Wikipedia:Reference desk/Archives/Mathematics/2018 September 8

= September 8 =

Hyperrectangle contains most of the cube, but not its vertices
Let $$n\in\mathbb N$$, and let $$p, q \in (0,1)$$, such that $$p<q$$.

Is there a hyperrectangle H satisfies:

1. $$\forall i\in\{1,\dots,n\}: H\supset \prod_{j=1,\dots,n} \left\{\begin{matrix} [p, q] & \text{if}~j=i\\{} [0,1] & \text{else} \end{matrix}\right.

$$

2. $$\{0,1\}^n\cap H=\emptyset$$

In 2D I was able to find such a hyperrectangle, but how about 3D and higher dimensions? Is there such a hyperrecangle? — Preceding unsigned comment added by 2.53.39.115 (talk) 07:45, 8 September 2018 (UTC)
 * First, the algebra is a bit simpler if you change the interval from [0, 1] to [-1, 1], just scale and translate accordingly. That said, I'm pretty sure that for n=3 the answer is no for any p and q. It is sufficient to show that if a rectangular cuboid H contains points (p, ±1, ±1), (±1, p, ±1), (±1, ±1, p), with -1<p<1 then H contains one of (±1, ±1, ±1). To prove this, let H be defined by
 * Lu ≤ u1x+u2y+u3z ≤ Mu
 * Lv ≤ v1x+v2y+v3z ≤ Mv
 * Lw ≤ w1x+w2y+w3z ≤ Mw
 * where u = (u1, u2, u3), v = (v1, v2, v3), w = (w1, w2, w3) are mutually orthogonal, non-zero vectors. Let
 * U = {(sign(u1), sign(u2), sign(u3)), (-sign(u1), -sign(u2), -sign(u3))} if all the ui are nonzero,
 * and
 * U = ∅ otherwise.
 * Similarly define V and W. The sets U, V and W have at most 2 elements each, and {(±1, ±1, ±1)} has eight elements, so the difference {(±1, ±1, ±1)}\(U∪V∪W) is non-empty, say
 * s = (s1, s2, s3)
 * is an element. We have s is one of the vertices of the cube, and I claim that s satisfies each of the three inequalities above. I'll prove the first inequality with the other two being similar. There are two possibilities, either some ui=0, or there is i so si and ui have the same sign, and j so that sj and uj have opposite signs. In the first case, take i=1, i=2 and i=3 being similar. Then u1=0 and note that
 * (p, s2, s3) ∈ H,
 * so
 * Lu ≤ u1p+u2s2+u3s3 ≤ Mu.
 * But u1=0 so
 * u1p+u2s2+u3s3 = u1s1+u2s2+u3s3
 * and
 * Lu ≤ u1s1+u2s2+u3s3 ≤ Mu.
 * For the second case, suppose u1 and s1 have the same sign, and u2 and s2 have opposite signs, the remaining possibilities being similar. Note that
 * (p, s2, s3) ∈ H
 * so
 * Lu ≤ u1p+u2s2+u3s3.
 * Also |p| < 1 so
 * u1p ≤ |u1p| < |u1| = u1s1
 * and therefore
 * Lu ≤ u1s1+u2s2+u3s3.
 * Also note that
 * (s1, p, s3) ∈ H
 * so
 * u1s1+u2p+u3s3 ≤ Mu.
 * Again |p|<1 so
 * -u2p ≤ |u2p| < |u2| = -u2s2
 * and
 * u2p ≥ u2s2
 * and therefore
 * u1s1+u2u2+u3s3 ≤ Mu.
 * This proves both parts of the first inequality, and as noted the other inequalities are similarly true, therefore
 * (s1, s2, s3) ∈ H.
 * This argument is rather fiddly and there were a couple previous versions where I discovered fatal flaws while writing them up; hopefully this version is correct. Note that I don't seem to use anywhere that u, v and w are orthogonal, so it looks like it remains true for parallelepipeds in general. The same argument works for n>3, the crucial fact needed being 2n<2n. So n=2 is actually the exceptional case. --RDBury (talk) 09:17, 9 September 2018 (UTC)


 * Actually there is a much more general theorem provided you apply some machinery of convex polytopes. Namely:
 * If P is a bounded convex polytope, Q a convex polytope which contains at least one point from every edge of P, and Q has fewer faces than P has vertices, then Q contains a vertex of P.
 * Proof: Suppose not. Every vertex of P is then excluded by some face of Q, in other words for each vertex v of Q there is a face of P, defined by L(x)=0 say, so that L(x)≤0 for x in Q and L(v)>0. There are more vertices of P than faces of Q, so two of the vertices must be excluded by the same face. In other words there is a linear function L so that L(x)≤0 for x in Q, and vertices v and w of P so that L(v)>0 and L(w)>0. If v and w are connected by and edge e, then L(y)>0 for all y in e, but Q contains a point of every edge of P and we have a contradiction. If L(v) is not the maximum value of L on P then apply the simplex algorithm to obtain an adjacent vertex u with L(u)>L(v)>0. This implies a contradiction as before, so L(v)=M, the maximum value of L on P. Similarly, L(w)=M. Then P∩{x: L(x)=M} is a polytope with at least two vertices, and therefore has an edge, which in fact is also an edge of P, again leading to a contradiction. There is a contradiction in every case so the initial assumption is incorrect and the theorem is true.
 * In the present case P is the hypercube with 2n vertices, and Q is a parallelepiped with 2n<2n faces. The simplest case is that if a triangle Q contains a point from every edge of a convex quadrilateral P, then Q must contain a vertex of P. This might be proven with just high school geometry but it would be a challenge. --RDBury (talk) 23:18, 9 September 2018 (UTC)

I like this theorem very much! Thank you! 2.53.159.234 (talk) —Preceding undated comment added 06:04, 11 September 2018 (UTC)


 * Notice however that a hyperrectangle (or even just a 2D-plane) that touches each face but doesn't contain any vertex does exist: After changing the interval from [0, 1] to [-1, 1] as RDBury sugggested above, in order to touch every face of the cube, you just need to take some hyperrectangele that contains the points $$\{(0,1,\dots,1),(0,-1,\dots,-1),(1,\dots,1,0),(-1,\dots,-1,0)\}$$. Actually, these points are linearly depndent, so you just need to to contain the points $$\{(0,1,\dots,1),(1,\dots,1,0)\}$$.
 * Hope this helps. עברית (talk) 17:42, 15 September 2018 (UTC)