## Le Monde puzzle [#1130]

Posted in Books, Kids, R, Statistics with tags , , , , , , , on February 7, 2020 by xi'an A two-player game as Le weekly Monde current mathematical puzzle:

Abishag and Caleb fill in alternance a row of N boxes in a row by picking one then two then three &tc. consecutive boxes. When a player is unable to find enough consecutive boxes, the player has lost. Who is winning when N=29? When N=30?

Using a basic recursive search for the optimal strategy, with the status of the row and the number of required boxes as entries,

f<-function(b=!1:N,r=0){
for(i in 1:(N-r)){
if(p<-!max(b[j<-i+r:0])){
q=b;q[j]=1
if(p<-!f(q,r+1))break}}
p}


returns Abishag as the winner for N=29 (as well as for N=1,2,7,…,13,19,…,29) and Caleb as the winner for N=30 (as well as for N=3,…,6,14,…,18). I am actually surprised that the recursion operates that deep, even though this means a √N depth for the recursion. While the code took too long to complete, the function operates for N=100. A side phenomenon is the apparent special case of N=47, which makes Abishag the looser, while N=46 and N=48 do not.This is an unusual pattern as otherwise (up to N=59), there are longer and longer stretches of adjacent wins and looses as N increases.

## shortened iterations [code golf]

Posted in Kids, pictures, Statistics, Travel with tags , , , , , , , , on October 29, 2019 by xi'an A codegolf lazy morning exercise towards finding the sequence of integers that starts with an arbitrary value n and gets updated by blocks of four as $a_{4k+1} = a_{4k} \cdot(4k+1)\\ a_{4k+2} = a_{4k+1} + (4k+2)\\ a_{4k+3} = a_{4k+2} - (4k+3)\\ a_{4k+4} = a_{4k+3} / (4k+4)$

until the last term is not an integer. While the update can be easily implemented with the appropriate stopping rule, a simple congruence analysis shows that, depending on n, the sequence is 4, 8 or 12 values long when $n\not\equiv 1(4)\\ n\equiv 1(4)\ \text{and}\ 3(n-1)+4\not\equiv 0(32)\\ 3(n-1)+4\equiv 0(32)$

respectively. But sadly the more interesting fixed length solution

~=rep #redefine function
b=(scan()-1)*c(32~4,8,40~4,1,9~3)/32+c(1,1,3,0~3,6,-c(8,1,9,-71,17)/8)
b[!b%%1] #keep integers only


ends up being longer than the more basic one:

a=scan()
while(!a[T]%%1)a=c(a,d<-a[T]*T,d+T+1,e<-d-1,e/((T<-T+4)-1))
a[-T]


where Robin’s suggestion of using T rather than length is very cool as T has double meaning, first TRUE (and 1) then the length of a…

## Le Monde puzzle [#1105]

Posted in Kids, R with tags , , , , , , on July 8, 2019 by xi'an Another token game as Le Monde mathematical puzzle:

Archibald and Beatrix play with a pile of n>100 tokens, sequentially picking m tokens from the pile with m being a prime number [including m=1] or a multiple of 6, the winner taking the last tokens. If Beatrix knows n and proposes to Archibald to start, what is the value of n?

Which cannot be solved in a few lines of R code:

k<-function(n)n<4||all(n%%2:ceiling(sqrt(n))!=0)||!n%%6
g=(1:3)
n=c(4,i<-4)
while(max(n)<101){
if(k(i)) g=c(g,i) else{
while(i%in%g)i=i+1;j=4;o=!j
while(!o&(j<i)){
o=(j%in%n)&k(i-j);j=j+1}
if(o) g=c(g,i) else n=c(n,i)}
i=i+1}


since it returned no unsuccessful value above 100! With 4, 8, 85, 95, and 99 as predecessors. A rather surprising outcome and a big gap that most certainly has a straightforward explanation! Or a lack of understanding from yours truly: this post appears after the solution was published in Le Monde and I am more bemused than ever since the losing numbers in the journal are given as 4, 8, 85, … 89, and 129. With the slight hiccup that 89 is a prime number…. The other argument in the solution that there can only be five such losers is well-taken since there are only five possible non-zero remainders in the division by 6.

## another attempt at code golf

Posted in Books, Kids, R with tags , , , on June 12, 2019 by xi'an I had another lazy weekend go at code golf, trying to code in the most condensed way the following task. Provided with a square matrix A of positive integers, keep iterating the steps

• take the highest square $$𝑥²$$ in A.
• find the smallest adjacent neighbour $$𝑛$$
• replace with x and n with nx

until no square is left (with neighbour defined as either horizontally or vertically and without wrapping around). While I managed a 217 bytes solution, compared with Robin’s 179b improvement, which remains surprising readable!, the puzzle offers two further questions:

1. is there a non-iterative way to find the final matrix B?
2. the puzzle assumes that A satisfies that at each step, the highest square and the smallest neighbour n will be unique, and that the sequence will not repeat forever. Is there a fool-proof way to check this is the case?

## an attempt at code golf

Posted in Kids, R with tags , , , , , , , , , on May 15, 2019 by xi'an Having discovered codegolf on Stack Exchange a few weeks ago, I spotted a few interesting puzzles since then but only got the opportunity at a try over a quiet and rainy weekend (and Robin being on vacation)! The challenge was to write an R code for deciding whether or not a given integer n is congruent or not, when congruent means that it is the surface of a rectangle triangle with all three sides rational. The question included a pointer to the Birch and Swinnerton-Dyer conjecture as a mean to check congruence although the real solution was provided by Tunnell’s Theorem, which states that n is congruent if and only if the number of integer solutions to 2x²+y²+8z²=n is twice as much as the number of integer solutions to 2x²+y²+32z²=n if n is odd and  the number of integer solutions to 8x²+y²+16z²=n is twice as much as the number of integer solutions to 8x²+y²+64z²=n if n is even. Although this is only true for squared-free integers. (I actually spent more time on figuring out the exact wording of the theorem than on optimising the R code!)

My original solution

p=function(n){
for (i in(n:2)^2)if(n%%i<1)n=n/i
if(n%%2){d=8;f=2;g=16}else{d=2;f=1;g=8}
A=0;b=(-n:n)^2
for(x in d*b)for(y in x+f*b)for(z in g*b)
A=A+(y+z==n)-2*(y+4*z==n)
A==0}


was quite naïve, as shown by the subsequent improvements by senior players, like the final (?) version of Guiseppe:

function(n){b=(-n:n)^2
for(i in b[b>0])n=n/i^(!n%%i)
P=2^(n%%2)
o=outer
!sum(!o(y<-o(8/P*b,2*b,"+")/P-n,z<-16/P*b,"+"),-2*!o(y,4*z,"+"))}


exhibiting a load of code golf tricks, from using an anonymous function to renaming functions with a single letter, to switching from integers to booleans and back with the exclamation mark.