and the Chinese Remainder Theorem, we can systematically count these failing elements and reach the desired conclusion.

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Chapter 10

Show that if \(\mathrm{PH}=\mathrm{PSPACE}\), then the polynomial time hierarchy has only finitely many distinct levels.

Chapter 10

Prove that if \(A\) is a regular language, a family of branching programs \(\left(B_{1}, B_{2}, \ldots\right)\) exists wherein each \(B_{n}\) accepts exactly the strings in \(A\) of length \(n\) and is bounded in size by a constant times \(n\).

Chapter 10

$$ \text { Prove that if } A \leq_{\mathrm{L}} B \text { and } B \text { is in NC, then } A \text { is in NC. } $$

Chapter 10

Define a ZPP-machine to be a probabilistic Turing machine that is permitted three types of output on each of its branches: accept, reject, and? A ZPP- machine \(M\) decides a language \(A\) if \(M\) outputs the correct answer on every input string \(w\) (accept if \(w \in A\) and reject if \(w \notin A\) ) with probability at least \(\frac{2}{3}\), and \(M\) never outputs the wrong answer. On every input, \(M\) may output? with probability at most \(\frac{1}{3}\). Furthermore, the average running time over all branches of \(M\) on \(w\) must be bounded by a polynomial in the length of \(w\). Show that RP $\cap \operatorname{coRP}=Z P P$, where ZPP is the collection of languages that are recognized by ZPP-machines.

Chapter 10

Let BPL be the collection of languages that are decided by probabilistic log space Turing machines with error probability \(\frac{1}{3}\). Prove that BPL \(\subseteq\) P.

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