## Le Monde puzzle [#1107]

Posted in Kids with tags , , , on July 8, 2019 by xi'an A light birthday problem as Le Monde mathematical puzzle:

Each member of a group of 35 persons writes down the number of those who share the same birth-month and the number of those who share the same birth-date [with them]. It happens that these 70 numbers include all integers from 0 to 10. Show that at least two people share a birth-day. What is the maximal number of people for this property to hold?

Which needs no R code since the result follows from the remark that the number of individuals sharing a birth-month with just one other, n¹, is a multiple of 2, the number of individuals sharing a birth-month with just two others, n², a multiple of 3, and so on. Hence, if no people share a birth-day, n¹,n²,…,n¹⁰>0 and

n¹+n²+…+n¹⁰ ≥ 2+3+…+11 = 6·11-1=65

which means that it is impossible that the 10 digits n¹,…,n¹⁰ are all positive. All the way up to 65 people. As an aside, no correction of the wrong solution to puzzle #1105 was published in the subsequent editions.

## Le Monde puzzle [#1081]

Posted in Books, Kids, R, Travel with tags , , , , on January 24, 2019 by xi'an A “he said-she said” Le Monde mathematical puzzle (again in the spirit of the famous Singapore high-school birthdate problem):

Abigail and Corentin are both given a positive integer, a and b, such that a+b is either 19 or 20. They are asked one after the other and repeatedly if they are sure of the other’s number. What is the maximum number of times they are questioned?

If Abigail is given a 19, b=1 necessarily. Hence if Abigail does not reply, a<19. This implies that, if Corentin is given b=1 or b=19, he can reply a+b=19 or a+b=20, necessarily. Else, 1<b<19 implies that, if a=1 or a=18, b=18 or b=2. And so on…which leads to a maximum of 20 questions, 10 for Abigail and 10 for Corentin. Here is my R implementation

az=bz=cbind(20-(1:19),19-(1:19))
qwz=0;at=TRUE;bt=FALSE
while ((max(az)>0)&(max(bz)>0)){
if (at){
for (i in 1:19){
if (sum(az[i,]>0)==2){
for (j in az[i,az[i,]>0]){
if (sum(bz[j,]==0)==2) az[i,]=rep(0,2)}}
if (sum(az[i,]>0)<2){
az[i,]=rep(0,2)}}}
if (bt){
for (i in 1:19){
if (sum(bz[i,bz[i,]>0]>0)==2){
for (j in bz[i,bz[i,]>0]){
if (sum(az[j,]==0)==2) bz[i,]=rep(0,2)}}
if (sum(bz[i,]>0)<2){ bz[i,]=rep(0,2)}}}
bt=!bt;at=!at;qwz=qwz+1}


## the beauty of maths in computer science [book review]

Posted in Books, Statistics, University life with tags , , , , , , , , , , , , , , , , , , , , on January 17, 2019 by xi'an CRC Press sent me this book for review in CHANCE: Written by Jun Wu, “staff research scientist in Google who invented Google’s Chinese, Japanese, and Korean Web search algorithms”, and translated from the Chinese, 数学之美, originating from Google blog entries. (Meaning most references are pre-2010.) A large part of the book is about word processing and web navigation, which is the author’s research specialty. And not so much about mathematics. (When rereading the first chapters to start this review I then realised why the part about language processing in AIQ sounded familiar: I had read it in the Beauty of Mathematics in Computer Science.)

In the first chapter, about the history of languages, I found out, among other things, that ancient Jewish copists of the Bible had an error correcting algorithm consisting in giving each character a numerical equivalent, summing up each row, then all rows, and  checking the sum at the end of the page was the original one. The second chapter explains why the early attempts at language computer processing, based on grammar rules, were unsuccessful and how a statistical approach had broken the blockade. Explained via Markov chains in the following chapter. Along with the Good-Turing [Bayesian] estimate of the transition probabilities. Next comes a short and low-tech chapter on word segmentation. And then an introduction to hidden Markov models. Mentioning the Baum-Welch algorithm as a special case of EM, which makes a return by Chapter 26. Plus a chapter on entropies and Kullback-Leibler divergence.

