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$f(n) = \Theta(g(n))$ and $g(n) = \Theta(h(n))$ implies $f(n) = \Theta(h(n))$ (by transitivity).

Now $f(n) = \Theta(h(n))$ implies $h(n) = \Theta(f(n))$ (symmetry property). So statement I is true.

Similarly in statement II, by transitivity, we get $f(n) = O(h(n))$ this implies $h(n) = \Omega(f(n))$, by transpose symmetry.

Statement III is false. consider $f(n) = n^2$ and $g(n)=4n^2$

Statement III would be true, if the conclusion was $f(n) = \Theta(g(n))$ or $g(n) = \Theta(f(n))$.

You can refer to these properties in CLRS.

Now $f(n) = \Theta(h(n))$ implies $h(n) = \Theta(f(n))$ (symmetry property). So statement I is true.

Similarly in statement II, by transitivity, we get $f(n) = O(h(n))$ this implies $h(n) = \Omega(f(n))$, by transpose symmetry.

Statement III is false. consider $f(n) = n^2$ and $g(n)=4n^2$

Statement III would be true, if the conclusion was $f(n) = \Theta(g(n))$ or $g(n) = \Theta(f(n))$.

You can refer to these properties in CLRS.

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