Rating:
# Random Encryption Fixed
Challenge description:
> We found the following file on a hacked terminal:
> ```
> import random
> flag = "flag{not_the_flag}" seeds = [] for i in range(0,len(flag)):
> seeds.append(random.randint(0,10000))
>
> res = [] for i in range(0, len(flag)):
> random.seed(seeds[i])
> rands = []
> for j in range(0,4):
> rands.append(random.randint(0,255))
>
> res.append(ord(flag[i]) ^ rands[i%4])
> del rands[i%4]
> print(str(rands))
>
> print(res) print(seeds)
>
> ```
>
> We also found sample output from a previous run:
>
> ```
> [249, 182, 79] [136, 198, 95] [159, 167, 6] [223, 136, 101] [66,
> 27, 77] [213, 234, 239] [25, 36, 53] [89, 113, 149] [65, 127, 119]
> [50, 63, 147] [204, 189, 228] [228, 229, 4] [64, 12, 191] [65, 176,
> 96] [185, 52, 207] [37, 24, 110] [62, 213, 244] [141, 59, 81] [166,
> 50, 189] [228, 5, 16] [59, 42, 251] [180, 239, 144] [13, 209, 132]
> [184, 161, 235, 97, 140, 111, 84, 182, 162, 135, 76, 10, 69, 246, 195,
> 152, 133, 88, 229, 104, 111, 22, 39] [9925, 8861, 5738, 1649, 2696,
> 6926, 1839, 7825, 6434, 9699, 227, 7379, 9024, 817, 4022, 7129, 1096,
> 4149, 6147, 2966, 1027, 4350, 4272]
> ```
## Solution
Here's my solution:
```
import random
flag = ""
res = [184, 161, 235, 97, 140, 111, 84, 182, 162, 135, 76, 10, 69, 246, 195, 152, 133, 88, 229, 104, 111, 22, 39]
seeds = [9925, 8861, 5738, 1649, 2696, 6926, 1839, 7825, 6434, 9699, 227, 7379, 9024, 817, 4022, 7129, 1096, 4149, 6147, 2966, 1027, 4350, 4272]
for i in range(0, len(seeds)):
random.seed(seeds[i])
rands = []
for j in range(0,4):
rands.append(random.randint(0,255))
flag += chr(rands[i%4] ^ res[i])
print( flag )
```
Since the seeds have been stored in `seeds`, we can reuse them in order to generate the same "random" numbers.
Also, XOR is reversible, so if:
```
res[i] = flag[i] ^ rands[i%4]
```
it means that:
```
flag[i] = rands[i%4] ^ res[i]
```
So, let's execute the script:
```
$ python3 solve.py
flag{Oppsie_LULZ_fixed}
```
