## Le Monde puzzle [#1020]

**A** collection of liars in this Le Monde mathematical puzzle:

*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.

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