# Thread: A i l m o p s t

1. ## A i l m o p s t

I just had a thought. Ed says the masking technique isnt known by anyone on the PLANET. Why does he use that word? PLA_i_N_t_E_x_T. Notice how the 9-letter word is disrupted by letters at positions 4,6 and 8? What if the mask is a similar sort of disruption based on the numeric sequence of ALMOST PI (hint revealed by the morse palindromes)? 3.142857142857... Then for every 27 letters in the pad we would have 6-letters to put to use. The repeating digits 142857 basically give us the following:
If the pad was simply the alphabet KRYPTOSABCDEFGHIJLMNQUVWXZK, the first, fourth, second, eighth, fifth, and seventh letters respectively would be KTSHNK. I figure the pad source needed for a K4 would need to be 97/6 or just over 16 alphabets in size, or 437 characters in length. How many characters are K1 and K2 togethter? I'd like to take a closer look at this application for ALMOST PI.

K1 is 63 letters and K2 is 373. That is 436 all together. Hmm. Im not sure if we need 437 or 436 because of my rounding off on the calculation but for the moment that sure is interesting to consider that K1 and K2 might together be enough to pad a 97-letter message.

2. So it turns out to be spot on. It's 16 rows of length 27 and a 17th row of just 3 letters more. I've picked out the letters that fit the ALMOST PI scheme.

Code:
```EMUFPHZLRFAXYUSDJKZLDKRNSHG
NFIVJYQTQUXQBQVYUVLLTREVJYQ
TMKYRDMFDVFPJUDEEHZWETZYVGW
HKK?QETGFQJNCEGGWHKKDQMCPFQ
ZDQMMIAGPFXHQRLGTIMVMZJANQL
VKQEDAGDVFRPJUNGEUNAQZGZLEC
GYUXUEENJTBJLBQCRTBJDFHRRYI
ZETKZEMVDUFKSJHKFWHKUWQLSZF
TIHHDDDUVH?DWKBFUFPWNTDFIYC
UQZEREEVLDKFEZMOQQJLTTUGSYQ
PFEUNLAVIDXFLGGTEZ?FKZBSFDQ
VGOGIPUFXHHDRKFFHQNTGPUAECN
UVPDJMQCLQUMUNEDFQELZZVRRGK
FFVOEEXBDMVPNFQXEZLGREDNQFM
PNZGLFLPMRJQYALMGNUVPDXVKPD
QUMEBEDMHDAFMJGZNUPLGEWJLLA
ETG

1   4 2    5       8      7
E   P Z    X       L      G
N   J Q    Q       L      Q
T   R M    P       W      W
H   Q T    N       K      Q
Z   M A    H       V      L
V   D G    P       A      C
G   U E    J       J      I
Z   Z M    K       K      F
T   D D    D       W      C
U   R E    F       L      Q
P   N A    F       F      Q
V   I U    D       T      N
U   J Q    M       L      K
F   E X    P       G      M
P   L L    Q       V      D
Q   B D    F       L      A
E```

3. EPZXLGNJQQLQTRMPWWHQTNKQZMAHVLVDGPACGUEJJIZZMKKFTD DDWCUREFLQPNAFFQVIUDTNUJQMLKFEXPGMPLLQVDQBDFLAE = 97 letters.

This could be the mask.

4. I observed that the 3 ?'s in K2 are spaced vertically the same as the Almost Pi series, 1 4 2.

5. I pounded away at some unmasking today, but didn't get anywhere.

Maybe I could apply 3.142587 better. I hadn't really tried applying the 3. Also I have to decide whether the first letter of the code block is where to start or do we start somewhere else and wrap around. Do we use the CT or the PT versions? How is that result applied to K4? As an OTP?

6. This time I acounted for the starting 3 from ALMOSTPI = 22/7 = 3.142857142857142857... = R.OYGBIVOYGBIVOYGBIV... (rainbow theme for visual tracking of the digits applied to the K1&K2 sequence of ciphertext.

7. How could these 98 letters mask K4?

Code:
```UFLFKRIVTUVEKYFVHZKQFJKMQMGFIJQEDFUGUXNTTHTKVUWQHHUHFDZEVDQUEUVDZBOGFHQUPDCQQVVOBMZDZGPRNXMEMDUSTG

?OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR```

8. If this works, then NYPVTTMZFPK relates to 1:1 to mask ZBOGFHQUPDC. That result then decodes by some process to BERLINCLOCK. ...if this works...

9. If I go with a one-time pad, and I consider the ?-mark may or may not be the first character of K4 with respect to lining up the mask, then I have to try an additional option, basicually dropping the starting letter U in the second case. I then have these two sets of letters lining up with the BERLINCLOCK crib:

(1) MZDBYACTUSM
(2) QXQJZYTPZEN

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