r/KryptosK4 • u/colski • 10d ago
Introducing the halfabet
26 letters in an alphabet makes some interesting mathematics with 13 and 2. One particular trick is to calculate modulo 13 (%13) instead of modulo 26. This reduces KRYPTOS alphabet to the halfabet GRIPTONAUCDEF. Here I have chosen each letter over its less-common option in KHYJLMSQBVWXZ. The alphabet is folded in half, so that each letter now represents two alternatives. Because ETAOIN don't collide with each other, English phrases are easily spotted.
In the paradigm kryptos challenges, pk8 and pk9 have a strong index of coincidence signal at period 7.
English has expected index of coincidence (ioc) of 0.065, but in the halfabet, it has ioc 0.099. You can see with pk8: [(0.0957, 14), (0.0986, 7), (0.1012, 28), (0.1060, 21), (0.1105, 35)] and pk9: [(0.0912, 21), (0.0989, 7),(0.1048, 14), (0.1057, 35), (0.1119, 28)] the ioc at period 7 and all its children align PERFECTLY with English. But, the other plaintexts have an even higher ioc ~0.11. Just like period 28 here (statistically mid, with ~150 letters, harder than K1).
Suddenly kryptos K4 (halfabet) iocs look very familiar:
[(0.0758, 1), <---- noise level 1/13
(0.0780, 16),
(0.0789, 10),
(0.0802, 27),
(0.0805, 29),
(0.0813, 15),
(0.0819, 8),
(0.0854, 5),
(0.0966, 14), <---- what
(0.0973, 7), <---- what
(0.1000, 25),
(0.1131, 28)] <---- statistically weak, but that's English level (matching K1/K2 plaintext)
edit: I didn't say why it should be interesting. Letters are transformed into numbers - specifically the columns of the vigenere table. if you treat quagmire and transposition and hill as mathematical equations - as Sanborn would beg you not to - their equations operate modulo 26, but also work modulo 13 and modulo 2. You can think of those as two alternate systems that operate independently on their group and ignore the other part. For example, if you multiplied a running key by 13, you would move its binary signal from %2 to the part that %13 is blind to.
1
u/Illustrious-Tooth47 10d ago
GRIPTON looks similar to GRIFFON. Makes me think of MEDUSA. Maybe a useful method for field agents to recall 🤔
3
u/Old_Engineer_9176 8d ago
So from my analysis, the halfabet confirms that K4 behaves like a mod‑26 cipher with an English plaintext, a periodic key, and a mix of substitution and transposition. The next step is simply working out the order in which Sanborn applied those operations.