EasyCTF_2018: Little Language

Category: Miscellanous Points: 250 Description:

I want root access to this special programming portal, and this file is my only clue. Maybe the password is inside? Even if it is, I'm not sure how to enter it. encrypted nc c1.easyctf.com 12480 Oh! Almost forgot... this might help.

Write-up

This challenge was fantastically annoying. Firstly, to start off with, the creator is @ztaylor54. A little googling leads us to his portfolio site. On there, you find a suspicious link and reference to 5ive.

screenshot.png

A little bit more googling and we land on this GitHub page for 5ive1terpreter. In this file, we see something highly familiar.

Stmt
  : Expr                                { NodeE $1 }
  | alive var '=' Expr                  { NodeLet $2 $4 }
  | unded var '=' Expr                  { NodeRecLet $2 $4 }
  | class cname "{" fields VList "..." MList "}"    { NodeClass $2 $5 $7 }
  | class cname extends cname "{" fields VList "..." MList "}"    {NodeClassExtends $2 $4 $7 $9 }

Going back to the files we got,

encrypted

Hmm... That certainly looks suspicious. Maybe the variables were renamed?

# nc c1.easyctf.com 12480
ctflang
commands begin with ":" (try :help)
global tag = 2
tag
2

Bingo! Now we know what sets the variable, let's try and look at the image above. What does that mean? Well, according to the PDF file, it seems to be some form of expression semantics. Let's get back to it later. To get the password, just strings.

# strings encrypted
[...]
zQYY
(1 ???99
%5)r
,..6
MbUU
<x0y
IEND
note: the password is l7&4C&Cg

Now, let's try to get the flag, which according to the expression semantic, is a function.

# nc c1.easyctf.com 12480
ctflang
commands begin with ":" (try :help)
global username = "root"
global password = "l7&4C&Cg"
flag
EasyCTF{5m4ll_573p_53m4n71c5_4r3_fun_r16h7?}

Therefore, the flag is EasyCTF{5m4ll_573p_53m4n71c5_4r3_fun_r16h7?}.

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