naive convex hull algorithms

I already understood how the well-known algorithms like Graham, Quickhull etc. work, but i have difficulties in understanding 2 naive versions of convex hull algorithms:

Let S={p1, ..., pn} a set of points and CH(S) the convex hull of S.


("any_sentence" is a quote from different course books.)

A1) "find the smallest subset R ⊂ S for CH(R) = CH(S). Since there are 2n subsets, the complexity is O(2n)."
Well my question is: since we don't know CH(S) - and indeed it's our goal to achieve CH(S) - , how can we check if CH(R) == CH(S)? This makes no sense. Pseudocode version:

foreach (s in S.subsets)
if (CH(R) == CH(S))
return R


A2) "Let p,q,r,s ∈ S."
"s ∈ CH(S) if s is NOT ∈ triangle(p,q,r)"
And here is the algorithm:

My question: Is the first part of the sentence not wrong? According to my opinion, it should be: Let p,q,r ∈ CH(S) and s ∈ S. (But we would have the same problem again, i.e. that we don't know at the beginning what CH(S) is.) My counter example why this algorithm wouldn't work:

at the start we check the triangle(1,2,3) and see that 4 is not in the triangle, and mark 4 as not an element of convex hull, which is FALSE.

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A1) You're right, the algorithm description sounds kind of weird. A naive algorithm that takes exponential time would rather be something like

foreach (R in S.subsets)
if (R.isConvexHull(S))
return R


However, the method for checking whether R is a convex hull needs $n$ time, so we obtain a runtime of $n2^n$.

A2) Your premise is missing some quantifiers. I think, what you wanted to say is

"$\forall s,$ if $\forall p,q,r \in S, s \not \in \triangle(p,q,r),$ then $s$ is on the convex hull."

Furthermore, go through your example again. You understood it wrong. If you check triangle (1,2,3), you see that 4 is not in the triangle. Hence, 4 might still be on the convex hull (but it doesn't have to). On the contrary, if you check the triangle (4,3,2), you see that 1 is on the interior. Therefore, 1 is definitely not on the convex hull.

This still does not give you a convex hull algorithm though, since the points still have to be sorted by the end. However, this is no problem, since the rest of the algorithm already needs $\mathcal{O}(n^4)$ time.

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