ISBA2020 program

The scheduled program for ISBA 2020 is now on-line. And full of exciting sessions, many with computational focus. With dear hopes that the nCo-2019 epidemics will have abated by then (and not solely for the sake of the conference, most obviously!). While early registration ends by 15 April, the deadline for junior travel support ends up this month. And so does the deadline for contributions.

Rashomon, plus 47 ronins, plus…

Another chance encounter (on Amazon) led me to read a graphical novel entitled Rashōmon, by Victor Santos. Which uses the same short stories from Ryūnosuke Akutagawa as Akira Kurosawa in his superlative film, if not with the same intensity. (The very first sentences are inspired from the first pages of the book, though.) And in a second part builds upon the tale of the 47 rônins which I read last summer in Koyasan. Plus a possible appearance of Miyamato Mushashi, the great 17th Century swordsman (depicted in two wonderful novels by Eiji Yoshikawa). While this is historically impossible, since Rashōmon takes place in the 12th Century and the 47 rônins acted in 1702, the theme cementing the story is the presence of a detective named Heigo Kobayashi, who “solves” both crimes but is nonetheless outsmarted by the novel “femme fatale”… Without a clear explanation as to how she did it.

While I found the rendering rather entertaining, with an original if convoluted drawing style, I was rather disappointed at the simplistic and Westernised adaptation of the subtle stories into a detective story. Calling upon (anachronic) ninjas as if the historical setting per se was not exotic enough. And the oddly modified role of the main female character into an Hammet-like heroin kills the ambivalence that is central to both Akutagawa’s and Kurosawa’s versions.

a very quick Riddle

A very quick Riddler’s riddle last week with the question

Find the (integer) fraction with the smallest (integer) denominator strictly located between 1/2020 and 1/2019.

and the brute force resolution

for (t in (2020*2019):2021){ 
   a=ceiling(t/2020)
   if (a*2019<t) sol=c(a,t)}

leading to 2/4039 as the target. Note that

Couplings and Monte Carlo [advanced graduate course at Dauphine by Pierre Jacob]

As a visiting professor at Paris-Dauphine next month, Pierre Jacob will give a series of lectures on coupling and Monte Carlo. Next month on Feb. 13, 14, 25 27, at Université Paris-Dauphine, the first two starting at 8:30 (room E) and the last two starting at 13:45 (room F and D201, respectively). Attendance is open to all and material will be made available on the lecture webpage.

my first parkrun [19:56,3/87,78.8%]

This morning, I had my first parkrun race in Gainesville, before heading back to Paris. (Thanks to Florence Forbes who pointed out this initiative to me.) Which reminded me of the race I ran in Helsinki a few years ago. Without the “self-transcendance” topping…! While the route was very urban, it was a fun opportunity to run a race with a few other runners. My time of 19.56 is not my best by far but, excuses, excuses, I was not feeling too well and the temperature was quite high (21⁰) and I finished in the first three runners, just seconds behind two young fellows who looked like they were still in high school.  (I am now holding the record of that race for my age group as well!) Anyway, this is a great way to join races when travelling and not worry about registration, certificates, &tc.

Parkrun also provides an age-grade adjusted ranking (78.8%), which is interesting but statistically puzzling as this is the ratio of one’s time over the fastest time (ever?) in the age x gender category. Given that fastest times are extreme, this depends on one individual and hence has a high variability. Especially in higher (meaning older!) veteran categories. A quantile in the empirical distribution would sound better. I came across this somewhat statistical analysis of the grade,

Le Monde puzzle [#1127]

A permutation challenge as Le weekly Monde current mathematical puzzle:

When considering all games between 20 teams, of which 3 games have not yet been played, wins bring 3 points, losses 0 points, and draws 1 point (each). If the sum of all points over all teams and all games is 516, was is the largest possible number of teams with no draw in every game they played?

The run of a brute force R simulation of 187 purely random games did not produce enough acceptable tables in a reasonable time. So I instead considered that a sum of 516 over 187 games means solving 3a+2b=516 and a+b=187, leading to 142 3’s to allocate and 45 1’s. Meaning for instance this realisation of an acceptable table of game results

games=matrix(1,20,20);diag(games)=0
while(sum(games*t(games))!=374){
  games=matrix(1,20,20);diag(games)=0
  games[sample((1:20^2)[games==1],3)]=0}
games=games*t(games)
games[lower.tri(games)&games]=games[lower.tri(games)&games]*
    sample(c(rep(1,45),rep(3,142)))* #1's and 3'
    (1-2*(runif(20*19/2-3)<.5)) #sign
games[upper.tri(games)]=-games[lower.tri(games)]
games[games==-3]=0;games=abs(games)

Running 10⁶ random realisations of such matrices with no constraint whatsoever provided a solution with] 915,524 tables with no no-draws, 81,851 tables with 19 teams with some draws, 2592 tables with 18 with some draws and 3 tables with 17 with some draws. However, given that 10*9=90 it seems to me that the maximum number should be 10 by allocating all 90 draw points to the same 10 teams, and 143 3’s at random in the remaining games, and I reran a simulated annealing version (what else?!), reaching a maximum 6 teams with no draws. Nowhere near 10, though!

an elegant sampler

Following an X validated question on how to simulate a multinomial with fixed average, W. Huber produced a highly elegant and efficient resolution with the compact R code

tabulate(sample.int((k-1)*n, s-n) %% n + 1, n) + 1

where k is the number of classes, n the number of draws, and s equal to n times the fixed average. The R function sample.int is an alternative to sample that seems faster. Breaking the outcome of

sample.int((k-1)*n, s-n)

as nonzero positions in an n x (k-1) matrix and adding a adding a row of n 1’s leads to a simulation of integers between 1 and k by counting the 1’s in each of the n columns, which is the meaning of the above picture. Where the colour code is added after counting the number of 1’s. Since there are s 1’s in this matrix, the sum is automatically equal to s. Since the s-n positions are chosen uniformly over the n x (k-1) locations, the outcome is uniform. The rest of the R code is a brutally efficient way to translate the idea into a function. (By comparison, I brute-forced the question by suggesting a basic Metropolis algorithm.)