From this table we pick out the letters B, E, F, O, R, T, X, as tentative letters of the key word on account of the variety of other letters with which they occur. As there are but two vowels for seven letters, we will add A to the list on account of its occurrences with B, D, E, R, and X. This leaves the letters for the bottom lines of the square as follows:
| . | . | . | . | . |
| . | . | . | C | D |
| G | H | IJ | K | L |
| M | N | P | Q | S |
| U | V | W | Y | Z |
Referring to the table again we find the most frequent combination to be EL, occurring 8 times, with no occurrence of LE. Now, TH is the commonest pair in plain text, and HT is not common. The fact that H occurs in the same horizontal line with L and that E and T are probably in the key, will lead us to put E in the first line over H and T in the first line over L, so as to make EL equal TH.
The next most frequent combination is PR occurring 7 times, with RP occurring twice. In the square as partially arranged, PR equals M_ or N_ or Q_ or I_. We may eliminate all these except N_, and this N_ could only be NO or NA, so that we will put, tentatively the R in the second line over H and the O and A in the same line over IJ. We have then:
| . | E | . | . | T |
| . | R | AO | C | D |
| G | H | IJ | K | L |
| M | N | P | Q | S |
| U | V | W | Y | Z |
Let us now check this by picking out the combinations beginning with EL and seeing if the table will solve them. We find, ELTV, ELAB, ELBXFZ, ELBXBT, ELAXCWBY, ELAXCWEQ, ELRH, ELBXFS. Now, on the assumption that the letter after EL represents E, we have it represented by A three times, B three times, R once and T once. This requires that A and B be put in the same horizontal line with E, since T is already there, and R is tentatively under E.
The combination ELTV now equals THEZ. If the T were moved one place to the left, it would be THEY, a more likely combination, but this requires the L to be moved one place to the left also, by putting I or K in the key word and taking out O, R or X and returning it to its place in the alphabetical sequence. The most frequent pairs containing O are B O six times, R O four times, and O X three times. Now these pairs equal respectively E N, E S and H E, if O is put between N and P in the fourth line. We will therefore cease to consider it as a letter of the the key word. The combination ELAB can only be THE_ on the assumption that A is the first letter to the right of E. The combination ELBX occurs three times. If it represents THE_, the B must be the first letter of the first line and the X must now be placed under E where the R was tentatively put. We can get THE_ out of ELRH by putting R in the first line or leaving it where it is, but the preponderance of the BX combination should suggest the former alternative.
A new square showing these changes will look like this:
| B | E | A | T | R |
| . | X | . | . | . |
| G | H | . | L | M |
| N | O | P | Q | S |
| U | V | W | Y | X |
As I put in the space under B will give the word BEATRIX and as a vowel is clearly necessary there, we will so use the IJ and leave K between H and L. This leaves C, D and F to be placed. It appeared at first that F was in the key but if it is in the second line, in proximity to the letters of the first line, it will give the same indications. Completing the square then, we have