Archive for mathematical puzzle

puzzles & riddles

Posted in Books, Kids, R, Statistics with tags , , , , , , , , on January 3, 2021 by xi'an

A rather simplistic game on the Riddler of 18 December:

…two players, each of whom starts with a whole number of points. Players take turns “attacking” each other, which involves subtracting their own number of points from their opponent’s until one of the players is out of points.

Easy to code in R:

g=function(N,M)ifelse(N<M,-g(M-N,N),1)

f=function(N,M=N){
  while(g(N,M)>0)M=M+1
  return(M)}

which converges to the separating ratio 1.618. If decomposing the actions until one player wins, one gets a sequence of upper and lower bounds associated with the Fibonacci sequence: 1⁻, 2⁺, 3/2⁻, 5/3⁺, 8/5⁻, &tc, converging to the “golden ratio” φ.

As an aside, I also solved a relatively quick codegolf challenge, where the question was to find the sum of all possible binary values from a bitmask. Meaning that for a binary input, e.g., 101X0XX0…01X, with some entries masked by X’s, one had to find the sum of all binary numbers compatible with the input. Which can be solved succinctly by counting the number of X’s, k, and adding the visible bits 2^k times and replacing the invisible ones by  2^{k-1}. With some help, and using 2 instead of X, my R code moved from 158 bytes to 50:

function(x)2^sum(a<-x>1)*rev(x/4^a)%*%2^(seq(x)-1)

Le Monde puzzle [#1164]

Posted in Books, Kids, R with tags , , , , , , , , , , , , , on November 16, 2020 by xi'an

The weekly puzzle from Le Monde is quite similar to older Diophantine episodes (I find myself impossible to point out):

Give the maximum integer that cannot be written as 105x+30y+14z. Same question for 105x+70y+42z+30w.

These are indeed Diophantine equations and the existence of a solution is linked with Bézout’s Lemma. Take the first equation. Since 105 and 30 have a greatest common divisor equal to 3×5=15, there exists a pair (x⁰,y⁰) such that

105 x⁰ + 30 y⁰ = 15

hence a solution to every equation of the form

105 x + 30 y = 15 a

for any relative integer a. Similarly, since 14 and 15 are co-prime,

there exists a pair (a⁰,b⁰) such that

15 a⁰ + 14 b⁰ = 1

hence a solution to every equation of the form

15 a⁰ + 14 b⁰ = c

for every relative integer c. Meaning 105x+30y+14z=c can be solved in all cases. The same result applies to the second equation. Since algorithms for Bézout’s decomposition are readily available, there is little point in writing an R code..! However, the original question must impose the coefficients to be positive, which of course kills the Bézout’s identity argument. Stack Exchange provides the answer as the linear Diophantine problem of Frobenius! While there is no universal solution for three and more base integers, Mathematica enjoys a FrobeniusNumber solver. Producing 271 and 383 as the largest non-representable integers. Also found by my R code

o=function(i,e,x){
  if((a<-sum(!!i))==sum(!!e))sol=(sum(i*e)==x) else{sol=0
    for(j in 0:(x/e[a+1]))sol=max(sol,o(c(i,j),e,x))}
  sol}
a=(min(e)-1)*(max(e)-1)#upper bound
M=b=((l<-length(e)-1)*prod(e))^(1/l)-sum(e)#lower bound
for(x in a:b){sol=0
for(i in 0:(x/e[1]))sol=max(sol,o(i,e,x))
M=max(M,x*!sol)}

(And this led me to recover the earlier ‘Og entry on the coin problem! As of last November.) The published solution does not bring any useful light as to why 383 is the solution, except for demonstrating that 383 is non-representable and any larger integer is representable.

Le Monde puzzle [#1158]

Posted in Books, Kids, R with tags , , , , , on November 10, 2020 by xi'an

A weekly puzzle from Le Monde on umbrella sharing:

Four friends, Antsa, Cyprien, Domoina and Fy, are leaving school to return to their common housing. It is raining and they only have one umbrella with only room for two. Given walking times, x¹, x², x³ and x⁴, find the fastest time by which all of the four will be home, assuming they all agree to come back with the umbrella if need be.

