| G | R | A | N | T | |
| G | A | B | C | D | E |
| R | F | G | H | I | K |
| A | L | M | N | O | P |
| N | Q | R | S | T | U |
| T | V | W | X | Y | Z |
Thus TG=E, AN=S, etc., as we have already found.
Case 9-b
Message
| 1950492958 | 3123252815 | 4418452815 | 2048115041 |
| 2252115345 | 5849134124 | 5028552526 | 5933195222 |
| 5245113215 | 6215584143 | 2861361265 | 2945565015 |
| 2342455850 | 6345542019 | 1550185311 | 2115415828 |
| 1124174553 | 4554205950 | 2552454132 | 1533492048 |
| 5018152364 |
An examination of the groups of two numerals each which make up this message, shows that we have 11 to 36 and 41 to 65 with eleven groups missing. Now the 11 to 36 combination is a very familiar one in numeral substitution ciphers (See Case 6-c) and it will be noted that 41 to 66 would give us a similar alphabet. Let us make a frequency table in this form:
| Group | Frequency | Group | Frequency |
| 11 | 11111 | 41 | 11111 |
| 12 | 1 | 42 | 1 |
| 13 | 1 | 43 | 1 |
| 14 | 44 | 1 | |
| 15 | 111111111 | 45 | 111111111 |
| 16 | 46 | ||
| 17 | 1 | 47 | |
| 18 | 111 | 48 | 11 |
| 19 | 111 | 49 | 111 |
| 20 | 1111 | 50 | 11111111 |
| 21 | 1 | 51 | |
| 22 | 11 | 52 | 1111 |
| 23 | 111 | 53 | 111 |
| 24 | 11 | 54 | 11 |
| 25 | 111 | 55 | 1 |
| 26 | 1 | 56 | 1 |
| 27 | 57 | ||
| 28 | 11111 | 58 | 11111 |
| 29 | 11 | 59 | 11 |
| 30 | 60 | ||
| 31 | 1 | 61 | 1 |
| 32 | 11 | 62 | 1 |
| 33 | 11 | 63 | 1 |
| 34 | 64 | 1 | |
| 35 | 65 | 1 | |
| 36 | 1 | 66 |
Each of these tables looks like the normal frequency table except for the position of 20 and 50 which should represent T, by all our rules, and should be apparently 30 and 60. But suppose we put the alphabet and corresponding numerals in this form:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0 | |
| 1 or 4 | A | B | C | D | E | F | G | H | I | J |
| 2 or 5 | K | L | M | N | O | P | Q | R | S | T |
| 3 or 6 | U | V | W | X | Y | Z |