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‎README.md

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- [pase.ca](pase.ca)
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- [hackcon.online](hackcon.online)
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- [ctf.sec.army](ctf.sec.army)
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- [ctf.redpwn.net](ctf.redpwn.net)
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Capturing flags for self-education in information security since July 2019.
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‎ctf.redpwn.net/README.md

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# ctf.redpwn.net
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## cryptography
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- [dunce-crypto](cryptography/dunce-crypto)
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- [super-hash](cryptography/super-hash)
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- [trinity](cryptography/trinity)
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## forensics
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- [msb](forensics/msb)
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## misc
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- [tux-trivia-show](misc/tux-trivia-show)
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## pwn
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- [babbypwn](pwn/babbypwn)
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- [bronze-ropchain](pwn/bronze-ropchain)
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- [dennis-says](pwn/dennis-says)
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- [penpal-world](pwn/penpal-world)
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- [rot26](pwn/rot26)
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- [zipline](pwn/zipline)
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## reverse-engineering
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- [generic-crackme](reverse-engineering/generic-crackme)
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- [generic-crackme-redux](reverse-engineering/generic-crackme-redux)
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‎ctf.redpwn.net/cryptography/dunce-crypto/README.md

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# Dunce Crypto
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## Cryptography - Points: 50
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> Written by: blevy
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>
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>
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>
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> Emperor Caesar encrypted a message with his record-breaking high-performance encryption method. You are his tax collector that he is trying to evade. Fortunately for you, his crown is actually a dunce cap.
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>
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>
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>
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> ```
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>
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> mshn{P_k0ua_d4ua_a0_w4f_tf_ahe3z}
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>
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> ```
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>
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Shift with key 7.
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flag: `flag{i_d0nt_w4nt_t0_p4y_my_tax3s}`

‎ctf.redpwn.net/cryptography/super-hash/README.md

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# Super Hash
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## Cryptography - Points: 50
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> Written by: NotDeGhost
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>
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>
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>
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> Does hashing something multiple times make it more secure? I sure hope so. I've hashed my secret ten times with md5! Hopefully this makes up for the fact that my secret is *really* short. Wrap the secret in `flag{}`.
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>
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>
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>
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> Note: Follow the format of the provided hash exactly
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>
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>
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>
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> Hash: `CD04302CBBD2E0EB259F53FAC7C57EE2`
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>
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Brute-force all printable characters, hash them 10 times and compare with the target hash.
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```python
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import hashlib
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import string
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target = 0xCD04302CBBD2E0EB259F53FAC7C57EE2
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for c in string.printable:
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m = hashlib.md5(c).hexdigest().upper()
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for i in range(9):
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m = hashlib.md5(m).hexdigest().upper()
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if int(m, 16) == target:
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print c
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break
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```
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flag: `flag{^}`

‎ctf.redpwn.net/cryptography/super-hash/sol.py

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import hashlib
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import string
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target = 0xCD04302CBBD2E0EB259F53FAC7C57EE2
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for c in string.printable:
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m = hashlib.md5(c).hexdigest().upper()
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for i in range(9):
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m = hashlib.md5(m).hexdigest().upper()
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if int(m, 16) == target:
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print c
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break

‎ctf.redpwn.net/cryptography/trinity/README.md

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# Trinity
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## Cryptography - Points: 77
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> Written by: blevy
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>
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>
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>
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> redpwn claims to have invented an unbreakable "trinity encoding."
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>
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> Security by obscurity!
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>
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>
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>
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> ```
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>
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> 1202010210201201021011200200021121112010202012010210102012102021000200121200210002021210112111200121200002111200121102000021211120010200212001020020102000212
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>
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> ```
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>
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>
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>
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> NB: This challenge does not use the flag{...} format. The flag is only lowercase letters.
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>
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>
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> Hint: The encoding is so simple, even your grandma knows about it!
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>
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Interpret the three numbers as morse code with the following mapping:
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'0': '.', '1': '-', '2': ' '
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The resulting morse code:
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- . .-. -. .- .-. -.-- .. ... -- --- .-. . .- .-. -.-. .- -. . -... ..- - .. -... . - -.-- --- ..- - .... --- ..- --. .... - --- ..-. .. - ..-. .. .-. ... -
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Decode it with some online decoder gives the flag.
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flag: `flag{TERNARYISMOREARCANEBUTIBETYOUTHOUGHTOFITFIRST}`

‎ctf.redpwn.net/cryptography/trinity/sol.py

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code = '1202010210201201021011200200021121112010202012010210102012102021000200121200210002021210112111200121200002111200121102000021211120010200212001020020102000212'
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mapping = {'0': '.', '1': '-', '2': ' '}
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morse = ''
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for c in code:
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morse += mapping[c]
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print morse

‎ctf.redpwn.net/forensics/msb/README.md

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# msb
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## Forensics - Points: 50
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> Written by: NotDeGhost
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>
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>
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> It's not LSB, its MSB!
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>
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> Red is Random,
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> Green is Garbage,
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> Blue is Boring.
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>
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> Hint: Only one channel is correct. Also, I like doing things top down.
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>
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> [lol.png](lol.png)
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>
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Flag is encoded in the MSB.
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flag: `flag{MSB_really_sucks}`

‎ctf.redpwn.net/forensics/msb/lol.png

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‎ctf.redpwn.net/forensics/msb/sol.py

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from PIL import Image
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def decode_binary_string(s):
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return ''.join(chr(int(s[i*8:i*8+8],2)) for i in range(len(s)//8))
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im = Image.open('lol.png')
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pix = im.load()
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x, y = im.size
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bits = ''
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for i in range(x):
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for j in range(y):
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r,g,b,a = pix[i,j]
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bits += str(b >> 7 & 1)
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print(decode_binary_string(bits))

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