A nonsensical string of letters to decipher:
b z z e a z d j j i z a i j a z z d d z z a d z e j z a d z j g a t e z a t z e d z z h a z z d
Don't be satisfied with a nonsensical string of words, which will be just halfway to the answer.
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My halfway answer:
-> bz ze az dj ji za ij az zd dz za dz ej za dz jg at ez at ze dz zh az zd
-> b e a d i a i a d d a d e a d g a e a e d h a d
-> b e a n s a s a d d a d o a d q u e u e d h a d
Possible nonsensical string:
"beans as add a do ad queued had"
My final answer:
-> be an sa sa dd ad oa dq ue ue dh ad
-> g o t t h e p u z z l e
@question_asker: I changed arbitrary letters (halfway answer), because:
-> j=10 (10th letter)
-> t=20 (20th letter)
-> z=26 (26th letter)
So I incremented every (non j,t,z) letter by its neighbouring indicating letter.
Examples:
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dj -> increment d (4th letter) by 10 -> n (14th letter)
ji -> increment i (9th letter) by 10 -> s (19th letter)
at -> increment a (1st letter) by 20 -> u (21th letter)
zd -> increment d (4th letter) by 26 -> d (4th letter) assuming round tripping
Second pass:b e a n s a s a d d a d o a d q u e u e d h a d 2 5 1 14 19 1 19 1 4 4 1 4 15 1 4 17 21 5 21 5 4 8 1 4 V V V V V V V V V V V V 7 15 20 20 8 5 16 21 26 26 12 5 g o t t h e p u z z l e
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5$\begingroup$ Looks like the right solution, but your explanation is rather confusing. Couldn't you just say you summed pairs of letters, modulo 26, and then did the same a second time? $\endgroup$– GentlePurpleRain ♦Apr 7 '16 at 13:12
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$\begingroup$ @GentlePurpleRain: You're right. That would be more clear. I gave your comment an upvote. My explanation just reflects the way I was thinking while solving the puzzle ;-) $\endgroup$– fondorApr 8 '16 at 6:50
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Partial Solution - Use Whatever Helps You
I am leaning heavily on the hint and taking a low-tech approach. I looked at the coded string and
pretty quickly saw a string of nonsense words. Presumably, this will take us halfway there. Here's what I see:
b z z e a z d j j = Bea sees (two letter z's) easy (ea z) DJ Jay
or you might render it "Busy, (bee zee) the (zee) easy DJ Jay
Next bit.
i z a i j a z z d d z = Is (iz) a jazzy (jaz zee), dizzy (dee's z) or possibly "I see (i zee)" or "I say (i z-ay)
Strings like "z a d" might be Sadie or (less likely) Zaydie (Yiddish for grandfather). There are places where d might be "Dee" as in a girl's name. Depending on how much latitude you are willing to allow for pronunciation, the ending might have something to do with "hazy day (h-ay-z zee dee)".
It looks to me as if something coherent might be made out of this. I leave it to abler thinkers to determine how. Or, I might be on entirely the wrong track. Happy puzzling!
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2$\begingroup$ Lovelee! After all, mairzee doats n dozee doats n lid-l lamzee divee. You discovered a layer of nonsense words that must've settled while this cipher was just sitting around. There is another layer elsewhere, alas, but at least you got rewarded for this nonsense. $\endgroup$– humnMar 19 '16 at 3:56
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Wrap-up: the making of Some string of letters
This is not a solution to the puzzle but provides notes from its poser. This type of answer has been approved by the community.
Caution: This post contains information about the solution.
(This post has been adapted from its original form, which consisted of just the first chart, presented as a hint.)
b z z e a z d j j i z a i j a z z d d z z a d z e j z a d z j g a t e z a t z e d z z h a z z d
The mystery encryption scheme was devised to require trivial effort to encode and decode, with a twist intended to defy most automated deciphering techniques. The twist is that...
...each original message letter can be encoded as 26 different encryption-letter pairs, so that a message with 200 letters can generally be encoded without repeated pairs. A longer message can deliberately repeat pairs for less common letters, such as consonants early in the alphabet, to serve as distractions from inevitably repeated pairs that will represent frequent letters, such as E and T.
The pairs used in the puzzle’s cipher were intentionally selected to provide clues.
As the accepted answer does not mention what led to solution, here is how clues were meant to work. Among other considerations, solving any letter-sequence cipher includes two initial steps.
