## Archive for competition

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

Posted in Statistics with tags competition, jatp, landscape photography, Old Man of Storr, Université Paris Dauphine, Vert le Futur on February 17, 2019 by xi'an## Le Monde puzzle [#1078]

Posted in Books, Kids, R with tags competition, Conservatoire National des Arts et Métiers, Le Monde, logic, mathematical puzzle, prizes, reward on November 29, 2018 by xi'an**R**ecalling 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 CEREMADE, competition, Dauphine, Le Monde, mathematical puzzle, optimisation, R, simulated annealing on November 22, 2018 by xi'an**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 [#1073]

Posted in Books, Kids, R with tags combinatorics, competition, Fibonacci, Le Monde, mathematical puzzle, matrix multiplication, R on November 3, 2018 by xi'an**A**nd 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

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 area of a triangle, brute-force solution, circumscribed circle, competition, discretisation, draw.circle, geometry, Heron's formula, Le Monde, mathematical puzzle, plotrix library, R on October 31, 2018 by xi'an**T**he 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.

**A**lthough 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 radi=sqrt(diz^2/2+(diz/sqrt(2)-10)^2)-10 if (abs(20-(diz/sqrt(2))-radi)<3e-3){ print(c(radi,diz));break()}}

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)){ radi=10*sqrt(2)-diz distA=sqrt((diz/sqrt(2)-4)^2+(diz/sqrt(2)-8)^2) if (abs(distA-radi)<4e-4){ print(c(radi,diz));break()}}

even though Heron’s formula was enough to produce the radius. In both approaches, this radius is 3.54, with the position of the centre O’at 10.6 from the lower left corner. When plotting these results, which showed consistency at this level of precision, I was reminded of the importance of plotting the points with the option “asp=1” in plot() as otherwise, draw.circle() does not plot the circles at the right location!

## Le Monde puzzle [#1071]

Posted in Books, Kids with tags affaire de logique, Alice and Bob, birthday problem, CEREMADE, competition, Dauphine, Le Monde, logic, mathematical puzzle, she said he said, simulated annealing, Singapore on October 18, 2018 by xi'an**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!):

Arwen and Brandwein are privately and respectively given the day and month of Caradoc’s birthday [in the Gregorian calendar] with the information that the month number is at least the day number. Arwen starts by stating she knows Brandwein cannot deduce the birthday, followed by Brandwein who says the same about Arwen. If this “she says he says” goes on for the largest possible number of steps before Arwen says she knows, when is Caradoc’s birthday? Arwen and Brandwein are later given two numbers corresponding to Deirdre’s birthday with no indication of which one is the day and which one is the month. They know both numbers end up with the same digit and that the month number is strictly less than the day number. Arwen states she does not know the date and she knows Brandwein cannot know either. Then Brandwein says he indeed does not the date but he knows whether he got the day or the month. This prompts Arwen to conclude she knows, then Brandwein to do the same. When is Deirdre’s birthday?

Since this was a fairly easy puzzle (and since I had spent too much time debugging the previous R code!), I did not try coding this one but instead drew the possibilities and remove the impossibilities on a blackboard. The first question is quite simple actually since the day numbers stand between 1 and 12 and that each “I cannot know” excludes one of the remaining endpoints, removing first excludes 1 from both lists, then 12, then 2, then …. 8, ending up with 7. And 07/07 as Caradoc’s birthday. The second case sees 13,…,20,23,…,30 eliminated from Arwen’s numbers, then 3,…,10 as well, which eliminates the same numbers from Brandwein’s possibilities. That he knows whether it is a month or a day leaves only 1,2,21,22,31 as possible numbers. Then Arwen’s certainty reduces her numbers to be 2, 21, 22, or 31, and since Brandwein is also sure, the only possible cases are (2,22) and (22,2). Meaning Deirdre’s birthday is on 22/02. I dunno if this symmetry was to be expected! (And I cannot fathom why this puzzle is awarded so many points, when compared with the others.)

## Le Monde puzzle [#1070]

Posted in Books, Kids, R, University life with tags CEREMADE, competition, Dauphine, dynamic programming, Le Monde, mathematical puzzle, optimisation, R on October 11, 2018 by xi'an**R**ewording 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.