Le Monde puzzle [#1086]

A terse Le Monde mathematical puzzle in the optimisation mode:

What is the maximal fraction of the surface of a triangle occupied by an inner triangle ABC where Abigail picks a summit A on a first side, Berenice B on a second side, and then Abigails picks C on the third side, towards Abigail maximising and Berenice minimising this surface?

Which I first tried to solve by pen & paper, completing another black block for the occasion, as coding the brute force R version sounded too painful:

leading me to conclude that, for a rectangle triangle (although the result sounds independent of this feature), the optimum was the middle triangle, weighting one-fourth of the original surface. Reprogramming the question in the plane to Angkor produced the same output, modulo my approximation of the triangle continuum with a 200×200/2grid:

Le Monde puzzle [#1087]

A board-like Le Monde mathematical puzzle in the digit category:

Given a (k,m) binary matrix, what is the maximum number S of entries with only one neighbour equal to one? Solve for k=m=2,…,13, and k=6,m=8.

For instance, for k=m=2, the matrix

is producing the maximal number 4. I first attempted a brute force random filling of these matrices with only a few steps of explorations and got the numbers 4,8,16,34,44,57,… for the first cases. Since I was convinced that the square k² of a number k previously exhibited to reach its maximum as S=k² was again perfect in this regard, I then tried another approach based on Gibbs sampling and annealing (what else?):

gibzit=function(k,m,A=1e2,N=1e2){
  temp=1 #temperature
  board=sole=matrix(sample(c(0,1),(k+2)*(m+2),rep=TRUE),k+2,m+2)
  board[1,]=board[k+2,]=board[,1]=board[,m+2]=0 #boundaries
  maxol=counter(board,k,m) #how many one-neighbours?
  for (t in 1:A){#annealing
    for (r in 1:N){#basic gibbs steps
      for (i in 2:(k+1))
        for (j in 2:(m+1)){
          prop=board
          prop[i,j]=1-board[i,j]
          u=runif(1)
          if (log(u/(1-u))<temp*(counter(prop,k,m)-val)){ 
             board=prop;val=counter(prop,k,m) 
             if (val>maxol){
               maxol=val;sole=board}}
      }}
    temp=temp*2}
  return(sole[-c(1,k+2),-c(1,m+2)])}

which leads systematically to the optimal solution, namely a perfect square k² when k is even and a perfect but one k²-1 when k is odd. When k=6, m=8, all entries can afford one neighbour exactly,

> gibzbbgiz(6,8)
[1] 48
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]    1    0    0    1    1    0    0    1
[2,]    1    0    0    0    0    0    0    1
[3,]    0    0    1    0    0    1    0    0
[4,]    0    0    1    0    0    1    0    0
[5,]    1    0    0    0    0    0    0    1
[6,]    1    0    0    1    1    0    0    1

but this does not seem feasible when k=6, m=7, which only achieves 40 entries with one single neighbour.

Le Monde puzzle [#1085]

A new Le Monde mathematical puzzle in the digit category:

Given 13 arbitrary relative integers chosen by Bo, Abigail can select any subset of them to be drifted by plus or minus one by Bo, repeatedly until Abigail reaches the largest possible number N of multiples of 5. What is the minimal possible value of N under the assumption that Bo tries to minimise it?

I got stuck on that one, as building a recursive functiion led me nowhere: the potential for infinite loop (add one, subtract one, add one, …) rather than memory issues forced me into a finite horizon for the R function, which then did not return anything substantial in a manageable time. Over the week and the swimming sessions, I thought of simplifying the steps, like (a) work modulo 5, (b) bias moves towards 1 or 4, away from 2 and 3, by keeping only one entry in 2 and 3, and all but one at 1 and 4, but could only produce five 0’s upon a sequence of attempts… With the intuition that only 3 entries should remain in the end, which was comforted by Le Monde solution the week after.

