It is clear from an examination of this table that we have to deal with a single alphabet but one in which the letters do not occur in their regular order.
We may assume that P and S are probably A and E, both on account of the frequency with which they occur and the variety of their prefixes and suffixes. If this is so, then B and H, are probably consonants and may represent R and N respectively. D and X are then vowels by the same method of analysis. Noting that HC occurs three times and taking H as N we conclude that C is probably T. Substitute these values in the last three words of the message because the letters assumed occur rather frequently there.
| PBPSZBSV | X | Y | X | SHC | D |
| I | I | I | |||
| ARAE_RE_ | _ | ENT | |||
| O | O | O |
Now Z is always prefixed by S and may be L. Taking X=I and D=O, (they are certainly vowels), V=G and Y=M, we have
| ARA EL REGIMIENTO |
Substituting these values in the rest of the message we have
| Q | DBYPBXH | YSOXPCP | YSHCSED | RBSZPTPB |
| _ | ORMARIN | ME_IATA | MENTE_O | _RELA_AR |
| BSCSB | PSHSZ | AJHCD | OSEXVHPODA | |
| RETER | AENEL | __NTO | _E_IGNA_O_ |
We may now take Q=F, O=D, E=S, R=B, T=C, A=P and J=U and the message is complete. We are assisted in our last assumption by noting that S=E and E=S, etc., and we may on that basis reconstruct the entire alphabet. The letters in parenthesis do not occur in the message but may be safely assumed to be correct.
| Ordinary | A | B | C | D | E | F | G | H | I | J | L | M | N | O | P | Q | R | S | T | U | V | X | Y | Z |
| Cipher | P | R | T | O | S | (Q) | (V) | N | (X) | (U) | (Z) | (Y) | (H) | D | A | F | B | E | C | J | G | I | M | L |
It is always well to attempt the reconstruction of the entire alphabet for use in case any more cipher messages written in it are received.——