Here are excellent latin squares of sizes 8,10,12,14 and 16 all so far as I can inform with out apparent symmetries. Sorry in regards to the formatting, at the least it is copy-n-paste pleasant (you would not wish to verify them by visible inspection anyway, I assume).

ABGFDHCE

BCHGEADF

HAFECGBD

CDAHFBEG

GHEDBFAC

DEBAGCFH

FGDCAEHB

EFCBHDGA

ABDGEIHCJF

BCEHFJIDAG

DEGJHBAFCI

HIADBFEJGC

EFHAICBGDJ

JACFDHGBIE

CDFIGAJEBH

IJBECGFAHD

GHJCAEDIFB

FGIBJDCHEA

ABDHCKILFEJG

BCEIDLJAGFKH

DEGKFBLCIHAJ

HIKCJFDGALEB

CDFJEAKBHGLI

LACGBJHKEDIF

EFHLGCADJIBK

KLBFAIGJDCHE

JKAELHFICBGD

FGIAHDBEKJCL

IJLDKGEHBAFC

GHJBIECFLKDA

ABDGKELJINFCMH

BCEHLFMKJAGDNI

DEGJNHAMLCIFBK

GHJMCKDBAFLIEN

KLNCGAHFEJBMID

EFHKAIBNMDJGCL

NACFJDKIHMEBLG

MNBEICJHGLDAKF

IJLAEMFDCHNKGB

FGILBJCANEKHDM

LMADHBIGFKCNJE

CDFIMGNLKBHEAJ

JKMBFNGEDIALHC

HIKNDLECBGMJFA

ABDGKPFMEONJHCLI

BCEHLAGNFPOKIDMJ

DEGJNCIPHBAMKFOL

GHJMAFLCKEDPNIBO

KLNAEJPGOIHDBMFC

PACFJOELDNMIGBKH

FGILPEKBJDCOMHAN

NOADHMCJBLKGEPIF

MNPCGLBIAKJFDOHE

HIKNBGMDLFEAOJCP

EFHKODJAICBNLGPM

LMOBFKAHPJIECNGD

JKMPDIOFNHGCALEB

CDFIMBHOGAPLJENK

OPBEINDKCMLHFAJG

IJLOCHNEMGFBPKDA

How they had been constructed and why this technique does not work for odd sizes:

If you change A with 0, B with 1 and many others. and for every domino as they sit within the latin squares compute the distinction of left and proper (prime and backside) fields modulo n, then you definitely’ll discover that these variations line up completely, i.e. all horizontal (vertical) dominos with distinction 2, say, i.e. 0:2 1:3 2:4 … (n-1):1 sit in the identical column (row) pair. The needed and sufficicient situation for this development to yield an ideal latin sq. is relatively easy: the set of diffences over all column (row) pairs should be {1,2…n-1} so all dominos are current and all sums over linear subsets of column pairs, i.e. all pairs of neighbors contained in a related block of columns (rows) should not be 0 modulo n in any other case in that row (column) a quantity happens at the least twice.

The latter criterion can also be the explanation we can’t apply this development to the odd case, as a result of the complete matrix may have column (row) sum 1+2+3+…+n-1 = n(n-1)/2 which for odd n is 0 mod n. Thus regardless of how the variations are organized, the primary and final variety of every row (column) will probably be similar which is, in fact, not allowed.

For even numbers as much as 16 we are able to discover preparations of variations satisfying the criterion simply by brute pressure (few seconds with completely unoptimized code). At 18 the pc takes longer than my consideration span. Also of notice is that 1,-2,3,-4… will work but when utilized to each rows and columns will create symmetries.