A 111111111111111111111111111 27 A 1 1 B 11 2 B 1111111 7 C 111111111 9 C D 1111111111 10 D E 1111111111111111111111111111 28 E 1 1 F 11 2 F 11111 5 G 111 3 G H 11 2 H 1 1 I 111111111111 12 I 111 3 J 1 1 J 1 1 L 1111111111 10 L 111 3 M 111111 6 M 1 1 N 111111111111 12 N 1 1 O 1111111111111111 16 O 111111 6 P 11111 5 P 111 3 Q 11 2 Q 111 3 R 111111111111111 15 R 11 2 S 11111111111111 14 S T 11111111 8 T U 1111111 7 U V 11 2 V X X 11 2 Y 11 2 Y Z 1 1 Z 1 1 Our first assumption might be that B = A and F = E but it is evident at once that in that case, S, T, U and V (equal to R, S, T and U) do not occur and a message even this short without R, S, T or U is practically impossible. By trying B = E we find that the two tables agree in a general way very well and this is all that can be expected with such a short message. The longer the message the nearer would its frequency table agree with the standard table. Note that if a cipher disk has been used, the alphabet runs the other way and we must count upward in working with a graphic table. Note also that if, in a fairly long message, it is impossible to coördinate the graphic table, reading either up or down, with the standard table and yet some letters occur much more frequently than others and some do not occur at all, we have a mixed alphabet to deal with. The example chosen for case 6-a is of this character. An examination of the frequency table given under that case shows that it bears no graphic resemblance to the standard table. However, as will be seen in case 7-b, the preparation of graphic tables enables us to state definitely that the same order of letters is followed in each of a number of mixed alphabets. General Remarks