A first intermede is provided by a chapter dedicated to the late Frederick Jelinek, the author’s mentor (including what I find a rather unfortunate equivalent drawn between the Nazi and Communist eras in Czechoslovakia, p.64). Chapter that sounds a wee bit too much like an extended obituary.

The next section of chapters is about search engines, with a few pages on Boolean logic, dynamic programming, graph theory, Google’s PageRank and TF-IDF (term frequency/inverse document frequency). Unsurprisingly, given that the entries were originally written for Google’s blog, Google’s tools and concepts keep popping throughout the entire book.

Another intermede about Amit Singhal, the designer of Google’s internal search ranking system, Ascorer. With another unfortunate equivalent with the AK-47 Kalashnikov rifle as “elegantly simple”, “effective, reliable, uncomplicated, and easy to implement or operate” (p.105). Even though I do get the (reason for the) analogy, using an equivalent tool which purpose is not to kill other people would have been just decent…

Then chapters on measuring proximity between news articles by (vectors in a 64,000 dimension vocabulary space and) their angle, and singular value decomposition, and turning URLs as long integers into 16 bytes random numbers by the Mersenne Twister (why random, except for encryption?), missing both the square in von Neumann’s first PRNG (p.124) and the opportunity to link the probability of overlap with the birthday problem (p.129). Followed by another chapter on cryptography, always a favourite in maths vulgarisation books (but with no mention made of the originators of public key cryptography, like James Hellis or the RSA trio, or of the impact of quantum computers on the reliability of these methods). And by an a-mathematic chapter on spam detection.

Another sequence of chapters cover maximum entropy models (in a rather incomprehensible way, I think, see p.159), continued with an interesting argument how Shannon’s first theorem predicts that it should be faster to type Chinese characters than Roman characters. Followed by the Bloom filter, which operates as an approximate Poisson variate. Then Bayesian networks where the “probability of any node is computed by Bayes’ formula” [not really]. With a slightly more advanced discussion on providing the highest posterior probability network. And conditional random fields, where the conditioning is not clearly discussed (p.192). Next are chapters about Viterbi’s algorithm (and successful career) and the EM algorithm, nicknamed “God’s algorithm” in the book (Chapter 26) although I never heard of this nickname previously.

The final two chapters are on neural networks and Big Data, clearly written later than the rest of the book, with the predictable illustration of AlphaGo (but without technical details). The twenty page chapter on Big Data does not contain a larger amount of mathematics, with no equation apart from Chebyshev’s inequality, and a frequency estimate for a conditional probability. But I learned about 23&me running genetic tests at a loss to build a huge (if biased) genetic database. (The bias in “Big Data” issues is actually not covered by this chapter.)

“One of my main objectives for writing the book is to introduce some mathematical knowledge related to the IT industry to people who do not work in the industry.”

To conclude, I found the book a fairly interesting insight on the vision of his field and job experience by a senior scientist at Google, with loads of anecdotes and some historical backgrounds, but very Google-centric and what I felt like an excessive amount of name dropping and of I did, I solved, I &tc. The title is rather misleading in my opinion as the amount of maths is very limited and rarely sufficient to connect with the subject at hand. Although this is quite a relative concept, I did not spot beauty therein but rather technical advances and trick, allowing the author and Google to beat the competition.

## surprises in probability [book review]

Posted in Books, Statistics, Travel with tags , , , , , , , , , on November 20, 2018 by xi'an

A very short book (128 pages, but with a very high price!) I received from CRC Press is Henk Tijms’ Surprises in Probability (Seventeen Short Stories). Henk Tijms is an emeritus professor of econometrics at the Vrije University in Amsterdam and he wrote these seventeen pieces either for the Dutch Statistical Society magazine or for a blog he ran for the NYt. (The video of A Night in Casablanca above is only connected to this blog through Chico mimicking the word surprise as soup+rice.)

