## Archive for latin square

## another Latin rectangle

Posted in pictures, University life with tags construction work, jatp, latin square, material sciences, multiple tests, Paris, péri, picture, Porte Dauphine, Université Paris Dauphine on June 17, 2021 by xi'an## genuine Latin~~square~~ rectangle

Posted in Books, pictures, Statistics, Travel, University life with tags art restoration, bioweapon, bug, Choi Seok-jeong, Firenze, Florence, Italia, latin square, Leonhard Euler, Medicis, NYT, sacrophagus, Serratia ficaria SH7 on June 9, 2021 by xi'an
## stained glass to go

Posted in pictures, University life with tags Caius & Gonville College, Cambridge, Cambridge colleges, controversies, design of experiments, DNA helix, eugenics, Francis Crick, latin square, Ronald Fisher, stained glass, Venn diagram on July 6, 2020 by xi'an## Le Monde puzzle [#1092]

Posted in Statistics with tags brute force, latin square, Le Monde, mathematical puzzle, optimisation, R, simulated annealing, sudoku on April 18, 2019 by xi'an**A** Latin square Le Monde mathematical puzzle that I found rather dreary:

A hidden 3×3 board contains all numbers from 1 to 9. Anselm wants to guess the board and makes two proposals. Berenicke tells him how many entries are in the right rows and colums for each proposal, along with the information that no entry is at the right location. Anselm deduces the right board.

Which I solved by brute force and not even simulated annealing, first defining a target

ordoku1=ordoku2=matrix(1,9,2) ordoku1[,1]=c(1,1,1,2,2,2,3,3,3) ordoku1[,2]=rep(1:3,3) ordoku2[,1]=c(3,2,3,1,2,3,2,1,1) ordoku2[,2]=c(2,2,3,2,3,1,1,3,1) fitz=function(ordo){ (sum(ordo[c(1,4,7),2]==1)==1)+(sum(ordo[c(2,5,8),2]==2)==1)+ (sum(ordo[c(3,6,9),2]==3)==0)+(sum(ordo[c(1,2,3),1]==1)==1)+ (sum(ordo[c(4,5,6),1]==2)==1)+(sum(ordo[c(7,8,9),1]==3)==2)+ (sum(ordo[c(6,7,9),2]==1)==2)+(sum(ordo[c(1,2,4),2]==2)==1)+ (sum(ordo[c(3,5,8),2]==3)==2)+(sum(ordo[c(4,8,9),1]==1)==1)+ (sum(ordo[c(7,2,5),1]==2)==1)+(sum(ordo[c(1,3,6),1]==3)==0)+ (!(0%in%apply((ordo-ordoku1)^2,1,sum)))+(!(0%in%apply((ordo-ordoku2)^2,1,sum))) }

on a 9×9 board entry reproducing all items of information given by Berenicke. If all constraints are met, the function returns 14. And then searched for a solution at random:

temp=1 randw=function(ordo){ for (t in 1:1e6){ chlg=sample(1:9,2) temp=ordo[chlg[1],] ordo[chlg[1],]=ordo[chlg[2],] ordo[chlg[2],]=temp if (fitz(ordo)==14){ print(ordo);break()}}}

which produces the correct board

4 3 5 6 7 1 9 2 8