Tags: misc
Rating:
# Seen
**Event:** Nullcon Goa HackIM 2026 CTF
**Category:** Misc
**Points:** 50
## Overview
The client-side checker in `index.html` decodes a long string of Unicode Variation Selectors into a byte array `t`, then validates your input with a simple PRNG-like update. Only the second half of `t` is actually compared, so the first half is noise.
## Variation Selectors to Bytes
The challenge uses the basic Variation Selectors block (`U+FE00...`). Two selectors encode one byte:
```js
const vs = 0xFE00;
for (let i = 0; i < s.length; i += 2) {
t.push(((s.charCodeAt(i) - vs) << 4) | (s.charCodeAt(i + 1) - vs));
}
```
So:
- Each selector encodes a nibble (`0..15`) with `codepoint - 0xFE00`.
- Two nibbles make one byte.
## Length Constraint
The checker enforces:
```
if (u.length * 2 != t.length) fail();
```
So the input length is exactly `t.length / 2`.
## Inversion
Only the low byte of the PRNG state is used. Reduced modulo 256:
- `gen & 0xFF = 0x48`
- `c & 0xFF = 0x48`
- `k & 0xFF = 0x83`
Let `G[i] = t[n + i]` (the second half of `t`). The per-byte inversion is:
```
input[i] = prev ^ cc ^ ((G[i] - kk) & 0xFF)
prev = G[i]
```
## Minimal Solver (Python)
```python
# Extract s from the HTML manually, paste it here:
s = "PASTE_THE_VARIATION_SELECTOR_STRING_HERE"
vs = 0xFE00
# Decode variation selectors to bytes
t = []
for i in range(0, len(s), 2):
b = ((ord(s[i]) - vs) << 4) | (ord(s[i+1]) - vs)
t.append(b)
n = len(t) // 2
G = t[n:]
prev = 0x48
cc = 0x48
kk = 0x83
u = []
for gi in G:
ui = prev ^ cc ^ ((gi - kk) & 0xFF)
u.append(ui)
prev = gi
flag = bytes(u).decode("utf-8")
print(flag)
```
## Flag
`ENO{W0W_1_D1DN'T_533_TH4T_C0M1NG!!!}`
