Col. 1. Col. 2. Col. 3. Col. 4. Col. 5. A 2 A 9 A 1 A 1 A 2 B— B 3 B 3 B— B 7 C 7 C 1 C 3 C 4 C— D 2 D 2 D 1 D— D 3 E 4 E— E 2 E 7 E— F 3 F— F 9 F 3 F 5 G 9 G— G 3 G 2 G 2 H 3 H 5 H 3 H 3 H 2 I 2 I 2 I 7 I 17 I 2 J 5 J 1 J 6 J— J 9 K 6 K 5 K— K 1 K 1 L— L 19 L 2 L 5 L 1 M— M— M 7 M 4 M 3 N 7 N 3 N 4 N— N 5 O 5 O— O 9 O 1 O— P 7 P 7 P 8 P 4 P— Q 5 Q— Q— Q 2 Q 6 R— R 1 R 1 R 6 R 1 S— S 8 S 6 S 12 S 7 T 7 T 3 T 5 T 1 T 14 U 7 U 3 U 6 U— U 1 V 5 V— V 2 V 5 V— W 3 W 4 W— W 5 W 7 X 2 X— X 4 X 8 X 6 Y 4 Y 5 Y— Y 3 Y 7 Z— Z 5 Z 3 Z— Z 3 Now, having erected these five enigmatical columns, Captain Hitt juggles them until he uncovers the hidden message, thus: “In the table for column 1 the letter G occurs 9 times,” he says with an air of a man having found something that is perfectly plain. “Let us consider it tentatively as E. “Then, if the cipher alphabet runs regularly and in the direction of the regular alphabet, C (7 times) is equal to A, and the cipher alphabet bears a close resemblance to the regular frequency table. Note that TUV (equal to RST) occurring respectively 7, 7, and 5 times and the non-occurrence of B, L, M, R, S, Z (equal to Z, J, K, P, Q, and X, respectively). “In the next table L occurs 19 times, and taking it for E with the alphabet running the same way, A is equal to H. The first word of our message, CT, thus becomes AM when deciphered with these two alphabets, and the first two letters of the key are CH. “Similarly in the third table we may take either F or O for E, but a casual examination shows that the former is correct and A is equal to B. “In the fourth table I is clearly E and A is equal to E. “The fifth table shows that T is equal to 14 and J is equal to 9. If we take J as equal to E then T is equal to O, and in view of the many Es already accounted for in the other columns this may be all right. It checks as correct if we apply the last three alphabets to the second word of our message, OSB, which deciphers NOW. Using these alphabets to decipher the whole message we find it to read: