**T**he riddle of this week is about an optimisation of positioning the four digits of a multiplication of two numbers with two digits each and is open to a coding resolution:

*Four digits are drawn without replacement from {0,1,…,9}, one at a time. What is the optimal strategy to position those four digits, two digits per row, as they are drawn, toward minimising the average product?*

Although the problem can be solved algebraically by computing **E**[X⁴|x¹,..] and **E**[X⁴X³|x¹,..] I wrote three R codes to “optimise” the location of the first three digits: the first digit ends up as a unit if it is 5 or more and a multiple of ten otherwise, on the first row. For the second draw, it is slightly more variable: with this R code,

second<-function(i,j,N=1e5){draw
drew=matrix(0,N,2)
for (t in 1:N)
drew[t,]=sample((0:9)[-c(i+1,j+1)],2)
conmean=(45-i-j)/8
conprod=mean(drew[,1]*drew[,2])
if (i<5){ #10*i
pos=c((110*i+11*j)*conmean,
100*i*j+10*(i+j)*conmean+conprod,
(100*i+j)*conmean+10*i*j+10*conprod)}else{
pos=c((110*j+11*i)*conmean,
10*i*j+(100*j+i)*conmean+10*conprod,
10*(i+j)*conmean+i*j+100*conprod)
return(order(pos)[1])}

the resulting digit again ends up as a unit if it is 5 (except when x¹=7,8,9, where it is 4) or more and a multiple of ten otherwise, but on the second row. Except when x¹=0, x²=1,2,3,4, when they end up on the first row together, 0 obviously in front.

For the third and last open draw, there is only one remaining random draw, which mean that the decision only depends on x¹,x²,x³ and **E**[X⁴|x¹,x²,x³]=(45-x¹-x²-x³)/7. Attaching x³ to x² or x¹ will then vary monotonically in x³, depending on whether x¹>x² or x¹<x²:

fourth=function(i,j,k){
comean=(45-i-j-k)/7
if ((i<1)&(j<5)){ pos=c(10*comean+k,comean+10*k)}
if ((i<5)&(j>4)){ pos=c(100*i*comean+k*j,j*comean+100*i*k)}
if ((i>0)&(i<5)&(j<5)){ pos=c(i*comean+k*j,j*comean+i*k)}
if ((i<7)&(i>4)&(j<5)){ pos=c(i*comean+100*k*j,j*comean+100*i*k)}
if ((i<7)&(i>4)&(j>4)){ pos=c(i*comean+k*j,j*comean+i*k)}
if ((i>6)&(j<4)){ pos=c(i*comean+100*k*j,j*comean+100*i*k)}
if ((i>6)&(j>3)){ pos=c(i*comean+k*j,j*comean+i*k)}
return(order(pos)[1])}

Running this R code for all combinations of x¹,x² shows that, except for the cases x¹≥5 and x²=0, for which x³ invariably remains in front of x¹, there are always values of x³ associated with each position.

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