# Extreme Point

In mathematics, an extreme point of a convex set S \displaystyle S in a real or complex vector space is a point in S \displaystyle S which does not lie in any open line segment joining two points of S . \displaystyle S. In linear programming problems, an extreme point is also called vertex or corner point of S . \displaystyle S. [1]

## extreme point

The perimeter of any convex polygon in the plane is a face of that polygon.[2] The vertices of any convex polygon in the plane R 2 \displaystyle \mathbb R ^2 are the extreme points of that polygon.

A closed convex subset of a topological vector space is called strictly convex if every one of its (topological) boundary points is an extreme point.[6] The unit ball of any Hilbert space is a strictly convex set.[6]

More generally, a point in a convex set S \displaystyle S is k \displaystyle k -extreme if it lies in the interior of a k \displaystyle k -dimensional convex set within S , \displaystyle S, but not a k + 1 \displaystyle k+1 -dimensional convex set within S . \displaystyle S. Thus, an extreme point is also a 0 \displaystyle 0 -extreme point. If S \displaystyle S is a polytope, then the k \displaystyle k -extreme points are exactly the interior points of the k \displaystyle k -dimensional faces of S . \displaystyle S. More generally, for any convex set S , \displaystyle S, the k \displaystyle k -extreme points are partitioned into k \displaystyle k -dimensional open faces.

I am trying to implement a 3D packing algorithm using the extreme point-based approach. The paper which introduced this approach can be seen here: Extreme Point-Based Heuristics for Three-Dimensional Bin Packing

(Note the 'i' at the end instead of the 'k'.)But again, what does it mean that an extremePoint lies on the side of an item? And which side does it mean? Any? Or a specific defined by the given parameter Xy (for example).

It's basically the conglomeration of X,Y,Z's of the current item being packed and the X,Y,Z's of every item already packed (since it's a loop), to get all new appropriate extreme points. Example...Yx is the x:(x + width) of the ith item, y:y + length of the item being packed, z of the item being packed. Xy is the x:(x + width) of the item being packed, y:(y + length) of the ith item, z of the item being packed.

What I am even more confused about is what the function CanTakeProjection does and what it needs the above mentioned symbols (Yx, Yz, ...) for? Also the explanation of the function didn't help me: "CanTakeProjection: function returning true if an EP k lie on the side of an item k" How should the extremePoint k not lie on a side of an item k ever? Or is this a typo and it should be like following: "CanTakeProjection: function returning true if an EP k lie on the side of an item i"

If you actually draw out the extreme points it makes sense. Sometimes an extreme point calculation won't be touching the new packed item at all. So that is not a valid extreme point. So it is correct being 'k'. Early in the paper, it refers to a Corner points algorithm, which tries a new item on each point. The more you pack the slower the algorithm gets. These guys realized that you don't really care about each point, just the furthest out in each axis direction, e.g. extreme point. So the CanTakeProjection is just saying, is the calculated point touching item k...we know it'll be touching item i.

A point in a convex set is called k extreme if and only if it is the interior point of a k-dimensional convex set within S, and it is not an interior point of a (k+1)- dimensional convex set within S. Basically, for a convex set S, k extreme points make k-dimensional open faces.

If you don't know calculus, then this is the way you may understand: notice that quadratic function is symmetrical by its extreme. The quadratic formula tells you the two roots are $x=\frac-b \pm \sqrtb^2-4ac2a$. They are symmetric by the extreme, so the extreme is at $x=-\fracb2a$.

We characterize the set of extreme points of monotonic functions that are either majorized by a given function f or themselves majorize f and show that these extreme points play a crucial role in many economic design problems. Our main results show that each extreme point is uniquely characterized by a countable collection of intervals. Outside these intervals the extreme point equals the original function f and inside the function is constant. Further consistency conditions need to be satisfied pinning down the value of an extreme point in each interval where it is constant. We apply these insights to a varied set of economic problems: equivalence and optimality of mechanisms for auctions and (matching) contests, Bayesian persuasion, optimal delegation, and decision making under uncertainty.

For a feasible solution $x\in\mathbbR^n$ to be an extreme point, there must be at least $n$ bounding hyperplanes of the feasible region that pass through $x$ (meaning at least $n$ constraints, including sign constraints, that are binding at $x$), and $n$ of those hyperplanes must have linearly independent normals. The constraints corresponding to those $n$ hyperplanes are satisfied as equalities (since they are binding) and have rank $n$ (since the constraints intersect only at one point, namely $x$).

