http://www.4cornersmedia.co.uk/hexagonalgeometry/newspherepacking.html

https://www.reddit.com/r/math/comments/72rkvk/calling_all_mathematicians_a_new_way_to_stack/

]]>But thinking through it was great! Thank you for your analysis.

Here was my writeup

—-

After the first round, every chick has a 1/4 probability to be unpecked. (1/2^2)

Of the unpecked chicks,

1/2 are immortal singletons, and 1/2 are in pairs.

So there is a

1/4 * 1/2 = 1/8 chance of immediately becoming a singleton unpecked chicken and therefore immortal, plus a 1/8th chance of ending up in a pair of unpecked chicks. (And a 6/8th chance of being pecked).

For each unpacked pair, there are four outcomes:

1/4: a pecks b

1/4: b pecks a

1/4: a pecks b and b pecks a

1/4: neither pecks the other.

Let the number of immortal chicks coming from a pair = u. Then,

u = 1/2 + 1/4u

3/4u = 1/2

u = 2/3

From above, the expected fraction of *chicks in pairs* is 1/8, so the fraction of *pairs* is 1/16 of the total chick count.

1/8+ 1/16(2/3) ~= .1667

]]>In the statement of the Hurwitz theorem, should it be \((\sqrt{5}b^2)^{-1}\)? ]]>

If a, b are relatively prime positive integers, and if \(|bx-a|<|dx-c|\) for all pairs c, d of relatively prime positive integers satisfying \(d\le b\) and \(c/d\ne a/b\), then a/b is called a best approximation of the 2nd kind (BA2) to x. Every BA2 to x is a BA1 to x, but not conversely. A fraction is a BA2 to x if and only if it is a convergent of the continued fraction for x.

A theorem of Hurwitz states that for every real irrational x there are infinitely many rational approximations a/b with \(|bx-a|<(\sqrt5 b)^{-1}\). So if you use \(|bx-a|^{-1}\) as a measure of the merit of the approximation a/b, you'll find that every real irrational has a sequence of approximations of unbounded merit. If you use \((b|bx-a|)^{-1}\), you'll find that each quadratic irrational has only approximations of bounded merit. The precise behavior of \(\log|b\pi-a|/\log b\) as a/b runs through approximations to pi is not known.

]]>As for the Carnival — of course, I’d be delighted. ]]>

P.S. May I include a link to this post on this month’s Aperiodical Carnival of Mathematics?

]]>`tanton :: [Rational]`

tanton = [1] ++ map (1+) (zipWith (/) (scanl1 (+) tanton) [1..])

]]>One nitpick, why not “ghost” the grid lines for all graphs as done in the “fraction surviving sequential pecking” graph?

]]>In the single-round process, every chick pecks, and so after that round every chick must have at least one pecked neighbor. On the basis of this fact I claim that the surviving population can never be greater than \(\frac{N}{2}\).

it should say:

In the single round process, all N chicks peck and every pecked chick is pecked by at most 2 other. Therefore there are at least \(\frac{N}{2}\) pecked chicks, and the surviving population can never be greater than \(\frac{N}{2}\).

The reason why the first block is wrong is that in configurations other than a circle it is indeed possible that everyone has a pecked neighbor and more than \(\frac{N}{2}\) survive, so the conclusion doesn’t follow from the argument.

*(A correction that doesn’t detract anything from the fascinating nerdiness of this full post)*

First, thanks so much for pointing out my black-and-white confusion. Should be fixed now.

Second, I am grateful to now have a probability model that works and whose derivation makes sense, at least for the one-round-of-pecking case. “. . . the pecking state of the chick under consideration is irrelevant; so let’s stop thinking about it” — that’s a step in the analysis that I would not have thought to take.

Just for the record, here’s a comparison of your odd and even distributions with the empirical data:

They both appear to be spot on.

]]>Red is observed data; purple is the graph of your PDF. I thought it was just missing a factor of 2, so I tried doubling the amplitude *(bright green)*, but that curve is a little off, too. The sum of \(p(k)\) for \(0 \le k \le 12\) is about 0.48. Perhaps I’ve misunderstood what you’re aiming at.

I’m also unsure how to deal with \((N/2)!\) when \(N\) is odd — gamma function?

]]>