Some ternary quasigroups over small sets.
Page one contains sections 1 through 4. Page two contains sections 5 through 10. Page three contains sections 11 and 12. This is page four, with sections 13 through 16. 
§13 Slices. If one input of a ternary operation is held constant while the other inputs vary, its behavior amounts to that of a binary operation, termed a slice. Because of cancellativity, each slice of a Latin cube must be a Latin square. The following tables give an example of TQG slicing, employing these symbols:
4:26230  

aaa=b  aab=d  aac=c  aad=a  aba=c  abb=a  abc=b  abd=d  aca=a  acb=c  acc=d  acd=b  ada=d  adb=b  adc=a  add=c 
baa=a  bab=c  bac=d  bad=b  bba=d  bbb=b  bbc=a  bbd=c  bca=b  bcb=d  bcc=c  bcd=a  bda=c  bdb=a  bdc=b  bdd=d 
caa=d  cab=b  cac=a  cad=c  cba=a  cbb=c  cbc=d  cbd=b  cca=c  ccb=a  ccc=b  ccd=d  cda=b  cdb=d  cdc=c  cdd=a 
daa=c  dab=a  dac=b  dad=d  dba=b  dbb=d  dbc=c  dbd=a  dca=d  dcb=b  dcc=a  dcd=c  dda=a  ddb=c  ddc=d  ddd=b 
O (a, q, r) = F (q, r)  F = 4::273  aa=b  ab=d  ac=c  ad=a  ba=c  bb=a  bc=b  bd=d  ca=a  cb=c  cc=d  cd=b  da=d  db=b  dc=a  dd=c  
O (b, q, r) = F (q, r)  F = 4::91  aa=a  ab=c  ac=d  ad=b  ba=d  bb=b  bc=a  bd=c  ca=b  cb=d  cc=c  cd=a  da=c  db=a  dc=b  dd=d  
O (c, q, r) = F (q, r)  F = 4::484  aa=d  ab=b  ac=a  ad=c  ba=a  bb=c  bc=d  bd=b  ca=c  cb=a  cc=b  cd=d  da=b  db=d  dc=c  dd=a  
O (d, q, r) = F (q, r)  F = 4::302  aa=c  ab=a  ac=b  ad=d  ba=b  bb=d  bc=c  bd=a  ca=d  cb=b  cc=a  cd=c  da=a  db=c  dc=d  dd=b 
O (p, a, r) = F (p, r)  F = 4::271  aa=b  ab=d  ac=c  ad=a  ba=a  bb=c  bc=d  bd=b  ca=d  cb=b  cc=a  cd=c  da=c  db=a  dc=b  dd=d  
O (p, b, r) = F (p, r)  F = 4::304  aa=c  ab=a  ac=b  ad=d  ba=d  bb=b  bc=a  bd=c  ca=a  cb=c  cc=d  cd=b  da=b  db=d  dc=c  dd=a  
O (p, c, r) = F (p, r)  F = 4::76  aa=a  ab=c  ac=d  ad=b  ba=b  bb=d  bc=c  bd=a  ca=c  cb=a  cc=b  cd=d  da=d  db=b  dc=a  dd=c  
O (p, d, r) = F (p, r)  F = 4::499  aa=d  ab=b  ac=a  ad=c  ba=c  bb=a  bc=b  bd=d  ca=b  cb=d  cc=c  cd=a  da=a  db=c  dc=d  dd=b 
O (p, q, a) = F (p, q)  F = 4::196  aa=b  ab=c  ac=a  ad=d  ba=a  bb=d  bc=b  bd=c  ca=d  cb=a  cc=c  cd=b  da=c  db=b  dc=d  dd=a  
O (p, q, b) = F (p, q)  F = 4::476  aa=d  ab=a  ac=c  ad=b  ba=c  bb=b  bc=d  bd=a  ca=b  cb=c  cc=a  cd=d  da=a  db=d  dc=b  dd=c  
O (p, q, c) = F (p, q)  F = 4::379  aa=c  ab=b  ac=d  ad=a  ba=d  bb=a  bc=c  bd=b  ca=a  cb=d  cc=b  cd=c  da=b  db=c  dc=a  dd=d  
O (p, q, d) = F (p, q)  F = 4::99  aa=a  ab=d  ac=b  ad=c  ba=b  bb=c  bc=a  bd=d  ca=c  cb=b  cc=d  cd=a  da=d  db=a  dc=c  dd=b 
If a BQG is further sliced, the result is a permutation.
