Quote:

Original post by Airo proposition 1, just means if i interpret correctly, that if you have the largest distance between two points on the convex hull, then this distance is the diameter of the encompassing sphere.

Quote:

proposition 2 specific, means that if you have the path like you describe, r is a maximum because it is the middle point on the path. If there would be another maximum, the convex hull would be different going through the other local maximum, this is a contradiction

Quote:

proposition 2 general, looks like some shift in point positions, but seems logical for a convex path.

Quote:

Original post by alvaro Quote: Original post by Airo proposition 1, just means if i interpret correctly, that if you have the largest distance between two points on the convex hull, then this distance is the diameter of the encompassing sphere. That's not what he said. The diameter of S* is by definition the lowest upper bound of the distances between two points of S*. If S* is compact (and it is if S is finite), then the maximum and the lowest upper bound are the same thing, so property one is guaranteed. Quote: proposition 2 specific, means that if you have the path like you describe, r is a maximum because it is the middle point on the path. If there would be another maximum, the convex hull would be different going through the other local maximum, this is a contradiction A maximum? A maximum of what? Am I missing something? Quote: proposition 2 general, looks like some shift in point positions, but seems logical for a convex path. Well, I guess I am confused by previous errors, like some compilers might say, but I am not even sure of what we are talking about. Is he trying to see if there are faster algorithms for computing the convex hull of sets S that satisfy some extra conditions? Or what?

Quote:

However, I'm having some trouble deciding which of these assumptions is true.

Quote:

[Proposition 1] If a point p in S* is furthest from q in S* and vice versa, then the segment pq forms the diameter of S*.

Quote:

[Proposition 2: Specific] Suppose points p, q, r are all in S*. If prq represents points on a boundary path along S*, and q is the first local maximum for p on the boundary path, then none of the points other than q along that path can be a local maximum for r.

Quote:

[Proposition 2: General] Suppose points p, q, r, s are all in S*. If prsq represents points on a boundary path along S* such that d[p,s] <= d[p,q], then d[r, s] <= d[r, q].

Quote:

Original post by kSquared [Proposition 2: Specific] Suppose points p, q, r are all in S*. If prq represents points on a boundary path along S*, and q is the first local maximum for p on the boundary path, then none of the points other than q along that path can be a local maximum for r.

 _________q_________
/                   \k          .
\_________p_______r_/           .

Quote:

[Proposition 2: General] Suppose points p, q, r, s are all in S*. If prsq represents points on a boundary path along S* such that d[p,s] <= d[p,q], then d[r, s] <= d[r, q].

Quote:

Original post by Dmytry Quote: Original post by kSquared [Proposition 2: Specific] Suppose points p, q, r are all in S*. If prq represents points on a boundary path along S*, and q is the first local maximum for p on the boundary path, then none of the points other than q along that path can be a local maximum for r. If i understand right, counterexample is: Draw such shape: _________q_________ / \k . \_________p_______r_/ . and k is placed on "\"

Quote:

You can place r and k that k is local maximum to r. Yet first (and only) local maximum to p is q.

Quote:

So i disproved 2nd and proved 3rd proposition...

Quote:

Original post by kSquared Quote: Original post by Dmytry Quote: Original post by kSquared [Proposition 2: Specific] Suppose points p, q, r are all in S*. If prq represents points on a boundary path along S*, and q is the first local maximum for p on the boundary path, then none of the points other than q along that path can be a local maximum for r. If i understand right, counterexample is: Draw such shape: _________q_________ / \k . \_________p_______r_/ . and k is placed on "\" I think you might have misread the statement. To disprove it, you'd have to show that when q is the local maximum for p, there exists a point other than q on the path which is the local maximum for r. Quote: You can place r and k that k is local maximum to r. Yet first (and only) local maximum to p is q. Umm... q is looking awfully like the local maximum for r in your diagram, not k. If anything, k looks like the local minimum!

Quote:

Quote: So i disproved 2nd and proved 3rd proposition... Also, since the third statement is a more general case of the second statement, it is impossible to prove the third statement without also proving the second. So I'm a little wary of your "proof" right now even though I can't see anything wrong with it at first glance.

Quote:

Original post by Dmytry ok, with second i was wrong because i misread it. But third proof is correct.

Quote:

Original post by kSquared Quote: Original post by Dmytry ok, with second i was wrong because i misread it. But third proof is correct. Ahh, okay, that cleared it up a bit. I was confused for a second. Now that you've proved the general one, you should be able to see why the second statement is a specific case, I think. Thanks for the help!