## Le Monde puzzle [#905]

Posted in Books, Kids, R, Statistics, University life with tags , , , on April 1, 2015 by xi'an

A recursive programming  Le Monde mathematical puzzle:

Given n tokens with 10≤n≤25, Alice and Bob play the following game: the first player draws an integer1≤m≤6 at random. This player can then take 1≤r≤min(2m,n) tokens. The next player is then free to take 1≤s≤min(2r,n-r) tokens. The player taking the last tokens is the winner. There is a winning strategy for Alice if she starts with m=3 and if Bob starts with m=2. Deduce the value of n.

Although I first wrote a brute force version of the following code, a moderate amount of thinking leads to conclude that the person given n remaining token and an adversary choice of m tokens such that 2m≥n always win by taking the n remaining tokens:

optim=function(n,m){

outcome=(n<2*m+1)
if (n>2*m){
for (i in 1:(2*m))
outcome=max(outcome,1-optim(n-i,i))
}
return(outcome)
}


eliminating solutions which dividers are not solutions themselves:

sol=lowa=plura[plura<100]
for (i in 3:6){
sli=plura[(plura>10^(i-1))&(plura<10^i)]
ace=sli-10^(i-1)*(sli%/%10^(i-1))
lowa=sli[apply(outer(ace,lowa,FUN="=="),
1,max)==1]
lowa=sort(unique(lowa))
sol=c(sol,lowa)}


> subs=rep(0,16)
> for (n in 10:25) subs[n-9]=optim(n,3)
> for (n in 10:25) if (subs[n-9]==1) subs[n-9]=1-optim(n,2)
> subs
[1] 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
> (10:25)[subs==1]
[1] 18


Ergo, the number of tokens is 18!

## Le Monde puzzle [#902]

Posted in Books, Kids, Statistics, University life with tags , , , , , , on March 8, 2015 by xi'an

Another arithmetics Le Monde mathematical puzzle:

From the set of the integers between 1 and 15, is it possible to partition it in such a way that the product of the terms in the first set is equal to the sum of the members of the second set? can this be generalised to an arbitrary set {1,2,..,n}? What happens if instead we only consider the odd integers in those sets?.

I used brute force by looking at random for a solution,

pb <- txtProgressBar(min = 0, max = 100, style = 3)
for (N in 5:100){
sol=FALSE
while (!sol){
k=sample(1:N,1,prob=(1:N)*(N-(1:N)))
pro=sample(1:N,k)
sol=(prod(pro)==sum((1:N)[-pro]))
}
setTxtProgressBar(pb, N)}
close(pb)


and while it took a while to run the R code, it eventually got out of the loop, meaning there was at least one solution for all n’s between 5 and 100. (It does not work for n=1,2,3,4, for obvious reasons.) For instance, when n=15, the integers in the product part are either 3,5,7, 1,7,14, or 1,9,11. Jean-Louis Fouley sent me an explanation:  when n is odd, n=2p+1, one solution is (1,p,2p), while when n is even, n=2p, one solution is (1,p-1,2p).

A side remark on the R code: thanks to a Cross Validated question by Paulo Marques, on which I thought I had commented on this blog, I learned about the progress bar function in R, setTxtProgressBar(), which makes running R code with loops much nicer!

For the second question, I just adapted the R code to exclude even integers:

while (!sol){
k=1+trunc(sample(1:N,1)/2)
pro=sample(seq(1,N,by=2),k)
cum=(1:N)[-pro]
sol=(prod(pro)==sum(cum[cum%%2==1]))
}


and found a solution for n=15, namely 1,3,15 versus 5,7,9,11,13. However, there does not seem to be a solution for all n’s: I found solutions for n=15,21,23,31,39,41,47,49,55,59,63,71,75,79,87,95…

## amazing Gibbs sampler

Posted in Books, pictures, R, Statistics, University life with tags , , , , , , on February 19, 2015 by xi'an

When playing with Peter Rossi’s bayesm R package during a visit of Jean-Michel Marin to Paris, last week, we came up with the above Gibbs outcome. The setting is a Gaussian mixture model with three components in dimension 5 and the prior distributions are standard conjugate. In this case, with 500 observations and 5000 Gibbs iterations, the Markov chain (for one component of one mean of the mixture) has two highly distinct regimes: one that revolves around the true value of the parameter, 2.5, and one that explores a much broader area (which is associated with a much smaller value of the component weight). What we found amazing is the Gibbs ability to entertain both regimes, simultaneously.

## Le Monde puzzle [#899]

Posted in Books, Kids, Statistics, University life with tags , , , , , on February 8, 2015 by xi'an

An arithmetics Le Monde mathematical puzzle:

For which n’s are the averages of the first n squared integers integers? Among those, which ones are perfect squares?

