Archive for The Riddler

a very quick Riddle

Posted in Books, Kids, pictures, R with tags , , , , , , on January 22, 2020 by xi'an

A very quick Riddler’s riddle last week with the question

Find the (integer) fraction with the smallest (integer) denominator strictly located between 1/2020 and 1/2019.

and the brute force resolution

for (t in (2020*2019):2021){ 
   a=ceiling(t/2020)
   if (a*2019<t) sol=c(a,t)}

leading to 2/4039 as the target. Note that

\dfrac{2}{4039}=\dfrac{1}{\dfrac{2020+2019}{2}}

riddle on a circle

Posted in Books, Kids, R, Travel with tags , , , , , , , on December 22, 2019 by xi'an

The Riddler’s riddle this week provides another opportunity to resort to brute-force simulated annealing!

Given a Markov chain defined on the torus {1,2,…,100} with only moves a drift to the right (modulo 100) and a uniformely random jump, find the optimal transition matrix to reach 42 in a minimum (average) number of moves.

Which I coded in my plane to Seattle, under the assumption that there is nothing to do when the chain is already in 42. And the reasoning that there is not gain (on average) in keeping the choice between right shift and random jump random.

dure=min(c(41:0,99:42),50)
temp=.01
for (t in 1:1e6){
  i=sample((1:100)[-42],1)
  dura=1+mean(dure)
  if (temp*log(runif(1))<dure[i]-dura) dure[i]=dura
  if(temp*log(runif(1))<dure[i]-(dura<-1+dure[i*(i<100)+1])) 
    dure[i]=dura 
  temp=temp/(1+.1e-4*(runif(1)>.99))}

In all instances, the solution is to move at random for any position but those between 29 and 41, for an average 13.64286 number of steps to reach 42. (For values outside the range 29-42.)

riddle by attrition

Posted in Books, Kids, R with tags , , , on December 2, 2019 by xi'an

The weekend riddle from The Riddler is rather straightforward [my wording and simplification]:

Construct a decimal number X between 0 and 1 by drawing the first digit a¹ uniformly over {0,1,…,9}, the second digit a² uniformly over {0,1,…,9}, &tc., until 0 is attained. What is the expectation of this random variable X?

Since each new digit has expectation half of the previous digit, the expectation is an infinite geometric series with common ratio 20⁻¹ and factor 9, leading to an expectation of 9/19. As verified with the following R code:

skr<-function(){
  a=9;b=0
  while((a<-sample(rep(0:a,2),1))>0)b=10*b+a
  while(b>=1)b=b/10
  return(b)}

Froebenius coin problem

Posted in pictures, R, Statistics with tags , , , , , , , , , , on November 29, 2019 by xi'an

A challenge from The Riddler last weekend came out as the classical Frobenius coin problem, namely to find the largest amount that cannot be obtained using only n coins of specified coprime denominations (i.e., with gcd equal to one). There is always such a largest value. For the units a=19 and b=538, I ran a basic R code that returned 9665 as the largest impossible value, which happens to be 19×538-538-19, the Sylvester solution to the problem when n=2. A recent paper by Tripathi (2017) manages the case n=3, for “almost all triples”, which decomposes into a myriad of sub-cases. (As an aside, Tripathi (2017) thanks a PhD student, Prof. Thomas W. Cusick, for contributing to the proof, which constitutes a part of his dissertation, but does not explain why he did not join as co-author.) The specific case when a=19, b=101, and c=538 suggested by The Riddler happens to fall in one of the simplest categories since, as ⌊cb⁻¹⌋ and ⌊cb⁻¹⌋ (a) are equal and gcd(a,b)=1 (Lemma 2), the solution is then the same as for the pair (a,b), namely 1799. As this was quite a light puzzle, I went looking for a codegolf challenge that addressed this problem and lo and behold! found one. And proposed the condensed R function

function(a)max((1:(b<-prod(a)))[-apply(combn(outer(a,0:b,"*"),sum(!!a))),2,sum)])

that assumes no duplicate and ordering in the input a. (And learned about combn from Robin.) It is of course very inefficient—to the point of crashing R—to look at the upper bound

