## Familiar (100 pts)
### Description
\-
### Solution
Diberikan kode berikut
```python
def encode(data):
charset = "!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~"
padd = "="
binstr = "".join(format(byte, "08b") for byte in data)
padding = (5 - len(binstr) % 5) % 5
binstr += "0" * padding
groups = [binstr[i:i+5] for i in range(0, len(binstr), 5)]
result = ""
for group in groups:
dec = int(group, 2)
result += charset[dec]
result += padd * (padding // 2)
return result
FLAG = "flag{fake_flag_dont_submit}"
print(encode(FLAG.encode()))
```
Kode tersebut melakukan konversi dari 8 bit ke 5 bit lalu nilai dari 5 bit dikonversikan ke nilai pada charset. Jadi untuk mendapatkan plaintextnya tinggal dapatkan nilai 5 bit (dari index) lalu knoversi ke 8 bit. Berikut solver yang kami gunakan
```python
charset = "!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~"
ct = "*&(&)<+$*\"$%+?_?:.,[;[+~+{](+`#%,|![{[*;.]^@}@,>'.:@)_\"<+.:?+`>$'\"#$#`=((|};=="
tmp_bin = ""
for i in ct:
ind = charset.index(i)
tmp_bin += bin(ind)[2:].rjust(5,"0")
flag = ""
for i in range(0,len(tmp_bin),8):
flag += chr(int(tmp_bin[i:i+8],2))
print(flag)
```
Flag : INTECHFEST{WhY\_W0ulD\_AnY0n3\_Us3\_Th1S\_Enc0D1nG?}
## Elysium (244 pts)
### Description
\-
### Solution
Diberikan kode berikut
```python
from Crypto.Util.number import bytes_to_long
from sage.all import *
def add(G, P):
return G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + P + P + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + \
G + G + P + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + P + G + G + P + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + \
G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + \
G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + \
G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + P + G + G
flag = open('flag.txt', 'rb').read()
p = 0xffffffff00000001000000000000000000000000ffffffffffffffffffffffff
K = GF(p)
a = 0xffffffff00000001000000000000000000000000fffffffffffffffffffffffc
b = 0x5ac635d8aa3a93e7b3ebbd55769886bc651d06b0cc53b0f63bce3c3e27d2604b
E = EllipticCurve(K, (a, b))
G = E.gens()[0]
m = bytes_to_long(flag)
P = E.lift_x(Integer(m))
Q = add(G, P)
print('Q:', Q)
```
Add menjalankan operasi 288\*G + 21\*P . Karena kita tahu nilai G maka kita bisa kurangkan saja Q dengan 288\*G untuk mendapatkan 21\*P. Pada ecc terdapat konsep public dan private key dalam bentuk Q = nP dimana Q merupakan public key, n private key, dan P adalah base point. Jadi untuk mendapatkan P disini kita bisa lakukan inverse nilai n dengan order dari Q lalu kalikan dengan Q atau dalam bentuk persamaan adalah P = 21^-1 \* Q. Berikut solver yang kami gunakan
```python
from Crypto.Util.number import *
def add(G, P):
return G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + P + P + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + \
G + G + P + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + P + G + G + P + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + G + \
G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + G + \
G + G + G + G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + G + \
G + G + G + G + G + P + G + G + G + G + G + G + G + G + G + G + P + G + G
p = 0xffffffff00000001000000000000000000000000ffffffffffffffffffffffff
K = GF(p)
a = 0xffffffff00000001000000000000000000000000fffffffffffffffffffffffc
b = 0x5ac635d8aa3a93e7b3ebbd55769886bc651d06b0cc53b0f63bce3c3e27d2604b
E = EllipticCurve(K, (a, b))
G = E.gens()[0]
Q = E(26326686390928441989926437302948364151298187886536227434090842323538336764500, 15057597490574272687879749163595226837809841897797118807290241444796596563842)
q = Q.order()
tmp = Q - 288*G
flag = inverse(21, q) * tmp
print(long_to_bytes(int(flag.xy()[0])))
```
Flag : INTECHFEST{ECC\_FUNd4m3nt4l}