Since $y_1 \in P$ and $y_2 \in P$, we get $b_i \ge a_i^Ty_1$ and $b_i \ge a_i^Ty_2$for all $i \in I$ and $a_i^Ty_1 = b_i$ and $a_i^Ty_2 = b_i$ for all $i \in E$.Let $i \in T$. Then $a_i^Tx = b_i \implies (1-\alpha)(b_i - a_i^Ty_1) + \alpha(b_i - a_i^Ty_2) = 0$.Hence, $a_i^Ty_1 = b_i$ and $a_i^Ty_2 = b_i$, so $a_i^T(y_1 - y_2) = 0$.Since this is true for all $i \in T$, we get $B(y_1 - y_2) = 0$.But $\rank(B) = n$, which implies that $y_1 = y_2$.This is a contradiction. Hence, $\xhat$ is an extreme point of $P$.

It is not required to specify the point on the vector when defining the direction.This function is usually used to find the bounding dimension of the body, especially when the body orientation is not aligned with global XYZ coordinate as it is not required to reorient the body to find its best fit bounding box.

The main question is what generality you want. As soon as you have just one strictly convex function in the space, any point where it attains its maximum on $K$ is an extreme point. If we were talking about separable normed spaces, the construction of such function would be trivial: $F(x)=\sum_j 2^-j(1+\x_j\)^-1\x-x_j\$ where $x_j$ is any countable dense set would work just fine. In a strictly convex normed space $F(x)=\x\$ would be an even simpler example. The problem is that you want it in an abstract locally convex topological linear space, so some form of AC seems, indeed, inavoidable.

The beef of the Krein-Milman theorem is the fact that each face of your compact convex set K has an extreme point; the statement about the (closed) convex hull then follows from a swift application of Hahn-Banach. Now notice that a face of K is itself compact and convex. So the difficulty of proving the Krein-Milman theorem is pretty much equal to the difficulty of proving that every compact convex set has an extreme point.

Proposition 1: Let $A\subseteq V$ be a non-empty compact convex set of an hausdorff locally convex semi-topological vector space (over some field which contains the reals topologically). Then $A$ has an extreme point.

Proof (Prop 1): If $A$ is a singleton, we are done. So suppose otherwise. First notice we can separate two points by a convex open set, $U$, yielding $U^-=U\cap A$ a convex proper subset open in $A$. So fix such a set $U^-$.

just suppose that point x is the point in the compact set that has the largest norm (Euclidean norm, L2). Let's now prove that it is actually a extreme point because otherwise we can choose x = 1/2 ( y + z) where y, z are points not identical to x. But because x has the largest norm, one can prove that under condition y and z has norm less or equal than x, but x is the middle point of the segment [y, z], it leads to the conclusion that x = y = z, which contradicts!

Recall from the Extreme Subsets and Extreme Points of a Set in a LCTVS page that if $X$ is a locally convex topological vector space and $K$ is a nonempty convex subset of $X$ then $E \subseteq K$ is said to be an extreme subset of $K$ if:

The Extreme Point bullet starts expanding with a much larger diameter than a traditional bullet, and this results in a quicker and more massive transfer of energy. Thanks to its original design and to its ballistic performances the Winchester Extreme Point delivers a higher knock down power than a traditional 180 gr soft point bullet

This paper presents a degenerate extreme point strategy for active set algorithms which classify linear constraints as either redundant or necessary. The strategy makes use of an efficient method for classifying constraints active at degenerate extreme points. Numerical results indicate that significant savings in the computational effort required to classify the constraints can be achieved.

Theorem 1 below is called the Extreme Value theorem. It describes a condition that ensures a function has both an absolute minimum and an absolute maximum. The theorem is important because it can guide our investigations when we search for absolute extreme values of a function.

This theorem says that a continuous function that is defined on a closed interval must have both an absolute maximum value and an absolute minimum value. It does not address how to find the extreme values.

One of the most useful results of calculus is that the absolute extreme values of a function must come from a list of local extreme values, and those values are easily found using the first derivative of the function.

It is clear from the definitions that for domains consisting of one or more intervals, any absolute extreme point must also be a local extreme point. So, absolute extrema can be found by investigating all local extrema. 041b061a72