I received at last my weekend edition of Le Monde and hence the solution proposed by the authors (Cohen and Busser) to the puzzle #14. They obtain a strategy that only requires at most 19 steps. The idea is to start with a first test, which gives a reference score S₀, and then work on groups of four questions, whose answers can be found in at most three steps. For instance, starting with x₁,…,x₂₅, the second test uses (1-x₁),(1-x₂),x₃,…,x₂₅, and changes the score S₀ by -2,2 or 0. In the first two cases, this determines y₁,y₂ and it suffices to use x₁,x₂,(1-x₃),(1-x₄),x₅…,x₂₅, to find y₃,y₄ in one or two steps. If the score S₀ does not change, considering x₁,(1-x₂),(1-x₃),(1-x₄),x₅…,x₂₅, and then maybe x₁,(1-x₂),x₃,(1-x₄),x₅…,x₂₅, produces again the value of y₁,…,y₄. If one repeats the algorithm one group of four after another, there are six such groups and the maximal number of step is

1+6*3=19

since the final answer y₂₅ is known by deduction from S₀. An additional improvement not mentioned in the journal is achieved in checking after any change whether or not the new score is equal to zero. (In both my solution and theirs, there is an extra step to propose the correct solution, which means in my case an exact average of steps equal to 25, by the geometric argument.)

For the current solution, here is an R code that evaluates the distribution of the number of steps:

rangep=rep(0,25)
for (t in 1:10^5){

s=2
for (j in 1:6){

y=sample(c(0,1),4,rep=T)
x=z=rep(1,4)
Delta0=sum(as.numeric(y!=x))

z[1:2]=0
Delta=sum(as.numeric(y!=z))

if (abs(Delta-Delta0)==2){
z=c(1,1,0,0)
Delta=sum(as.numeric(y!=z))
s=s+2+(abs(Delta-Delta0)!=2)

}else{
z=c(1,0,0,0)
Delta=sum(as.numeric(y!=z))
s=s+2+(abs(Delta-Delta0)!=3)
}
}
rangep[s]=rangep[s]+1
}

The fit by a binomial is rather poor, but this is not surprising given the two-stage decision. In any case, this does better than my earlier solution!