2014 Olympic CTF 200

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그래서 무작정 base64 디코더를 돌려봤다. 그런데 아무것도 나오지 않았다. (사실 생각해보면 당연한거다.) 그래서 한줄한줄 디코더를 돌려봤다.

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IHdyaXRpbmcgaGlkZGVuIG1lc3NhZ2VzIGluIHN1Y2ggYSB3YXkgdGhhdCBubyBvbmV=

이 문장은 Steganography is the art and science of writing hidden messages in such a way that no one 이러한 평문이였다. 

그리고 LCBhcGFydCBmcm9tIHRoZSBzZW5kZXIgYW5kIGludGVuZGVkIHJlY2lwaWVudCwgc3VzcGU= 이 문장을 디코딩하니 , apart from the sender and intended recipient, suspe 이러한 평문이 나왔다.

Y3RzIHRoZSBleGlzdGVuY2Ugb2YgdGhlIG1lc3M=

이 문장은 cts the existence of the mess 이러한 평문이다. 이렇게 한줄한줄 해보니 어느정도는 된다. 대충 몇번 돌려보니 스테가노그래피 뭐라뭐라한다..

그래서 스테가노그래피의 원리를 생각하며 이 암호문에서 어떻게 flag를 뽑아 올 수 있을까 고민을 많이 했다. 그러다가 RFC4648에 관련된 문서를 읽었고. 여기서 힌트를 얻었다. RFC4648은 처음봐서 이해하고 스크립트를 짜는데 꽤 시간이 오래걸렸다. 대충 보면 base64는 인코딩할때 마지막 바이트가 변경되는 경우가 있다. 대충 패턴을 보면  텍스트(3x8=24비트)의 3글자는 6비트 인코딩 문자 중 4개로 변환 > 긴 텍스트는 3char 당 한 그룹으로 나누고 앞에 말한 것이 각 그룹에 적용> 마지막 그룹에 2char이 포함된 경우 8비트 문자 중 2개(2x8=16)는 6비트 암호화 문자 중 3개(3x6=18)로 표현되고 '='(padding) 문자를 추가  원래는 16비트인데 인코딩된 텍스트는 18비트로 2개의 비트가 남는다. > 그 2비트가 스테가노그래피를 위한 거겠지!!!!!

8비트마다 스테가노그래피를 위한 남는 비트를 합치면 flag가 나온다.

def get_base64_diff_value(a, b):
 base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
 result = 0
 for i in xrange(len(a)):
  if a[i] != b[i]:
   return abs(base64chars.index(a[i]) - base64chars.index(b[i]))
 return result
def exploit():
 with open('olyctf.txt', 'rb') as f:
  file_lines = f.readlines()
 bin_str = ''
 for line in file_lines:
  a_line = line.replace('\n', '')
  b_line = line.replace('\n', '').decode('base64').encode('base64').replace('\n', '')
  diff = get_base64_diff_value(a_line, b_line)
  pads_num = a_line.count('=')
  if diff:
   bin_str += bin(diff)[2:].zfill(pads_num * 2)
  else:
   bin_str += '0' * pads_num * 2
 res_str = ''
 for i in xrange(0, len(bin_str), 8):
  res_str += chr(int(bin_str[i:i+8], 2))
 print (res_str)
exploit()

Base_sixty_four_point_five