An easy R code, for instance

n=10^3
car=as.integer(as.integer(1:n)^2)
sumcar=as.integer((cumsum(car)%/%as.integer(1:n)))
diff=as.integer(as.integer(cumsum(car))-as.integer(1:n)*sumcar)
print((1:n)[diff==00])


which produces 333 values

  [1]   1   5   7  11  13  17  19  23  25  29  31  35  37  41  43  47  49  53
[19]  55  59  61  65  67  71  73  77  79  83  85  89  91  95  97 101 103 107
[37] 109 113 115 119 121 125 127 131 133 137 139 143 145 149 151 155 157 161
[55] 163 167 169 173 175 179 181 185 187 191 193 197 199 203 205 209 211 215
[73] 217 221 223 227 229 233 235 239 241 245 247 251 253 257 259 263 265 269
[91] 271 275 277 281 283 287 289 293 295 299 301 305 307 311 313 317 319 323
[109] 325 329 331 335 337 341 343 347 349 353 355 359 361 365 367 371 373 377
[127] 379 383 385 389 391 395 397 401 403 407 409 413 415 419 421 425 427 431
[145] 433 437 439 443 445 449 451 455 457 461 463 467 469 473 475 479 481 485
[163] 487 491 493 497 499 503 505 509 511 515 517 521 523 527 529 533 535 539
[181] 541 545 547 551 553 557 559 563 565 569 571 575 577 581 583 587 589 593
[199] 595 599 601 605 607 611 613 617 619 623 625 629 631 635 637 641 643 647
[217] 649 653 655 659 661 665 667 671 673 677 679 683 685 689 691 695 697 701
[235] 703 707 709 713 715 719 721 725 727 731 733 737 739 743 745 749 751 755
[253] 757 761 763 767 769 773 775 779 781 785 787 791 793 797 799 803 805 809
[271] 811 815 817 821 823 827 829 833 835 839 841 845 847 851 853 857 859 863
[289] 865 869 871 875 877 881 883 887 889 893 895 899 901 905 907 911 913 917
[307] 919 923 925 929 931 935 937 941 943 947 949 953 955 959 961 965 967 971
[325] 973 977 979 983 985 989 991 995 997


which are made of all odd integers that are not multiple of 3. (I could have guessed the exclusion of even numbers since the numerator is always odd. Why are the triplets excluded, now?! Jean-Louis Fouley gave me the answer: the sum of squares is such that

$\frac{1+2^2+\cdots+m^2}{m}=\frac{m(m+1)(2m+1)}{6m}=\frac{(m+1)(2m+1)}{6}$

and hence m must be odd and 2m+1 a multiple of 3, which excludes multiples of 3.)

The second part is as simple:

sole=sumcar[(1:n)[diff==0]]
scar=as.integer(as.integer(sqrt(sole))^2)-sole
sum(scar==0)


with the final result

> sum(scar==0)
[1] 2
> ((1:n)[diff==0])[scar==0]
[1] 1 337


since  38025=195² is a perfect square. (I wonder if there is a plain explanation for that result!)

## Le Monde puzzle [#887]

Posted in Books, Kids, R, Statistics with tags , , , on November 15, 2014 by xi'an

A simple combinatorics Le Monde mathematical puzzle:

N is a golden number if the sequence {1,2,…,N} can be reordered so that the sum of any consecutive pair is a perfect square. What are the golden numbers between 1 and 25?

Indeed, from an R programming point of view, all I have to do is to go over all possible permutations of {1,2,..,N} until one works or until I have exhausted all possible permutations for a given N. However, 25!=10²⁵ is a wee bit too large… Instead, I resorted once again to brute force simulation, by first introducing possible neighbours of the integers

  perfect=(1:trunc(sqrt(2*N)))^2
friends=NULL
le=1:N
for (perm in 1:N){
amis=perfect[perfect>perm]-perm
amis=amis[amis<N]
le[perm]=length(amis)
friends=c(friends,list(amis))
}


and then proceeding to construct the permutation one integer at time by picking from its remaining potential neighbours until there is none left or the sequence is complete

orderin=0*(1:N)
t=1
orderin[1]=sample((1:N),1)
for (perm in 1:N){
friends[[perm]]=friends[[perm]]
[friends[[perm]]!=orderin[1]]
le[perm]=length(friends[[perm]])
}
while (t<N){
if (length(friends[[orderin[t]]])==0)
break()
if (length(friends[[orderin[t]]])>1){
orderin[t+1]=sample(friends[[orderin[t]]],1)}else{
orderin[t+1]=friends[[orderin[t]]]
}
for (perm in 1:N){
friends[[perm]]=friends[[perm]]
[friends[[perm]]!=orderin[t+1]]
le[perm]=length(friends[[perm]])
}
t=t+1}


and then repeating this attempt until a full sequence is produced or a certain number of failed attempts has been reached. I gained in efficiency by proposing a second completion on the left of the first integer once a break occurs:

while (t<N){
if (length(friends[[orderin[1]]])==0)
break()
orderin[2:(t+1)]=orderin[1:t]
if (length(friends[[orderin[2]]])>1){
orderin[1]=sample(friends[[orderin[2]]],1)}else{
orderin[1]=friends[[orderin[2]]]
}
for (perm in 1:N){
friends[[perm]]=friends[[perm]]
[friends[[perm]]!=orderin[1]]
le[perm]=length(friends[[perm]])
}
t=t+1}


