**A** Le Monde mathematical puzzle that reminds me of an earlier one:

*Given ten tokens with different unknown weights, and a scale that can rank three tokens at a time, starting with ranking three tokens, what is the minimum number of actions necessary to rank the ten of them if (a) one token at a time is added, (b) one or two tokens are added? If no restriction is imposed on the token introduction, is there a more efficient strategy?*

**I**t indeed relates to earlier posts on sorting and ternary sorting. Checking further on StackOverflow I found this reply to a question about ternary sorting:

Average number of comparisons:

```
in ternary search = ((1/3)*1+(2/3)*2)*ln(n)/ln(3)~1.517*ln(n)
in binary search = 1*ln(n)/ln(2)~1.443*ln(n)
```

Worst number of comparisons:

```
in ternary search = 2*ln(n)/ln(3)~1.820*ln(n)
in binary search = 1*ln(n)/ln(2)~1.443*ln(n)
```

albeit with no reference. So this somewhat answers part (c) of the question. (If asymptotically since this does not work for n=10! Even for n=100, it is way too large.) Looking for a solution to part (a), I looked at the performances of a dyadic sorting algorithm, partitioning recursively the already sorted part of the sample into three parts to locate each time the proper position of the new token. This led me to the following code

rang=function(x,ranj,taken){
mag=max(ranj)-min(ranj)+1
i1=ranj[1]-1+trunc(mag/3)
i2=ranj[1]-1+trunc(2*mag/3)
locrk=rank(c(taken[c(i1,i2)],x))
if (locrk[3]==1) return(ifelse(i1>3,
1+rang(x,ranj=1:(i1-1),taken),
1+(i1>1)))
if (locrk[3]==2) return(ifelse(i2-i1>4,
1+rang(x,ranj=(i1+1):(i2-1),taken),ranj=1:(i1-1),taken),
1+(i1>1)))
if (locrk[3]==3) return(ifelse(mag-i2>2,
1+rang(x,ranj=(i2+1):mag,taken),ranj=1:(i1-1),taken),
1+(i1>1)))
}
ternone=function(toke){
toke[1:3]=sort(toke[1:3])
counter=1
for (ul in (4:length(toke))){
counter=counter+rang(toke[ul],1:(ul-1),toke)
toke[1:ul]=sort(toke[1:ul])
}
return(counter)
}
ternone(sample(1:10))

which produces a minimum of eight (of course!) and a maximum of 20 uses of the scale (based on 10⁶ random permutations of (1,…,10)). Unsurprisingly, the solution proposed in *Le Monde* the week after does better as it obtains 16 as the worst case figure. Repeating the experiment with n=100 values, the maximum was 303 (with a mean of 270 and a minimum of 240 ternary comparisons).

**M**oving to an insertion two tokens at a time, I tested a scheme where two new tokens were tested against the current median, then the median of one half, and so on until they split on both sides of this median. Here is the corresponding R code

ring=function(x,y,ranj,taken){
mag=max(ranj)-min(ranj)+1
i1=ranj[1]-1+trunc(mag/2)
locrk=rank(c(taken[i1],x,y))
if (locrk[1]==3) return(ifelse(i1>2,
1+ring(x,y,ranj=1:(i1-1),taken),
1+(i1==2)))
if (locrk[1]==2) return(ifelse(mag>4,
1+rang(min(x,y),ranj=min(ranj):(i1-1),taken)
+rang(max(x,y),(i1+1):max(ranj),taken),1))
if (locrk[1]==1) return(ifelse(mag-i1>2,
1+ring(x,y,ranj=(i1+1):mag,taken),
1+(mag-i1>1)))
}
terntwo=function(toke){
toke[1:3]=sort(toke[1:3])
counter=1+rang(toke[4],1:3,toke)
toke[1:4]=sort(toke[1:4])
for (ul in -1+2*(3:(length(toke)/2))){
counter=counter+ring(toke[ul],toke[ul+1],1:(ul-1),toke)
toke[1:ul]=sort(toke[1:ul])
ul=ul+1
}
return(counter)
}

leading to a value of 13.

**T**his Feb. 19 issue of Le Monde Science&Médecine leaflet also contains a two page report on growing psychological work-related issues like stress, burn-out and depression, in research labs, mostly connected with the increase in administrative duties and grant writing, duties most of us have not been trained for. There is also a short column by Etienne Gys telling about the short lived claim by Moukhtarbaï Otelbaev to have solved the Navier-Stokes problem. Until Terry Tao wrote a paper stating why the proof could not work.

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