Number of Cubies problem

[Herbert Kociemba]

Oo, nice find. Wolfram Alpha helpfully points out that that function is equivalent to
\( K(x) = \frac{\sin(\pi x)}{\pi - \pi x} \)

Be careful, equivalent except for K(1), which is the most important function value. Here you get a division by zero error.

You must define K(x)= Sin(pi*x)/(pi - pi*x) for x<>1 and K(x)= 1 for x=1

But in this way you get a case distinction between x=1 and x >1 which is funny if we remember what this thread is all about.

 

[qqwref]

That's true, but the limit at x=1 is 1, which is formal enough for me. In WolframAlpha, if you ask to evaluate that function at x=1, it simply spits back a 1 without even warning you about the denominator. (Incidentally, if gamma(2-x) is undefined for most positive integers, is it still alright to divide by it?)

 

[Herbert Kociemba]

That's true, but the limit at x=1 is 1, which is formal enough for me. In WolframAlpha, if you ask to evaluate that function at x=1, it simply spits back a 1 without even warning you about the denominator. (Incidentally, if gamma(2-x) is undefined for most positive integers, is it still alright to divide by it?)

I think indeed this discussion is very academic, nevertheless an expression like 0/0 is meaningless per se, whereas in the extendend complex plane (see for example http://en.wikipedia.org/wiki/Riemann_sphere) Gamma(2-x) is well defined with Gamma(2-x)=ComplexInfinity for x=2,3,4... and hence 1/Gamma[2-x]=0 is defined too.
But I will not insist on all this, I just wanted to have a bulletproof definition where no one could say that implicitly it includes a case distinction.

 

[m1r0]

I don't know if this has been already resolved, but wouldn't (x^3-[x-1]^3)+([x-1]^3-[x-2]^3)*|[sin(pi y)/(pi - pi y) at y=x]-1| work? (27-8)+(8-1)*|0-1|= 26 for a 3 by 3. And for a 1 by 1, (1-0)+(0-1)*|1-1|=1
I'm not sure that this is correct, because I am slightly basing this off of the post by qqwref. Yes, and I do know this is an old discussion.

Edit: Also, this was kind've thrown together in a minute, not a very elegant situation, just using a little limit problem to detect if x is one, it is not very elegant at all.

 

[AndersB]

Hmm, I think this has been made way more complex then it actually is.

I like to approach the problem like this: For example in a 7x7, when you remove all invisible cubies, you remove a 5x5. That means 7x7x7-5x5x5=the number of cubies.
For all cubes it would look like this, if x is the sidelength: x^3-(x-2)^3 = number of cubies. Am i not right?

This does not apply to the cubes smaller then a 2x2, but i think it does'nt really matter...

 

[vcuber13]

but thats the whole challenge