Asymptotic. If f(n) = theta(g(n)) and g(n) = theta(h(n)), then why h(n) = theta(f(n))

Question

it is f(n)=theta(h(n)) as theta is transitive. But Can any one explain why h(n)=theta(f(n)).

This question appears to be off-topic because it is about computer science. Consider migrating it to http://cs.stackexchange.com/ — Timothy Shields, Dec 22 '13 at 20:22
All of the growth notations are transitive. Theta is the only one that is also symmetric (and tilde, if that counts). — Teepeemm, Dec 22 '13 at 20:31

score 5 · Accepted Answer · edited Nov 06 '20 at 13:55

5

Expanding the Big-O notation by its definition usually makes things easy.

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edited Nov 06 '20 at 13:55

Angel Politis

answered Dec 22 '13 at 21:38

Timothy

putting H(n)<=G(n)/s.. cool..and easy ..thanks @Skyler.Is it why the theta is symmetric .. – Xax Dec 24 '13 at 07:01
very nice explanation. – choudhury smrutiranjan parida Sep 01 '17 at 06:04

score 1 · Answer 2 · answered Dec 22 '13 at 20:31

1

If k1.h(n) <= f(n) <= k2.h(n) for large n, then (1/k2)f(n) <= h(n) <= (1/k1)f(n).

answered Dec 22 '13 at 20:31

Paul Hankin

score 0 · Answer 3 · answered Dec 22 '13 at 20:33

0

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))

answered Dec 22 '13 at 20:33

phimuemue

@MartijnPieters Sorry, I didn't pay attention at this very moment. Can I reject it now? – phimuemue Jan 21 '14 at 11:44
1

Nope; we had to clean up the tag to remove the spam. If you are not paying attention, why review at all? – Martijn Pieters Jan 21 '14 at 11:45

3 Answers3