The definition of a Latin cube used throughout this report is consistent with what most algebraists use: each slice is a Latin square. Yet a caution is in order because, according to Keedwell and Dénes, statisticians sometimes define the Latin cube more loosely for use in their sampling design. If the size of a statisticians' Latin cube is n × n × n, then each twodimensional slice will as above contain n instances of each of the n elements — but statisticians might omit the requirement that each twodimensional slice be a Latin square in its own right. Consequently, a onedimensional slice might not be a permutation.
Below is such a cube, shown as four layers. The elements are represented by a, b, c, and d, while the smaller characters x, y, and z suggest the directions in which nonpermutative onedimensional slices lie.
b  c y  a  d y  d  a  c  b  a x  b x  d x  a x  c x  d x y  b x  c x y  
a  d y  b  c y  c  b  d  a  d  a  c  b  b  c y  a  d y  
d  b y z  c  a y z  b  d z  a  c z  c x  d x z  b x  c x z  a x  a x y z  d x  b x y z  
c  b y z  d  a y z  a  c z  b  d z  b  c z  a  d z  d  a y z  c  b y z 
§14 Fuzziness. An analogue of fuzzy logic can be used to extend the crisp ternary semigroups discussed above. The usual approach to fuzzy logic requires selecting a tnorm and tconorm from among many possibilities, and the following example opts for probabilistic norms because they are among the easiest to calculate. (How they are probabilistic will be explained at the end of this section.)
Each data item will be an ordered quadruple (because C = 4) of nonnegative real numbers, for example ⟨ 0.29, 0.13, 0.26, 0.32 ⟩. Components are written with subscripts: x = ⟨ x_{0}, x_{1}, x_{2}, x_{3} ⟩. A quadruple is complete when the sum of its four elements is precisely one. Meanwhile, an incomplete quadruple, whose sum is less than one, can appear only as an intermediate result in calculations.
Addition is defined in parallel: ⟨ x_{0}, x_{1}, x_{2}, x_{3} ⟩ + ⟨ y_{0}, y_{1}, y_{2}, y_{3} ⟩ = ⟨ x_{0} + y_{0}, x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3} ⟩. For a real number t and the set elements a, b, c and d, these multiplications are defined:
ta = ⟨ t, 0, 0, 0 ⟩
tb = ⟨ 0, t, 0, 0 ⟩
tc = ⟨ 0, 0, t, 0 ⟩
td = ⟨ 0, 0, 0, t ⟩
Hence ⟨ x_{0}, x_{1}, x_{2}, x_{3} ⟩ = x_{0}a + x_{1}b + x_{2}c + x_{3}d.
