Insomni'Hack Teaser 2023 - Artscii (Misc 200)

21/01 - 22/01/2023 (24hr)

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Description:

Can you read the flag?

generate.py
output.txt

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Solution

The flag is encoded in the output-82af45781b1a0057f2bf9b69d1702976928383941b7f9651de8e2c819935bcf2.txt file:

##   ##  #     ####  #    #  ##  
### ###  # #  # ## ####  ##### ##
#######  ##    ## ##    ###     #
#### ##  ##     ## #    ##   #   
#    ##  ##       ##    ##   #  #
#    ##  # #  # # ##### #### ####
##  ###  #      ## ###   ##  #   
##   ##  ##     ####      # ##   
##   ##     #      #   ##  #### ##   #  ##   ##   ##  
### ###        #  ##    #   ##  ##   #   #   ### ###  
#######        #    #####  ###  ##   #       #######  
#### ##   # ##     ## #     #   ##   #  ###  #### ##  
#    ##   #  ##      ##     ######   #####   #    ##  
#    ##  ### ##      ##     ##  ##   # ##    #    ##  
##  ###  #### ##    ####   #### ##  ### ##   ##  ###  
##   ##  ##   ##     ##    ##   ##  ## ###   ##   ##  
   #     #        ##  #     # # #   ###   # #    ##    ## #   #  # #    #       ## ####  
      #  #  ####   #   ##  #  ###   # ##  # ##   #     ####   #  #      #  #### #  # ##  
      #      # #  # #  ##  # #  #   #   # #  #   # #   # # #  #             # # ## # ##  
 # ##       #      #  ###   # # #   ###   #  # ####     ## #  #  # # #     #    #  #  #  
 #  ##    ### # #  ###  #   # # #   # #   #    ##      ########   ##    #   # # #   ###  
### ##    ### # #  ##      ##  ##     ### #    ##          ## #   ##    #   # # #  # ##  
#### ##    #      #### ##  ## #      ##         # #  ####### ###   #      #     ## ####  
##   ##           ##   ##           # # #      ##    ######## ## ###            ##   ##

The code is in the generate-5612300a7e5a0bf0120c06ab3ccbc3cb0003e209d1e6667d70bf0797d82a307c.py file:

import re
import art

with open("flag.txt") as f:
    flag = f.readline()
    assert(flag[0:4]=="INS{" and flag[-1]=="}")
    content = flag[4:-1]
    assert(re.search(r'^[A-Z1-9_]*$', content))
    assert(content.count("_") == 2)
    content = content.replace("_","\n")


def mergeLines(line1,line2):
    line = list(map(lambda xy: " " if xy[0] == xy[1] else "#", zip(line1, line2)))
    return ''.join(line)


def mergeText(text1, text2):
    a = text1.split("\n")
    b = text2.split("\n")
    c = []
    for j in range(25):
        c.append(mergeLines(a[j],b[j]))
    return '\n'.join(c) 


i = 0
while i<3:
    text = art.text2art(content, font="rnd-medium", chr_ignore=False)
    if re.search(r'^[ #\n]*$', text) and text.count('\n') == 24:
        if i == 0:
            res = text
        else:
            res = mergeText(res,text)
        i = i+1

print(res)

Cracking the Flag

The first task is to find which 3 fonts were used to generate the flag. From the output we can easily tell that the first line is MISC (or M1SC or MI5C or M15C).

Therefore, we brute force all font combinations (there are ~100 fonts, so it doesn't take long) and check which combination produces the desired output. After ~1 minute we get the correct fonts:

    future_2
    green_be
    z-pilot

Then we move on to the second line. We crack the flag character by character. We brute force each character and we check how many columns matches with the target output. The character that matches the most columns is the correct one. We repeated the same process for the 3rd line of the output and get the flag.

For more details, please refer to artscii_crack.py file.

So, the flag is: INS{MISC_MAYHEM_A7R93Y4E7H}

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Original writeup (https://github.com/ispoleet/ctf-writeups/tree/master/insomnihack_2023/Artscii).