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The convergents of 2x

Take some real number !!\alpha!! and let its convergents be !!c_0, c_1, c_2, \ldots!!. Now consider the convergents of !!2\alpha!!. Sometimes they will include !!2c_0, 2c_1, 2c_2, \ldots!!, sometimes only some of these.

For example, the convergents of !!\pi!! and !!2\pi!! are

$$ \begin{array}{rlc} \pi & \approx & \color{darkblue}{3},&&& \color{darkblue}{\frac{22}{7}}, & \color{darkblue}{\frac{333}{106}}, && \color{darkblue}{\frac{355}{113}}, & \color{darkblue}{\frac{103993}{33102}}, && \frac{104348}{33215}, & \color{darkblue}{\frac{208341}{66317}}, & \ldots \\ 2\pi & \approx & \color{darkblue}{6}, & \frac{19}{3}, & \frac{25}{4}, & \color{darkblue}{\frac{44}{7}}, & \color{darkblue}{\frac{333}{53}}, & \frac{377}{60}, & \color{darkblue}{\frac{710}{113}}, & \color{darkblue}{\frac{103393}{16551}}, & \frac{312689}{49766}, && \color{darkblue}{\frac{416682}{66317}}, & \ldots \end{array}
$$

Here are the analogous lists for !!\frac{1+\sqrt{5}}2!! and !!1+\sqrt5!!:

$$ \begin{array}{rlc} \frac12\left({1+\sqrt{5}}\right)& \approx & 1, & 2, & \color{darkblue}{\frac32}, & \frac53, & \frac85, & \color{darkblue}{\frac{13}8}, & \frac{21}{13}, & \frac{34}{21}, & \color{darkblue}{\frac{55}{34}}, & \frac{89}{55}, & \frac{144}{89}, & \color{darkblue}{\frac{233}{144}}, & \frac{377}{233}, &\frac{610}{377} , & \color{darkblue}{\frac{987}{610} }, & \ldots \\ 1+\sqrt{5} & \approx & & & \color{darkblue}{3}, &&& \color{darkblue}{\frac{13}4}, &&& \color{darkblue}{\frac{55}{17}}, &&& \color{darkblue}{\frac{233}{72}}, &&& \color{darkblue}{\frac{987}{305}}, & \ldots \end{array} $$

This time all the convergents in the second list are matched by convergents in the first list. The number !!\frac{1+\sqrt5}{2}!! is notorious because it's the real number whose convergents converge the most slowly. I'm surprised that !!1+\sqrt5!! converges so much more quickly; I would not have expected the factor of 2 to change the situation so drastically.

I haven't thought about this at all yet, but it seems to me that a promising avenue would be to look at what Gosper's algorithm would do for the case !!x\mapsto 2x!! and see what simplifications can be done. This would probably produce some insight, and maybe a method for constructing a number !!\alpha!! so that all the convergents of !!2\alpha!! are twice those of !!\alpha!!.

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“Forensic” doesn't mean what I thought it did

Last week at work we released bad code, which had somehow survived multiple reviews. I was very interested in finding out how this happened, dug into the Git history to find out, and wrote a report. Originally I titled the report something like “Forensic analysis of Git history” (and one of my co-workers independently referred to the investigation as forensic) but then I realized I wasn't sure what “forensic” meant. I looked it up, and learned it was the wrong word.

A forensic analysis is one performed in the service of a court or court case. “Forensic” itself is from Latin forum, which is a public assembly place where markets were held and court cases were heard.

Forensic medicine is medicine in service of a court case, for example to determine a cause of death. For this reason it often refers to a postmortem examination, and I thought that “forensic” meant a postmortem or other retrospective analysis. That was the sense I intended it. But no. I had written a postmortem analysis, but not a forensic one.

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A simple dice-throwing game that seems hard to play

I ran into a fun math problem yesterday, easy to ask, easy to understand, but somewhat open-ended and seems to produce fairly complex behavior. It might be a good problem for a bright high school student to tinker with.

Consider the following one-player game. You start with a total of n points. On each turn, you choose to throw either a four-, six-, or eight-sided die, and then subtract the number thrown from your point total. The game continues until your total reaches zero (and you win) or goes below zero (and you lose).

This game seems surprisingly difficult to analyze. The computer analysis is quite easy, but what I mean is, if someone comes to you offering to pay you a dollar if you can win starting with !!n=9!! points, and it would be spoilsportish to say “just wait here for half an hour while I write this computer program”, what's your good move?

Is there even a way to make an educated guess, short of doing a full analysis? The !!n≤4!! strategy is obvious, but even for !!n=5!! you need to start calculating: rolling the d4 is safe. Rolling the d6 gives you a chance of wiping out, but also a chance of winning instantly; is that an improvement? (Spoiler: it is, quite substantially so! Your chance of winning increases from !!36\%!! to !!40.7\%!!.)