To establish the fuzzy operation, multiply quadruples x, y and z:
(x_{0}a + x_{1}b + x_{2}c + x_{3}d) (y_{0}a + y_{1}b + y_{2}c + y_{3}d) (z_{0}a + z_{1}b + z_{2}c + z_{3}d)
Then select a crisp operation O, and apply the distributive law numerous times to obtain the fuzzy version of operation O. When three of the set members are "multiplied", take it to mean applying the operation O to them. For instance, the "product" of b, d, and a would be O (b, d, a) or more concisely Obda :
(x_{0}y_{0}z_{0})Oaaa +  (x_{0}y_{0}z_{1})Oaab +  (x_{0}y_{0}z_{2})Oaac +  (x_{0}y_{0}z_{3})Oaad + 
(x_{0}y_{1}z_{0})Oaba +  (x_{0}y_{1}z_{1})Oabb +  (x_{0}y_{1}z_{2})Oabc +  (x_{0}y_{1}z_{3})Oabd + 
(x_{0}y_{2}z_{0})Oaca +  (x_{0}y_{2}z_{1})Oacb +  (x_{0}y_{2}z_{2})Oacc +  (x_{0}y_{2}z_{3})Oacd + 
(x_{0}y_{3}z_{0})Oada +  (x_{0}y_{3}z_{1})Oadb +  (x_{0}y_{3}z_{2})Oadc +  (x_{0}y_{3}z_{3})Oadd + 
(x_{1}y_{0}z_{0})Obaa +  (x_{1}y_{0}z_{1})Obab +  (x_{1}y_{0}z_{2})Obac +  (x_{1}y_{0}z_{3})Obad + 
(x_{1}y_{1}z_{0})Obba +  (x_{1}y_{1}z_{1})Obbb +  (x_{1}y_{1}z_{2})Obbc +  (x_{1}y_{1}z_{3})Obbd + 
(x_{1}y_{2}z_{0})Obca +  (x_{1}y_{2}z_{1})Obcb +  (x_{1}y_{2}z_{2})Obcc +  (x_{1}y_{2}z_{3})Obcd + 
(x_{1}y_{3}z_{0})Obda +  (x_{1}y_{3}z_{1})Obdb +  (x_{1}y_{3}z_{2})Obdc +  (x_{1}y_{3}z_{3})Obdd + 
(x_{2}y_{0}z_{0})Ocaa +  (x_{2}y_{0}z_{1})Ocab +  (x_{2}y_{0}z_{2})Ocac +  (x_{2}y_{0}z_{3})Ocad + 
(x_{2}y_{1}z_{0})Ocba +  (x_{2}y_{1}z_{1})Ocbb +  (x_{2}y_{1}z_{2})Ocbc +  (x_{2}y_{1}z_{3})Ocbd + 
(x_{2}y_{2}z_{0})Occa +  (x_{2}y_{2}z_{1})Occb +  (x_{2}y_{2}z_{2})Occc +  (x_{2}y_{2}z_{3})Occd + 
(x_{2}y_{3}z_{0})Ocda +  (x_{2}y_{3}z_{1})Ocdb +  (x_{2}y_{3}z_{2})Ocdc +  (x_{2}y_{3}z_{3})Ocdd + 
(x_{3}y_{0}z_{0})Odaa +  (x_{3}y_{0}z_{1})Odab +  (x_{3}y_{0}z_{2})Odac +  (x_{3}y_{0}z_{3})Odad + 
(x_{3}y_{1}z_{0})Odba +  (x_{3}y_{1}z_{1})Odbb +  (x_{3}y_{1}z_{2})Odbc +  (x_{3}y_{1}z_{3})Odbd + 
(x_{3}y_{2}z_{0})Odca +  (x_{3}y_{2}z_{1})Odcb +  (x_{3}y_{2}z_{2})Odcc +  (x_{3}y_{2}z_{3})Odcd + 
(x_{3}y_{3}z_{0})Odda +  (x_{3}y_{3}z_{1})Oddb +  (x_{3}y_{3}z_{2})Oddc +  (x_{3}y_{3}z_{3})Oddd 
That expression being lengthy, a numerical example is in order. Let:
x = ⟨ 0.16, 0.23, 0.31, 0.30 ⟩
y = ⟨ 0.22, 0.34, 0.08, 0.36 ⟩
z = ⟨ 0.13, 0.29, 0.05, 0.53 ⟩
O = 4:14046
