GROUP ixvakiaxcf: or total symmetry 1451 



Table IV — Transmission Matrix for 



T = S (3, 5, 6, 7) = S2,z(xi , X2 , .1-3) 



X3 X2 Xi 



3 Oil 



5 10 1 



6 110 



7 111 



example, the transmission of Table IV equals 1 whenever two or 

 three relays are operated. For such transmissions any permutation of 

 the variables leaves the transmission unchanged. These transmissions 

 are called totally symmetric. They are usually written in the form, 

 T = Soi , a«---a„X^i , X2 , ••• Xn), whcrc thc transmission is to equal 

 1 only ^^•ilen exactly ai or a-z or • • • or Um of the variables Xi , x^ • ■ • Xn 

 are equal to one. The transmission of Table IV can be written as 

 'S2,3(.i"i , x<i , Xz). This definition of symmetric transmissions can be gen- 

 eralized by allowing some of the variables {xi , X2 , ■ ■ • .r„) to be primed. 

 Thus the transmission S^ixi , X2 , Xz) will equal 1 only when Xi = X2 = 

 X:i = 1 or 0^1 = x-i = 1 and X2 = 0. It is useful to know when a trans- 

 mission is totally symmetric since special design techniques exist for 

 such functions.'* 



It is possible to determine whether a transmission is totally symmetric 

 from its matrix. Unless all columns of the standard matrix derived from 

 the transmission matrix have the same weight, the transmission cannot 

 possibly be totally symmetric. If all columns do have equal weights, the 

 rows should be partitioned into groups of rows which all have the same 

 weight. Whether the transmission is totally symmetric can now be de- 

 termined by inspection. If there is a row of weight k; that is, a row which 

 contains k I's, then every possible row of weight k must also be included 

 in the matrix. This means that there must be nCk rows of weight k where 

 71 is the number of columns (variables).* If any possible row of weight k 

 was not included then the corresponding k literals could be set equal to 

 1 without the transmission being equal to 1. This contradicts the defi- 

 nition of a totally symmetric transmission. In Table V(b) there are 4 

 rows of weight 1 and 1 row of weight 4. Since 4C1 = 4 and 4C4 = 1 this 

 transmission is totally symmetric and can be written as Si,i(xi , X2', Xz , 

 Xi). The number of rows of weight 1 in Table V(d) is 2 and since iCi = 4 

 this transmission is not totally symmetric. 



A difficulty arises if all columns of a transmission matrix contain equal 



* nCk is the binomial coefficient -. " ,, , 



(n — K)\li\ 



