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Algorithm Growth Functions?

Well, I have two questions: -

  • If f (n) is a function whose growth rate will be found, then for all three notations g (n) will be the same as for f (n) = O (g (n)), and similarly for omega and theta?

  • The theta designation is “omega and o”, if in some case, if the functions oh and omega are different, then how do we find the theta function there? Thank:)

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O, & Theta; and & Omega; notation is an interconnected, but very different concept. O-notation expresses an asymptotic upper bound for the growth rate of a function; he says that a function is ultimately bounded above by some constant multiple of some other function. Ω The designation is similar, but gives a lower bound. & Theta; the notation gives an asymptotic bounded boundary — for sufficiently large inputs, the algorithm grows at a rate bounded by a constant, multiple function, both from above and below.

If f (n) = O (g (n)), then it is not necessarily true that f (n) = & Omega; (g (n)) or f (n) =? (g (n)). For example, 1 = O (n), but 1? Ne & Omega; (n) because n grows strictly faster than 1.

, f (n) = O (g (n)) & Omega; (h (n)), g (n) & ne; h (n), j (n) , f (n) = & Theta; (j (n)). g (n) = & Theta; (h (n)), , f (n) = & Theta; (g (n)), , the & Theta; .

, !

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f (n) = O (g (n)) , n > N = > | f (n) | ≤C | g (n) | N C.

f (n) = & Omega; (g (n)) , n > N = > | f (n) | ≥C | g (n) | N C.

f (n) = & Theta (g (n)) , f (n) = O (g (n)) f (n) = & Omega; (g (n)).

, f ag , f (n) = & Theta; (g (n)), , g "" (.. - n ^ r * Log (n ) ^ s). , f (n) = cos (n) ² * n + sin (n) ² * n², f (n) = O (n²) f (n) = & Omega; (n), "" g , f (n) = & Theta; (g (n)).

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