URL: http://score.quals.seccon.jp/question/
Type: crypto
Solution: SECCON{6in_tex7}
Description
Reverseit http://files.quals.seccon.jp/Reverseit
First we try to identify the type of the given file :
$ file Reverseit Reverseit: data $ hexdump -C Reverseit 00000000 9d ff 70 0d b6 da fc 93 72 63 28 22 22 bd d2 18 |..p.....rc(""...| ... 000013b0 02 02 02 02 02 02 02 02 02 02 02 02 02 02 02 02 |................| * 00001ba0 02 02 02 02 02 02 02 02 02 02 02 02 02 e3 16 47 |...............G| ... 00001e00 84 00 84 00 10 10 10 00 64 94 64 a4 01 00 0e ff |........d.d.....| 00001e10 8d ff |..|
As the challenge name told us, we first try to reverse the bytes in the file:
00000000 ff 8d ff 0e 00 01 a4 64 94 64 00 10 10 10 00 84 |.......d.d......| ...
No success… Let’s try next taking into account high and low nibbles:
import sys def swap_nibbles(byte): return ((byte<<4) | (byte>>4)) & 0xff i=open(sys.argv[1], 'rb') o=open(sys.argv[1] + '.rev', 'wb') o.write(''.join(map(chr, map(swap_nibbles, map(ord, i.read()[::-1]))))) i.close() o.close()
$ python rev_bytes.py Reverseit $ file Reverseit.rev Reverseit.rev: JPEG image data, JFIF standard 1.01 $ hexdump -C Reverseit.rev 00000000 ff d8 ff e0 00 10 4a 46 49 46 00 01 01 01 00 48 |......JFIF.....H| ...
Let’s have a look at the image :
So, we also have to reverse the image:
import sys from PIL import Image i = Image.open(sys.argv[1]) o = Image.new(i.mode, i.size) idata, odata = i.load(), o.load() for y in range(i.size[1]): for x in range(i.size[0]): odata[x, y] = idata[i.size[0]-x-1, y] o.save(sys.argv[1] + '.jpg')
$ python revh_image.py Reverseit.rev
We then obtain the following image which contains the flag SECCON{6in_tex7}:
Programming is fun but lazyness is better! You can do the same with the following commands:
$ hexdump -ve '1/1 "%.2x"' Reverseit | rev | xxd -r -p > Reverseit.rev $ convert -flop Reverseit.rev Reverseit.rev.jpg