A recursive R function produces the solution

bez=function(starz=rexp(4),finiz=rep(0,4),rtrn=F){
  if((!rtrn)&(sum(starz>0)==2)){return(max(starz))
    }else{
      tim=1e6
      if(rtrn){
        for(i in (1:4)[finiz>0]){
          nstart=starz;nstart[i]=finiz[i]
          nfini=finiz;nfini[i]=0
          targ=finiz[i]+bez(nstart,nfini,FALSE)
          if(targ<tim){tim=targ}} 
          }else{
          for(i in (1:4)[starz>0])
          for(j in (1:4)[starz>0]){
            if(i!=j){
              nstar=starz;nstar[i]=nstar[j]=0
              nfini=finiz;nfini[i]=starz[i];nfini[j]=starz[j]
              targ=max(starz[i],starz[j])+bez(nstar,nfini,TRUE)
              if (targ<tim){tim=targ}
            }}}
      return(tim)}

which gives for instance

> bez()
[1] 3.297975
> bez(1:4)
[1] 11
> bez(rep(3,4))
[1] 15

asymmetric information

Posted in Kids, R with tags , , , , , on November 4, 2020 by xi'an

The Riddler of 16 October had the following puzzle:

Take a real number θ uniformly distributed over (0,100). Among three players, the winner is whoever guessed the closest price without going over θ. In the event all guesses exceeded θ, the contestant with the lowest (and therefore closest) guess is declared the winner. The second player knows the first player’s guess and the third player knows both other guesses. What is the optimal guess for the first player, assuming all players maximise their probability of winning?

Looking at the optimal solution z for the third player leads to six possible choices, depending on the connection between the other guesses, x and y. Which translates in the R code

topz=function(x,y){
  if((2*y>=x)&(y>=1-x))  z=y-.001
  if(max(4*y,1+y)<=2*x)  z=y+.001
  if((2*x<=1+y)&(x<=1-y))z=x+.001
  z}
  
third=function(x,y) ifelse(y<x,topz(x,y),topz(y,x))

For there, the optimal choice y for the second player follows and happens on a boundary of one of the six regions, which itself returns that the optimal choice for the first player is x=2/3, leading to equal chances of winning (although there is some uncertainty on the boundaries). It is thus feasible to beat the asymmetric information. The picture above was my attempt at representing the probabilities of gain for all three players, some of the six regions being clearly visible, with first axis being x and second being y [and z is one of x⁻,x⁺,y⁻,y⁺]. The R code is too pedestrian to be reproduced!

Le Monde puzzle [#1159]

Posted in Books, Kids, R with tags , , , , , , on October 6, 2020 by xi'an

The weekly puzzle from Le Monde is quite similar to #1157:

Is it possible to break the ten first integers, 1,…,10, into two groups such that the sum over the first group is equal to the product over the second? Is it possible that the second group is of cardinal 4? of cardinal 3?

An exhaustive R search returns 3 solutions by

library(R.utils)
bitz<-function(i)
  c(rev(as.binary(i)),rep(0,10))[d<-1:10]
for (i in 1:2^10)
  if (sum(d[!!bitz(i)])==prod(b<-d[!bitz(i)])) print(b)
[1]  1  4 10 #40
[1] 6 7 #42
[1] 1 2 3 7 #42

which brings a positive reply to the question. Moving from N=10 to N=19 produces similar results

[1]  1  9 18 #162
[1]  2  6 14 #168
[1]  1  3  4 14 #168
[1]  1  2  7 12 #168

with this interesting pattern of only two acceptable products, but I am obviously unable to run the same code for N=49, which is the subsidiary question to the puzzle. Turning to a more conceptual (!) approach, over a long insomnia bout (!!) and a subsequent run, I realised that if there are three terms, x¹,x² and x³, in the second group, they need satisfy

x¹x²x³+x¹+x²+x³=½N(N+1)

and if in addition one of them is equal to 1, x¹ say, this equation simplifies into

(x²+1)(x³+1)=½N(N+1)

which always leads to a solution, as e.g. for N=49,

x¹=1, x²=24 and x³=48.

A brute-force search also led to a four term solution in that case

x¹=1, x²=7, x³=10 and x⁴=17.