-3

I believe we have to use a variant of master theorem...can someone suggest me how to find complexity of such equation which doesn't directly fit into Masters theorem.

-1

I'd first put parentheses around the expressions to make clear what the right side actually is. Once it is clear that it is (n ^ 2.93) * ((log n)^93) = o (n^3) the solution is easy, but rather pointless: If log n is the natural logarithm, then for n ≥ 8 or so no computer in the world will ever solve the problem. If log n is base 10 logarithm, then the same is true for n = 100.

If you graph it and conclude that (n ^ 2.93) * ((log n)^93) is always greater than n^3: If log is the natural logarithm, then n^3 is larger when x is around 10^5445.

PS. codesInChaos is apparently not very well versed with the use of little-o. It is very common to use = and not "element of" with big-O / little-o notation, and anything that grows slower than n^3 is indeed o (n^3).

  • Shouldn't your = be a ? – CodesInChaos Mar 17 '16 at 10:58
  • It's common to use = with big O / little o. – gnasher729 Mar 17 '16 at 11:30
  • 1
    Sounds like a horrible idea. Especially in this case, since o(n^2.93 * (log n)^93) ⊂ o(n^3) and thus o(n^2.93 * (log n)^93) ≠ o(n^3). – CodesInChaos Mar 17 '16 at 11:35

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