## Bernoulli factory in the Riddler

Posted in Books, Kids, R, Statistics with tags , , , , , , , , , , on December 1, 2020 by xi'an

“Mathematician John von Neumann is credited with figuring out how to take a p biased coin and “simulate” a fair coin. Simply flip the coin twice. If it comes up heads both times or tails both times, then flip it twice again. Eventually, you’ll get two different flips — either a heads and then a tails, or a tails and then a heads, with each of these two cases equally likely. Once you get two different flips, you can call the second of those flips the outcome of your “simulation.” For any value of p between zero and one, this procedure will always return heads half the time and tails half the time. This is pretty remarkable! But there’s a downside to von Neumann’s approach — you don’t know how long the simulation will last.” The Riddler

The associated riddle (first one of the post-T era!) is to figure out what are the values of p for which an algorithm can be derived for simulating a fair coin in at most three flips. In one single flip, p=½ sounds like the unique solution. For two flips, p²,(1-p)^2,2p(1-p)=½ work, but so do p+(1-p)p,(1-p)+p(1-p)=½, and the number of cases grows for three flips at most. However, since we can have 2³=8 different sequences, there are 2⁸ ways to aggregate these events and thus at most 2⁸ resulting probabilities (including 0 and 1). Running a quick R code and checking for proximity to ½ of any of these sums leads to

[1] 0.2062997 0.7937005 #p^3
[1] 0.2113249 0.7886753 #p^3+(1-p)^3
[1] 0.2281555 0.7718448 #p^3+p(1-p)^2
[1] 0.2372862 0.7627143 #p^3+(1-p)^3+p(1-p)^2
[1] 0.2653019 0.7346988 #p^3+2p(1-p)^2
[1] 0.2928933 0.7071078 #p^2
[1] 0.3154489 0.6845518 #p^3+2p^2(1-p)
[1] 0.352201  0.6477993 #p^3+p(1-p)^2+p^2(1-p)
[1] 0.4030316 0.5969686 #p^3+p(1-p)^2+3(1-p)p^2
[1] 0.5


which correspond to

1-p³=½, p³+(1-p)³=½,(1-p)³+(1-p)p²=½,p³+(1-p)³+p²(1-p),(1-p)³+2(1-p)p²=½,1-p²=½, p³+(1-p)³+p²(1-p)=½,(1-p)³+p(1-p)²+p²(1-p)=½,(1-p)³+p²(1-p)+3p(1-p)²=½,p³+p(1-p)²+3(p²(1-p)=½,p³+2p(1-p)²+3(1-p)p²=½,p=½,

(plus the symmetric ones), leading to 19 different values of p producing a “fair coin”. Missing any other combination?!

Another way to look at the problem is to find all roots of the $2^{2^n}$ equations

$a_0p^n+a_1p^{n-1}(1-p)+\cdots+a_{n-1}p(1-p)^{n-1}+a_n(1-p)^n=1/2$

where

$0\le a_i\le{n \choose i}$

(None of these solutions is rational, by the way, except p=½.) I also tried this route with a slightly longer R code, calling polyroot, and finding the same 19 roots for three flips, [at least] 271 for four, and [at least] 8641 for five (The Riddler says 8635!). With an imprecision in the exact number of roots due to rather poor numerical rounding by polyroot. (Since the coefficients of the above are not directly providing those of the polynomial, I went through an alternate representation as a polynomial in (1-p)/p, with a straightforward derivation of the coefficients.)

## visitors allowed in Svalbard

Posted in Statistics with tags , , , , , , , , , , , , , , , on November 8, 2020 by xi'an

## from Svalbard [with snow]

Posted in Statistics with tags , , , , , , , , , , , , on April 25, 2020 by xi'an

## ABC in Svalbard [news #1]

Posted in Mountains, pictures, Running, Statistics, Travel, University life with tags , , , , , , , , , , , , , , , , , , , , on March 23, 2020 by xi'an

We [Julien and myself] are quite pleased to announce that

• the scientific committee for the workshop has been gathered
• the webpage for the workshop is now on-line (with a wonderful walrus picture whose author we alas cannot identify)
• the workshop is now endorsed by both IMS and ISBA, which will handle registration (to open soon)
• the reservation of hotel rooms will be handled by Hurtigruten Svalbard through the above webpage (this is important as we already paid deposit for a certain number of rooms)
• we are definitely seeking both sponsors and organisers of mirror workshops in more populated locations

As an item of trivia, let me recall that Svalbard stands for the archipelago, while Spitsbergen is the name of the main island, where Longyearbyen is located. (In Icelandic, Svalbarði means cold rim or cold coast.)

## 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…