Homemade_Keeloq_II

Flag: ECW{w0w_Th4t_w4s_sUp3r_s3cUr3_53217}

Challenge

5KB

astek_HomemadeKeeloq_II.zip

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Description

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Your mission : Find the next key value (again?!), more information given in the source.

Flag is in the format ECW{w0w_Th4t_w4s_sUp3r_s3cUr3_NEXTKEY}. NEXTKEY being the next 16 bit key in decimal format.

Challenge made by:

Solution

Ce challenge est la suite de Homemade_Keeloq_I. On réutilise le même parser et on obtient :

{'fixed': '1', 'vlow': '0', 'button_status_1': '1110', 'serial': 184232904, 'button_status_2': '0010', 'discrimination': 746, 'counter': 210}
{'fixed': '1', 'vlow': '0', 'button_status_1': '1110', 'serial': 184232900, 'button_status_2': '0001', 'discrimination': 746, 'counter': 10629}
{'fixed': '1', 'vlow': '0', 'button_status_1': '1110', 'serial': 184232904, 'button_status_2': '0010', 'discrimination': 746, 'counter': 40800}
{'fixed': '1', 'vlow': '0', 'button_status_1': '1110', 'serial': 184232900, 'button_status_2': '0001', 'discrimination': 746, 'counter': 25187}
{'fixed': '1', 'vlow': '0', 'button_status_1': '1110', 'serial': 184232904, 'button_status_2': '0010', 'discrimination': 746, 'counter': 29326}

C’est un polynôme interpolé sur des points consécutifs, on peut utiliser les différences finie :

x	y	Δ¹	Δ²	Δ³	Δ⁴
0	210
1	10629	10629−210=10419
2	40800	40800−10629=30171	30171−10419=19752
3	25187	25187−40800=−15613≡49923	49923−30171=19752	0
4	29326	29326−25187=4139	4139−49923=−45784≡19752	0	0

Les différences stabilisent à Δ⁴=0 → degré 3 polynôme

Pour points uniformes,

$$f(n) = y_0 + n\,\Delta^1_0 + \frac{n(n - 1)}{2!}\,\Delta^2_0 + \frac{n(n - 1)(n - 2)}{3!}\,\Delta^3_0 + \cdots$$

Les valeurs :

• Δ¹₀ = 10419

• Δ²₀ = 19752

• Δ³₀ = 0

$$f(x) = 210 + 10419 \, \frac{x}{1!} + 19752 \, \frac{x \, (x - 1)}{2!} \pmod{65536}$$

En python ça nous donne :

code = lambda x: int((210+n*10419+((x*(x-1)) / 2)*19752) % 2**16)
for i in range(6):
    print(code(i))

210
10629
40800
25187
29326
53217