## 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.

## Egyptian fractions [Le Monde puzzle #922]

Posted in Books, Kids, R with tags , , , , , , , on July 28, 2015 by xi'an

For its summer edition, Le Monde mathematical puzzle switched to a lighter version with immediate solution. This #922 considers Egyptian fractions which only have distinct denominators (meaning the numerator is always 1) and can be summed. This means 3/4 is represented as ½+¼. Each denominator only appears once. As I discovered when looking on line, a lot of people are fascinated with this representation and have devised different algorithms to achieve decompositions with various properties. Including Fibonacci who devised a specific algorithm called the greedy algorithm in 1202 in the Liber Abaci. In the current Le Monde edition, the questions were somewhat modest and dealt with the smallest decompositions of 2/5, 5/12, and 50/77 under some additional constraint.

Since the issue was covered in so many places, I just spent one hour or so constructing a basic solution à la Fibonacci and then tried to improve it against a length criterion. Here are my R codes (using the numbers library):

osiris=function(a,b){
#can the fraction a/b be simplified
diva=primeFactors(a)
divb=primeFactors(b)
divc=c(unique(diva),unique(divb))
while (sum(duplicated(divc))>0){
n=divc[duplicated(divc)]
for (i in n){a=div(a,i);b=div(b,i)}
diva=primeFactors(a)
divb=primeFactors(b)
divc=c(unique(diva),unique(divb))
}
return(list(a=a,b=b))
}


presumably superfluous for simplifying fractions

horus=function(a,b,teth=NULL){
#simplification
anubis=osiris(a,b)
a=anubis$a;b=anubis$b
#decomposition by removing 1/b
isis=NULL
if (!(b %in% teth)){
a=a-1
isis=c(isis,b)
teth=c(teth,b)}
if (a>0){
#simplification
anubis=osiris(a,b)
bet=b;a=anubis$a;b=anubis$b
if (bet>b){ isis=c(isis,horus(a,b,teth))}else{
# find largest integer
k=ceiling(b/a)
while (k %in% teth) k=k+1
a=k*a-b
b=k*b
isis=c(isis,k,horus(a,b,teth=c(teth,k)))
}}
return(isis)}


which produces a Fibonacci solution (with the additional inclusion of the original denominator) and

nut=20
seth=function(a,b,isis=NULL){
#simplification
anubis=osiris(a,b)
a=anubis$a;b=anubis$b
if ((a==1)&(!(b %in% isis))){isis=c(isis,b)}else{
ra=hapy=ceiling(b/a)
if (max(a,b)<1e5) hapy=horus(a,b,teth=isis)
k=unique(c(hapy,ceiling(ra/runif(nut,min=.1,max=1))))
propa=propb=propc=propd=rep(NaN,le=length((k %in% isis)))
bastet=1
for (i in k[!(k %in% isis)]){
propa[bastet]=i*a-b
propb[bastet]=i*b
propc[bastet]=i
propd[bastet]=length(horus(i*a-b,i*b,teth=c(isis,i)))
bastet=bastet+1
}
k=propc[order(propd)[1]]
isis=seth(k*a-b,k*b,isis=c(isis,k))
}
return(isis)}


which compares solutions against their lengths. When calling those functions for the three fractions above the solutions are

> seth(2,5)
[1] 15 3
> seth(5,12)
[1] 12  3
> seth(50,77)
[1]   2 154   7


with no pretension whatsoever to return anything optimal (and with some like crashes when the magnitude of the entries grows, try for instance 5/121). For this latest counter-example, the alternative horus works quite superbly:

> horus(5,121)
[1] 121 31 3751 1876 7036876