Restructuring

DAMCTF2021/haveibeenpwned/README.md

@@ -27,6 +27,7 @@ The server creates a list of $x,y$ points where $x$ is md5(pw) and $y$ is a rand
 Then the points are interpolated together using lagrange interpolation
 Notice that both polynomials have the private keys = $P(md5(pw_i))$ where $P$ is the polynomial and can be substituted in to the curve equation
 $P_y(md5(pw_i))^2 = P_x(md5(pw_i))^3 + 486662 * P_x(md5(pw_i))^2 + P_x(md5(pw_i)) \mod{2^{255} - 19}$
+## Finding the hashes
 To recover the hashes we seed to find for what n $P_x(n)^3 + 486662 * P_x(n)^2 + P_x(n) - P_y(n)^2 \equiv 0 \mod{2^{255} - 19}$
 ```python
 pol = xPoly^3 + 486662 * xPoly^2 + xPoly - yPoly^2
@@ -41,7 +42,9 @@ def md5(msg: bytes) -> int:
 	h.update(msg)
 	return int.from_bytes(h.digest(), 'little')
 ```
-Here we see that the `md5()` function uses little endianess (ew) so we have to convert the hashes using `int(r).to_bytes(16, 'little').hex()`, the we can throw this into an online md5 lookup table (or hashcat if you are feeling fancy) and find
+Here we see that the `md5()` function uses little endianess (ew) so we have to convert the hashes using `int(r).to_bytes(16, 'little').hex()`.
+## Finishing up
+Now we can throw this into an online md5 lookup table (or hashcat if you are feeling fancy) and find
 ```
 481cf17a2f3a0cfafcedaa0dbea09fdc DDBBFF
 476e90c539400f0e2988b32db64ae6bf honolulu-lulu