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'm having troubles with searching for analytical solution of following problem. Let we work in 3-D space and have the set of points (uniform net at cube's facets):

($-1,\hspace{2mm} -1+j*h,\hspace{2mm} -1+k*h$),

($1,\hspace{2mm} -1+j*h,\hspace{2mm} -1+k*h$),

($-1+i*h,\hspace{2mm} -1,\hspace{2mm} -1+k*h$),

($-1+i*h,\hspace{2mm} 1,\hspace{2mm} -1+k*h$),

($-1+i*h,\hspace{2mm} -1+j*h,\hspace{2mm} -1$),

($-1+i*h,\hspace{2mm} -1+j*h,\hspace{2mm} 1$),

where $h$ is const, $i = 0,1,2,...,2/h, \hspace{2mm} j = 0,1,2,...,2/h,\hspace{2mm} k = 0,1,2,...,2/h$.

Let's pass in review the foolowing vectors: $(0,0,0)$ is the beginning of each vector and the end of the vector is points that was written above.

It's necessary to find formula for mimimal positive (nonnegative) value of cosine of the angle between two vectors $(a,b,c)$ and $(d,e,f)$, where $|a-d| \leq h, |b-e| \leq h, |c-f| \leq h$ (for example, $a$ can be $d$, $d+h$ or $d-h$ only).

Const $h$ is such than $2/h$ is integer always.

And I repeat again, I'm interested in analytical solution.

share|improve this question
    
You can factor $h$ out of $f$ entirely. You can write $f$ as $\frac{<(i,j,k),(a,b,c)>}{||(i,j,k)|| \; ||(a,b,c)||}$. So it comes down to minimizing the cosine of the angle between $(i,j,k)$ and $(a,b,c)$. –  copper.hat May 8 '12 at 16:40
    
Of course, $f$ is not defined at $0$. I would suspect that choosing something like $(i,j,k) = (1,0,0)$ and $(a,b,c) = (0,1,0)$ would give you the minimum which I suspect is $0$. –  copper.hat May 8 '12 at 16:56
add comment

2 Answers

  1. You should not change the problem statement so much that the existing answer no longer connects to it.

  2. The problem boils down to minimizing the angle between $\mathbf v$ and $\mathbf v+\mathbf w$ where $\mathbf w$ is restricted to have coordinates $\pm h$ or $0$. On geometric grounds, we should seek the shortest $\mathbf v$ and longest $\mathbf w$. Such as: $\mathbf v = (1,0,0)$ and $\mathbf w = (-h,h,h)$.

share|improve this answer
add comment

The analytical solution is $0$.

You can factor $h$ out of the $f$ to get $$f((i,j,k),(a,b,c)) = \frac{<(i,j,k),(a,b,c)>}{||(i,j,k)|| \; ||(a,b,c)||}$$ Since all of $(i,j,k),(a,b,c)$ are non-negative, then $f \geq 0$ also.

Since $f((1,0,0), (0,1,0)) = 0$, this is a solution to the problem.

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.