I have some questions about Strong Induction where the inductive procedure isn't entirely clear to me. I will use a specific example to demonstrate and present my attempt at a proof with questions following.

Suppose we have a recurrence relation $T(n)$ defined as follows \begin{align*} T(0) &= 0 \\ T(n) &= T\left(\left\lfloor n/3\right\rfloor\right) + T\left(\left\lfloor n/5\right\rfloor\right) + T\left(\left\lfloor n/7\right\rfloor\right) + n, \enspace n > 0 \end{align*}

Use Strong Induction to prove that $T(n) \in O(n)$.


$\underline{\text{Base Case}}$: By definition of BigOh, we want to show that $\exists c > 0 \enspace \text{s.t.} \enspace T(n) \le cn, \enspace \forall n \ge n_0$. Since $T(0) = 0$, this holds for any choice of $c$ with $n=0$ and $n_0 = 0$.

$\underline{\text{Inductive Hypothesis (I.H.)}}$: Suppose there exists such a $c$ for all $0 \le n < k$.

$\underline{\text{Inductive Step}}$: We need to show that $T(n) \in O(n)$ for $n = k$.

Since $\lfloor k/3 \rfloor,\lfloor k/5 \rfloor,\lfloor k/7 \rfloor < k$, it follows by the I.H. that $T(\lfloor k/3 \rfloor) \le ck/3$, $T(\lfloor k/5 \rfloor) \le ck/5$, and $T(\lfloor k/7 \rfloor) \le ck/7$.

Then using the definition of $T(n)$ we can say that \begin{align*} T(k) &\le \frac{ck}{3} + \frac{ck}{5} + \frac{ck}{7} + k\\ &\le ck\left(\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{c}\right) \end{align*} To prove that $T(n) \in O(n)$ for $n=k$, we need to show that this quantity is less than $ck$ as follows \begin{align*} ck\left(\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{c}\right) &\le ck \\ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{c} &\le 1 \\ \frac{71}{105} + \frac{1}{c} &\le 1 \\ \frac{1}{c} &\le \frac{34}{105} \\ c &\le \frac{105}{34} \end{align*}

So for any $c \ge 105/34$, $T(n) \in O(n)$ for $n = k$, and thus $T(n) \in O(n)$.


  1. What is the particular reason that strong induction is necessary for this proof? Is it because $T(n)$ is defined in terms of three previous recursive instances with floor functions that don't follow the same $+1$ ordering as the natural numbers? Or does it relate to proving asymptotic complexity where a claim $P(n)$ involves $\forall n \ge n_0$ for each $n$?
  2. Is my base case sufficient? Since BigOh typically applies to asymptotic limits, I have trouble understanding how to prove BigOh for a base case when $n$ takes on a specific value.
  3. Are there any issues with my attempted proof? Don't hold back, I would appreciate some formal criticism.
  • 1
    $\begingroup$ I understand that the last inequality is with $\ge$, isn't it? $\endgroup$
    – ajotatxe
    Jul 19 '15 at 18:56
  • $\begingroup$ @ajotatxe, yes, sorry for the confusion. I am not (intentionally) asserting that the steps I used to find c ≤ 105/34 prove that c ≥ 105/34. I'm using the value I found as a choice of c that works for the BigOh notation, and so anything greater should also work. $\endgroup$ Jul 19 '15 at 20:20

Strong induction is necessary for this proof because weak induction isn't enough to solve the problem. Specifically, weak induction looks at $T(n-1)$, but you need to plug different values into $T$, so you need to assume the claim is true for more cases.

The base case is enough, you showed that $T(0)\leq c\cdot 0$.

Your attempted proof is backwards (it's more like scratch work). In your "proof," you found that $c\geq\frac{105}{34}$ (YOU FORGOT TO FLIP THE INEQUALITY). Your work helped you to figure out what $c$ should be. Now, you have to go back and start with an appropriate $c$ value and prove that that particular value for $c$ works. (You could round $c$ up to the nearest integer).

Therefore, start the entire proof over using $c=\frac{105}{34}$ (or any integer greater than this quotient) and prove that for the base case $T(0)\leq c\cdot 0$. Also, prove that $T(n)\leq cn$ by doing your work in the opposite direction.

  • $\begingroup$ Can you elaborate on your first point? Specifically, can you address the ordering difference between weak induction with T(n-1) vs. strong induction with T(n/3) + T(n/3) + T(n/7)? Can you also address if it matters that the claim to prove involves a range of values for n greater than a minimum n0 for each choice of n? $\endgroup$ Jul 19 '15 at 19:57
  • $\begingroup$ The difference in the types of induction are in the assumptions. Weak induction assumes the truth of one smaller case, while strong induction assumes the truth of all smaller cases. $\endgroup$ Jul 19 '15 at 20:58
  • $\begingroup$ When you invert fractions, the inequality reverses. For example, $2\leq 3$, but $\frac12\geq\frac13$. Take $2\leq3$, and divide through by $2\cdot 3$. $\endgroup$ Jul 19 '15 at 20:59

The "danger" in this exercise is in writing the proof in such a way that $c$ depends on $n$. But you haven't done this. I think that it is ok.

But to make things clearer, you can stablish from the beginning a value for $c$, say $c=4$. Your original work is useful as a draft, to find this $c$.


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