# Solution to SECCON 2017 Vigenere3d

Vigenere3d

crypto

100

#### Writeup

We got a python program:

``````import sys
def _l(idx, s):
return s[idx:] + s[:idx]
def main(p, k1, k2):
s = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz_{}"
t = [[_l((i+j) % len(s), s) for j in range(len(s))] for i in range(len(s))]
i1 = 0
i2 = 0
c = ""
for a in p:
c += t[s.find(a)][s.find(k1[i1])][s.find(k2[i2])]
i1 = (i1 + 1) % len(k1)
i2 = (i2 + 1) % len(k2)
return c
print main(sys.argv[1], sys.argv[2], sys.argv[2][::-1])
``````

and some example output:

``````\$ python Vigenere3d.py SECCON{**************************} **************
POR4dnyTLHBfwbxAAZhe}}ocZR3Cxcftw9
``````

Lets call the first argument “flag” and the second “seed”.

Line 2 in the main function creates a three dimensional lookup table, and as seen in the first line in the for loop, the lookups are based on flag string, the seed string and the seed string inversed.

Since we know 7 characters of the flag, we can calculate the key string like this:

``````    plain = "SECCON{"
encry = "POR4dnyTLHBfwbxAAZhe}}ocZR3Cxcftw9"
#print seed
for a in range(0, 7):
for d in range(0, len(t[s.find(plain[a])])):
for e in range(0, len(t[s.find(plain[a])][s.find(plain[a])])):
if t[s.find(plain[a])][d][e] == encry[a] and s[d] == 'A':
print "%i %c %c" % (a, s[d], s[e])
``````

which gives us the output:

``````0 A _
1 A K
2 A P
3 A 2
4 A Z
5 A a
6 A _
``````

Once we know that the seed is AAAAAAA_aZ2PK_ we can calculate the flag with:

``````    #print flag
for a in range(0, len(p)):
for d in range(0, len(s)):
if t[d][s.find(k1[i1])][s.find(k2[i2])] == encry[a]:
print "%i %c" % (a, s[d])

i1 = (i1 + 1) % len(k1)
i2 = (i2 + 1) % len(k2)
``````

Flag was: SECCON{Welc0me_to_SECCON_CTF_2017}