# CyberHeroines 2023 ## Shannon Kent > [Shannon Kent](https://en.wikipedia.org/wiki/Shannon_M._Kent): A specialist > in cryptologic warfare and fluent in seven languages, Senior Chief Shannon > Kent served multiple tours in Iraq, participating in numerous special > operations that contributed to the capture of hundreds of enemy insurgents. > She paved the way for greater inclusion of women in Special Operations > Forces and was one of the first women to pass the Naval Special Warfare > Direct Support Course. Kent was killed in action in Syria on Jan. 16, 2019, > and posthumously promoted to senior chief petty officer. - [Navy.mil > Reference](https://www.navy.mil/Women-In-the-Navy/Past/Display-Past-Woman- > Bio/Article/2959760/senior-chief-shannon-kent/) > > Chal: Take the time to learn more about [Senior Chief Petty Officer Shannon > Kent](https://www.youtube.com/watch?v=IlM-5FK0TL4) and unlock the > cryptographic puzzle hosted at `0.cloud.chals.io 27572` > > Hint: Strongly recommended you build your solution in provided docker > container. > > Author: [TJ](https://www.tjoconnor.org/) > > [`distrib.tar.gz`](https://raw.githubusercontent.com/D13David/ctf- > writeups/main/cyberheroines23/crypto/shannon_kent/distrib.tar.gz) Tags: _crypto_ ## Solution For this challenge we get a `Dockerfile` and a python script with only some imports. The files are ment for local exploit development and the imports really are a hint to find the solution. When we connect to the service we get some information: ```bash $ nc 0.cloud.chals.io 27572 -------------------------------------------------------------------------------- Flag Storage Safe v0.1337 -------------------------------------------------------------------------------- WWWWWWWWWW WNXK000000000000KXNW WNK0O0KKXNNWWWWNNXKK000KNW WX0O0XWW WWWWWWWW WWX0O0XW WXOOKNW WWXK0000000000KXW WNKOOXW WKk0NW WNKOO0KXNWWWWNXK0OOKNW WN0kKW W0kKW NKO0XW WX0OKN WKk0W WKkKW NOOXW WXOON WKkKW NOON W0kXW WXk0W NOON WXk0W Xk0W W0kX W0kX WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WNXOkKW WKkOKKKKKKKKKKKKKKKKKKKKKKOkKW WKkOXNW WX0O00KN NKKKKKKKKKKKKKKKKKKKKKKKKKKN NK00O0XW W0kKNW WWWWWWNNNNNNWWWWWWWWWWWWNNNNNNWWWWWW WNKk0W Xk0W NKOkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkOKN W0kX WKkKWWNOddddddddddddddddddddddddddddddddddddddONWWKkKW WKkKWWXkddddddddddddddddddddddddddddddddddddddkX WKkKW WKkKW XkddddddddddddddxO000000OxddddddddddddddkX WKkKW WKkKW Xkddddddddddddx0XW WX0xddddddddddddkX WKkKW WKkKWWXkdddddddddddxKW WKxdddddddddddkXWWKkKW WKkKW XkdddddddddddON NOdddddddddddkXWWKkKW WKkKW XkdddddddddddkX XkdddddddddddkXWWKkKW WKkKW XkdddddddddddxONW WNOxdddddddddddkXWWKkKW WKkKW XkdddddddddddddkKW WKkdddddddddddddkXWWKkKW WKkKW XkddddddddddddddkXW WXkddddddddddddddkXWWKkKW WKkKW XkddddddddddddddkXW WXkddddddddddddddkXWWKkKW WKkKW Xkddddddddddddddx0NW WN0xddddddddddddddkXWWKkKW WKkKW Xkdddddddddddddddxk0KK0kxdddddddddddddddkXWWKkKW WKkKW XkddddddddddddddddddddddddddddddddddddddkXWWKkKW WKkKW NOddddddddddddddddddddddddddddddddddddddONWWKkKW Xk0WWWNKOkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkOKNW W0kX W0k0NWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW WWN0k0W WX0O00KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK00O0XW WNXKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKXNW -------------------------------------------------------------------------------- <<< Tying a Secure Knot Around the File <<< Started: 2023-09-10 09:57:51 -------------------------------------------------------------------------------- <<< XORing flag with random 6 byte key <<< GZIPing result (with gzip.compress with max compression) <<< XORing result with random 10 byte key <<< ZIPping result (with zipfile.ZipFile compression=ZIP_DEFLATED) <<< Flag