Archive for Alice and Bob

Bayesian Adversarial Privacy [v2]

Posted in Books, Statistics, University life with tags , , , , , , , , , , , , , , , , , , , , , , , , on September 11, 2026 by xi'an

We have just reposted our paper Bayesian Adversarial Privacy on arXiv to reflect the revision we wrote in the past months, to address the (quite sensible) comments from the reviewers. Interestingly the discussants of my Akaike lecture made similar points. The main changes are in explaining more clearly the nature of the combined loss, with Antoine coming up with the use of illuminating R-U map representations, in enlarging the references to other approaches, in stressing that Eve was an Alice’s construct rather than a genuine adversary, but still integrating the case of “multiple Eves”, in mellowing our criticisms of DP, and in expanding the conclusion with limitations and extensions subsections.

Bayesian, adversarial, oceanic, privacy

Posted in Books, Statistics, University life with tags , , , , , , , , , , , , , , , , , , , , , , , on March 6, 2026 by xi'an

We just arXived a new paper on Bayesian privacy! We meaning Cameron Bell, Antoine Luciano, Timothy Johnston and myself, as members of my ERC OCEAN lab at PariSanté and Paris Dauphine. While sharing the same ground as my recent paper with James Bailie, Joshua Bon and Judith Rousseau, this one is definitely more mainstream Bayesian in that the entire decision process falls under the Bayesian hat, with the ultimate decision being the choice of the release mechanism by the data holder (or hoarder!). To rationalise this decision process, we break the framework as resulting from the actions of three actors, namely the data holder, Alice, the data scientist, Bob, and the eavesdropper. Eve. (As in my earlier posts on solving Le Monde’s math puzzles, we could have used pronouns from other cultures, but I feared this would have confused some of the readers. Incidentally, I found out that the earliest use of the first two pronouns was within the groundbreaking cryptography 1977 paper of Rivest, Shamir and Adleman, bringing the RSA algorithm to the World! With Eve appearing in an early, highly-cited privacy paper by Montréal’s Bennett, Brassard, and (unconnected to me!) Robert, in 1988.)

We thus consider a Bayesian setting in which, given data x, held by Alice, inference is to be performed by Bob on a parameter θ. Performing such inference requires Alice releasing information derived from x, which may contain sensitive content, exploited by Eve. Our approach is to compare Alice’s release mechanisms according to both the quality of inference on θ (from Bob’s viewpoint) and the privacy leakage regarding x (sought by Eve and dreaded by Alice). To formalise this evaluation, we posit that Alice refers to a loss function that is a linear combination of Bob’s and Eve’s losses, the weight on Eve’s loss being then negative. (An alternative to be considered in future work is Alice using a ratio of Bob’s and Eve’s losses, possibly set to different powers, the rationale being that a zero loss for Eve is intolerable for Alice.) As in Bayesian experimental design, a prior on the data is necessary for Eve to infer on the hidden data based on the release mechanism and released output and for Alice to evaluate the risk of said release mechanism . (They may differ, as long as they are both made public.) To calibrate Alice’s loss, we opted for a balance that returns the same risk for a full data release and a total lack of release. In specific, informed, settings, other weights could be chosen. While finding the optimal release strategy is impossible but for highly discrete settings, the framework obviously allows for the ranking of natural strategies like insufficient statistics and synthetic datasets. Comments welcome!

Le Monde puzzle [#1130]

Posted in Books, Kids, R, Statistics with tags , , , , , , , on February 7, 2020 by xi'an

A two-player game as Le weekly Monde current mathematical puzzle:

Abishag and Caleb fill in alternance a row of N boxes in a row by picking one then two then three &tc. consecutive boxes. When a player is unable to find enough consecutive boxes, the player has lost. Who is winning when N=29? When N=30?

