adv’surd! [jatp]

AI for good

Last 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]

MUDAM

As our son is doing an internship in Luxembourg City this semester, we visited him last weekend and took the opportunity to visit the Museum of Modern Art (or MUDAM) there. The building itself is quite impressive, inserted in the walls of the 18th Century Fort Thüngen designed by Vauban, with a very luminous and airy building designed by Ming Pei. The main exhibit at the MUDAM is a coverage of the work on Su-Mei Tse, an artist from Luxembourg I did not know but whom vision I find both original and highly impressive, playing on scales and space, from atoms to planets… With connections to Monet’s nympheas. And an almost raw rendering of rock forms that I appreciate most particularly!

The bottom floor also contains an extensive display of the political drawings of Ad Reinhardt, who is more (?) famous for his black-on-black series…

splitting a field by annealing

A 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…]