A CLASS OF BINARY SIGNALING ALPHABETS 211 



T. If it does, we say T satisfies the tth-place parity check; otherwise T 

 fails the zth-place parity check. When a set of parity check rules (9) is 

 giN'cii, we can associate an (n — /i^-place binary sequence, R{T), with 

 each element T of 5„. We examine each check place of T in order starting 

 with the (k -\- 1 )-st place of T. We write a zero if a place of T satisfies 

 the parity check; we write a one if a place fails the parity check. The re- 

 sultant sequence of zeros and ones, written from left to right is R(T). 

 We call R(T) the parity check sequence of T. Example: with the parity 

 rules 04 = 02 -j- 03 , 05 = Oi -j- 02 -j- c^s used to define the (5, 3)-alphabet 

 in the examples of Theorem 3, we find i?(11000) = 10 since the sum of 

 the entries in the second and third places of 11001 is not the entry of 

 the fourth place and since the sum of Oi = 1, 02 = 1, and 03 = is 

 = 05 . 



Theorem 4- Let I, A2 , • • • ^^^ be an {n, /c)-alphabet. Let R{T) be the 

 parity check sequence of an element T of B„ formed in accordance with 

 the parity check rules of the (n, /c) -alphabet. Then R(Ti) = R(T2) if 

 and only if Ti and T2 lie in the same row of array (4). The coset leaders 

 can be ordered so that R{Si) is the binary symbol for the integer i — 1. 



As an example of Theorem 4 consider the (4, 2)-alphabet shown with 

 its cosets below 



The parity check rules for this alphabet are 03 = oi , 04 = Oi -j- ^2 • 

 Every element of the second row of this array satisfies the parity check 

 in the third place and fails the parity check in the 4th place. The parity 

 check sequence for the second row is 01. The parity check for the third 

 row is 10, and for the fourth row 11. Since every letter of the alphabet 

 satisfies the parity checks, the parity check sequence for the first row is 

 00. We therefore make the following association between parity check 

 sequences and coset leaders 



00 -^ 0000 = Si 



01 -^ 0100 = S2 



10 -^ 0010 = S, 



11 -^ 1000 = ^4 



1.9 INSTRUMENTING A GROUP ALPHABET 



Proposition 4 attests to the ease of the encoding operation involved 



