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Assuming the upper bound on the maximum value of the $n$ input elements is $5n$ (as written in the question) then your solution works. Their solution works too but it is overcomplicated. If the maximum value of the $n$ input elements is at most $n^5$, then your solution would require time $\Omega(n^5)$. In general, if all $n$ input element are written in ...


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The statement is incorrect*. Consider that $$ RE = \{L(M) \mid M \text{ is a turing machine} \} $$ Thus $$\{ \langle M \rangle \mid M \text{ is a turing machine}\} \subseteq L \implies S_L =\{\ L(M) \mid \langle M \rangle \in L \} = RE$$ Therefore if I set $$L = \{ \langle M \rangle \mid M \text{ is a turing machine}\} \cup \{ \#\langle M \rangle \mid M \...


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The second idea seems to guide you on the right track. You can use the fact that all multiplies of each prime cannot be co-prime. So, for 14, as an example, you can remove all multiples of 2 and seven from the list. Hence, you can remove all multiplies of 2 ($\frac{10^6}{2}$), plus all odd multiplies of 7 ($\frac{10^6}{7\times 2}$). Hence, you need to check ...


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