The author mentions that the book can be useful for teachers and indeed this is a collection of surprising probability results, surprising in the sense that the numerical probabilities are not necessarily intuitive. Most illustrations involve betting of one sort or another,  with only basic (combinatorial) probability distributions involved. Readers should not worry about even this basic probability background since most statements are exposed without a proof. Most examples are very classical, from the prisoner’s problem, to the Monty Hall paradox, to the birthday problem, to Benford’s distribution of digits, to gambler’s ruin, gambler’s fallacy, and the St Petersbourg paradox, to the secretary’s problem and stopping rules. The most advanced notion is the one of (finite state) Markov chains. As martingales are only mentionned in connection with pseudo-probabilist schemes for winning the lottery. For which (our very own!) Jeff Rosenthal makes an appearance, thanks to his uncovering of the Ontario Lottery scam!

“In no other branch of mathematics is it so easy for experts to blunder as in probability theory.”  Martin Gardner

A few stories have entries about Bayesian statistics, with mentions made of the O.J. Simpson, Sally Clark and Lucia de Berk miscarriages of justice, although these mentions make the connection most tenuous. Simulation is also mentioned as a manner of achieving approximations to more complex probabilities. But not to the point of discussing surprises about simulation, which could have been the case with the simulation of rare events.

Ten most beautiful probability formulas (Story 10) reminded me of Ian Steward 17 formulas that changed the World. Obviously at another scale and in a much less convincing way. To wit, the Normal (or Gauss) density, Bayes’ formula, the gambler’s ruin formula, the squared-root formula (meaning standard deviation decreases as √n), Kelly’s betting formula (?), the asymptotic law of distribution of prime numbers (??), another squared-root formula for the one-dimensional random walk, the newsboy formula (?), the Pollaczek-Khintchine formula (?), and the waiting-time formula. I am not sure I would have included any of these…

All in all this is a nice if unsurprising database for illustrations and possibly exercises in elementary probability courses, although it will require some work from the instructor to link the statements to their proof. As one would expect from blog entries. But this makes for a nice reading, especially while traveling and I hope some fellow traveler will pick the book from where I left it in Mexico City airport.

## Le Monde puzzle [#1071]

Posted in Books, Kids with tags , , , , , , , , , , , on October 18, 2018 by xi'an A “he said she said” Le Monde mathematical puzzle sixth competition problem that reminded me of the “Singapore birthday problem” (nothing to do with the original birthday problem!):

Arwen and Brandwein are privately and respectively given the day and month of Caradoc’s birthday [in the Gregorian calendar] with the information that the month number is at least the day number. Arwen starts by stating she knows Brandwein cannot deduce the birthday, followed by Brandwein who says the same about Arwen. If this “she says he says” goes on for the largest possible number of steps before Arwen says she knows, when is Caradoc’s birthday? Arwen and Brandwein are later given two numbers corresponding to Deirdre’s birthday with no indication of which one is the day and which one is the month. They know both numbers end up with the same digit and that the month number is strictly less than the day number. Arwen states she does not know the date and she knows Brandwein cannot know either. Then Brandwein says he indeed does not the date but he knows whether he got the day or the month. This prompts Arwen to conclude she knows, then Brandwein to do the same. When is Deirdre’s birthday?

Since this was a fairly easy puzzle (and since I had spent too much time debugging the previous R code!), I did not try coding this one but instead drew the possibilities and remove the impossibilities on a blackboard. The first question is quite simple actually since the day numbers stand between 1 and 12 and that each “I cannot know” excludes one of the remaining endpoints, removing first excludes 1 from both lists, then 12, then 2, then …. 8, ending up with 7. And 07/07 as Caradoc’s birthday. The second case sees 13,…,20,23,…,30 eliminated from Arwen’s numbers, then 3,…,10 as well, which eliminates the same numbers from Brandwein’s possibilities. That he knows whether it is a month or a day leaves only 1,2,21,22,31 as possible numbers. Then Arwen’s certainty reduces her numbers to be 2, 21, 22, or 31, and since Brandwein is also sure, the only possible cases are (2,22) and (22,2). Meaning Deirdre’s birthday is on 22/02. I dunno if this symmetry was to be expected! (And I cannot fathom why this puzzle is awarded so many points, when compared with the others.)

