Archive for Le Monde

Johnnysteria

Posted in Statistics with tags , , , , , , on December 6, 2017 by xi'an

Le Monde puzzle [#1029]

Posted in Books, Kids, R with tags , , on November 22, 2017 by xi'an

A convoluted counting Le Monde mathematical puzzle:

A film theatre has a waiting room and several projection rooms. With four films on display. A first set of 600 spectators enters the waiting room and vote for their favourite film. The most popular film is projected to the spectators who voted for it and the remaining spectators stay in the waiting room. They are joined by a new set of 600 spectators, who then also vote for their favourite film. The selected film (which may be the same as the first one) is then shown to those who vote for it and the remaining spectators stay in the waiting room. This pattern is repeated for a total number of 10 votes, after which the remaining spectators leave. What are the maximal possible numbers of waiting spectators and of spectators in a projection room?

A first attempt by random sampling does not produce extreme enough events to reach those maxima:

wm=rm=600 #waiting and watching
for (v in 1:V){
 film=rep(0,4) #votes on each fiLm
 for (t in 1:9){
  film=film+rmultinom(1,600,rep(1,4))
  rm=max(rm,max(film))
  film[order(film)[4]]=0
  wm=max(wm,sum(film)+600)}
 rm=max(rm,max(film)+600)}

where the last line adds the last batch of arriving spectators to the largest group of waiting ones. This code only returns 1605 for the maximal number of waiting spectators. And 1155 for the maximal number in a projection room.  Compared with the even separation of the first 600 into four groups of 150… I thus looked for an alternative deterministic allocation:

wm=rm=0
film=rep(0,4)
for (t in 1:9){
 size=sum(film)+600
 film=c(rep(ceiling(size/4),3),size-3*ceiling(size/4))
 film[order(film)[4]]=0
 rm=max(rm,max(film)+600)
 wm=max(wm,sum(film)+600)}

which tries to preserve as many waiting spectators as possible for the last round (and always considers the scenario of all newcomers backing the largest waiting group for the next film). The outcome of this sequence moves up to 1155 for the largest projection audience and 2264 for the largest waiting group. I however wonder if splitting into two groups in the final round(s) could even increase the size of the last projection. And indeed halving the last batch into two groups leads to 1709 spectators in the final projection. With uncertainties about the validity of the split towards ancient spectators keeping their vote fixed! (I did not think long enough about this puzzle to turn it into a more mathematical problem…)

While in Warwick, I reconsidered the problem from a dynamic programming perspective, always keeping the notion that it was optimal to allocate the votes evenly between some of the films (from 1 to 4). Using the recursive R code

optiz=function(votz,t){
  if (t==9){ return(sort(votz)[3]+600)
  }else{
    goal=optiz(sort(votz)+c(0,0,600,-max(votz)),t+1)
    goal=rep(goal,4)
    for (i in 2:4){
      film=sort(votz);film[4]=0;film=sort(film)
      size=sum(film[(4-i+1):4])+600
      film[(4-i+1):4]=ceiling(size/i)
      while (sum(film[(4-i+1):4])>size) film[4]=film[4]-1
      goal[i]=optiz(sort(film),t+1)}
    return(max(goal))}}    

led to a maximal audience size of 1619. [Which is also the answer provided by Le Monde]

Le Monde puzzle [#1028]

Posted in Books, Kids with tags , , , on November 16, 2017 by xi'an

Back to standard Le Monde mathematical puzzles (no further competition!), with this arithmetic one:

While n! cannot be a squared integer for n>1, does there exist 1<n<28 such that 28(n!) is a square integer? Does there exist 1<n,m<28 such that 28(n!)(m!) is a square integer? And what is the largest group of distinct integers between 2 and 27 such that the product of 28! by their factorials is a square?

The fact that n! cannot be a square follows from the occurrence of single prime numbers in the resulting prime number decomposition. When considering 28!, there are several single prime numbers like 17, 19, and 23, which means n is at least 23, but then the last prime in the decomposition of 28! being 7 means this prime remains alone in a product by any n! when n<28. However, to keep up with the R resolution tradition, I started by representing all integers between 2 and 28 in terms of their prime decomposition:

primz=c(2,3,5,7,11,13,17,19,23)
dcmpz=matrix(0,28,9)
for (i in 2:28){
 for (j in 1:9){
    k=i
    while (k%%primz[j]==0){ 
      k=k%/%primz[j];dcmpz[i,j]=dcmpz[i,j]+1}}
}

since the prime number factorisation of the factorials n! follows by cumulated sums (over the rows) of dcmpz, after which checking for one term products

fctorz=apply(dcmpz,2,cumsum)
for (i in 23:28)
  if (max((fctorz[28,]+fctorz[i,])%%2)==0) print(i)

and two term products

for (i in 2:28)
for (j in i:27)
 if (max((fctorz[28,]+fctorz[i,]+fctorz[j,])%%2)==0) 
  print(c(i,j))

is easy and produces i=28 [no solution!] in the first case and (i,j)=(10,27) in the second case. For the final question,  adding up to twelve terms together still produced solutions so I opted for the opposite end by removing one term at a time and

for (a in 2:28)
  if (max(apply(fctorz[-a,],2,sum)%%2)==0) print(a)

exhibited a solution for a=14. Meaning that

2! 3! …. 13! 15! …. 28!

is a square.

