Expanding the Big-O notation by its definition usually makes things easy.
Asymptotic. If f(n) = theta(g(n)) and g(n) = theta(h(n)), then why h(n) = theta(f(n))
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20-09-2022 - |
سؤال
it is f(n)=theta(h(n)) as theta is transitive. But Can any one explain why h(n)=theta(f(n)).
المحلول
نصائح أخرى
If k1.h(n) <= f(n) <= k2.h(n) for large n, then (1/k2)f(n) <= h(n) <= (1/k1)f(n).
That's basically simply because f(n) € theta(h(n))
is equivalent to h(n) € theta(f(n))
because of the following:
f(n) € O(h(n)) => h(n) € Omega(f(n))
f(n) € Omega(h(n)) => f(n) € O(h(n))
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