Le Monde puzzle [#1083]

A Le Monde mathematical puzzle that seems hard to solve without the backup of a computer (and just simple enough to code on a flight to Montpellier):

Given the number N=2,019, find a decomposition of N as a sum of non-trivial powers of integers such that (a) the number of integers in the sum is maximal or (b) all powers are equal to 4.  Is it possible to write N as a sum of two powers?

It is straightforward to identify all possible terms in these sums by listing all powers of integers less than N

pool=(1:trunc(sqrt(2019)))^2
for (pow in 3:11)
  pool=unique(c(pool,(2:trunc(2019^(1/pow)))^pow))

which leads to 57 distinct powers. Sampling at random from this collection at random produces a sum of 21 perfect powers:

 1+4+8+9+16+25+27+32+36+49+64+81+100+121+125+128+144+169+196+243+441

But looking at the 22 smallest numbers in the pool of powers leads to 2019, which is a sure answer. Restricting the terms to powers of 4 leads to the sequence

1⁴+2⁴+3⁴+5⁴+6⁴ = 2019

And starting from the pools of all possible powers in a decomposition of 2019 as the sum of two powers shows this is impossible.

MASH in Le Monde

Le Monde puzzle [#1081]

A “he said-she said” Le Monde mathematical puzzle (again in the spirit of the famous Singapore high-school birthdate problem):

Abigail and Corentin are both given a positive integer, a and b, such that a+b is either 19 or 20. They are asked one after the other and repeatedly if they are sure of the other’s number. What is the maximum number of times they are questioned?

If Abigail is given a 19, b=1 necessarily. Hence if Abigail does not reply, a<19. This implies that, if Corentin is given b=1 or b=19, he can reply a+b=19 or a+b=20, necessarily. Else, 1<b<19 implies that, if a=1 or a=18, b=18 or b=2. And so on…which leads to a maximum of 20 questions, 10 for Abigail and 10 for Corentin. Here is my R implementation

az=bz=cbind(20-(1:19),19-(1:19))
qwz=0;at=TRUE;bt=FALSE
while ((max(az)>0)&(max(bz)>0)){
 if (at){ 
  for (i in 1:19){ 
   if (sum(az[i,]>0)==2){
   for (j in az[i,az[i,]>0]){ 
     if (sum(bz[j,]==0)==2) az[i,]=rep(0,2)}}
   if (sum(az[i,]>0)<2){ 
    az[i,]=rep(0,2)}}} 
  if (bt){ 
   for (i in 1:19){ 
    if (sum(bz[i,bz[i,]>0]>0)==2){
     for (j in bz[i,bz[i,]>0]){ 
      if (sum(az[j,]==0)==2) bz[i,]=rep(0,2)}}
     if (sum(bz[i,]>0)<2){ bz[i,]=rep(0,2)}}}
  bt=!bt;at=!at;qwz=qwz+1}

y a plus de mouchoirs au bureau des pleurs

Once the French government started giving up to some requests of the unstructured “gilets jaunes” protesters, it was like a flood or flush gate had opened and every category was soon asking for a rise (in benefits) and a decrease (in taxes) or the abolition of a recent measure (like the new procedure for entering university after high school). As an illustration, I read a rather bemusing tribune in Le Monde from a collective of PhD students against asking non-EU students (including PhD students) to pay fees to study in French universities. This may sound a bit of a surrealistic debate from abroad, but the most curious point in the tribune [besides the seemingly paradoxical title of students against Bienvenue En France!] is to argue that asking these students to pay the intended amount would bring their net stipends below the legal minimum wage, considering that they are regular workers… (Which is not completely untrue when remembering that in France the stipends are taxed for income tax and retirement benefits and unemployment benefits, meaning that a new PhD graduate with no position can apply for these benefits.) It seems to me that the solution adopted in most countries, namely that the registration fees are incorporated within the PhD grants, could apply here as well… The other argument that raising these fees from essentially zero to 3000 euros is going to stop bright foreign students to do their PhD in France is not particularly strong when considering that the proportion of foreign students among PhD students here is slightly inferior to the proportion in the UK and the US, where the fees are anything but negligible, especially for foreign students.