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I have to find the lower bound of the following recursion:

$A_1 = C_1 = p_1$, $B_1 = D_1 = 1-p_1$, $F_k = A_k + B_k$. Evaluate $F_n$. \begin{align} A_{k+1} &= (A_k + 2C_k) p_{k+1} + (1-p_k) p_{k+1}, \\ B_{k+1} &= (B_k + 2D_k) (1-p_{k+1}) + p_k (1-p_{k+1}), \\ C_{k+1} &= C_k p_{k+1} + (1-p_k) p_{k+1}, \\ D_{k+1} &= D_k (1-p_{k+1}) + p_k (1-p_{k+1}). \end{align}

In this case all numbers of the form $p_i$ are given inputs (there are $n$ of them).

I have to find the lower bound of this algorithm.

Since it's runtime is just $O(n)$ (right?), I know for sure that the maximum possible value of the lower bound is $\Omega(n)$. How would I go about proving/disproving that $\Omega(n)$ truly is the minimum bound and that $\Omega(n-1)$ is not/is attainable?

The only main method I know for proving stuff like this is an Adversary Argument but after a couple hours of thinking, I still haven't been able to come up with an argument to prove what I want.

Any helps/tips/suggestions would be greatly appreciated.

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There's quite a lot of confusion here:

  1. $\Omega(n)$ and $\Omega(n-1)$ are absolutely the same.
  2. You don't an adversary argument to show that this algorithm runs in $\Theta(n)$. It follows immediately from the description of the argument.
  3. There is a difference between analyzing the running time of a particular algorithm and showing that every algorithm that correctly solves a problem must run in at least so many steps.

In order to show that any algorithm computing $F_n$ correctly must run in time $\Omega(n)$, we will show that any such algorithm must read all inputs $p_1,\ldots,p_n$. The starting point is the following observation:

The correct output is different for the instances $p_1=\cdots=p_n=0$ and $p_1=\cdots=p_{i-1}=0,p_i=1,p_{i+1}=\cdots=p_n=0$.

Given that, we can use a simple adversary argument to show that any algorithm correctly computing the function must read all $p_j$. Indeed, run the given algorithm, answering $0$ whenever it queries the value of some $p_j$. If it announces the answer without querying, say, $p_i$, then it cannot tell apart the two instances above, and in particular, its answer will be wrong on at least one of them.

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  • $\begingroup$ I have a couple of questions: 1. why does proving that the algorithm needs all $n$ inputs mean that it runs in $\Omega(n)$ time? 2. The problem asks "Give the best lower bounds you can on a runtime of such an algorithm." Would that mean for all of the possible algorithms, I need to come up with the shortest one out of all of them? 3. If it asks for the best possible lower bounds, would arguing a lower bound of say $\Omega{5n}$ be considered as "better" than $\Omega{n}$ since technically, it is a more tight bound? $\endgroup$ – Arjun the Mathematician Nov 20 at 15:03
  • $\begingroup$ I'm not sure you have a very good grasp of the meaning of "$\Omega(n)$". We have several reference questions covering asymptotic notations. You can start here. $\endgroup$ – Yuval Filmus Nov 20 at 15:34

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