Archive for the Books Category

Le Monde puzzle [#1009]

Posted in Books, Kids with tags , , , , on May 26, 2017 by xi'an

An incomprehensible (and again double) Le Monde mathematical puzzle (despite requests to the authors! The details in brackets are mine.):

  1. A [non-circular] chain of 63 papers clips can be broken into sub-chains by freeing one clip [from both neighbours] at a time. At a given stage, considering the set of the lengths of these sub-chains, the collection of all possible sums of these lengths is a subset of {1,…,63}. What is the minimal number of steps to recover the entire set {1,…,63}?  And what is the maximal length L of a chain of paper clips that allows this recovery in 8 steps?
  2.  A tri-colored chain of 200 paper clips starts with a red, a blue and a green clip. Removing one clip every four clips produces a chain of 50 removed clips identical to the chain of 50 first clips of the original chain and a chain of remaining 150 clips identical to the 150 first clips of the original chain. Deduce the number of green, red, and blue clips.

The first question can be easily tackled by random exploration. Pick one number at random between 1 and 63, and keep picking attached clips until the set of sums is {1,…,63}. For instance,

rebreak0]
 sumz=cumsum(sample(difz))
 for (t in 1:1e3)
  sumz=unique(c(sumz,cumsum(sample(difz))))
 if (length(sumz)<63)
    brkz=rebreak(sort(c(brkz,sample((1:63)[-brkz],1))))
 return(brkz)}

where I used sampling to find the set of all possible partial sums. Which leads to a solution with three steps, at positions 5, 22, and 31. This sounds impossibly small but the corresponding lengths are

1 1 1 4 8 16 32

from which one can indeed recover by summation all numbers till 63=2⁶-1. From there, a solution in 8 steps can be found by directly considering the lengths

1 1 1 1 1 1 1 1 9 18=9+8 36=18+17+1 72 144 288 576 1152 2303

whose total sum is 4607. And with breaks

10 29 66 139 284 573 1150 2303

The second puzzle is completely independent. Running another R code reproducing the constraints leads to

tromcol=function(N=200){
  vale=rep(0,N)
  vale[1:3]=1:3
  while (min(vale)==0){
    vale[4*(1:50)]=vale[1:50]
    vale[-(4*(1:50))]=vale[1:150]}
  return(c(sum(vale==1),sum(vale==2),sum(vale==3)))}

and to 120 red clips, 46 blue clips and 34 green clips.

ARS: when to update?

Posted in Books, Kids, Statistics, University life with tags , , , , , on May 25, 2017 by xi'an

An email I got today from Heng Zhou wondered about the validity of the above form of the ARS algorithm. As printed in our book Monte Carlo Statistical Methods. The worry is that in the original version of the algorithm the envelope of the log-concave target f(.) is only updated for rejected values. My reply to the question is that there is no difference in the versions towards returning a value simulated from f, since changing the envelope between simulations does not modify the accept-reject nature of the algorithm. There is no issue of dependence between the simulations of this adaptive accept-reject method, all simulations remain independent. The question is rather one about efficiency, namely does it pay to update the envelope(s) when accepting a new value and I think it does because the costly part is the computation of f(x), rather than the call to the piecewise-exponential envelope. Correct me if I am wrong!

Manchester, United we stand!

Posted in Books, Kids with tags , , , , , on May 24, 2017 by xi'an

This is the place
In the north-west of England. It’s ace, it’s the best
And the songs that we sing from the stands, from our bands
Set the whole planet shaking.
Our inventions are legends. There’s nowt we can’t make, and so we make brilliant music
We make brilliant bands

We make goals that make souls leap from seats in the stands
And we make things from steel
And we make things from cotton
And we make people laugh, take the mick summat rotten
And we make you at home
And we make you feel welcome and we make summat happen
And we can’t seem to help it
And if you’re looking from history, then yeah we’ve a wealth

But the Manchester way is to make it yourself.
And make us a record, a new number one
And make us a brew while you’re up, love, go on
And make us feel proud that you’re winning the league
And make us sing louder and make us believe that this is the place that has helped shape the world

And so this is the place now with kids of our own. Some are born here, some drawn here, but they all call it home.
And they’ve covered the cobbles, but they’ll never defeat, all the dreamers and schemers who still teem through these streets.
Because this is a place that has been through some hard times: oppressions, recessions, depressions, and dark times.
But we keep fighting back with Greater Manchester spirit. Northern grit, Northern wit, and Greater Manchester’s lyrics.

