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MIT 6.189 A Gentle Introduction to Programming Using Python (OCW)
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OpenStudy (anonymous):
how do u find size search. why prof. John G mentioned 12345/.01^2= 26? help. thanks a million
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OpenStudy (anonymous):
Found it! I knew I wasn't listening carefully enough. it's Log2(N) where N is the number of possible answers, which is the upper bound divided by the epsilon squared. Verified here: http://www.wolframalpha.com/input/?i=log2%28123450000%29
OpenStudy (anonymous):
oohh wow thanks a million! 'heart'
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