Given a set $\textbf y\subset\mathbb R^2$, let $y = (y_1,y_2), y'=(y'_1,y'_2)\in\textbf y$ be elements of that set, let $\alpha_{min}\in\mathbb R$, $\alpha_{min}<1$, $\alpha_{max}\in\mathbb R$, $\alpha_{max}>0$, $\alpha_{min}\leq\alpha_{max}$, and let:
$$\begin{align}y\ I\text{-dominates }y' \text{ iff }&\begin{cases} \forall\alpha\in\{\alpha_{min},\alpha_{max}\}, \alpha y_1 + (1-\alpha)y_2\leq\alpha y'_1 + (1-\alpha)y'_2\\ \exists\alpha\in\{\alpha_{min},\alpha_{max}\}, \alpha y_1 + (1-\alpha)y_2 < \alpha y'_1 + (1-\alpha)y'_2 \end{cases} \\ y\text{ Pareto-dominates }y'\text{ iff } &\begin{cases} y\neq y'\\ y_1\leq y'_1\\ y_2\leq y'_2 \end{cases}\end{align}$$
Call $y\ I$-optimal iff $\nexists y'\in\textbf{y}, y'\ I$-dominates $y$ and call it Pareto-optimal iff $\nexists y'\in\textbf{y}, y'$ Pareto-dominates $y$. Call $NI$ the set of all $I$-optimal elements of $\textbf y$ and $ND$ the set of all Pareto-optimal elements of $\textbf y$. It's clear that $NI\subseteq ND$; to prove that, suppose $y\not\in ND$, which implies that there's some $y'$ that Pareto-dominates $y$, and show that $y'$ also $I$-dominates $y$ and thus $y\not\in NI$, so $y\not\in ND\rightarrow y\not\in NI$ and modus ponens to modus tollens, the theorem is proven.
With this out of the way: I have an algorithm for finding the Pareto-optimal elements of a finite set $\textbf y$ that runs in $O(n\log n)$ where $n = |\textbf y|$. I need to reduce the problem $\Pi$ of finding $I$-optimal elements of that set to a problem $\Pi'$ of finding Pareto-optimal elements of the set, in order to build an algorithm to find the $I$-optimal elements. But I have no clue how to do this! My first instinct would be replacing $(y_1,y_2)$ with $(\alpha y_1, (1-\alpha)y_2)$ and running the same algorithm, but I really don't think this is it.
This is for a school project which can be found here (link in French), and that's really all the information I have and what's asked of me. What do I do?
I figured it out.
For each vector, I was supposed to replace $(y_1, y_2)$ with $(\alpha_{min}y_1 + (1-\alpha_{min})y_2, \alpha_{max}y_1 + (1-\alpha_{max})y_2)$, and then apply the same algorithm.