Archive for Luxembourg
adv’surd! [jatp]
Posted in Statistics with tags advertising, COVID-19, Didier Raoult, kiosque, Luxembourg, Paris, Pétain, radio, RTL, street advertising on September 3, 2021 by xi'anAI for good
Posted in pictures, Statistics, Travel with tags AI for good, Amnesty International, Julien Cornebise, Luxembourg on December 26, 2017 by xi'anLast week, I had a quick chat in front of the Luxembourg gardens with Julien Cornebise and he told me about the AI for Good Foundation with whom he was going to work through Element AI, after doing volunteer work with Amnesty International. Great!
champagne, not Guinness??? [jatp]
Posted in Statistics with tags champagne, Guinness, jatp, Luxembourg, W. Gosset on October 29, 2017 by xi'ansplitting a field by annealing
Posted in Kids, pictures, R, Statistics with tags Autrans, Luxembourg, mathematical puzzle, Palais Ducal, R, random walk, simulated annealing, The Riddler, Vercors on October 18, 2017 by xi'anA recent riddle [from The Riddle] that I pondered about during a [long!] drive to Luxembourg last weekend was about splitting a square field into three lots of identical surface for a minimal length of separating wire… While this led me to conclude that the best solution was a T like separation, I ran a simulated annealing R code on my train trip to
AutransValence, seemingly in agreement with this conclusion.I discretised the square into n² units and explored configurations by switching two units with different colours, according to a simulated annealing pattern (although unable to impose connectivity on the three regions!):
partz=matrix(1,n,n) partz[,1:(n/3)]=2;partz[((n/2)+1):n,((n/3)+1):n]=3 #counting adjacent units of same colour nood=hood=matrix(4,n,n) for (v in 1:n2) hood[v]=bourz(v,partz) minz=el=sum(4-hood) for (t in 1:T){ colz=sample(1:3,2) #picks colours a=sample((1:n2)[(partz==colz[1])&(hood<4)],1) b=sample((1:n2)[(partz==colz[2])&(hood<4)],1) partt=partz;partt[b]=colz[1];partt[a]=colz[2] #collection of squares impacted by switch nood=hood voiz=unique(c(a,a-1,a+1,a+n,a-n,b-1,b,b+1,b+n,b-n)) voiz=voiz[(voiz>0)&(voiz<n2)] for (v in voiz) nood[v]=bourz(v,partt) if (nood[a]*nood[b]>0){ difz=sum(nood)-sum(hood) if (log(runif(1))<difz^3/(n^3)*(1+log(10*rep*t)^3)){ el=el-difz;partz=partt;hood=nood if (el<minz){ minz=el;cool=partz} }}}
(where bourz computes the number of neighbours), which produces completely random patterns at high temperatures (low t) and which returns to the T configuration (more or less):if not always, as shown below:
Once the (a?) solution was posted on The Riddler, it appeared that one triangular (Y) version proved better than the T one [if not started from corners], with a gain of 3% and that a curved separation was even better with an extra gain less than 1% [solution that I find quite surprising as straight lines should improve upon curved ones…]