## [un]solved riddles

Posted in Books, Kids, R with tags , , on July 4, 2017 by xi'an

On the Riddler of last week, first a birthday puzzle:

Given a group of 23 persons, what is the probability of observing three pairs of identical birthdays?

which can be found by a quick simulation as

ave=0
for (t in 1:1e6){
dupz=dates[duplicated(sample(1:365,23,rep=TRUE))]
ave=ave+as.integer((length(dupz)==3)&
(length(unique(dupz))==3))}}
ave/M


returning a value of 0.0183, but which combinatoric resolution I could not fully fathom without a little help from a friend (-ly blog). I had the multinomial coefficient ${23\choose 2\ 2\ 2}$

for the allocation of the 23 persons to one of the three pairs or none, as well as the probability $\dfrac{365\times\cdots\times(365-(23-4)+1)}{365^{23}}$

but I had forgotten the 3! in the denominator for the permutations of the three pairs, which leads again to ${23 \choose 2\ 2\ 2}\dfrac{365\times\cdots\times(365-(23-4)+1)}{3!\times365^{23}} = 0.0183$

A question that also led to an unsolved question: in the even this probability was much smaller, what is an easy way into constructing a more efficient importance sampler?

The second riddle was just as easy to code in R:

A game of tag goes by the following rules: (i) anyone untagged can tag anyone untagged; (ii) anyone tagged by a player tagged gets untagged; (iii) the winner is the last untagged player. What is the expected number of runs for N players?

The outcome of

game=function(N=12){
a=rep(0,N);T=0
while (sum(a==0)>1){
ij=sample((1:N)[a==0],2)
a[ij]=ij;a[a==ij]=0
T=T+1}
return(T)}


leads to an average value of $2^{N-1}-1$

but I had no clear quick explanation for the doubling phenomenon. Until I picked a pen and a sheet of paper and drew the last steps of the game: to decrease down to 1, the size of the untagged survivors has to get through …,3,2 and each time the eliminated player needs to have tagged no other player since otherwise the population grows again. This has to apply all the way to the second round, where N-1 players remain and the one tagged needs to be anyone but the one who tagged the first one. And so on…

## sampling by exhaustion

Posted in Books, Kids, R, Statistics with tags , , , , on November 25, 2016 by xi'an

The riddle set by The Riddler of last week sums up as follows:

Within a population of size N, each individual in the population independently selects another individual. All individuals selected at least once are removed and the process iterates until one or zero individual is left. What is the probability that there is zero individual left? While I cannot see a clean analytical solution to this problem, it reminds me of an enveloppe-versus-letter (matches) problem I saw in graduate school. Indeed, the expected number of removed (or selected) individuals is given by $N\left\{1-\frac{N-2}{N-1}\right\}^{N-1}$

which is equivalent to (1-e⁻¹)N for N large, meaning that the population decreases by an average factor of e⁻¹ at each round. And that it takes on average approximately log(N) iterations to reach a single individual. A simulation of the first probabilities of ending with one individual led me to the above curve, which wiggles in an almost periodic way around the probability ½, equal to the average of those probabilities. Using the R code

rad=function(N){#next population size
ut=sample(rep(2:N,2),1)
for (i in 2:N)#sampling
ut=c(ut,sample(rep((1:N)[-i],2),1))
return(N-length(unique(ut))}
sal=rep(0,m);sal=1
for (N in 3:M){
prop=0;
for (t in 1:T){#one single step $p_n(m)={n\choose m}\sum_{v=}^{n-m} (-1)^v {n-m\choose v} \left(1-\frac{m+v}{n}\right)^n$