Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I think I know how to do this problem but just want to check that my reasoning is correct. "Let $U$ denote the unit cube in $\mathbb{R}^n$, and let $D$ be a given diagonal of $U$. How many other diagonals of $U$ are perpendicular to $D$?" I am unsure how to proceed with the 1 and 0 dimensional case, but in 2 dimensions that answer is 1. I conjecture that for an n-dimensional cube, if the diagonal traverses $m_o$, an odd number of dimensions, the number is zero, and if it traverses $m_e$, an even number of dimensions the number is the binomial coefficient $\binom{m_e}{m_e/2}$. We write the m dimensional diagonal as $(d_1,d_2,...d_m)$, with $d_i=\pm1$. By rotation, all diagonals are similar to $(1,1,..1)$ m times, so we need only find the solution for that particular diagonal. We look for another m dimensional vector such that its dot product with $(1,1,...1)$ is zero. For m odd, this is clearly impossible. For m even, half of the coordinates must be negative 1's and the other positive ones, so you have $\binom{m_e}{m_e/2}$ options. Does this reasoning make sense? Also, what should I do for the 0 and 1 dimensional cases? Sorry, wasn't sure what to tag this as either. Thanks.

share|improve this question
    
You can get $\binom nm$ using \binom nm, without the spurious vinculum that you get with \frac. –  joriki Aug 7 '11 at 19:06
    
You can also get ${n \choose m}$ using {n \choose m}. –  anon Aug 7 '11 at 19:41
    
I think you should clarify your question OP: of an $n$-dimensional hypercube and a diagonal of one of its $m$-dimensional hypercells, you're looking for the number of diagonals of likewise $m$-dimensional hypercells that are orthogonal with the given one. If that's the case, your reasoning is sound (I understand it clearly). A $0$-dimensional "diagonal" is just the point $\vec{0}$, so this problem is ill-defined there. A $1$-dimensional diagonal is an edge, and the # of orthogonal edges equals the number of edges of a $(n-1)$-dimensional hypercube. –  anon Aug 7 '11 at 19:55
    
You have to divide your answer by 2 since the negative of a diagonal vector give a vector in the opposite direction that occupies the same line as the original vector. For example, (-1,1) and (1,-1) are the same diagonal in a square. –  neelp Aug 13 '12 at 21:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.