232 THE BELL SYSTEM TECHNICAL JOURNAL, JANUARY 1956 



An alphabet with one or more elements of weight 1 must have an 

 «2 ^ 36, for there are nine elements of weight 2 which cannot possibly 

 be coset leaders. To see this, suppose (without loss of generality) that 

 the alphabet contains the letter 1. The elements 12, 13, 14, • • • 1 10 can- 

 not possibly be coset leaders since the product of any one of them with 

 the letter 1 yields an element of weight 1 . 



An alphabet with one or more elements of weight 2 must have an 

 ai S 37. Suppose for example, the alphabet contained the letter 12. 

 Then 13 and 23 must be in the same coset, 14 and 24 must be in the 

 same coset, ■ • • , 1 10 and 2 10 must be in the same coset. There are at 

 least eight elements of weight two which are not coset leaders. 



Each element of weight 3 in the alphabet prevents three elements of 

 weight 2 from being coset leaders. For example, if the alphabet contains 

 123, then 12, 13, and 23 cannot be coset leaders. We say that the three 

 elements of weight 2 are "blocked" by the letter of weight 3. Suppose an 

 alphabet contains at least three letters of weight three. There are several 

 cases: (A) if three letters have no numerals in common, e.g., 123, 456, 

 789, then nine distinct elements of weight 2 are blocked and a-2 S 36; 

 (B) if no two of the letters have more than a single numeral in common, 

 e.g., 123, 345, 789, then again nine elements of weight 2 are blocked and 

 a-2 ^ 36; and (C) if two of the letters of weight 3 have two numerals in 

 common, e.g., 123, 234, then their product is a letter of weight 2 and l)y 

 the preceding paragraph ao ^ 37. If an alphabet contains exactly two 

 elements of weight 3 and no elements of weight 2, the elements of weight 



3 block six elements of weight 2 and 0:2 ^ 39. 

 The preceding argument shows that to be better than the exhibited 



alphabet a (10, 4)-alphabet with letters of weight 3 must have just one 

 such letter. A similar argument (omitted here) shows that to be better 

 than the exhibited alphabet, a (10, 4)-alphabet cannot contain more 

 than one element of weight 4. Furthermore, it is easily seen that an 

 alphabet containing one element of weight 3 and one element of weight 



4 must have an ao ^ 39. 

 The only new contenders for best (10, 4)-alphabet are, therefore, 



alphabets with a single letter other than / of weight less than 5, and this 

 letter must have weight 3 or 4. Application of Propositions 6 and 7 show 

 that the only possible weights for alphabets of this sort are: 35 6 and 



5 46' where 5' means seven letters of weight 5, etc. We next show that 

 there do not exist (10, 4)-alphabets having these weights. 



Consider first the suggested alphabet with weights 35 6'. As explained 

 in Section 2.9, from such an alphabet we can construct a matrix repre- 

 sentation of ('4 having the character x(/) = 10, one matrix of trace 4, 



