10
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First, hide the text, come on, no spoilers here;

Second, bash at it, you have nothing to fear;

Third, pair up and basic pigpen, flagging every time;

Fourth, use the force, wait, no, rather a rhyme?

Fifth, flip from left to right, come on, you're almost done;

Sixth, where is number six, is it a free one?

Last, come on, you're almost there, have chemistry with me;

Finally, the answer is lflflflfljffljljljlflffflfljljljljmflflflfljljffljlflflflfmflfljljljljfflflfljljljmfljljljlflfljljljljljmfljljljljljfflfljljljlj

Hint 1:

Bash at, at bash, that's a clue for five less three;
but you still have to clear them chronologically.

Hint 2:

okay, okay, a clue for number one,
partial answers in the comments when you're still not done;
furthermore, a clue for one plus two,
the pigpen shows the pattern of the flags for you

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  • $\begingroup$ first thoughts ROT13: Lbh zragvba cvtcra fb V'z nffhzvat fbzr xvaq bs cvtcra pvcure? V nyfb abgvprq gung gur bayl 5 yrggref hfrq va gur "nafjre" ner W,V,Y,Z (juvpu ner neenatrq ba n djregl xrlobneq va n phefbe-yvxr funcr) naq S. Gurl erzvaq zr n ybg bs hc,qbja,yrsg,evtug naq sver xrlf lbh zvtug cynl n tnzr jvgu. Be, vg pbhyq or gb qb jvgu znexvat bhg gur funcr bs gur cvtcra pvcure? $\endgroup$ – Astralbee Dec 6 '18 at 14:38
  • $\begingroup$ i love this, and even more when i see the answer. +1 to both and a star to the question! $\endgroup$ – Omega Krypton Jul 11 at 9:37
10
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First, let me say thank you for this great riddle.

My answer is

Binary

Here are my steps:

1. First, hide the text, come on, no spoilers here;

Explanation:

First thoughts were the spoiler Tag >! but we got Hint2:

okay, okay, a clue for number one,
partial answers in the comments when you're still not done;

How are we hiding text in comments? We use Rot13, a Ceasar Cipher with Shift 13!

Lets do it:

We start decrypting:
rot13(lflflflfljffljljljlflffflfljljljljmflflflfljljffljlflflflfmflfljljljljfflflfljljljmfljljljlflfljljljljljmfljljljljljfflfljljljlj) =
ysysysysywssywywywysysssysywywywywzsysysysywywssywysysysyszsysywywywywssysysywywywzsywywywysysywywywywywzsywywywywywssysywywywyw

2. Second, bash at it, you have nothing to fear;

Explanation:

First thoughts were some characters used in bash-scripts #! but we got Hint1:

Bash at, at bash, that's a clue for five less three;
but you still have to clear them chronologically.


So we know the next Cipher: Atbash (A->Z , B->Y ... Z->A)

Lets do it:

We continue decrypting:
atbash(ysysysysywssywywywysysssysywywywywzsysysysywywssywysysysyszsysywywywywssysysywywywzsywywywysysywywywywywzsywywywywywssysywywywyw) =
bhbhbhbhbdhhbdbdbdbhbhhhbhbdbdbdbdahbhbhbhbdbdhhbdbhbhbhbhahbhbdbdbdbdhhbhbhbdbdbdahbdbdbdbhbhbdbdbdbdbdahbdbdbdbdbdhhbhbdbdbdbd

3. Third, pair up and basic pigpen, flagging every time;

Explanation:

a) Pair Up: Take two letters at once, makes 64 pairs
b) basic pigpen: in pigpen cipher we place the letters into something like #
c) Lookup the pairs in the pigpen structure, interpreting the positions as 2 flags (semaphore cipher)
d) Write the Flags as letters