(An alternative would have been to complete left and right by squared numbers taken at random…) The result of running this program showed there exist permutations with the above property for N=15,16,17,23,25,26,…,77.  Here is the solution for N=49:

25 39 10 26 38 43 21 4 32 49 15 34 30 6 3 22 42 7 9 27 37 12 13 23 41 40 24 1 8 28 36 45 19 17 47 2 14 11 5 44 20 29 35 46 18 31 33 16 48

As an aside, the authors of Le Monde puzzle pretended (in Tuesday, Nov. 12, edition) that there was no solution for N=23, while this sequence

22 3 1 8 17 19 6 10 15 21 4 12 13 23 2 14 11 5 20 16 9 7 18

sounds fine enough to me… I more generally wonder at the general principle behind the existence of such sequences. It sounds quite probable that they exist for N>24. (The published solution does not bring any light on this issue, so I assume the authors have no mathematical analysis to provide.)

## The winds of Winter [Bayesian prediction]

Posted in Books, Kids, R, Statistics, University life with tags , , , , , , , , , , on October 7, 2014 by xi'an

A surprising entry on arXiv this morning: Richard Vale (from Christchurch, NZ) has posted a paper about the characters appearing in the yet hypothetical next volume of George R.R. Martin’s Song of ice and fire series, The winds of Winter [not even put for pre-sale on amazon!]. Using the previous five books in the series and the frequency of occurrence of characters’ point of view [each chapter being told as from the point of view of one single character], Vale proceeds to model the number of occurrences in a given book by a truncated Poisson model,

$x_{it} \sim \mathcal{P}(\lambda_i)\text{ if }|t-\beta_i|<\tau_i$

in order to account for [most] characters dying at some point in the series. All parameters are endowed with prior distributions, including the terrible “large” hyperpriors familiar to BUGS users… Despite the code being written in R by the author. The modelling does not use anything but the frequencies of the previous books, so knowledge that characters like Eddard Stark had died is not exploited. (Nonetheless, the prediction gives zero chapter to this character in the coming volumes.) Interestingly, a character who seemingly died at the end of the last book is still given a 60% probability of having at least one chapter in  The winds of Winter [no spoiler here, but many in the paper itself!]. As pointed out by the author, the model as such does not allow for prediction of new-character chapters, which remains likely given Martin’s storytelling style! Vale still predicts 11 new-character chapters, which seems high if considering the series should be over in two more books [and an unpredictable number of years!].

As an aside, this paper makes use of the truncnorm R package, which I did not know and which is based on John Geweke’s accept-reject algorithm for truncated normals that I (independently) proposed a few years later.

## a weird beamer feature…

Posted in Books, Kids, Linux, R, Statistics, University life with tags , , , , , , , , , , , , on September 24, 2014 by xi'an

As I was preparing my slides for my third year undergraduate stat course, I got a weird error that got a search on the Web to unravel:

! Extra }, or forgotten \endgroup.
\endframe ->\egroup
\begingroup \def \@currenvir {frame}
l.23 \end{frame}
\begin{slide}
?


which was related with a fragile environment

\begin{frame}[fragile]
\frametitle{simulation in practice}
\begin{itemize}
\item For a given distribution $F$, call the corresponding
pseudo-random generator in an arbitrary computer language
\begin{verbatim}
> x=rnorm(10)
> x
[1] -0.021573 -1.134735  1.359812 -0.887579
[7] -0.749418  0.506298  0.835791  0.472144
\end{verbatim}
\item use the sample as a statistician would
\begin{verbatim}
> mean(x)
[1] 0.004892123
> var(x)
[1] 0.8034657
\end{verbatim}
to approximate quantities related with $F$
\end{itemize}
\end{frame}\begin{frame}


but not directly the verbatim part: the reason for the bug was that the \end{frame} command did not have a line by itself! Which is one rare occurrence where the carriage return has an impact in LaTeX, as far as I know… (The same bug appears when there is an indentation at the beginning of the line. Weird!) [Another annoying feature is wordpress turning > into &gt; in the sourcecode environment…]