Step 1. Convert each letter into its alphabetic position number.
Step 2. Count how many times each letter/number appears.
A chart that combines these two steps for the cipher in question reveals an odd distribution.
z
z
z
z
z
z
z
z
z
a z
a d z
a d z
a d j z
a d e j z
a d e j z
a d e i j t z
letter a b d e g h i j t z
alphabetic # 1 2 3 4 5 6 7 8 9 10 . . . . 20 26
count 8 1 7 4 1 1 2 5 2 17
:.................: :...........:
Almost all letters are The only outliers
early in the alphabet. are 20th and last.
The noticeable clustering was meant as a compelling clue while a subtler clue is that the total number of letters is 48, which reeks of doubling, and that the letter counts divide into neat halves.
letter a b d e g h i j t z
count 8 1 7 4 1 1 2 5 2 17
:...............: :...............................:
The left portion The right portion also
includes 24 letters. includes 24 letters.
Compare all that to how this distribution could have easily been made more cryptic.
e f g h i j k l m n o p q r s t u v w x y z
letter a b c d e f g h i j k l m n o p q r s t u v w x y z
alphabetic # 1 2 3 4 5 6 7 8 9 10 . . . . . . . 26
count 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
The actual distribution was meant to draw attention...
...to the letters j, t and z, along with their suggestive alphabet positions of 10th, 20th and last.
The next two steps were meant to follow naturally.
Step 3. Highlight the interesting letters.
zz z jj z j zz zz z jz zj t z tz zz zz b ea d i ai a dd ad e ad ga e a ed ha d
A very even dispersion emerges, meant to strongly suggest...
...letter pairing, which further reveals that each pair includes exactly one j, t, or z.
| z|z | z| j|j |z | j| z|z | z|z | z| j|z | z|j | t| z| t|z | z|z | z|z | |b | e|a |d | i| a|i |a | d|d | a|d |e | a|d | g|a |e |a | e|d | h|a | d|
Step 4. Tally...
...the letter pairs.
z a z d z a z d z a z d z a z d z e j i t a z a z d z e j d j e j g j i t a z a z b z d z e z h 10 4 10 5 10 7 10 9 20 1 26 1 26 2 26 4 26 5 26 8
This was meant to be the “Aha!” moment, where the distribution can be recognized as having characteristics of a typical letter-count distribution, especially if...
...z = 26 is taken as z = 0 (zero)
z a z d z a z d z a z d z a z d z e z a z d z e j i t a z a z b z d z e z h j d j e j g j i t a 0 1 0 2 0 4 0 5 0 8 10 4 10 5 10 7 10 9 20 1 1 2 4 5 8 14 15 17 19 21 a b d e h n o q s u
This may indeed represent a 24-letter message as it has 10 unique letters, including 4 vowels.
Step 5. Rewrite the cipher...
. ...in terms of these pairs.
bz ze az dj ji za ij az zd dz za dz ej za dz jg at ez at ze dz zh az zd b e a n s a s a d d a d o a d q u e u e d h a d beans as add a do ad queued had
Terse as the puzzle statement is, it does say:
Don’t be satisfied with a nonsensical string of words, which will be just halfway to the answer.
This validates the progress so far and leads to completion by repeating steps 4 and 5. Unlike the initial letter distribution with deliberate clues, the chart at this stage would draw attention to nothing useful.
a d
a d
a d
a d e
a d e s u
letter a b d e h n o q s u
alphabetic # 1 2 4 5 8 14 15 17 19 21
count 6 1 6 3 1 1 1 1 2 2
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1$\begingroup$ Wait, I just noticed you're the OP... So is this a hint, or do you not know the solution yourself? $\endgroup$– AlconjaMar 18 '16 at 23:52
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1$\begingroup$ Yeah, @Alconja, both a hint and a way to reduce false starts, not to mention a way to encourage any starts at all after a week $\endgroup$– humnMar 19 '16 at 0:15
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1$\begingroup$ This may be hinting at the fact that each pair of adjacent letters in the original jumble has either a "j", a "t" or a "z". $\endgroup$– ffaoApr 5 '16 at 17:45
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$\begingroup$ @humn I've paired them up but not found anything yet; also, not sure I get what's special about j, t, and z (aside from the fact that they're in each pair, as mentioned, and thus comprise exactly half of the letters) $\endgroup$ Apr 6 '16 at 12:51