As a starting-point the cipher-man assumed that the text was in English because he got it out of an English newspaper, but he did not stop there. He checked it from a negative view-point by finding the letter _w_ in it, which does not occur in the Latin languages, and by finding that the last fifteen words of the message had from two to four letters each, which would have been impossible in German. Then he proceeds to analyze. The message has 108 groups that are presumably words, and there are 473 letters in it. This makes an average of 4.4 letters to the group, whereas one versed in the art normally expects about five. There are ninety vowels of the AEIOU group and seventy-eight letters JKQXZ. Harking back to that first statement of percentages, it is certain that this is a substitution cipher because the percentage does not check with the transposition averages. The canny man with the sharp pencil then looks for recurring groups and similar groups in his message and he finds that they are: AIIWG AII BKSM BKAI CT CTWY CTW DLMMJXL DL ESF ESBP FJNVX FJHVSI NPSI NPUXQG OSB OSY ROSB OL OL PORTELOGJ PO SQ SQA TP TP TLBWTPZ TLFNS TLFTWMC UVZUA UVD UV SMKUL XMKUHW YJL YJVTI. Passing along by the elimination route he refers to his frequency tables to see how often the same letters occur, and he finds that they are all out of proportion, and he can proceed to hunt the key for several alphabets. He factors the recurring groups like a small boy doing a sum in arithmetic when he wants to find out how many numbers multiplied by each other will produce a larger one. The number of letters between recurring groups and words is counted and dissected in this wise: AII AII 45, which equals 3x3x5 BK BK 345, which equals 23x3x5 CT CT 403, no factors CTW CTW 60, which equals 2x2x2x5 DL DL 75, which equals 3x5x5 ES ES 14, which equals 2x7 FJ FJ 187, no factors NP NP 14, which equals 2x7 OL OL 120, which equals 2x2x2x3x5 OS OS 220, which equals 11x2x2x5 OSB OSB 465, which equals 31x3x5 PO PO 105, which equals 7x3x5 SQ SQ 250, which equals 2x5x5x5 TLF TLF 80, which equals 2x2x2x2x5 TP TP 405, which equals 3x3x3x3x5 UV UV 115, which equals 23x5 XMKU XMKU 120, which equals 2x2x2x3x5 UV UV 73, no factors YJ YJ 85, which equals 17x5 Now the man who is doing the studying takes a squint at this result and he sees that the dominant factor all through the case is the figure 5, so he is reasonably sure that five alphabets were used, and that the key-word had, therefore, five letters, so he writes the message in lines of five letters each and makes a frequency table for each one of the five columns he has formed, and he gets the following result: