## Le Monde puzzle [#1070]

Posted in Books, Kids, R, University life with tags , , , , , , , on October 11, 2018 by xi'an

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]

Posted in Books, Kids, R with tags , , , , , , , on September 26, 2018 by xi'an

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){
y=x%%(10^dig)
while (keep&(y>1e3)){
y=y%%(10^trunc(log(y,10)))
if (keep) solz=x}}}}
if (solz<10^dig) break()
topsol=max(solz)}


## Le Monde puzzle [#1066]

Posted in Books, Kids, R with tags , , , , , , , on September 13, 2018 by xi'an

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!

## journée algorithmes stochastiques à Dauphine vendredi

Posted in Statistics, University life with tags , , , , , , , , , , on November 28, 2017 by xi'an

A final reminder (?) that we hold a special day of five talks around stochastic algorithms at Dauphine this Friday, from 10:00 till 17:30. Attendance is free, coffee and tea are free (while they last!), come and join us!

## methods for quantifying conflict casualties in Syria

Posted in Books, Statistics, University life with tags , , , , , , , , , , on November 3, 2014 by xi'an

On Monday November 17, 11am, Amphi 10, Université Paris-Dauphine,  Rebecca Steorts from CMU will give a talk at the GT Statistique et imagerie seminar:

Information about social entities is often spread across multiple large databases, each degraded by noise, and without unique identifiers shared across databases.Entity resolution—reconstructing the actual entities and their attributes—is essential to using big data and is challenging not only for inference but also for computation.

In this talk, I motivate entity resolution by the current conflict in Syria. It has been tremendously well documented, however, we still do not know how many people have been killed from conflict-related violence. We describe a novel approach towards estimating death counts in Syria and challenges that are unique to this database. We first introduce computational speed-ups to avoid all-to-all record comparisons based upon locality-sensitive hashing from the computer science literature. We then introduce a novel approach to entity resolution by discovering a bipartite graph, which links manifest records to a common set of latent entities. Our model quantifies the uncertainty in the inference and propagates this uncertainty into subsequent analyses. Finally, we speak to the success and challenges of solving a problem that is at the forefront of national headlines and news.

This is joint work with Rob Hall (Etsy), Steve Fienberg (CMU), and Anshu Shrivastava (Cornell University).

[Note that Rebecca will visit the maths department in Paris-Dauphine for two weeks and give a short course in our data science Master on data confidentiality, privacy and statistical disclosure (syllabus).]

## vector quantile regression

Posted in pictures, Statistics, University life with tags , , , , , , , on July 4, 2014 by xi'an

My Paris-Dauphine colleague Guillaume Carlier recently arXived a statistics paper entitled Vector quantile regression, co-written with Chernozhukov and Galichon. I was most curious to read the paper as Guillaume is primarily a mathematical analyst working on optimisation problems like optimal transport. And also because I find quantile regression difficult to fathom as a statistical problem. (As it happens, both his co-authors are from econometrics.) The results in the paper are (i) to show that a d-dimensional (Lebesgue) absolutely continuous random variable Y can always be represented as the deterministic transform Y=Q(U), where U is a d-dimensional [0,1] uniform (the paper expresses this transform as conditional on a set of regressors Z, but those essentially play no role) and Q is monotonous in the sense of being the gradient of a convex function,

$Q(u) = \nabla q(u)$ and $\{Q(u)-Q(v)\}^\text{T}(u-v)\ge 0;$

(ii) to deduce from this representation a unique notion of multivariate quantile function; and (iii) to consider the special case when the quantile function Q can be written as the linear

$\beta(U)^\text{T}Z$

where β(U) is a matrix. Hence leading to an estimation problem.

While unsurprising from a measure theoretic viewpoint, the representation theorem (i) is most interesting both for statistical and simulation reasons. Provided the function Q can be easily estimated and derived, respectively. The paper however does not provide a constructive tool for this derivation, besides indicating several characterisations as solutions of optimisation problems. From a statistical perspective, a non-parametric estimation of  β(.) would have useful implications in multivariate regression, although the paper only considers the specific linear case above. Which solution is obtained by a discretisation of all variables and  linear programming.