# Title:_ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣ ⁣

Key:_

Message:_  ⁣  ⁣  ⁣  ⁣  ⁣  ⁣ 　  ⁣ 　  ⁣ 　  ⁣ 　  ⁣ 　  ⁣ 　  ⁣ 　 ⁣ 　  ⁣ 　  ⁣ 　 　⁣   ⁣   　⁣   ⁣  ⁣  ⁣  ⁣ 　⁣  ⁣  ⁣  ⁣ 　⁣  ⁣  ⁣  ⁣  ⁣  ⁣  ⁣  ⁣  ⁣  ⁣ 　⁣  ⁣  ⁣  ⁣  ⁣  ⁣  ⁣ 　⁣  ⁣  ⁣  ⁣  ⁣  ⁣

Hint{

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• The hint seems to be code written in Whitespace. – Deusovi Apr 4 '18 at 18:57

I will continue where wolfram42 left off:

Instead of substituting the key by "0123456789ABC", we can use "0123456789BCD" to get the following:

 55 73 89 42 24 24 2DB4 2DB5 2DB6 2DB7 2DB8 2DB9 2DB 2DBB 2DBC 2DBD 307 303D 469 34 24 65 7D 85 24 74 7D 85 7C 72 24 84 76 73 24 71 7D 82 82 73 71 84 24 6D 7C 83 87 73 82 34
If we interpret these as base-14 numbers, and then take the corresponding characters, we get:
 Key:        ɉ  ​ɓ ͭ. You found the correct answer.
This probably means that the puzzle is solved. In order to explain the first part of the answer, we can look at the base 14 numbers corresponding to the part after "Key: " and before the period:
 24 2DB4 2DB5 2DB6 2DB7 2DB8 2DB9 2DB 2DBB 2DBC 2DBD 307 303D 469
and we can compare this to the base 14 representation of the characters in the real key:
 24 2DB4 2DB5 2DB6 2DB7 2DB8 2DB9 2DBA 2DBB 2DBC 3007 303D 469A
These two keys are really similar, in the key in the answer there are some A's missing (since these can't be represented), the character 2DBD is added (which should probably correspond to the A), and there is a 0 missing (maybe just a mistake).
The only unsolved mystery is the title.

The source for the puzzle contains a bunch of different blank and non-printable Unicode characters.

Title:_U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+2009U+2063U+2009U+2063U+2008U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063U+20U+2063

Key:_U+20U+2002U+2003U+2004U+2005U+2006U+2007U+2008U+2009U+200aU+202fU+205fU+3000

Message:_U+2006U+2006U+2063U+2008U+2004U+2063U+2009U+200aU+2063U+2005U+2003U+2063U+2003U+2005U+2063U+2003U+2005U+2063U+2003U+3000U+202fU+2005U+2063U+2003U+3000U+202fU+2006U+2063U+2003U+3000U+202fU+2007U+2063U+2003U+3000U+202fU+2008U+2063U+2003U+3000U+202fU+2009U+2063U+2003U+3000U+202fU+200aU+2063U+2003U+3000U+202fU+2063U+2003U+3000U+202fU+202fU+2063U+2003U+3000U+202fU+205fU+2063U+2003U+3000U+202fU+3000U+2063U+2004U+20U+20U+2008U+2063U+2004U+20U+2004U+3000U+2063U+2005U+2007U+200aU+2063U+2004U+2005U+2063U+2003U+2005U+2063U+2007U+2006U+2063U+2008U+3000U+2063U+2009U+2006U+2063U+2003U+2005U+2063U+2008U+2005U+2063U+2008U+3000U+2063U+2009U+2006U+2063U+2008U+205fU+2063U+2008U+2003U+2063U+2003U+2005U+2063U+2009U+2005U+2063U+2008U+2007U+2063U+2008U+2004U+2063U+2003U+2005U+2063U+2008U+2002U+2063U+2008U+3000U+2063U+2009U+2003U+2063U+2009U+2003U+2063U+2008U+2004U+2063U+2008U+2002U+2063U+2009U+2005U+2063U+2003U+2005U+2063U+2007U+3000U+2063U+2008U+205fU+2063U+2009U+2004U+2063U+2009U+2008U+2063U+2008U+2004U+2063U+2009U+2003U+2063U+2004U+2005