[ex?] Paris-Saclay univerXity

Posted in Kids, pictures, University life with tags , , , , , , , , , on November 5, 2017 by xi'an

In the plane to Warwick last Tuesday, I read a fairly long and pessimistic article in Le Monde about the future of the Paris-Saclay university. This debate presumably makes no sense outside French circles, for it relates to the century-old opposition between universities and grandes écoles, these selective engineering and business schools that operate independently from the university structure. In the sense that the selection and the schooling of their students is completely separated, meaning these students may never attend a university program if they choose to do so. But not so independently in terms of hiring part-time professors from universities and sharing resources for their Master programs. And depending on the same agencies (like CNRS) for funding their research program.

Anyway, the core message of this article was that the influence of former students from Polytechnique [aka X] in the high administration is such that they can prevent the integration of the different engineer schools on the Saclay plateau into the intended superstructure of the Paris-Saclay university. And turn this somewhat megalomanic Paris-Saclay project initiated by Nicolas Sarkozy into a mere geographical superposition of separate institutions, with very unequal State funding, perpetuating the two speed regime for public higher education… And a ever more confusing international image that will not help an inch moving up the Shanghai ranking (a major reason for Sarkozy pushing this project). Very French indeed!

Le Monde [last] puzzle [#1026]

Posted in Books, Kids, R with tags , , , , , on November 2, 2017 by xi'an

The last and final Le Monde puzzle is a bit of a disappointment, to wit:

A 4×4 table is filled with positive and different integers. A 3×3 table is then deduced by adding four adjacent [i.e. sharing a common corner] entries of the original table. Similarly with a 2×2 table, summing up to a unique integer. What is the minimal value of this integer? And by how much does it increase if all 29 integers in the tables are different?

For the first question, the resulting integer writes down as the sum of the corner values, plus 3 times the sum of the side values, plus 9 times the sum of the 4 inner values [of the 4×4 table]. Hence, minimising the overall sum means taking the inner values as 1,2,3,4, the side values as 5,…,12, and the corner values as 13,…,16. Resulting in a total sum of 352. As checked in this computer code in APL by Jean-Louis:

This configuration does not produce 29 distinct values, but moving one value higher in one corner does: I experimented with different upper bounds on the numbers and 17 always provided with the smallest overall sum, 365.

firz=matrix(0,3,3)#second level
thirz=matrix(0,2,2)#third level
for (t in 1:1e8){
flor=matrix(sample(1:17,16),4,4)
for (i in 1:3) for (j in 1:3)
firz[i,j]=sum(flor[i:(i+1),j:(j+1)])
for (i in 1:2) for (j in 1:2)
thirz[i,j]=sum(firz[i:(i+1),j:(j+1)])
#last
if (length(unique(c(flor,firz,thirz)))==29)
solz=min(solz,sum(thirz))}

and a further simulated annealing attempt did not get me anywhere close to this solution.

Le Monde puzzle [poll]

Posted in Books, Kids with tags , , on November 1, 2017 by xi'an

As the 25 Le Monde mathematical puzzles have now been delivered (plus the extraneous #1021), the journal is asking the players for their favourites, in order to separate ex-aequos. For readers who followed the entire sequence since puzzle #1001, what are your favourite four puzzles? (No more than four votes!)

publish or perish [or move to .005]

Posted in Books, pictures, Statistics, University life with tags , , , , on October 24, 2017 by xi'an

A series of articles in the Sciences et Médecine part of Le Monde reproduced coverages found elsewhere on the debates running within the scientific community on improving the quality of scientific papers. Through reproducible experiments and conclusions. And on using new bounds for the p-value, the solution to all woes! The article borrows a lot from the Nature proposal [discussed quite a lot here in the past weeks] and does not provide particularly insightful views. It however contains a coverage (rightmost columns) on a peer community approach called PubPeer, which was launched by two neuroscientists, Brandon Stell and Boris Barbour, both at CNRS, towards sharing comments on published papers. Mostly to criticise the methodology used in these papers. Or to point out multiple usages of the same graphs. Or doctoring of pictures. In the vast majority of cases, the papers are in biology and the comments not addressed by the authors of the papers. (With this exception of a discussion of the Nature paper covering the call for new bounds on p-values. Nature paper that had the appealing feature of calling for an end to `one-size-fits-all’ thresholds.) Creating a platform for discussing papers from a journal is already hard enough (as shown with the closure of Series B’log!), hence running a global discussion forum for all journals sounds hard to manage and foster. By which I mean it is difficult to fathom the impact of the discussions on the published papers and the journals where they are published, given the reticence of said journals to engage into reassessments of published papers…