Reply To: Challenge 10B
@tahausman, are you suggesting that Cipher Challenge create its own YouTube channel and post tutorials on cipher-cracking? I myself cannot contribute, because my ugly face breaks any camera lens that points at it.
What I can do is offer you this explanation of how @AspiringPenguin figured it out. It is not how Harry expected the cipher to be solved, so it gives a different and interesting perspective. Here’s how s/he did it:
First, notice that the characters come in pairs, and the first of each pair is from A23456789XJQK, while the second is from CDHS. Remember all the stuff about playing cards at the beginning of the season? These character sets should pop out as playing cards, if we take the reasonable assumption that X = 10. OK, so they are all playing cards. In the Bridge ordering of the deck, cards run AC, 2C, …, XC, JC, QC, KC, AD, 2D, …, KD, AH, …, KH, AS, …, KS. Number them from 0 to 51. Now the ciphertext is a long list of numbers.
Next, notice that once again things come in pairs. The first number of each pair is in the range 0-35, while the second is in the range 36-51. That second range has length 16, so let’s subtract 36 from each of them to get numbers in the range 0-15.
At this point, the Penguin notices that numbers in the range 0-35 are two-digit numbers when expressed in base 6, and numbers 0-15 are two-digit numbers when expressed in base 4. For each pair of numbers, replace them with their base-6 and base-4 equivalents; call them AB and CD for the heck of it, where A and B are in the range 0-5, and C and D in the range 0-3.
Now, we must remember that 24 is one of Harry’s favo(u)rite numbers. And an alphabet with 24 letters is usually big enough, if the plaintext does not contain J and Z, for example. Well, 24 = 6 x 4. So we can shuffle the digits A, B, C, and D to make two numbers in the range 0-23. There are several ways to do this, but the one that puts the pairs in the right order and lines up the alphabets so that the keywords stand out is this:
X = B + 6xC
Y = D + 4xA
Do this calculation for each pair of numbers AB(base6) and CD(base4), and replace them with X and Y. Or replace X and Y with letters (A=0, B=1, …, Z=25). Either way, what remains is a polyalphabetic substitution cipher with period 2. This can be solved by hill-climbing or by clever trial-and-error.
When you are finished, you will find that the two substitution keys are gibberish, but if you invert them, you get
SHADOWBCEFGIKLMNPQRTUVXYJZ
FULHEARTBCDGIKMNOPQSVWXYJZ
Lop off the JZ at the ends because you need 24-letter keys for this cipher. The keywords are now evident.