Tony Walsh

end of a long era [1982-2017]

Posted in Books, pictures, Running, University life with tags , , , , , , , , , , , on May 23, 2017 by xi'an

This afternoon I went to CREST to empty my office there from books and a few papers (like the original manuscript version of Monte Carlo Statistical Methods). This is because the research centre, along with the ENSAE graduate school (my Alma mater), is moving to a new building on the Saclay plateau, next to École Polytechnique. As part of this ambitious migration of engineering schools from downtown Paris to a brand new campus there. Without getting sentimental about this move, it means leaving the INSEE building in Malakoff, on the outskirts of downtown Paris, which has been an enjoyable part of my student and then academic life from 1982 till now. And also leaving the INSEE Paris Club runners! (I am quite uncertain about being as active at the new location, if only because going there by bike is a bit more of a challenge. To be addressed anyway!) And I left behind my accumulation of conference badges (although I should try to recycle them for the incoming BNP 11 in Paris!).

Bacon in the Library [jatp]

Posted in Books, Kids, pictures, Travel, University life with tags , , , , , , , on May 23, 2017 by xi'an

The Riddle of the Sands [not a book review]

Posted in Books, Kids, pictures, Travel with tags , , , , , on May 22, 2017 by xi'an

Visiting Dublin last weekend led me to learn of the sad end of the author of The Riddle of the Sands, (Robert) Erskine Childers.  To my surprise, I indeed found out when reading about the Irish Civil War of the early 20’s that he was executed by a firing squad as a member of the anti-Treaty Sinn Féin. What could have led the author of the role model of classical spy novels, The Riddle of the Sands, to this tragical ending?! While his book was immensely popular in Britain, to the point of impacting the preparations for war in the years before WWI, and while he served as instructor of pilots towards a possible attack of Germany through the very Frisian islands appearing in the novel, he turned progressively towards Irish nationalism, smuggling weapons on his own boat, and opposing the treaty with Britain to  the point of joining the anti-treaty Sinn Féin. When a pistol was found at his home, he was sentenced to death and executed two days later.

continental divide

Posted in Books, Kids, pictures, R with tags , , , , on May 19, 2017 by xi'an

While the Riddler puzzle this week was anticlimactic,  as it meant filling all digits in the above division towards a null remainder, it came as an interesting illustration of how different division is taught in the US versus France: when I saw the picture above, I had to go and check an American primary school on-line introduction to division, since the way I was taught in France is something like that

with the solution being that 12128316 = 124 x 97809… Solved by a dumb R exploration of all constraints:

for (y in 111:143)
for (z4 in 8:9)
for (oz in 0:999){
  z=oz+7e3+z4*1e4
  x=y*z
  digx=digits(x)
  digz=digits(z)
  if ((digz[2]==0)&(x>=1e7)&(x<1e8)){ 
   r1=trunc(x/1e4)-digz[5]*y 
   if ((digz[5]*y>=1e3)&(digz[4]*y<1e4) &(r1>9)&(r1<100)){ 
    r2=10*r1+digx[4]-7*y 
    if ((7*y>=1e2)&(7*y<1e3)&(r2>=1e2)&(r2<1e3)){     
     r3=10*r2+digx[3]-digz[3]*y 
     if ((digz[3]*y>=1e2)&(digz[3]*y<1e3)&(r3>9)&(r3<1e2)){
       r4=10*r3+digx[2]
       if (r4<y) solz=rbind(solz,c(y,z,x))
  }}}}

Looking for a computer-free resolution, the constraints on z exhibited by the picture are that (a) the second digit is 0 and the fourth digit is 7.  Moreover, the first and fifth digits are larger than 7 since y times these digits is a four-digit number. Better, since the second subtraction from a three-digit number by 7y returns a three-digit number and the third subtraction from a four-digit number by ny returns a two-digit number, n is larger than 7 but less than the first and fifth digits. Ergo, z is necessarily 97809! Furthermore, 8y<10³ and 9y≥10³, which means 111<y<125. Plus the constraint that 1000-8y≤99 implies y≥112. Nothing gained there! This leaves 12 values of y to study, unless there is another restriction I missed…