Reply To: Challenge 10B
@cat_robin the important pattern to notice was that there are 36 cards in the first of each pair, and 16 cards in the second.
Through comparing the IoC of ciphertext letter pairs to plaintext letter pairs (of some large plaintext you already have), and noticing repeats in the ciphertext, and that the 36/16 pattern occurs every pair, you can guess that each pair of cards is a substitution for a pair of plaintext letters.
If the cipher used a 26 letter alphabet, there would be up to 26×26 = 676 different letter pairs. But 36×16 = 576, so we observe far fewer pairs of cards. Since 36 and 16 are squares, 576 must be a square: 24×24. This is a pretty solid indication of a 24 letter alphabet.
24 = 6×4 (so, number of rows/cols of the 36 square × number of rows/cols of the 16 square), and each pair of cards contains one card from the 36 and one from the 16; a simple way to address pairs with this setup is a four square (rectangle?) cipher, using the 36 and 16 squares with two 6×4 rectangles containing the plaintext alphabets.