A scheme’s running time is: O((1/E)^2*n^2). Is this a fully polynomial-time approximation scheme? Explain why?

Asked Nov 05 '22 at 23:08

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I am looking for an answer to this question: A scheme’s running time is: O((1/E)^2*n^2). Is this a fully polynomial-time approximation scheme? Explain why?

asked Nov 05 '22 at 23:08

Majid

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A scheme’s running time is: O((1/E)^2*n^2). Is this a fully polynomial-time approximation scheme? Explain why?

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