For the purposes of this short delve into codebreaking I have, erm, borrowed part of the Jellyneo Book of Ages.
FIIRPBK FP 7 HFKA B7OQE C7BOFB, TEL IFSBP FK JBOFABII. 7Q LKB QFJB PEB IFSBA FK C7BOFBI7KA, 8RQ IBCQ CLO OB7PLKP RKHKLTK. PEB EBIMBA ABCBKA JBOFABII AROFKD QEBFO T7OP, 7KA E7P OBJ7FKBA QEBOB FK EBO DI7AB PFK9B. FIIRPBK PBKAP J7KV VLRKD KBLMBQP LRQ LK NRBPQP, 7IQELRDE KL LKB HKLTP TE7Q PEB MI7KP QL AL TFQE 7II QEB FQBJP PEB’P 9LIIB9QBA.
DGGPN9I’N BM95O9NO AJ9 DN ECP8JM5, 5GOCJPBC IJ JI9 N5Q9 OC9 ORJ JA OC9H FIJRN RC5O C5KK9I98 69OR99I OC9H – NJH9 9Q9I N5T OC9T R9M9 JI79 AMD9I8N.
BG MA7 63KD7LM 837KB7 93F7, BEENL7G 4753F7 IHLL7LL76 4R MA7 63KD7LM 837KB7 38M7K MA7 P7K7ENI7 DBG9 LMHE7 A7K 5A3KF (PAB5A IKHM75M76 A7K 8KHF MA7 63KD7LM 837KB7). MHK 3G6 KH47KM3 K75HO7K76 MA7 5A3KF, 3G6 BG MA3GDL, BEENL7G 93O7 MA7F MA7 837KB7 HK4, PAB5A 53G 5NM MAKHN9A F39B53E 43KKB7KL.
The most basic approach it is to look for letters on their own, they must be I or A. This only works because the encrypted data has not been reformatted into same sized blocks so we know these are likely to be correct. Here, we know I is unlikely since we also know the context – its written about someone not by them. So that would make the 7 an A.
Another easy approach is Frequency Analysis. E is the most common letter in the English Language, so it is likely, assuming they have not encoded spaces or added extra letters, that the most common one in the code is E. This approach would then move onto letters such as T and the remaining vowels. It also works to pick out the uncommon ones. In the First block we can also see N appears just once, so is likely to be Z, Q, X or J. If R turns out to be U, then this N would likely be a Q.
When you input the first block cypher into the second or third block you will find it doesn’t produce the desired result. This indicates that the code has changed, so you have to start again. And unfortunately there are no lone letters in the 2nd block to find A with.
The addition of numbers to this code actually makes no difference using these methods, where as if you were to try and find the key it could since you wouldn’t know where the numbers were in the rest of the array.
All else failing, you could just go for a random approach. But at this late stage you probably wont find it in time. Btw, Black-bag & rubber-hose cryptanalysis aren’t going to work, so dont even try them. 😄
And before you go running off to put 7=A into the first line, I didn’t encrypt the data above the same way.