Then the sum becomes:
(0.16 × 0.22 × 0.13)b +  (0.16 × 0.22 × 0.29)a +  (0.16 × 0.22 × 0.05)c +  (0.16 × 0.22 × 0.53)d + 
(0.16 × 0.34 × 0.13)a +  (0.16 × 0.34 × 0.29)b +  (0.16 × 0.34 × 0.05)d +  (0.16 × 0.34 × 0.53)c + 
(0.16 × 0.08 × 0.13)d +  (0.16 × 0.08 × 0.29)c +  (0.16 × 0.08 × 0.05)a +  (0.16 × 0.08 × 0.53)b + 
(0.16 × 0.36 × 0.13)c +  (0.16 × 0.36 × 0.29)d +  (0.16 × 0.36 × 0.05)b +  (0.16 × 0.36 × 0.53)a + 
(0.23 × 0.22 × 0.13)a +  (0.23 × 0.22 × 0.29)b +  (0.23 × 0.22 × 0.05)d +  (0.23 × 0.22 × 0.53)c + 
(0.23 × 0.34 × 0.13)b +  (0.23 × 0.34 × 0.29)a +  (0.23 × 0.34 × 0.05)c +  (0.23 × 0.34 × 0.53)d + 
(0.23 × 0.08 × 0.13)c +  (0.23 × 0.08 × 0.29)d +  (0.23 × 0.08 × 0.05)b +  (0.23 × 0.08 × 0.53)a + 
(0.23 × 0.36 × 0.13)d +  (0.23 × 0.36 × 0.29)c +  (0.23 × 0.36 × 0.05)a +  (0.23 × 0.36 × 0.53)b + 
(0.31 × 0.22 × 0.13)c +  (0.31 × 0.22 × 0.29)d +  (0.31 × 0.22 × 0.05)b +  (0.31 × 0.22 × 0.53)a + 
(0.31 × 0.34 × 0.13)d +  (0.31 × 0.34 × 0.29)c +  (0.31 × 0.34 × 0.05)a +  (0.31 × 0.34 × 0.53)b + 
(0.31 × 0.08 × 0.13)b +  (0.31 × 0.08 × 0.29)a +  (0.31 × 0.08 × 0.05)c +  (0.31 × 0.08 × 0.53)d + 
(0.31 × 0.36 × 0.13)a +  (0.31 × 0.36 × 0.29)b +  (0.31 × 0.36 × 0.05)d +  (0.23 × 0.36 × 0.53)c + 
(0.30 × 0.22 × 0.13)d +  (0.30 × 0.22 × 0.29)c +  (0.30 × 0.22 × 0.05)a +  (0.30 × 0.22 × 0.53)b + 
(0.30 × 0.34 × 0.13)c +  (0.30 × 0.34 × 0.29)d +  (0.30 × 0.34 × 0.05)b +  (0.30 × 0.34 × 0.53)a + 
(0.30 × 0.08 × 0.13)a +  (0.30 × 0.08 × 0.29)b +  (0.30 × 0.08 × 0.05)d +  (0.30 × 0.08 × 0.53)c + 
(0.30 × 0.36 × 0.13)b +  (0.30 × 0.36 × 0.29)a +  (0.30 × 0.36 × 0.05)c +  (0.30 × 0.36 × 0.53)d 
After combining similar terms, the result is exactly ⟨ 0.246512, 0.255600, 0.249264, 0.248624 ⟩.
To see this operation's probabilistic behavior, consider an example with many zeros:
x = ⟨ 0.16, 0.84, 0.00, 0.00 ⟩
y = ⟨ 0.66, 0.34, 0.00, 0.00 ⟩
z = ⟨ 0.75, 0.25, 0.00, 0.00 ⟩
O = 4:14046, as before
Now substitute as in the first example:
(0.16 × 0.66 × 0.75)b +  (0.16 × 0.66 × 0.25)a + 
(0.16 × 0.34 × 0.75)a +  (0.16 × 0.34 × 0.25)b + 
(0.84 × 0.66 × 0.75)a +  (0.84 × 0.66 × 0.25)b + 
(0.84 × 0.34 × 0.75)b +  (0.84 × 0.34 × 0.25)a 
Many terms have dropped out because of the zeros in the inputs, and the result is ⟨ 0.5544, 0.4456, 0.0000, 0.0000 ⟩.
