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

Natum

Scott Sisson’s ABC seminar in Paris [All about that Bayes]

On the “All about that Bayes” seminar tomorrow (Tuesday 21 at 3p.m., room 42, AgroParisTech, 16 rue Claude Bernard, Paris 5ième), Scott Sisson, School of Mathematics and Statistics at UNSW, and visiting Paris-Dauphine this month, will give a talk on

Approximate posteriors and data for Bayesian inference

Abstract
For various reasons, including large datasets and complex models, approximate inference is becoming increasingly common. In this talk I will provide three vignettes of recent work. These cover a) approximate Bayesian computation for Gaussian process density estimation, b) likelihood-free Gibbs sampling, and c) MCMC for approximate (rounded) data.

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. 12, 25, 26, 27, at Université Paris-Dauphine, all starting at 13:45 (room yet to be announced). Attendance is open to all and material will be made available on the lecture webpage.

maths & AI

夢幻花 [Dream flower]

Another Japanese mystery novel by Higashino Keigo, which I read in French under the title La fleur de l´illusion [on a sunny Sunday afternoon, under my fig tree] and enjoyed both for its original, convoluted (and mostly convincing) plot and for the well-rendered interaction between the young protagonists. And also for having a few connections with my recent trip, from one protagonist studying nuclear physics at the University of Osaka to a visit to the back country of Katsuura. (The author himself graduated from Osaka Prefecture University with a Bachelor of Engineering degree.) Spoiler warning: the only annoying part of the plot was the resolution of the mystery via a secret society run by a few families of civil servants, which as always sounds to me like a rather cheap way out. But not enough to ruin the entire novel.

 

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!