Le Monde puzzle [#1073]

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

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]

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
  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]

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]

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]

A purely (?) algorithmic Le Monde mathematical puzzle

For the table below, what is the minimal number of steps required to reach equal entries when each step consists in adding ones to three entries sitting in a L, such as (7,11,12) or (5,6,10)? Same question for the inner table of four in yellow.

For the inner table, this is straightforward as there are four possible L’s, three equations like 6+n⁶=7+n⁷,  and two degrees of freedom leading to a unique entry of N=13 (impossible!)  or 16 (feasible). Hence adding 10 L’s. For the entire table, summing up all entries after completion leads to 16N, which is also equal to 1+3+3+…+16+M, where M is the number of added L’s, itself equal to 138+3O, if O denotes the number of ones added. Hence M can only take the values 18, 21, … It took me quite a while to R code an approach to complete the table into 16 18’s, as my versions of simulated annealing did not seem to converge. In the end, I used a simplified version where the table was completed by multinomial draws, like a M(17;3⁻¹,3⁻¹,3⁻¹) for the upper left corner, corresponding to random draws of one of the 36 available L’s, which should be used 50 times in total, and then augmented or reduced of one L depending on the value at a randomly selected entry. Leading to the result

> aneal(grid=c(1,3,3:13,15,15,16),maxT=1e5)
 [1] 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18

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 [#1067]

The second Le Monde mathematical puzzle in the new competition is sheer trigonometry:

When in the above figures both triangles ABC are isosceles and the brown segments are all of length 25cm, find the angle in A and the value of DC², respectively.

This could have been solved by R coding the various possible angles of the three segments beyond BC until the isosceles property is met, but it went much faster by pen and paper, leading to an angle of π/9 in the first case and a square of 1250 in the second case. The third puzzle is basic arithmetic that only seems solvable by enumeration…