is secure: 504b0304140000000800394f2a57106bae6a3e0000003900000008000000646174612e62696e013900c6ff1585e970e97250e4c1b30b2ce1ad09f3cd7746c4f60b81d758631996f8b69fbf0947a126140f2a5332f44fd542fed16938f78fda0952f6e1ad504b01021403140000000800394f2a57106bae6a3e00000039000000080000000000000000000000800100000000646174612e62696e504b0506000000000100010036000000640000000000 ``` The reciepe how the flag was *secured* is printed here also a starting time. First I thought we had to retreive the correct seed to generate the same random values for `xor`. But this was not the case, the length of the random byte keys are also a hint. In both cases we know the exact amount of information the reconstruct the keys. For the first round we know the flag starts with the six chars `chctf{`. And for the second round we have all the information the build the 10 byte [gzip header](https://docs.fileformat.com/compression/gz/). Let's get cracking, the first round is easy: convert the hex-string to bytes and unzip the one file from this buffer: ```python #<<< ZIPping result (with zipfile.ZipFile compression=ZIP_DEFLATED) zip_buffer = bytes.fromhex(flag) with ZipFile(BytesIO(zip_buffer), 'r') as file: print(file.namelist()) buffer1 = file.read("data.bin") ``` For the second part we know it's a gzipped buffer. So it must start with a gzip header which is 10 bytes long and exactly the length of our xor key (perfect isn't it)? ``` Offset Size Value Description 0 2 0x1f 0x8b Magic number 2 1 Compression method 3 1 Flags 4 4 32 bit timestamp 8 1 Compression flags 9 1 Operating system ID ``` From this we can retrieve the xor key and then retrieve the gzipped buffer. ```python #<<< XORing result with random 10 byte key t = datetime(2023, 9, 10, 9, 57, 51) ts = pack(" Original writeup (https://github.com/D13David/ctf- writeups/blob/main/cyberheroines23/crypto/shannon_kent/README.md).# CyberHeroines 2023 ## Shannon Kent > [Shannon Kent](https://en.wikipedia.org/wiki/Shannon_M._Kent): A specialist > in cryptologic warfare and fluent in seven languages, Senior Chief Shannon > Kent served multiple tours in Iraq, participating in numerous special > operations that contributed to the capture of hundreds of enemy insurgents. > She paved the way for greater inclusion of women in Special Operations > Forces and was one of the first women to pass the Naval Special Warfare > Direct Support Course. Kent was killed in action in Syria on Jan. 16, 2019, > and posthumously promoted to senior chief petty officer. - [Navy.mil > Reference](https://www.navy.mil/Women-In-the-Navy/Past/Display-Past-Woman- > Bio/Article/2959760/senior-chief-shannon-kent/) > > Chal: Take the time to learn more about [Senior Chief Petty Officer Shannon > Kent](https://www.youtube.com/watch?v=IlM-5FK0TL4) and unlock the > cryptographic puzzle hosted at `0.cloud.chals.io 27572` > > Hint: Strongly recommended you build your solution in provided docker > container. > > Author: [TJ](https://www.tjoconnor.org/) > > [`distrib.tar.gz`](https://raw.githubusercontent.com/D13David/ctf- > writeups/main/cyberheroines23/crypto/shannon_kent/distrib.tar.gz) Tags: _crypto_ ## Solution For this challenge we get a `Dockerfile` and a python script with only some imports. The files are ment for local exploit development and the imports really are a hint to find the solution. When we connect to the service we get some information: ```bash $ nc 0.cloud.chals.io 27572 \-------------------------------------------------------------------------------- Flag Storage Safe v0.1337 \-------------------------------------------------------------------------------- WWWWWWWWWW WNXK000000000000KXNW WNK0O0KKXNNWWWWNNXKK000KNW WX0O0XWW WWWWWWWW WWX0O0XW WXOOKNW WWXK0000000000KXW WNKOOXW WKk0NW WNKOO0KXNWWWWNXK0OOKNW