If the probabilities of three independent events are 0.16, 0.66 and 0.75, then the probability of zero or any two (i.e. an even number) of them occurring is 0.5544, and the probability of any one or three (an odd number) occuring is 0.4456.
§15A. Authors discussing binary quasigroups often introduce two kinds of inverse to the principal operation.
Select as the obverse a cancellative binary operation F (p, q), but employ instead the centered dot notation p · q. Based on the following relation:
The names of these three operations are · ⧸ ⧹ They might be pronounced "dot", "foreslash", and "backslash". Because of cancellativity, each is guaranteed to exist and be unique.
Researchers in universal algebra might prefer to work in the opposite direction, postulating these three operations to define a quasigroup, and leaving cancellativity as a consequence. This approach calls for a set S of values, and the three binary operations from above. Necessary and sufficient is that the four following identities be satisfied for all p, q ∈ S:
(p · q) ⧸ q = p  p ⧹ (p · q) = q  
(p ⧸ q) · q = p  p · (p ⧹ q) = q 
The structure { S · ⧸ ⧹ } so assembled is sometimes called a primitive quasigroup, or an equasigroup (from "equation quasigroup"). Some writers hold that a primitive quasigroup is indeed a fullyfledged quasigroup, while others say that a primitive quasigroup is merely isomorphic to a quasigroup. Whatever the classification, each inverse operation engenders a quasigroup to precisely the same extent as the obverse does.
§15B. With that precedent in place, ternary operations can now be addressed.
Select as the obverse a cancellative ternary operation O (p, q, r), employing the twodot notation p · q · r. Based on the following relation:
The names of the four operations are ·· ⧸⧸ ⧹⧸ ⧹⧹ Because of cancellativity, each is guaranteed to exist and be unique.
Observe that the notation is p = s ⧸ r ⧸ q rather than p = s ⧸ q ⧸ r. The distinction is critical because these operations are in general noncommutative. Here is a rationale for the notation:
In step 2, the expressions p · q and s ⧸ r do not actually mean anything; they merely convey part of the rationale for the notation.
Also observe that the symbol ⧸⧹ is not defined.
With a set S, the universal algebra specification will have five components: { S ·· ⧸⧸ ⧹⧸ ⧹⧹ }. The operations will need to satisfy six identities:
(p · q · r) ⧸ r ⧸ q = p  p ⧹ (p · q · r) ⧸ r = q  q ⧹ p ⧹ (p · q · r) = r  
(p ⧸ q ⧸ r) · r · q = p  p · (p ⧹ q ⧸ r) · r = q  q · p · (p ⧹ q ⧹ r) = r 
As in the binary case, each inverse operation leads to as much a quasigroup as the obverse.
§15C. Many investigations of quasigroups examine only one obverse operation at a time. On the other hand, this report frequently examines the interaction of multiple obverse operations, for instance the distributive law shown in section 2D. This makes disambiguation sometimes necessary. Recall:
Thus one can write without confusion mixedoperation expressions such as the following:
O (p ⧹ N (q ⧸ r ⧸ s) ⧸ M (t · u · v))
Such a notation also allows various identities to be concisely expressed. Define the following names for certain operations:


This set of eight operations is closed under inversion as follows:
O  (p · q · r)  = N  (p ⧸ q ⧸ r)  = M  (p ⧹ q ⧸ r)  = K  (p ⧹ q ⧹ r) 
N  (p · q · r)  = O  (p ⧸ q ⧸ r)  = L  (p ⧹ q ⧸ r)  = J  (p ⧹ q ⧹ r) 
M  (p · q · r)  = I  (p ⧸ q ⧸ r)  = O  (p ⧹ q ⧸ r)  = L  (p ⧹ q ⧹ r) 
K  (p · q · r)  = J  (p ⧸ q ⧸ r)  = I  (p ⧹ q ⧸ r)  = O  (p ⧹ q ⧹ r) 
L  (p · q · r)  = H  (p ⧸ q ⧸ r)  = N  (p ⧹ q ⧸ r)  = M  (p ⧹ q ⧹ r) 
J  (p · q · r)  = K  (p ⧸ q ⧸ r)  = H  (p ⧹ q ⧸ r)  = N  (p ⧹ q ⧹ r) 
I  (p · q · r)  = M  (p ⧸ q ⧸ r)  = K  (p ⧹ q ⧸ r)  = H  (p ⧹ q ⧹ r) 
H  (p · q · r)  = L  (p ⧸ q ⧸ r)  = J  (p ⧹ q ⧸ r)  = I  (p ⧹ q ⧹ r) 
This illustrates how inversion does not produce genuinely new operations. Still, inversion can nonetheless shed light on how one operation relates to another. (Fortunately, there is no analogue to the divisionbyzero inconvenience.)
