## Calanques, calanques [aka 40/40]

Posted in Mountains, pictures, Running, Travel, University life with tags , , , , , , , , , , , on May 17, 2019 by xi'an

## 50/50 photography competition [another public image]

Posted in Statistics with tags , , , , , on February 17, 2019 by xi'an ## Le Monde puzzle [#1078]

Posted in Books, Kids, R with tags , , , , , , on November 29, 2018 by xi'an Recalling Le Monde mathematical puzzle  first competition problem

Given yay/nay answers to the three following questions about the integer 13≤n≤1300 (i) is the integer n less than 500? (ii) is n a perfect square? (iii) is n a perfect cube?  n cannot be determined, but it is certain that any answer to the fourth question (iv) are all digits of n distinct? allows to identify n. What is n if the answer provided for (ii) was false.

When looking at perfect squares less than 1300 (33) and perfect cubes less than 1300 (8), there exists one single common integer less than 500 (64) and one single above (729). Hence, it is not possible that answers to (ii) and (iii) are both positive, since the final (iv) would then be unnecessary. If the answer to (ii) is negative and the answer to (iii) is positive, it would mean that the value of n is either 512 or 10³ depending on the answer to (i), excluding numbers below 500 since there is no unicity even after (iv). When switching to a positive answer to (ii), this produces 729 as the puzzle solution.

Incidentally, while Amic, Robin, and I finished among the 25 ex-aequos of the competition, none of us reached the subsidiary maximal number of points to become the overall winner. It may be that I will attend the reward ceremony at Musée des Arts et Métiers next Sunday.

## Le Monde puzzle [#1075]

Posted in Books, Kids, R with tags , , , , , , , on November 22, 2018 by xi'an

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 [#1073]

Posted in Books, Kids, R with tags , , , , , , on November 3, 2018 by xi'an And here is Le Monde mathematical puzzle  last competition problem

Find the number of integers such that their 15 digits are all between 1,2,3,4, and the absolute difference between two consecutive digits is 1. Among these numbers how many have 1 as their three-before-last digit and how many have 2?

Combinatorics!!! While it seems like a Gordian knot because the number of choices depends on whether or not a digit is an endpoint (1 or 4), there is a nice recurrence relation between the numbers of such integers with n digits and leftmost digit i, namely that

n⁴=(n-1)³, n³=(n-1)²+(n-1)⁴, n²=(n-1)²+(n-1)¹, n¹=(n-1)²

with starting values 1¹=1²=1³=1⁴=1 (and hopefully obvious notations). Hence it is sufficient to iterate the corresponding transition matrix $M= \left(\begin{matrix}0 &1 &0 &0\\1 &0 &1 &0\\0 &1 &0 &1\\0 &0 &1 &0\\\end{matrix} \right)$

on this starting vector 14 times (with R, which does not enjoy a built-in matrix power) to end up with

15¹=610, 15²= 987, 15³= 987, 15⁴= 610

which leads to 3194 different numbers as the solution to the first question. As pointed out later by Amic and Robin in Dauphine, this happens to be twice Fibonacci’s sequence.

For the second question, the same matrix applies, with a different initial vector. Out of the 3+5+5+3=16 different integers with 4 digits, 3 start with 1 and 5 with 2. Then multiplying (3,0,0,0) by M¹⁰ leads to 267+165=432 different values for the 15 digit integers and multiplying (0,5,0,0) by M¹⁰ to. 445+720=1165 integers. (With the reassuring property that 432+1165 is half of 3194!) This is yet another puzzle in the competition that is of no special difficulty and with no further challenge going from the first to the second question…

## Le Monde puzzle [#1072]

Posted in Books, Kids, R with tags , , , , , , , , , , , on October 31, 2018 by xi'an The penultimate Le Monde mathematical puzzle  competition problem is once again anti-climactic and not worth its points:

For the figure below [not the original one!], describing two (blue) half-circles intersecting with a square of side 20cm, and a (orange) quarter of a circle with radius 20cm, find the radii of both gold circles, respectively tangent to both half-circles and to the square and going through the three intersections. Although the problem was easily solvable by some basic geometry arguments, I decided to use them a minima and resort to a mostly brute-force exploration based on a discretisation of the 20×20 square, from which I first deduced the radius of the tangent circle by imposing (a) a centre O on the diagonal (b) a distance from O to one (half-)circle equal to the distance to the upper side of the square, leading to a radius of 5.36 (and a centre O at a distance 20.70 from the bottom left corner):

diaz=sqrt(2)*20
for (diz in seq(5/8,7/8,le=1e4)*diaz){#position of O


In the case of the second circle I first deduced the intersections of the different circles by sheer comparison of black (blue!) pixels, namely A(4,8), B(8,4), and C(10,10), then looked at the position of the centre O’ of the circle such that the distances to A, B, and C were all equal:

for (diz in seq(20*sqrt(2)-20,10*sqrt(2),le=1e4)){
distA=sqrt((diz/sqrt(2)-4)^2+(diz/sqrt(2)-8)^2) A “he said she said” Le Monde mathematical puzzle sixth competition problem that reminded me of the “Singapore birthday problem” (nothing to do with the original birthday problem!):