$$T(N) = \begin{cases}1 & \text{if } N = 1\\ T(\varphi(N)) + \lg(\varphi(N))^3 & \text{otherwise} \end{cases}$$

where $\varphi(N)$ is Euler's totient function.

Can I somehow express $\varphi(N)$ as $N/b$, so I can apply the Master Theorem and resolve this recurrence?

You may assume $\varphi(N) = (p-1)(q-1)$, if it's easier that way. You may also assume, if it helps, that $p$, $q$ are safe primes, that is, $p = 2p' + 1$ and $q = 2q' + 1$. (Assume anything that makes the problem easier. For instance, you can replace the function $\lg^3(\varphi(N))$ with any other that makes the problem easier, but do so only as a last resort.)


1 Answer 1


You can not apply the master theorem directly. However, you can play with your expression a bit to get an upper bound on which you can then apply the master theorem.

First, show that $\phi(\phi(n)) < n/2$. This can be done as such:

Let $n = \prod_{i=1}^rp_i^{k_i}$ be the prime factorisation of $n$ ($p_i$ prime, $k_i>0$)

  • Suppose $n$ is even. Then $\phi(n) = n\prod_{i=1}^r(1-\frac{1}{p_i}) \leq n(1-\frac{1}{2}) \leq n/2.$ Thus $\phi(\phi(n)) < n/2$.
  • Suppose $n$ is odd and $n > 1$. Then $\phi(n) = \prod_{i=1}^r (p_i-1)p_i^{k_i-1}$ is even and smaller than $n$. By the previous result $\phi(\phi(n)) < n/2$.

So we get the desired result.

Now suppose $n\geq2$. You can write: $$T(n) = T(\phi(n))+\log(\phi(n))^3 = T(\phi(\phi(n)))+\log(\phi(\phi(n)))^3 + \log(\phi(n))^3.$$ $$T(n) \leq T(n/2) + 2\log(n)^3.$$

Now you can apply the master theorem here to get: $$T(n) = O(log(n)^4).$$

  • $\begingroup$ Are you sure the answer is $O(\log^4 n)$? Looking at Cormen's book, I understood $a = 2$, $b = 2$ and case 1 applies because $2 \log^3 n \in O(n^{(\log_2 2) - e}) = O(n)$, so the answer would be $\Theta(n^{\log_2 2}) = \Theta(n)$. Where am I wrong? How did you get the 4th power? $\endgroup$
    – R. Chopin
    Aug 25, 2019 at 19:04
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    $\begingroup$ You have a=1 and not 2, so you can use case 2 of the more precise formulation given here : en.wikipedia.org/wiki/Master_theorem_(analysis_of_algorithms). Alternatively, you can just do a prove by induction without using the master theorem, it will yield the same result quite easily. $\endgroup$
    – Tassle
    Aug 25, 2019 at 20:03
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    $\begingroup$ Notice also that because we work on an upper bound of $T$ and not the exact formula, we can no longer deduce lower bounds on the complexity via the master theorem, thus we can only get $T(n)=O(\text{something})$ and no longer $T(n)=\Theta(\text{something})$. $\endgroup$
    – Tassle
    Aug 25, 2019 at 20:09
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    $\begingroup$ Oh I see, no worries :) And yep, $O(n)$ for $a=2$ seems right to me. $\endgroup$
    – Tassle
    Aug 25, 2019 at 21:28
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    $\begingroup$ Oh I'm sorry, I just realized that my upper bound doesn't work out that way any more if $a = 2$. In fact, the new upper bound would be $T(n) \leq 4T(n/2) + ...$, which leads to $O(n^2)$. But regarding your point, yes, in the first case of the master theorem growth can be arbitrarily fast in $F(n)$. Try it out with $F(n) = 0$ and you'll get, hum, infinite growth in $F(n)$, whatever that means $\endgroup$
    – Tassle
    Aug 25, 2019 at 22:14

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