The exercise is

Given a set of point $S$ and a point $p$. Decide in $O(n)$ time if $p$ is a vertex of convex polygon formed from points of $S$.

The problem is I am a little bit confused with time complexity $O(n)$. The more naive solution would be to construct convex polygon in $O(n\log n)$ and test if $p$ is one of the vertices.

  • $\begingroup$ That problem statement is wrong. Given $p$ and $S$, it suffices to pick two points from $S$ to form a triangle, which is a convex polygon. This is $O(1)$. $\endgroup$
    – user16034
    Apr 6, 2021 at 19:31

3 Answers 3


Hint: the point $p$ is a vertex of the convex hall iff there are two half-lines from it such that all points fall inside the angle created by them.

Here is an idea for an algorithm based on this hint:

Design an algorithm with two variables $q$ and $r$ (input points). The algorithm will examine each of the input points and will update $q$ and $r$ so that all points checked so far fall inside the wedge $\angle qpr$.

  • $\begingroup$ @JeffE, is there a mistake in my idea? You can see it if you hover the mouse over the part below the second paragraph. $\endgroup$
    – Kaveh
    May 20, 2012 at 18:42
  • $\begingroup$ I figured it out. (I had a different solution in mind.) $\endgroup$
    – JeffE
    May 20, 2012 at 18:43
  • $\begingroup$ @JeffE, I should have written an in place of the, will fix it in a second. :) $\endgroup$
    – Kaveh
    May 20, 2012 at 18:46
  • 2
    $\begingroup$ @JeffE, I am curious, what was your idea? $\endgroup$
    – com
    May 25, 2012 at 5:49

The problem is to find a line that passes through $p$ and has all the other points in $S$ on one side. This is a two-dimensional linear-programming problem, so it can be solved in $O(n)$ time using textbook geometric algorithms. But let me describe a self-contained solution.

To simplify notation, translate all the points so that $p$ is the origin $(0,0)$, and let $Q = S\setminus\{p\}$. Then we want to determine if there is a real number $m$ such that either (1) $y < mx$ for all $(x,y)\in Q$ or (2) $y > mx$ for all $(x,y)\in Q$. In the first case, $p$ is a vertex of the upper hull of $S$; in the second case, $p$ is a vertex of the lower hull of $S$. I'll describe an algorithm for the first case; the other case is symmetric.

If any point in $Q$ lies directly above $p$ (that is, if any point in $Q$ has coordinates $(0,y)$ for some $y>0$), then $p$ cannot lie on the upper hull. It is easy to check this condition in $O(n)$ time.

So assume no points in $Q$ lie directly above $p$. The $y$-axis splits $Q$ into two subsets $L$ (left) and $R$ (right). Points in $L$ have negative $x$-coordinates, and points in $R$ has positive $x$-coordinates. (Points directly below $p$ don't matter; just ignore them.) Let $$ m_L = \min_{(x,y)\in L} \frac{y}{x}, \quad M_R = \max_{(x,y)\in R} \frac{y}{x}, \quad \text{and} \quad m = \frac{m_L + M_R}{2}. $$ Now there are three cases to consider:

  • If $m_L < M_R$, then every point in $Q$ lies strictly below the line $y = mx$, so $p$ is a vertex of the upper hull.

  • If $m_L = M_R$, then the line $y=mx$ passes through a point in $L$ and a point in $R$, and no point in $Q$ is strictly above that line. So $p$ lies on an edge of the upper hull, but it is not a vertex.

  • If $m_L > M_R$, then at least one point in $L$ and at least one point in $R$ lie strictly above the line $y=mx$. So $p$ lies strictly below the upper hull.

It is easy to compute $m_L$ and $M_R$ in $O(n)$ time. We don't actually need to compute $m$.


I think that the accepted answer is too sketchy, and the answer by Jeff Erickson is too complex - so my two cents are below.

For all points $q \in S$ calculate an oriented angle between the basis vector $e_x=(1,0)$ (or any other non-null vector - it's your choice) and the vector $\vec {pq}$. These angles will be in the $[0, 2\pi)$ semi-closed interval. Find minimum $\alpha_{min}$ and maximum $\alpha_{max}$ for all such angles. The point $p$ belongs to the convex hull of the set $S$ if and only if the difference $\alpha_{max}-\alpha_{min} < \pi$. Obviously all that can be done in $O(n)$ time.


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