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I've been learning master theorem in school now and have learnt how to apply it to a number of recurrence relations. However one of my assignments has the following recurrence relation:

T(n) = T(n-2) + n^2

Am I wrong in thinking that I cannot apply Master Theorem to this equation? If it can be applied, I am unsure how, as the variable T(n-2) gives no value for "b" which would usually be used to solve with master theorem.

If it can be solved with Master Theorem, how would it be solved? If it cannot be, how would one go about solving this specific recurrence relation?

Thank you in advance for any help and clarification!

keenns
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    Correct: the master theorem does not apply because the recurrence is not of the form $T(n) = aT(n/b) + f(n)$. For how to solve it, see our reference question. I recommend just expanding a few terms of the sum and seeing what happens. – David Richerby Jun 24 '16 at 23:55

4 Answers4

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Think about what the derivative of T is. If the difference between T (n) and T (n-2) is $n^2$ then the derivative T' (n) is about $n^2 / 2$ and T (n) is about $2/3 · n^3$.

gnasher729
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Don't use the Master theorem for this recurrence. Instead, use guess-and-prove, expand a few times and spot the pattern, or express it as a summation.

Community
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D.W.
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If in the form:

$T(n) = aT(n-b)+f(n) ,\hspace{0.5cm} if \hspace{0.5cm}n>1,\hspace{0.5cm}a>0,\hspace{0.5cm}b>0,\hspace{0.5cm}k>=0$

and if $f(n)$ is $O(n^k)$ then

$T(n)=O(n^k) \hspace{0.5cm} if \hspace{0.5cm} a<1$

$T(n)=O(n^{k+1}) \hspace{0.5cm} if \hspace{0.5cm} a=1$

$T(n)=O(n^k.a^{n/b}) \hspace{0.5cm} if \hspace{0.5cm} a>1$

Alwyn Mathew
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  • Citation or proof needed. 2) Why not more precise solutions? $\Theta$s should be easy, but we really want explicit, precise solutions here.
    • – Raphael Jun 26 '16 at 00:43
  • In expanded form, this may be suitable for this question or even the reference question. – Raphael Jun 26 '16 at 00:45