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I got the following from an ARG (alternate reality game) website called terminal00:

Msp  n.rf  rmu.pbq  rfc  rpsrfyyy  gr  f.u  ,ccI  ,jmuI  .u.wz    q.kc  .q  kw  umpbqy
Rfc  ugIbyyy  ir  gq  ecrrgIe  qrpmIecpy
,rq  F  ´.k  j.hb  Irq  fqp  qo.´bpxxx  Vbpy  qebobyp  pIjbqefkd  qeboby  .iIkd  qeb  tfkapx
Qeb  qtI  ´roobkqpy  qebvyob  gIfkfkdxxx
Jfqqckz  qfbpb  .oc  ksj,bcoqxyx  Wbqxyx  recoc  .pb  Irk´coq  qfbpb.  .  aIbbz  mcof.np?
Ecocb  Gzji  n.qp  gq  .imke  qm  vmry
Ehybo/  Qfvy  Mkcy  XbpIz  Kgkcy  LfIbz  Cmrpx  wkjb 

The original html-comment clue:

<!--Msp n.rf bccncpyyy gr f.u ,ccl ,jmul .u.wz q.kc .q kw umpbqy-->

It looks like it uses some sort of ROT2, ROT3 shift to encode the first 1-2 and 3-4 lines respectively. The last three lines contain letters that repeat, indicating there is some sort of mono-alphabetic cipher. The solution should be a URL or part of a URL. The rest I know is speculative.

EDIT: Added the html-comment clue

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  • 2
    $\begingroup$ Welcome to Puzzling SE! $\endgroup$ – North Mar 21 '18 at 15:05
  • $\begingroup$ reddit.com/r/codes/comments/85ed8t/cipher_riddle $\endgroup$ – somebody Mar 21 '18 at 22:12
  • 1
    $\begingroup$ @somebody that is my post. I wanted to keep things free from my assumptions. Nonetheless I'm stuck at the same place where other have too (so far). $\endgroup$ – What Mar 21 '18 at 22:30
  • $\begingroup$ @SimpleMath well your assumptions so far look completely accurate, we just need to find out what to do from there. e.g. the / looks kinda suspicious, and so does the last line with all those 4/5 letter words) $\endgroup$ – somebody Mar 22 '18 at 1:10
  • $\begingroup$ Thanks to Pex_Juv on reddit we now know that there is an elliptical rot cipher. That is changing the rot between 2 and 3 for each letter. $\endgroup$ – What Mar 22 '18 at 16:44
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As you stated, ROT2 and ROT3 help a little bit

ROT2:

Our p.th tow.rds the truthaaa it h.w ,eeK ,lowK .w.yb s.me .s my wordsa
The wiKdaaa kt is gettiKg stroKgera
,ts H ´.m l.jd Kts hsr sq.´drzzz Xdra sgdqdar rKldsghmf sgdqda .kKmf sgd vhmcrz
Sgd svK ´tqqdmsra sgdxaqd iKhmhmfzzz

ROT3:

Pvs q.ui upx.set uif usvuibbb ju i.x ,ffL ,mpxL .x.zc t.nf .t nz xpsetb
Uif xjLebbb lu jt hfuujLh tuspLhfsb
,ut I ´.n m.ke Lut its tr.´esaaa Yesb therebs sLmething thereb .lLng the windsa
The twL ´urrentsb theybre jLiningaaa

Mixing them together give us:

Our p.th tow.rds the truth it h.w ,eeK ,lowK .w.y s.me .s my words
The wiKd kt is gettiKg stroKger
,ut I ´.n m.ke Lut its tr.´es Yes theres sLmething there .lLng the winds
The twL ´urrents theyre jLining

The next step is replacing some letters with each others :

K => n
L => o
´=> c
.=>a
,=>b

this give us:

Our path towards the truth it haw been blown away same as my words
The wind kt is getting stronger
but I can make out its traces Yes theres something there along the winds
The two currents theyre joining

but that doesn't work for the last lines, must be another encryption

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  • $\begingroup$ The letters are replaced with the character three chars up $\endgroup$ – somebody Mar 21 '18 at 22:15
2
+50
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Robert Garrett have already posted a similar answer on reddit. However, he didn't explain how he got it.

After a while it becomes obvious that letter 'I' in the original code is actually 'l' lost during some data operation. So, we replace it as is.

Now for the first part.

