Filed under: Statistics Tagged: AI, artificial intelligence, Cédric Villani, French parliement, French politics, Le Monde, singularity, superintelligence, weapons of math destruction ]]>

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*A most democratic electoral system allows every voter to have at least one representative by having each of the N voters picking exactly m candidates among the M running candidates and setting the size n of the representative council towards this goal, prior to the votes. If there are M=25 candidates, m=10 choices made by the voters, and n=10 representatives, what is the maximal possible value of N? And if N=55,555 and M=33, what is the minimum value of n for which m=n is always possible?
*

**I** tried a brute force approach by simulating votes from N voters at random and attempting to find the minimal number of councillors for this vote, which only provides an upper bound of the minimum [for one vote], and a lower bound in the end [over all votes]. Something like

for (i in 1:N) votz[i,]=sample(1:M,n) #exploration by majority remz=1:N;conz=NULL while (length(remz)>0){ seatz=order(-hist(votz[remz,], breaks=(0:M)+0.5,plot=FALSE)$density)[1] conz=c(conz,seatz);nuremz=NULL for (v in remz) if (!(seatz%in%votz[v,])) nuremz=c(nuremz,v) remz=nuremz} solz=length(conz) #exploration at random kandz=matrix(0,N,M) for (i in 1:N) kandz[i,votz[i,]]=1 for (t in 1:1e3){ #random choice of councillors zz=sample(c(0,1),M,rep=TRUE) while (min(kandz%*%zz)!=1) zz=sample(c(0,1),M,rep=TRUE) solz=min(solz,sum(zz)) #random choice of remaining councillor per voter remz=1:N;conz=NULL while (length(remz)>0){ seatz=sample(votz[remz[1],],1) conz=c(conz,seatz);nuremz=NULL for (i in remz) if (!(seatz%in%votz[i,])) nuremz=c(nuremz,i) remz=nuremz} solz=min(solz,length(conz))} maxz=max(solz,maxz)}

which leads to a value near N=4050 for the first question, with 0% confidence… Obviously, the problem can be rephrased as a binary integer linear programming problem of the form

where A is the NxM matrix of votes and c is the vector of selected councillors. But I do not see a quick way to fix it!

Filed under: Books, Kids, R Tagged: binary integer linear programming, competition, Le Monde, linear programming, mathematical puzzle, R ]]>

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*A circle of 16 liars and truth-tellers is such that everyone states that their immediate neighbours are both liars. How many liars can there be?**A circle of 12 liars and truth-tellers is such that everyone state that their immediate neighbours are one liar plus one truth-teller. How many liars can there be?**A circle of 8 liars and truth-tellers is such that four state that their immediate neighbours are one liar plus one truth-teller and four state that their immediate neighbours are both liars . How many liars can there be?*

**T**hese questions can easily be solved by brute force simulation. For the first setting, using 1 to code truth-tellers and -1 liars, I simulate acceptable configurations as

tabz=rep(0,16) tabz[1]=1 #at least one tabz[2]=tabz[16]=-1 for (i in 3:15){ if (tabz[i-1]==1){ tabz[i]=-1}else{ if (tabz[i+1]==-1){ tabz[i]=1}else{ if (tabz[i+1]==1){ tabz[i]=-1}else{ if (tabz[i-2]==-1){ tabz[i]=1}else{ tabz[i]=sample(c(-1,1),1) }}}}}

which produces 8, 9, and 10 as possible (and obvious) values.

The second puzzle is associated with the similar R code

tabz=sample(c(-1,1),12,rep=TRUE) rong=FALSE while (!rong){ for (i in sample(12)){ if (tabz[i-1+12*(i==1)]*tabz[i%%12+1]==-1){ tabz[i]=1}else{ tabz[i]=sample(c(-1,1),1)} } rong=TRUE for (i in (1:12)[tabz==1]) rong=rong&(tabz[i-1+12*(i==1)]*tabz[i%%12+1]==-1) if (rong){ for (i in (1:12)[tabz==-1]) rong=rong&(tabz[i-1+12*(i==1)]*tabz[i%%12+1]!=-1) }}

with numbers of liars (-1) either 12 (obvious) or 4.

The final puzzle is more puzzling in that figuring out the validating function (is an allocation correct?) took me a while, the ride back home plus some. I ended up with the following code that samples liars (-1) and thruth-seekers (1) at random, plus forces wrong and right answers (in 0,1,2) on these, and check for the number of answers of both types:

rong=FALSE while (!rong){ tabz=sample(c(-1,1),8,rep=TRUE) #truth tabz[1]=1;tabz[sample(2:8,1)]=-1 tt=(1:8)[tabz==1];lr=(1:8)[tabz==-1] statz=rep(0,8) #stmt statz[tt]=(tabz[tt-1+8*(tt==1)]*tabz[tt%%8+1]==-1)+ 2*(tabz[tt-1+8*(tt==1)]+tabz[tt%%8+1]==-2) #answering 0 never works statz[lr]=2*(tabz[lr-1+8*(lr==1)]*tabz[lr%%8+1]==-1)+ (tabz[lr-1+8*(lr==1)]+tabz[lr%%8+1]==-1)+ sample(c(1,2),8,rep=TRUE)[lr]* (tabz[lr-1+8*(lr==1)]+tabz[lr%%8+1]==1) rong=(sum(statz==1)==4)&(sum(statz==2)==4)}

with solutions 3, 4, 5 and 6.

Filed under: Books, Kids, R Tagged: competition, Le Monde, liar puzzle, mathematical puzzle ]]>

- Professor in Statistics, Biostatistics or Data Science at U de M, deadline October 30th, 2017, a requirement being proficiency in the French language;
- Tenure-Track Professorship in Statistics at Harvard University, Department of Statistics, details there.

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The current paper by

The next part of the paper compares this approach with seven other solutions found in the literature, from Matthew Stephens’ (2000) to our permutation reordering. Which does pretty well in terms of MSE in the simulation study (see the massive Table 3) while being much cheaper to implement than the proposed pivotal relabelling (Table 4). And which, contrary to the authors’ objection, *does not require* the precise computation of the MAP since, as indicated in our paper, the relative maximum based on the MCMC iterations can be used as a proxy. I am thus less than convinced at the improvement brought by this alternative…

Filed under: Statistics Tagged: Bayesian Essentials with R, finite mixtures, label switching, relabelling, ResearchGate ]]>

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