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🔒 UMDCTF 2023 / ✅-crypto-ekans
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Sutx BOT 04/28/2023 3:00 PM
Ekans - 500 points
Category: Crypto Description: Ekans is literally just "snake" backwards. If it works for pokemon naming, it works for cryptography, right? Author: aparker Files:Tags: No tags.
04/28/2023 3:00 PM
Sutx pinned a message to this channel. 04/28/2023 3:00 PM
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Sutx BOT 04/28/2023 3:31 PM
@kanon wants to collaborate 🤝
15:34
@Violin wants to collaborate 🤝
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Sutx BOT 04/28/2023 4:50 PM
@unpickled admin bot wants to collaborate 🤝
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unpickled admin bot 04/28/2023 4:50 PM
(will prob afk this till i finish my hw but)
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Sutx BOT 04/28/2023 8:43 PM
@fleming wants to collaborate 🤝
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Sutx BOT 04/28/2023 11:11 PM
@layka_ wants to collaborate 🤝
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Sutx BOT 04/29/2023 9:02 AM
@Utaha wants to collaborate 🤝
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Utaha 04/29/2023 9:13 AM
I guess the original decryption would be (V - B' * s)
09:13
I don't really know the name of encryption, but the only issue is probably the fact that s, e is related
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Utaha
used /ctf solve
Sutx BOT 04/29/2023 9:39 AM
✅ Challenge solved.
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Utaha 04/29/2023 9:39 AM
basically solve linear system to get s and win
09:40
A = ... B = ... B_prime = ... V = ... DIM = 256 q = 15901 for i in range(DIM): A[i][DIM-1-i] += 1 A = matrix(GF(q), A) B = vector(GF(q), [x[0] for x in B]) s = ~A * B B_prime = matrix(GF(q), B_prime) pt = vector(GF(q), [x[0] for x in V]) - B_prime * s flag = "" for i in range(DIM): tmp = int(pt[i]) tmp = (tmp + q // 32) // (q // 16) tmp %= 16 flag += hex(tmp)[2:] flag = [int(flag[i:i+2], 16) for i in range(0, len(flag), 2)] print(bytes(flag))
Exported 14 message(s)