Fri, 04 Oct 2019
Addenda to recent articles 201910
Several people have written in with helpful remarks about recent posts:
Regarding online tracking of legislation:
- Ed Davies directed my attention to www.legislation.gov.uk, an official organ of the British government, which says:
The aim is to publish legislation on this site simultaneously, or at least within 24 hours, of its publication in printed form.
M. Davies is impressed. So am I. Here is the European Union (Withdrawal) Act 2018.
Nik Clayton showed me the @unitedstates project, whcih among other things has a program that attempts to parse the United States Code and produce a tree-structured output.
Twitter user
@sunk818pointed out How I changed the law with a GitHub pull request. The title oversells what actually happened: the codified law contained a minor technical error, which could be corrected without requiring any actual legislation. But the takeaway is that the District of Columbia officially publishes its law to a Github repo.
This then led me to Standardizing the World’s Legislative Information — One hackathon at a time on the LII's VOXPOPULII blog.
(Reminder to readers: I do not normally read Twitter, and it is not a reliable way to contact me.)
Regarding the mysteriously wide letter ‘O’ on the Yeadon firehouse. I had I had guessed that it was not in the same family as the others, perhaps because the original one had been damaged. I asked Jonathan Hoefler, a noted font expert; he agreed.
But one reader, Steve Nicholson, pointed out that it is quite common, in Art Deco fonts, for the ‘O’ to be circular even when that makes it much wider than the other letters. He provided ten examples, such as Haute Corniche.
I suggested this to M. Hoefler, but he rejected the theory decisively:
True; it's a Deco mannerism to have 'modulated capitals'… . But this isn't a deco font, or a deco building, and in any case it would have been HIGHLY unlikely for a municipal sign shop to spec something like this for any purpose, let alone a firehouse. It's a wrong sort O, probably installed from the outset.
(The letter spacing suggests that this is the original ‘O’.)
Several people wrote to me about the problem of taking half a pill every day, in which I overlooked that the solution was simply the harmonic numbers.
Robin Houston linked to this YouTube video, “the frog problem”, which has the same solution, and observed that the two problems are isomorphic, proceeding essentially as Jonathan Dushoff does below.
Shreevatsa R. wrote a long blog article detailing their thoughts about the solution. I have not yet read the whole thing carefully but knowing M. Shreevatsa, it is well worth reading. M. Shreevatsa concludes, as I did, that a Markov chain approach is unlikely to be fruitful, but then finds an interesting approach to the problem using probability generating functions, and then another reformulating it as a balls-in-bins problem.
Jonathan Dushoff sent me a very clear and elegant solution and kindly gave me permission to publish it here:
The first key to my solution is the fact that you can add expectations even when variables are not independent.
In this case, that means that each time we break a pill we can calculate the probability that the half pill we produce will "survive" to be counted at the endpoint. That's the same as the expectation of the number of half-pills that pill will contribute to the final total. We can then just add these expectations to get the answer! A little counter-intuitive, but absolutely solid.
The next key is symmetry. If I break a half pill and there are !!j!! whole pills left, the only question for that half pill is the relative order in which I pick those !!j+1!! objects. In particular, any other half pills that exist or might be generated can be ignored for the purpose of this part of the question. By symmetry, any of these !!j+1!! objects is equally likely to be last, so the survival probability is !!\frac1{j+1}!!.
If I start with !!n!! pills and break one, I have !!n-1!! whole pills left, so the probability of that pill surviving is !!\frac1n!!. Going through to the end we get the answer:
$$\frac1n + \frac1{n-1} + \ldots + 1.$$
I have gotten feedback from several people about my Haskell type constructor clutter, which I will write up separately, probably, once I digest it.
Thanks to everyone who wrote in, even people I forgot to mention above, and even to the Twitter person who didn't actually write in.
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