## parking riddle

Posted in Books, Kids, pictures, R, Statistics, Travel with tags , , , , , , on October 23, 2020 by xi'an

The Riddler of this week had a quick riddle: if one does want to avoid parallel parking a car over a six spot street, either the first spot is available or two consecutive spots are free. What is the probability this happens with 4 other cars already parked (at random)?

While a derivation by combinatorics easily returns 9/15 as the probability to fail to park, a straightforward R code does as well

l=0
for(t in 1:1e6){
k=sort(sample(0:5,4))
l=l+1*(!!k[1]|3%in%diff(k)|!k[4]%%3)}

since

> round(choose(6,2)*F/1e6)
[1] 10

## Le Monde puzzle [#1121]

Posted in Books, Kids with tags , , , , on December 17, 2019 by xi'an

A combinatoric puzzle as Le weekly Monde current mathematical puzzle:

A class of 75<n<100 students is divided at random into two groups of sizes a and b=n-a, respectively, such that the probability that two particular students Ji-ae and Jung-ah have a probability of exactly 1/2 to be in the same group. Find a and n.

(with an original wording mentioning an independent allocation to the group!). Since the probability to be in the same group (under a simple uniform partition distribution) is

$\frac{a(1-1)}{n(n-1)}+\frac{b(b-1)}{n(n-1)}$

it is sufficient to seek by exhaustion values of (a,b) such that this ratio is equal to ½. The only solution within the right range is then (36,45) (up to the symmetric pair). This can be also found by seeking integer solutions to the second degree polynomial equation, namely

$b^\star=\left[ 1+2a\pm\sqrt{1+8a}\right]/2 \in \mathbb N$

## numbers of numbers with numbers of their numbers

Posted in Statistics with tags , , , on April 6, 2019 by xi'an

A funny riddle from The Riddler where the number of numbers that indicate the numbers of their numbers is requested. By which one means numbers like 22 (two 2’s) and 21322314 (two 1’s, three 2’s, two 3’s and one 4), the convention being that “numbers consist of alternating tallies and numerals”. And that numerals are “tallied in increasing order”. A reasoning based on the number of 1’s, from zero (where 22 seems to be the only possibility) to six (where 613223141516171819 is the only case) leads to a total of 59 cases (unless zero counts as an extra case) and a brute force R exploration returns the same figure:

for (t in 1:1e5){
az=sample((0:6),9,rep=TRUE)
count=rep(TRUE,9)
for (i in 1:9) count[i]=(sum(az==i)+(az[i]>0))==az[i]
nit=0
while ((min(count)==0)&(nit<1e2)){
j=sample((1:9)[!count],1)
az[j]=(sum(az==j)+(az[j]>0))
nit=nit+1}
if (min(count)==1) solz=unique.matrix(rbind(solz,az),mar=1)}

The solution published on The Riddler differs because it also includes numbers with zeros. Which I find annoying to the extreme because if zeroes are allowed then every digit not in the original solution should appear as multiplied by 0, which is self-contradictory… For instance 22 should be 012203…09, except then there is one 1, one 3, and so on.

## 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 [#1029]

Posted in Books, Kids, R with tags , , on November 22, 2017 by xi'an

A convoluted counting Le Monde mathematical puzzle:

A film theatre has a waiting room and several projection rooms. With four films on display. A first set of 600 spectators enters the waiting room and vote for their favourite film. The most popular film is projected to the spectators who voted for it and the remaining spectators stay in the waiting room. They are joined by a new set of 600 spectators, who then also vote for their favourite film. The selected film (which may be the same as the first one) is then shown to those who vote for it and the remaining spectators stay in the waiting room. This pattern is repeated for a total number of 10 votes, after which the remaining spectators leave. What are the maximal possible numbers of waiting spectators and of spectators in a projection room?

A first attempt by random sampling does not produce extreme enough events to reach those maxima:

wm=rm=600 #waiting and watching
for (v in 1:V){
for (t in 1:9){
film=film+rmultinom(1,600,rep(1,4))
rm=max(rm,max(film))
film[order(film)[4]]=0
wm=max(wm,sum(film)+600)}
rm=max(rm,max(film)+600)}

where the last line adds the last batch of arriving spectators to the largest group of waiting ones. This code only returns 1605 for the maximal number of waiting spectators. And 1155 for the maximal number in a projection room.  Compared with the even separation of the first 600 into four groups of 150… I thus looked for an alternative deterministic allocation:

wm=rm=0
film=rep(0,4)
for (t in 1:9){
size=sum(film)+600
film=c(rep(ceiling(size/4),3),size-3*ceiling(size/4))
film[order(film)[4]]=0
rm=max(rm,max(film)+600)
wm=max(wm,sum(film)+600)}

which tries to preserve as many waiting spectators as possible for the last round (and always considers the scenario of all newcomers backing the largest waiting group for the next film). The outcome of this sequence moves up to 1155 for the largest projection audience and 2264 for the largest waiting group. I however wonder if splitting into two groups in the final round(s) could even increase the size of the last projection. And indeed halving the last batch into two groups leads to 1709 spectators in the final projection. With uncertainties about the validity of the split towards ancient spectators keeping their vote fixed! (I did not think long enough about this puzzle to turn it into a more mathematical problem…)

While in Warwick, I reconsidered the problem from a dynamic programming perspective, always keeping the notion that it was optimal to allocate the votes evenly between some of the films (from 1 to 4). Using the recursive R code

optiz=function(votz,t){
if (t==9){ return(sort(votz)[3]+600)
}else{
goal=optiz(sort(votz)+c(0,0,600,-max(votz)),t+1)
goal=rep(goal,4)
for (i in 2:4){
film=sort(votz);film[4]=0;film=sort(film)
size=sum(film[(4-i+1):4])+600
film[(4-i+1):4]=ceiling(size/i)
while (sum(film[(4-i+1):4])>size) film[4]=film[4]-1
goal[i]=optiz(sort(film),t+1)}
return(max(goal))}}

led to a maximal audience size of 1619. [Which is also the answer provided by Le Monde]