This is inspired by a recent question.

Given a positive integer $n\in\mathbb{N}$, is there a setting of finitely many points and a designated "starting point" $s$ in $\mathbb{R}^2$ such that the nearest-neighbor algorithm (described below) gives a tour that is $n$ times longer than the optimal solution starting at $s$?

Starting at $s$, pick the nearest neighbor not visited so far as the next node to visit.

EDIT: If the answer is no, what is the maximum value that the ratio $r$ of "nearest neighbor trip" vs "best trip" can take?

This is inspired by a recent question.

Given a positive integer $n\in\mathbb{N}$, is there a setting of finitely many points and a designated "starting point" $s$ in $\mathbb{R}^2$ such that the nearest-neighbor algorithm (described below) gives a tour that is $n$ times longer than the optimal solution starting at $s$?

Starting at $s$, pick the nearest neighbor not visited so far as the next node to visit.

EDIT: If the answer is no, what is the maximum value that the ratio $r$ of "nearest neighbor trip" vs "best trip" can take?

This is inspired by a recent question.

Given a positive integer $n\in\mathbb{N}$, is there a setting of finitely many points and a designated "starting point" $s$ in $\mathbb{R}^2$ such that the nearest-neighbor algorithm (described below) gives a tour that is $n$ times longer than the optimal solution starting $s$?

Starting at $s$, pick the nearest neighbor not visited so far as the next node to visit.

This is inspired by a recent question.

Given a positive integer $n\in\mathbb{N}$, is there a setting of finitely many points and a designated "starting point" $s$ in $\mathbb{R}^2$ such that the nearest-neighbor algorithm (described below) gives a tour that is $n$ times longer than the optimal solution starting at $s$?

Starting at $s$, pick the nearest neighbor not visited so far as the next node to visit.

EDIT: If the answer is no, what is the maximum value that the ratio $r$ of "nearest neighbor trip" vs "best trip" can take?