With the game as described, and optimal play, the probability of winning approaches !!45.66\%!! as the number of points increases, and the strategy is not simple: the best strategy for !!n≤20!! uses the d4 in 13 cases, the d6 in 4 cases, and the d8 in 3 cases:

$$\begin{array}{rcl} n & \text{Best play} & \text{Win %} \\ \hline 1 & 4\quad & 25.00\% \\ 2 & 4\quad & 31.25 \\ 3 & 4\quad & 39.06 \\ 4 & 4\quad & 48.83 \\ \hline 5 & 6 & 40.69\% \\ 6 & 6 & 47.47 \\ 7 & 4\quad & 44.01 \\ 8 & \quad8 & 47.04 \\ \hline 9 & 4\quad & 44.80\% \\ 10 & 4\quad & 45.83 \\ 11 & 4\quad & 45.42 \\ 12 & 4\quad & 45.77 \\ \hline 13 & 6 & 45.48 \% \\ 14 & \quad8 & 45.73 \\ 15 & 4\quad & 45.60 \\ 16 & \quad8 & 45.71 \\ \hline 17 & 4\quad & 45.63 \% \\ 18 & 4\quad & 45.67 \\ 19 & 4\quad & 45.65 \\ 20 & 6 & 45.67 \end{array} $$

It seems fairly clear (and not hard to prove) that when the die with fewest sides has !!d!! sides, the good numbers of points are multiples of !!d!!, with !!kd+1!! somewhat worse, and then !!kd+2, kd+3, \ldots!! generally better and better to the next peak at !!kd+d!!. But there are exceptions: even if !!d!! is not the smallest die, if you have a !!d!!-sided die, it is good to have !!d!! points, and when you do you should roll the !!d!!-sided die.

I did get a little more insight after making the chart above and seeing the 4-periodicity. In a comment on my Math SE post I observed:

There is a way to see quickly that the d4 is better for !!n=7!!. !!n=1!! is the worst possible position. !!n=2,3,!! and !!4!! are increasingly good; !!4!! is best because you can't lose and you might win outright. After that !!5!! is bad again, but not as bad as !!1!!, with !!6,7,8!! increasingly good. The pattern continues this way, with !!4k−3,4k−2,4k−1,4k!! being increasingly good, and then !!4k+1!! being worse again but better than !!4k−3!!. For !!n=7!!, the d6 allows one to land on !!\{1,2,3,4,5,6\}!!, and the d4 on !!\{3,4,5,6\}!!. But !!1!! is worse than !!5!! and !!2!! is worse than !!6!!, so prefer the d4.

The d4-d6-d8 case is unusually confusing, because for example it's not clear whether from 12 points you should throw d4, hoping to land on 8, or d6, hoping to land on 6. (I haven't checked but I imagine the two strategies perform almost equally well; similarly it probably doesn't matter much if you throw the d4 or the d6 first from !!n=10!!.)

That the d6 is best for !!n=13!! is very surprising to me.

Why !!45.66\%!!? I don't know. With only one die, the winning probability for large !!n!! converges to !!\frac2{n+1}!! which I imagine is a fairly straightforward calculation (but I have not done it). For more than one die, it seems much harder.

Is there a way to estimate the winning probability for large !!n!!, given the list of dice? Actually yes, a little bit: the probability of winning with just a d4 is !!\frac 25!!, and the d6 and d8 can't hurt, so we know the chance of winning with all three dice available will be somewhat more than !!40\%!!, as it is. The value of larger dice falls off rapidly with the number of sides, so for example with d4+d6 the chance of winning increases from !!40\%!! to almost !!45\%!!, and adding the d8 only nudges this up to !!45.66\%!!.

The probability of winning with a d2 is !!\frac 23!!, and if you have a d3 also the probability goes up to !!\frac 34!!, which seems simple enough, but if you add a d4 instead of the d3 instead it goes to !!68.965\%!!, whatever that is. And Dfan Schmidt tells me that d3 + d4 converges to !!\frac{512}{891}!!.

I wrote it up for Math StackExchange but nobody has replied yet.

Here's Python code to calculate the values. Enjoy.

[ Addendum: Michael Lugo points out that the d2+d4 probability (“!!68.965\%!!, whatever that is”) is simply !!\frac{20}{29}!!, and gives some other similar results. One is that d3+d4+d5 has a winning probability of !!\frac{16}{27}!!; the small denominator is surprising. ]

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Examples of dummy pronouns

Katara is interested in linguistics. When school was over for the year and she had time to think about things, I gave her all my old linguistics books. The other day for some reason I mentioned to her that I had known people who were engaged in formal research on the problem of how to get a computer to know what a pronoun referred to, and that this is very difficult.

(I once had a co-worker who claimed that it was simple: the pronoun always refers back to the nearest noun. It wasn't hard to go back in his Slack history and find a counterexample he had uttered a few minutes before.)

Today I wanted to tell Katara about dummy pronouns, which refer to nothing at all. I intended to send her the example from Wiktionary:

it is good to know that you are okay

I started my message:

Here's an interesting example of how hard it can be to find what a pronoun refers to

Then I realized I no longer needed the example.

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Duckface in German

In English, this is called duckface:

In German, I've learned, it's Schlauchbootlippen.

Schlauch is “tube”. A Schlauchboot is a tube-boat — an inflatable rubber dingy. Schlauchbootlippen means dinghy-lips.

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