• The title contains some too. – gnovice Apr 3 '18 at 19:04
• Ah that was my problem. I made a false assumption. – user47403 Apr 3 '18 at 19:07
• @thecoder16 I clicked on "Edit", copy/pasted into a text file, and then wrote a throw-away Qt program to read in characters one at a time and output the U+nnnn versions. – Daniel Schepler Apr 3 '18 at 19:10
• I was just thinking - I wouldn't be at all surprised if the answer turns out to be a quotation relating to "space". :) – Daniel Schepler Apr 3 '18 at 20:28
• at first I thought it was morse code with 2 different sizes of space, but it looks like it's more deep than that – Florian Bourse Apr 4 '18 at 8:31

# Also a partial answer, based on Daniel Schepler :

EDIT : I changed some characters to help the reading

Let's replace all characters with visibles one, following this table :

U+3000 = A
U+2063 = -
U+205F = C
U+202F = D
U+200A = E
U+2009 = F
U+2008 = G
U+2007 = H
U+2006 = I
U+2005 = J
U+2004 = K
U+2003 = L
U+2002 = M
U+20 = N

Let's apply this to our riddle :

Title : N-N-N-N-N-N-N-F-F-G-N-N-N-N-N-N-N-N-N-

Key : NMLKJIHGFEDCA

Here is some observations :

There is a - (U+2063) between each other characters in the title.
There is a lot of N (U+20) in the title, compare to the other 2 crypted strings
There is almost every character in the key, except for - (U+2063)
The following "-LJ" and "-LAD" are very common in the first part of the message
- (U+2063) seems to come up aproximately every 3 characters in the message, except for the precedent part
There is exactly 10 characters with the unicode U+200X. Maybe digits from 0 to 9 ?

So taking the hint as help:

It is composed of spaces and tabs. If we interpret a space as 0 and a tab as 1 we can convert to binary. The sequential increasing of values seem to imply this is the intended choice.

...--....        000110000 -> 0x30 -> '0'
.....--...-    00000110001 -> 0x31 -> '1'
.....--..-.    00000110010 -> 0x32 -> '2'
.....--..--    00000110011 -> 0x33 -> '3'
.....--.-..    00000110100 -> 0x34 -> '4'
.....--.-.-    00000110101 -> 0x35 -> '5'
.....--.--.    00000110110 -> 0x36 -> '6'
.....--.---    00000110111 -> 0x37 -> '7'
.....---...    00000111000 -> 0x38 -> '8'
.....---..-    00000111001 -> 0x39 -> '9'
.....--....-  000001100001 -> 0x61 -> 'a'
.....--...-.  000001100010 -> 0x62 -> 'b'
.....--...--  000001100011 -> 0x63 -> 'c'
..                      00

There are 14 values, which is the same as the 13 values in the key plus 1 null character (or seperator) which appears in the message and title. EDIT: To continue where I left off: If we do a substitution of these values directly with those in the key we end up with:
Key:_0123456789ABC
Title:_0 0 0 0 0 0 0 8 8 7 0 0 0 0 0 0 0 0 0
Message:_55 73 89 42 24 24 2CA4 2CA5 2CA6 2CA7 2CA8 2CA9 2CA 2CAA 2CAB 2CAC 3007 303C 469 34 24 65 7C 85 24 74 7C 85 7B 72 24 84 76 73 24 71 7C 82 82 73 71 84 24 6C 7B 83 87 73 82 34
The spaces are the null character that does not appear in the key. Technically the letters should be lower case, but I am interpreting them as hex values. Converting directly to utf-8 encoding does not reveal much. Perhaps a Vignette cipher can help? The number 24 appears many times and there is the series that goes 2ca4 2ca5 2ca6.. which seems interesting as well.

• Thank you for the edits, I had to go in a hurry and didn't have time to double check it. I added an additional step in my solution. – wolfram42 Apr 4 '18 at 18:31