WN0kKW W0kKW NKO0XW WX0OKN WKk0W WKkKW NOOXW WXOON WKkKW NOON W0kXW WXk0W NOON WXk0W Xk0W W0kX W0kX WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WKkKW WNXOkKW WKkOKKKKKKKKKKKKKKKKKKKKKKOkKW WKkOXNW WX0O00KN NKKKKKKKKKKKKKKKKKKKKKKKKKKN NK00O0XW W0kKNW WWWWWWNNNNNNWWWWWWWWWWWWNNNNNNWWWWWW WNKk0W Xk0W NKOkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkOKN W0kX WKkKWWNOddddddddddddddddddddddddddddddddddddddONWWKkKW WKkKWWXkddddddddddddddddddddddddddddddddddddddkX WKkKW WKkKW XkddddddddddddddxO000000OxddddddddddddddkX WKkKW WKkKW Xkddddddddddddx0XW WX0xddddddddddddkX WKkKW WKkKWWXkdddddddddddxKW WKxdddddddddddkXWWKkKW WKkKW XkdddddddddddON NOdddddddddddkXWWKkKW WKkKW XkdddddddddddkX XkdddddddddddkXWWKkKW WKkKW XkdddddddddddxONW WNOxdddddddddddkXWWKkKW WKkKW XkdddddddddddddkKW WKkdddddddddddddkXWWKkKW WKkKW XkddddddddddddddkXW WXkddddddddddddddkXWWKkKW WKkKW XkddddddddddddddkXW WXkddddddddddddddkXWWKkKW WKkKW Xkddddddddddddddx0NW WN0xddddddddddddddkXWWKkKW WKkKW Xkdddddddddddddddxk0KK0kxdddddddddddddddkXWWKkKW WKkKW XkddddddddddddddddddddddddddddddddddddddkXWWKkKW WKkKW NOddddddddddddddddddddddddddddddddddddddONWWKkKW Xk0WWWNKOkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkOKNW W0kX W0k0NWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW WWN0k0W WX0O00KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK00O0XW WNXKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKXNW \-------------------------------------------------------------------------------- <<< Tying a Secure Knot Around the File <<< Started: 2023-09-10 09:57:51 \-------------------------------------------------------------------------------- <<< XORing flag with random 6 byte key <<< GZIPing result (with gzip.compress with max compression) <<< XORing result with random 10 byte key <<< ZIPping result (with zipfile.ZipFile compression=ZIP_DEFLATED) <<< Flag is secure: 504b0304140000000800394f2a57106bae6a3e0000003900000008000000646174612e62696e013900c6ff1585e970e97250e4c1b30b2ce1ad09f3cd7746c4f60b81d758631996f8b69fbf0947a126140f2a5332f44fd542fed16938f78fda0952f6e1ad504b01021403140000000800394f2a57106bae6a3e00000039000000080000000000000000000000800100000000646174612e62696e504b0506000000000100010036000000640000000000 ``` The reciepe how the flag was *secured* is printed here also a starting time. First I thought we had to retreive the correct seed to generate the same random values for `xor`. But this was not the case, the length of the random byte keys are also a hint. In both cases we know the exact amount of information the reconstruct the keys. For the first round we know the flag starts with the six chars `chctf{`. And for the second round we have all the information the build the 10 byte [gzip header](https://docs.fileformat.com/compression/gz/). Let's get cracking, the first round is easy: convert the hex-string to bytes and unzip the one file from this buffer: ```python #<<< ZIPping result (with zipfile.ZipFile compression=ZIP_DEFLATED) zip_buffer = bytes.fromhex(flag) with ZipFile(BytesIO(zip_buffer), 'r') as file: print(file.namelist()) buffer1 = file.read("data.bin") ``` For the second part we know it's a gzipped buffer. So it must start with a gzip header which is 10 bytes long and exactly the length of our xor key (perfect isn't it)? ``` Offset Size Value Description 0 2 0x1f 0x8b Magic number 2 1 Compression method 3 1 Flags 4 4 32 bit timestamp 8 1 Compression flags 9 1 Operating system ID ``` From this we can retrieve the xor key and then retrieve the gzipped buffer. ```python #<<< XORing result with random 10 byte key t = datetime(2023, 9, 10, 9, 57, 51) ts = pack(" Original writeup (https://github.com/D13David/ctf- writeups/blob/main/cyberheroines23/crypto/shannon_kent/README.md).