Le Monde puzzle [#1075]

A Le Monde mathematical puzzle from after the competition:

A sequence of five integers can only be modified by subtracting an integer N from two neighbours of an entry and adding 2N to the entry.  Given the configuration below, what is the minimal number of steps to reach non-negative entries everywhere? Is this feasible for any configuration?

As I quickly found a solution by hand in four steps, but missed the mathematical principle behind!, I was not very enthusiastic in trying a simulated annealing version by selecting the place to change inversely proportional to its value, but I eventually tried and also obtained the same solution:

      [,1] [,2] [,3] [,4] [,5]
   -3    1    1    1    1
    1   -1    1    1   -1
    0    1    0    1   -1
   -1    1    0    0    1
    1    0    0    0    0

But (update!) Jean-Louis Fouley came up with one step less!

      [,1] [,2] [,3] [,4] [,5]
   -3    1    1    1    1
    3   -2    1    1   -2
    2    0    0    1   -2
    1    0    0    0    0

The second part of the question is more interesting, but again without a clear mathematical lead, I could only attempt a large number of configurations and check whether all admitted “solutions”. So far none failed.

Le Monde puzzle [#1070]

Rewording Le Monde mathematical puzzle  fifth competition problem

For the 3×3 tables below, what are the minimal number of steps to move from left to rights when the yellow tokens can only be move to an empty location surrounded by two other tokens?

In the 4×4 table below, there are 6 green tokens. How many steps from left to right?

Painful and moderately mathematical, once more… For the first question, a brute force simulation of random valid moves of length less than 100 returns solutions in 4 steps:

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
     1    1    1    0    0    1    0    0    1
     1    0    1    0    1    1    0    0    1
     0    0    1    1    1    1    0    0    1
     0    0    1    1    0    1    0    1    1
     0    0    1    0    0    1    1    1    1

But this is not an acceptable move because of the “other” constraint. Imposing this constraint leads to a solution in 9 steps, but is this the lowest bound?! It actually took me most of the weekend (apart from a long drive to and from a short half-marathon!) to figure out a better strategy than brute force random exploration: the trick I eventually figured out is to start from the finishing (rightmost) value F of the grid and look at values with solutions available in 1,2,… steps. This is not exactly dynamic programming, but it keeps the running time under control if there is a solution associated with the starting (leftmost) value S. (Robin proceeded reverse-wise, which in retrospect is presumably equivalent, if faster!) The 3×3 grid has 9 choose 5, ie 126, possible configurations with 5 coins, which means the number of cases remains under control. And even so for the 4×4 grid with 6 coins, with 8008 configurations. This led to a 9 step solution for n=3 and the proposed starting grid in yellow:

[1] 1 1 1 0 0 1 0 0 1
[1] 1 1 0 0 1 1 0 0 1
[1] 1 1 0 1 1 0 0 0 1
[1] 0 1 0 1 1 1 0 0 1
[1] 0 1 1 1 0 1 0 0 1
[1] 1 1 1 1 0 0 0 0 1
[1] 0 1 1 1 1 0 0 0 1
[1] 0 0 1 1 1 0 0 1 1
[1] 0 0 1 1 0 0 1 1 1
[1] 0 0 1 0 0 1 1 1 1

and a 19 step solution for n=4:

[1] 1 0 0 0 0 1 0 0 0 1 1 1 0 0 0 1
[1] 1 0 0 0 0 1 0 0 0 0 1 1 0 0 1 1
[1] 1 0 0 0 0 1 1 0 0 0 1 1 0 0 1 0
[1] 1 1 0 0 0 1 1 0 0 0 0 1 0 0 1 0
[1] 1 1 0 0 0 0 1 1 0 0 0 1 0 0 1 0
[1] 1 1 1 0 0 0 1 1 0 0 0 0 0 0 1 0
[1] 1 0 1 1 0 0 1 1 0 0 0 0 0 0 1 0
[1] 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 0
[1] 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 0
[1] 1 0 0 1 1 1 1 0 0 0 0 0 0 0 1 0
[1] 0 0 0 1 1 1 1 0 0 0 1 0 0 0 1 0
[1] 0 0 0 1 1 1 0 0 0 1 1 0 0 0 1 0
[1] 0 0 0 1 1 1 0 0 1 1 0 0 0 0 1 0
[1] 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 0
[1] 0 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0
[1] 0 0 0 1 1 1 0 0 1 0 0 0 1 0 1 0
[1] 0 0 0 1 0 1 0 0 1 1 0 0 1 0 1 0
[1] 0 0 0 1 0 1 0 0 0 1 1 0 1 0 1 0
[1] 0 0 0 1 0 1 1 0 0 1 1 0 1 0 0 0