§16A Commutative complementation. This is best explained with examples. Define names for six particular operations:


None of those six operations has any commutativity in its own right, but each has other operations as commutative complements as listed below. These six operations form a closed set under commutative complementation.
commutative complements of O, N, M, L, K, J  

sinisterior:  O (p, q, r) = N (q, p, r)  exterior:  O (p, q, r) = M (r, q, p)  dexterior:  O (p, q, r) = L (p, r, q) 
M (p, q, r) = K (q, p, r)  N (p, q, r) = J (r, q, p)  N (p, q, r) = K (p, r, q)  
L (p, q, r) = J (q, p, r)  L (p, q, r) = K (r, q, p)  M (p, q, r) = J (p, r, q)  
rcyclical:  O (p, q, r) = J (r, p, q)  qcyclical:  O (p, q, r) = K (q, r, p)  
N (p, q, r) = M (r, p, q)  N (p, q, r) = L (q, r, p)  
M (p, q, r) = L (r, p, q)  M (p, q, r) = N (q, r, p)  
L (p, q, r) = N (r, p, q)  L (p, q, r) = M (q, r, p)  
K (p, q, r) = O (r, p, q)  K (p, q, r) = J (q, r, p)  
J (p, q, r) = K (r, p, q)  J (p, q, r) = O (q, r, p) 
To set up another example, define names for three more operations:
X = 4:35451  Y = 4:35465  Z = 4:36041 
Each of the them is commutative (with itself) in exactly one way: X sinisterior, Y exterior, and Z dexterior. They too form a closed set under commutative complementation:
commutative complements of X, Y, Z  

sinisterior:  X (p, q, r) = X (q, p, r)  exterior:  Y (p, q, r) = Y (r, q, p)  dexterior:  Z (p, q, r) = Z (p, r, q) 
Y (p, q, r) = Z (q, p, r)  Z (p, q, r) = X (r, q, p)  X (p, q, r) = Y (p, r, q)  
rcyclical:  X (p, q, r) = Z (r, p, q)  qcyclical:  X (p, q, r) = Y (q, r, p)  
Y (p, q, r) = X (r, p, q)  Y (p, q, r) = Z (q, r, p)  
Z (p, q, r) = Y (r, p, q)  Z (p, q, r) = X (q, r, p) 
§16B. One might hope for associative complementation, but the situation becomes murkier. Define these:
F = 3::1  G = 3::2  H = 3::5 
It turns out that F has more than one associative complement:
F (F (p, q), r) = G (p, G (q, r)) = H (p, H (q, r))
Also, the distinction between operations G and H is lost when each is associated on the right. Meanwhile, 3::4 has no associative complement, and 3::0 has only itself.
Page one contains sections 1 through 4.
Page two contains sections 5 through 10. Page three contains sections 11 and 12. This is page four, with sections 13 through 16. 
directory of files  

numbered pages  operation charts 
page one (§1 – §4): page two (§5 – §10): page three (§11 – §12) page four (§13 – §16) 
binary: ternary: 