Using a basic recursive search for the optimal strategy, with the status of the row and the number of required boxes as entries,

f<-function(b=!1:N,r=0){
  for(i in 1:(N-r)){
    if(p<-!max(b[j<-i+r:0])){
      q=b;q[j]=1
      if(p<-!f(q,r+1))break}}
  p}

returns Abishag as the winner for N=29 (as well as for N=1,2,7,…,13,19,…,29) and Caleb as the winner for N=30 (as well as for N=3,…,6,14,…,18). I am actually surprised that the recursion operates that deep, even though this means a √N depth for the recursion. While the code took too long to complete, the function operates for N=100. A side phenomenon is the apparent special case of N=47, which makes Abishag the looser, while N=46 and N=48 do not.This is an unusual pattern as otherwise (up to N=59), there are longer and longer stretches of adjacent wins and looses as N increases.

Le Monde puzzle [#1115]

Posted in Kids, R with tags , , , , , on October 28, 2019 by xi'an

A two-person game as Le weekly Monde current mathematical puzzle:

Two players Amaruq and Atiqtalik are in a game with n tokens where Amaruq chooses a number 1<A<10 and then Atiqtalik chooses a different 1<B<10, and then each in her turn takes either 1, A or B tokens out of the pile.The player taking the last token wins. If n=150, who between Amaruq and Atiqtalik win if both are acting in an optimal manner? Same question for n=210.

The run of a brute force R code like

B=rep(-1,200);B[1:9]=1
for (i in 10:200){
    v=matrix(-2,9,9)
    for (b in 2:9){
       for (a in (2:9)[-b+1])
       for (d in c(1,a,b)){
        e=i-d-c(1,a,b)
        if (max(!e)){v[a,b]=max(-1,v[a,b])}else{
         if (max(e)>0) v[a,b]=max(v[a,b],min(B[e[which(e>0)]]))}}
     B[i]=max(B[i],min(v[v[,b]>-2,b]))}

always produces 1’s in B, which means the first player wins no matter… I thus found out (from the published solution) that my interpretation of the game rules were wrong. The values A and B are fixed once for all and each player only has the choice between withdrawing 1, A, and B on her turn. With the following code showing that Amaruq looses both times.

B=rep(1,210)
for(b in(2:9))
 for(a in(2:9)[-b+1])
  for(i in(2:210)){
   be=-2
   for(d in c(1,a,b)){
    if (d==i){best=1}else{
      e=i-d-c(1,a,b)
      if (max(!e)){be=max(-1,be)}else{
       if (max(e)>0)be=max(be,min(B[e[which(e>0)]]))}}}
   B[i]=be}

Le Monde puzzle [#1105]

Posted in Kids, R with tags , , , , , , on July 8, 2019 by xi'an

Another token game as Le Monde mathematical puzzle:

Archibald and Beatrix play with a pile of n>100 tokens, sequentially picking m tokens from the pile with m being a prime number [including m=1] or a multiple of 6, the winner taking the last tokens. If Beatrix knows n and proposes to Archibald to start, what is the value of n?

Which cannot be solved in a few lines of R code:

k<-function(n)n<4||all(n%%2:ceiling(sqrt(n))!=0)||!n%%6
g=(1:3)
n=c(4,i<-4)
while(max(n)<101){
  if(k(i)) g=c(g,i) else{
  while(i%in%g)i=i+1;j=4;o=!j
  while(!o&(j<i)){ 
    o=(j%in%n)&k(i-j);j=j+1}
  if(o) g=c(g,i) else n=c(n,i)}
  i=i+1}

since it returned no unsuccessful value above 100! With 4, 8, 85, 95, and 99 as predecessors. A rather surprising outcome and a big gap that most certainly has a straightforward explanation! Or a lack of understanding from yours truly: this post appears after the solution was published in Le Monde and I am more bemused than ever since the losing numbers in the journal are given as 4, 8, 85, … 89, and 129. With the slight hiccup that 89 is a prime number…. The other argument in the solution that there can only be five such losers is well-taken since there are only five possible non-zero remainders in the division by 6.