The first resolution takes less than a minute and the second one just a few minutes (or less than my short morning run!). Surprisingly, using this approach does not require more work, which makes me wonder at the solution Le Monde journalists will propose. Given the (misguided) effort put into my resolution, seeing a larger number of points for this puzzle is no wonder.

Le Monde puzzle [#1068]

And here is the third Le Monde mathematical puzzle  open for competition:

Consider for this puzzle only integers with no zero digits. Among these an integer x=a¹a²a³… is refined if it is a multiple of its scion, the integer defined as x without the first digit, y=a²a³…. Find the largest refined integer x such the sequence of the successive scions of x with more than one digit is entirely refined. An integer x=a¹a²a… is distilled if it is a multiple of its grand-scion, the integer defined as x without the first two digits, z=a³… Find the largest distilled integer x such the sequence of the successive scions of x with more than two digits is entirely distilled.

Another puzzle amenable to a R resolution by low-tech exploration of possible integers, first by finding refined integers, with  no solution between 10⁶ and 10⁸ [meaning there is no refined integer larger than 10⁶] and then checking which refined integers have refined descendants, e.g.,

raf=NULL
for (x in (1e1+1):(1e6-1)){
  y=x%%(10^trunc(log(x,10)))
  if (y>0){
    if (x%%y==0)
      raf=c(raf,x)}}

that returns 95 refined integers. And then

for (i in length(raf):1){
  gason=raf[i]
  keep=(gason%in%raf)
  while (keep&(gason>100)){
    gason=gason%%(10^trunc(log(gason,10)))
    keep=keep&(gason%in%raf)}
  if (keep) break()}}

that returns 95,625 as the largest refined integer with the right descendance. Rather than finding all refined integers at once, going one digit at a time and checking that some solutions have proper descendants is actually faster.

Similarly, running an exploration code up to 10⁹ produces 1748 distilled integers, with maximum 9,977,34,375, so it is unlikely this is the right upper bound but among these the maximum with the right distilled descendants is 81,421,875. As derived by

rad=(1:99)[(1:99)%%10>0]
for (dig in 2:12){
 for (x in ((10^dig+1):(10^{dig+1}-1))){
  y=x%%(10^{dig-1})
  if (y>0){
   if (x%%y==0){
    if (min(as.integer(substring(x,seq(nchar(x)),seq(nchar(x)))))>0){
     rad=c(rad,x)
     y=x%%(10^dig)
     keep=(y%in%rad)
     while (keep&(y>1e3)){
       y=y%%(10^trunc(log(y,10)))
       keep=keep&(y%in%rad)}
     if (keep) solz=x}}}}
 if (solz<10^dig) break()
 topsol=max(solz)}

Le Monde puzzle [#1066]

Recalling Le Monde mathematical puzzle  first competition problem

For the X table below, what are the minimal number of lights that are on (green) to reach the minimal and maximal possible numbers of entries (P) with an even (P as pair) number of neighbours with lights on? In the illustration below, there are 16 lights on (green) and 31 entries with an even number of on-neighbours.