Let's imagine it's all a single code, without any stretches. Then (as we have seen rot-2 approximation) '.' is definitely 'a' and 'y' is definitely '.'. Why is so? Well, my assumption is, the rotation wheel contains not only alphabet, but also punctuation. Basing on found terms, the wheel is "abcdefghijklmnopqrstuvwxyz.,", giving us, as it is:
Our path towards the truth... it haw been blown away, same as my words.
The wind... kt is getting stronger.

That's as close as I can get, while remaining logical. As for the second part,

The rotation wheel becomes 'abcdefghijklmnopqrstuvwxyz.,'', with apostrophe. Why? Either to confuse us, or to add the apostrophe in the process. Or, maybe, we still want 'a' to be '.'. With that and rot-3, the string transforms to
But I can make out its traces... Yes, there,s something there, along the winds.
The two currents, they,re joining...

While I tried to find explanation for these mistakes, I still see none - no modified cipher, for instance. Perhaps, it's one of your fancy code inside a code thing - and mistakes actully mean something else. I don't know.

Now the third code is a mix of first to 'rot'-s. Probably. I'll show you how I applied it:

"222 2222 2222222 222 22222222 22 222 2222 22222 22222 2222 22 22 222222",
"222 2222222 22 22 2222222 222222222",
"333 3 333 3333 333 333 333333333 3333 3333333 333333333 333333 33333 333 333333",
"333 333 333333333 3333333 3333333333",
"2323232 32323 232 32323232323 232323 23232 323 2323232 323232 3 23232 3232323?",
"32323 2323 2323 23 23232 32 3232",
"32323/ 2323 2323 23232 32323 23232 32323 2323"
Codes are changed each letter, without taking spaces into account - so I believe, the last '/' sign is not taken into account. Anyway, that didn't give me a lot.

This is what a result looks like:

Listen, there are numbeers... Yes... there are numcers therea a code, perhaps?
Heree I´ll pass it along to you.
Hj,dr/ Six, One, Zero, Nine, Nine, Four. ynle

Now, the last string again, in two rot-s:

Gj.dq/ Shx. Ome. Zdrn, Mime. Nhnd, Eotrz ymld
Hk,er/ Tiy, Pnf, .eso´ Njnf, Oioe´ Fpus. znme
in case you want to find a meaningful combination.

And the whole thing:

Our path towards the truth... it haw been blown away, same as my words.
The wind... kt is getting stronger.
but I can make out its traces...
Yes, there,s something there, along the winds.
The two currents, they,re joining...
Listen, there are numbeers... Yes... there are numcers therea a code, perhaps?
Heree I´ll pass it along to you.
Hj,dr/ Six, One, Zero, Nine, Nine, Four. ynle

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  • $\begingroup$ I do see the imperfections / confusions as mistakes from the puzzle-creators side. The puzzle has no consistency behind it and is rather random in nature and badly executed. It is more tedious guesswork then actual problem solving. It seems to me that the author makes the assumption that the function used and the searched inverse function are actually easy to find even without the key. $\endgroup$ – What Mar 29 '18 at 21:07
  • $\begingroup$ I have granted you the bounty even though it does not contain the fully decoded message it is a step closer then the most upvoted post at the time of writing this comment. $\endgroup$ – What Mar 30 '18 at 22:29
1
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Puttting Rot2 seems to give you this following solution:

Our p.th tow.rds the truthaaa it h.w ,eeK ,lowK .w.yb s.me .s my wordsa
The wiKdaaa kt is gettiKg stroKgera
,ts H ´.m l.jd Kts hsr sq.´drzzz Xdra sgdqdar rKldsghmf sgdqda .kKmf sgd vhmcrz
Sgd svK ´tqqdmsra sgdxaqd iKhmhmfzzz
Lhssemb shdrd .qe mul,deqszaz Ydszaz tgeqe .rd Ktm´eqs shdrd. . cKddb oeqh.pr?
Geqed Iblk p.sr is .komg so xota
Gjadq/ Shxa Omea ZdrKb Mimea NhKdb Eotrz ymld

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  • $\begingroup$ Whoops my spoiler is not working $\endgroup$ – North Mar 21 '18 at 15:06
  • $\begingroup$ Using return for line breaks breaks the spoilers, try using <br> instead $\endgroup$ – Fifth_H0r5eman Mar 21 '18 at 15:10
  • $\begingroup$ @Fifth_H0r5eman ... please use two spaces instead because that's the sane thing to do $\endgroup$ – somebody Mar 21 '18 at 15:24
  • $\begingroup$ @Somebody, I find two spaces hard to track during edits personally, but yes that also works. $\endgroup$ – Fifth_H0r5eman Mar 21 '18 at 15:25

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