Lets do it:

a) bh bh bh bh bd hh bd bd bd bh bh hh bh bd bd bd bd ah bh bh bh bd bd hh bd bh bh bh bh ah bh bd bd bd bd hh bh bh bd bd bd ah bd bd bd bh bh bd bd bd bd bd ah bd bd bd bd bd hh bh bd bd bd bd
This gives 4 distinct pairs to decode: bh, bd, hh, ah
b) create the basic pigpen (we only need the first #, cause our letters are a,b,d and h):
Basic PigPen structure
c) place the four pairs in the pigpen as follows
PigPen bh PigPen bd PigPen hh PigPen ah
d) lookup the flag-codes
bh = flag bh = D
bd = flag bd = P
hh = flag hh = space/break _
ah = flag ah = C

which finally gives us DDDDP_PPPDD_DPPPPCDDDPP_PDDDDCDPPPP_DDPPPCPPPDDPPPPPCPPPPP_DPPPP

4. Fourth, use the force, wait, no, rather a rhyme?

Explanation

My first thought was a rhyme on "the force" is "semaphores" so I missed the real step because I used semaphores already in the last step
Then I tried "force" -> "morse" as pointed out by @oryxandcake in a comment
Well, how to do some morse code on DDDDP_PPPDD....?
D could be Dash, P could be Point, _ and C are separators for chars and words

Lets do it:

Convert DDDDP_PPPDD_DPPPPCDDDPP_PDDDDCDPPPP_DDPPPCPPPDDPPPPPCPPPPP_DPPPP to morse code:
----. ...-- -..../---.. .----/-.... --.../...--...../..... -....
I used " " for "__" and "/" for "C"
the blocks of 5 Symbols (_ and .) point us to numbers, so it decode as: 936/81/67/35/56

5. Fifth, flip from left to right, come on, you're almost done;

Explanation:

Easy one, just flip it around

Lets do it:

flip(936/81/67/35/56) = 65/53/76/18/639

6. Sixth, where is number six, is it a free one?

Explanation:

Kick out those sixes

Lets do it:

65/53/76/18/639 -> 5/53/7/18/39

7. Last, come on, you're almost there, have chemistry with me;

Explanation:

Lookup those numbers in the periodic table

Lets do it:

5 -> Boron -> B
53 -> Iodine -> I
7 -> Nitrogen -> N
18 -> Argon -> Ar
39 -> Yttrium -> Y
--> Binary

8. Finally, the answer is lflflflfljffljljljlflffflfljljljljmflflflfljljffljlflflflfmflfljljljljfflflfljljljmfljljljlflfljljljljljmfljljljljljfflfljljljlj

Explanation:

Replace the string with our decoded one and check if it rhymes

Lets do it:

Finally, the answer is binary
rhymes with the ending of line 7 (have chemistry with me)
Looks correct.

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  • $\begingroup$ This is insanely brilliant. Well done!! $\endgroup$ – Mohirl Jul 11 at 15:57
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Partial:

First, hide the text, come on, no spoilers here;

The symbols used to create spoilers here are >!

Second, bash at it, you have nothing to fear;

In bash scripts, they often start with #!

Third, pair up and basic pigpen, flagging every time;

The symbols > and # look like they're in pigpen form, maybe the ! means to add the dots to them. Assuming # means all of the letters in the # format, this translates to: XJKLMNOPQR

Fourth, use the force, wait, no, rather a rhyme?

Thanks to @oryxandcake, this step involves using morse code (a rhyme of force).

Fifth, flip from left to right, come on, you're almost done;

Pretty self explanatory to just reverse the order of what you have.

Sixth, where is number six, is it a free one?

Not sure without getting this far, perhaps replacing a 6 with 31

Last, come on, you're almost there, have chemistry with me;

Maybe, if step 6 uses numbers, we'd have numbers now, and this clue means to get the chemical symbols with atomic numbers that we have and it will form a word.

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  • $\begingroup$ Second should be #! not !#? Though it's not just bash, a lot of scripting languages (python, perl, etc.) make use of this construct $\endgroup$ – ace_HongKongIndependence Dec 6 '18 at 16:52
  • $\begingroup$ @ace you're right, I've only ever seen it used with bash, will edit it in $\endgroup$ – AHKieran Dec 7 '18 at 8:34
  • 1
    $\begingroup$ For the fourth step... rot13(zbefr)? $\endgroup$ – oryxandcake Dec 7 '18 at 11:04
  • $\begingroup$ @oryxandcake oooh yes $\endgroup$ – AHKieran Dec 7 '18 at 14:48

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