As suggested last week, this was amenable to a R resolution by low-tech simulated annealing although the number of configurations was not that large when accounting for symmetries. The R code is a wee bit long for full reproduction here but it works on moving away from a random filling of this cross by 0’s and 1’s, toward minimising or maximising the number of P’s, this simulated annealing loop being inserted in another loop recording the minimal number of 1’s in both cases. A first round produced 1 and 44 for the minimal and maximal numbers of P’s, respectively, associated with at least 16 and 3 1’s, respectively, but a longer run exhibited 45 for 6 1’s crossing one of the diagonals of the X, made of two aligned diagonals of the outer 3×3 tables. (This was confirmed by both Amic and Robin in Dauphine!) The next [trigonometry] puzzle is on!

optimal approximations for importance sampling

“…building such a zero variance estimator is most of the times not practical…”

As I was checking [while looking at Tofino inlet from my rental window] on optimal importance functions following a question on X validated, I came across this arXived note by Pantaleoni and Heitz, where they suggest using weighted sums of step functions to reach minimum variance. However, the difficulty with probability densities that are step functions is that they necessarily have a compact support, which thus make them unsuitable for targeted integrands with non-compact support. And making the purpose of the note and the derivation of the optimal weights moot. It points out its connection with the reference paper of Veach and Guibas (1995) as well as He and Owen (2014), a follow-up to the other reference paper by Owen and Zhou (2000).

Bayesian gan [gan style]

In their paper Bayesian GANS, arXived a year ago, Saatchi and Wilson consider a Bayesian version of generative adversarial networks, putting priors on both the model and the discriminator parameters. While the prospect seems somewhat remote from genuine statistical inference, if the following statement is representative

“GANs transform white noise through a deep neural network to generate candidate samples from a data distribution. A discriminator learns, in a supervised manner, how to tune its parameters so as to correctly classify whether a given sample has come from the generator or the true data distribution. Meanwhile, the generator updates its parameters so as to fool the discriminator. As long as the generator has sufficient capacity, it can approximate the cdf inverse-cdf composition required to sample from a data distribution of interest.”

I figure the concept can also apply to a standard statistical model, where x=G(z,θ) rephrases the distributional assumption x~F(x;θ) via a white noise z. This makes resorting to a prior distribution on θ more relevant in the sense of using potential prior information on θ (although the successes of probabilistic numerics show formal priors can be used on purely numerical ground).

The “posterior distribution” that is central to the notion of Bayesian GANs is however unorthodox in that the distribution is associated with the following conditional posteriors

where D(x,θ) is the “discriminator”, that is, in GAN lingo, the probability to be allocated to the “true” data generating mechanism rather than to the one associated with G(·,θ). The generative conditional posterior (1) then aims at fooling the discriminator, i.e. favours generative parameter values that raise the probability of wrong allocation of the pseudo-data. The discriminative conditional posterior (2) is a standard Bayesian posterior based on the original sample and the generated sample. The authors then iteratively sample from these posteriors, effectively implementing a two-stage Gibbs sampler.

“By iteratively sampling from (1) and (2) at every step of an epoch one can, in the limit, obtain samples from the approximate posteriors over [both sets of parameters].”

What worries me about this approach is that  just cannot work, in the sense that (1) and (2) cannot be compatible conditional (posterior) distributions. There is no joint distribution for which (1) and (2) would be the conditionals, since the pseudo-data appears in D for (1) and (1-D) in (2). This means that the convergence of a Gibbs sampler is at best to a stationary σ-finite measure. And hence that the meaning of the chain is delicate to ascertain… Am I missing any fundamental point?! [I checked the reviews on NIPS webpage and could not spot this issue being raised.]

are profile likelihoods likelihoods?!

A recent arXived paper by Oliver J. Maclaren is asking this very question. And argue for a positive answer. One of the invoked sources is Murray Aitkin’s integrated likelihood book, which I criticised here and elsewhere. With the idea of the paper being that

“….there is an appropriate notion of integration over variables that takes likelihood functions to likelihood functions via maximization.”

Hmm…. The switch there is to replace addition with maximisation, probability with possibility, and… profile likelihood as marginal possibility under this new concept. I just do not see how adapting these concepts for the interpretation of the profile likelihood makes the latter more meaningful, since it still overwhelmingly does not result from a distribution density at an observed realisation of a random variable. This reminds me a paper I refereed quite a long while ago where the authors were using Schwarz’ theory of distributions to expand the notion of unbiasedness. With unclear consequences.