\prod_{i=1}^n a_i \ \ \ \ \ \ \ (1)

for the Frobenius number since

\min_{(i,j);\text{gcd}(a_i,a_j)=1} (a_i-1)(a_j-1)\ \ \ \ \ \ \ (2)

is already an upper bound, by Sylvester’s formula. But coding (2) would alas take much more space…

riddle of the seats

Posted in Statistics with tags , , , on November 8, 2019 by xi'an

An arithmetic quick riddle from The Riddler:

If an integer n is a multiple of every integer between 1 and 200, except for two consecutive ones, find those consecutive integers.

Since the highest power of 2 less than 200 is 2⁷=128 and since 127 is a prime number, the number

2^6\times \prod_{i=0,i\ne 63}^{99} (2i+1)

should work in that it contains all odd integers but 127, and all even numbers, but 128. Of course a smaller number that avoids duplicates by only considering the 44 primes other than 127 and 2 to a power that keep them less than 200 is also valid. Which gives a number of the order of 1.037443 10⁸⁵.

a non-riddle

Posted in Books, Kids, R with tags , , on July 12, 2019 by xi'an

Unless I missed a point in the last riddle from the Riddler, there is very little to say about it:

Given N ocre balls, N aquamarine balls, and two urns, what is the optimal way to allocate the balls to the urns towards drawing an ocre ball with no urn being empty?

Both my reasoning and a two line exploration code led to having one urn with only one ocre ball (and no acquamarine ball) and all the other balls in the second urn.

odz<-function(n,m,t) 2*m/n+(t-2*m)/(t-n)
probz=matrix(0,trunc(N/2)-1,N-1)
for (n in 1:(N-1))
  for (m in 1:(trunc(N/2)-1))
    probz[m,n]=odz(n,m,N)

riddles on Egyptian fractions and Bernoulli factories

Posted in Books, Kids, R with tags , , , , , , , , , , , , , on June 11, 2019 by xi'an

Two fairy different riddles on the weekend Riddler. The first one is (in fine) about Egyptian fractions: I understand the first one as

Find the Egyptian fraction decomposition of 2 into 11 distinct unit fractions that maximises the smallest fraction.

And which I cannot solve despite perusing this amazing webpage on Egyptian fractions and making some attempts at brute force  random exploration. Using Fibonacci’s greedy algorithm. I managed to find such decompositions

2 = 1 +1/2 +1/6 +1/12 +1/16 +1/20 +1/24 +1/30 +1/42 +1/48 +1/56

after seeing in this short note

2 = 1 +1/3 +1/5 +1/7 +1/9 +1/42 +1/15 +1/18 +1/30 +1/45 +1/90

And then Robin came with the following:

2 = 1 +1/4 +1/5 +1/8 +1/10 +1/12 +1/15 +1/20 +1/21 +1/24 +1/28

which may prove to be the winner! But there is even better:

2 = 1 +1/5 +1/6 +1/8 +1/9 +1/10 +1/12 +1/15 +1/18 +1/20 +1/24

The second riddle is a more straightforward Bernoulli factory problem:

Given a coin with a free-to-choose probability p of head, design an experiment with a fixed number k of draws that returns three outcomes with equal probabilities.

For which I tried a brute-force search of all possible 3-partitions of the 2-to-the-k events for a range of values of p from .01 to .5 and for k equal to 3,4,… But never getting an exact balance between the three groups. Reading later the solution on the Riddler, I saw that there was an exact solution for 4 draws when

p=\frac{3-\sqrt{3(4\sqrt{9}-6)}}{6}

Augmenting the precision of my solver (by multiplying all terms by 100), I indeed found a difference of

> solver((3-sqrt(3*(4*sqrt(6)-9)))/6,ba=1e5)[1]
[1] 8.940697e-08

which means an error of 9 x 100⁻⁴ x 10⁻⁸, ie roughly 10⁻¹⁵.