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Divide and Rule: DiFA  - Division Property Based Fault Attacks on PRESENT and GIFT

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Applied Cryptography and Network Security (ACNS 2023)

Part of the book series: Lecture Notes in Computer Science ((LNCS,volume 13905))

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Abstract

The division property introduced by Todo in Crypto 2015 is one of the most versatile tools in the arsenal of a cryptanalyst which has given new insights into many ciphers primarily from an algebraic perspective. On the other end of the spectrum we have fault attacks which have evolved into the deadliest of all physical attacks on cryptosystems. The current work aims to combine these seemingly distant tools to come up with a new type of fault attack. We show how fault invariants are formed under special input division multi-sets and are independent of the fault injection location. It is further shown that the same division trail can be exploited as a multi-round Zero-Sum distinguisher to reduce the key-space to practical limits. As a proof of concept division trails of PRESENT and GIFT are exploited to mount practical key-recovery attacks based on the random nibble fault model. For GIFT-64, we are able to recover the unique master-key with 30 nibble faults with faults injected at rounds 21 and 19. For PRESENT-80, DiFA reduces the key-space from \(2^{80}\) to \(2^{16}\) with 15 faults in round 25 while for PRESENT-128, the unique key is recovered with 30 faults in rounds 25 and 24. This constitutes the best fault attacks on these ciphers in terms of fault injection rounds. We also report an interesting property pertaining to fault induced division trails which shows its inapplicability to attack GIFT-128. Overall, the usage of division trails in fault based cryptanalysis showcases new possibilities and reiterates the applicability of classical cryptanalytic tools in physical attacks.

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Correspondence to Anup Kumar Kundu .

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Appendices

A Results on GIFT-128

The balanced bits from 5-round property is given below. Using a random nibble fault we get a total 80 balanced bits after 5 rounds. However the balanced bits are divided into two sets. If the active nibble belongs to \(\{N_0,...,N_{15}\}\) we get the following balanced bits

  • \(\mathcal {A}\) = 0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 16, 17, 18, 20, 21, 22, 24, 25, 28, 29, 32, 33, 34, 36, 37, 38, 40, 41, 44, 45, 48, 49, 50, 52, 53, 54, 56, 57, 60, 61, 64, 65, 66, 68, 69, 70, 72, 73, 76, 77, 80, 81, 82, 84, 85, 86, 88, 89, 92, 93, 96, 97, 98, 100, 101, 102, 104, 105, 108, 109, 112, 113, 114, 116, 117, 118, 120, 121, 124, 125.

If the active nibble belongs to \(\{N_{16},...,N_{31}\}\) we get the following balanced bits

  • \(\mathcal {B}\) = 0, 1, 4, 5, 8, 9, 10, 12, 13, 14, 16, 17, 20, 21, 24, 25, 26, 28, 29, 30, 32, 33, 36, 37, 40, 41, 42, 44, 45, 46, 48, 49, 52, 53, 56, 57, 58, 60, 61, 62, 64, 65, 68, 69, 72, 73, 74, 76, 77, 78, 80, 81, 84, 85, 88, 89, 90, 92, 93, 94, 96, 97, 100, 101, 104, 105, 106, 108, 109, 110, 112, 113, 116, 117, 120, 121, 122, 124, 125, 126.

While the intersection of these two sets contains the following 64 balanced bits

  • \(\mathcal {A} \cap \mathcal {B}\) = {0, 1, 4, 5, 8, 9, 12, 13, 16, 17, 20, 21, 24, 25, 28, 29, 32, 33, 36, 37, 40, 41, 44, 45, 48, 49, 52, 53, 56, 57, 60, 61, 64, 65, 68, 69, 72, 73, 76, 77, 80, 81, 84, 85, 88, 89, 92, 93, 96, 97, 100, 101, 104, 105, 108, 109, 112, 113, 116, 117, 120, 121, 124, 125 }.

Thus we have 64 balanced bits after 5 round in fault invariant setting, i.e., any random active nibble at the input results to 64 output bits. Moreover, input of each sbox in the 6-th round contains two balanced bits, namely the first and second input bit.

B Sbox Division Trail Algorithm

figure i

C PRESENT Partial Subkeys Recovery of the Last Two Rounds

Here we discuss partial recovery of the last two round keys. To recover the partial subkeys of the last two rounds we use our random nibble fault distinguisher for 4 and 5 rounds. For this we construct an active nibble at the input of \(R^{25}\) by injecting 15 faults. From Table 3, we can observe that all the 64 bits at \(R^{28}\) and the first 4 bits of \(R^{29}\) become balanced. Now take all possible key values of the nibbles \(N^{30}_j\) for \(j\in \{0, 4, 8, 12\}\) and partially decrypt the corresponding nibbles of the ciphertext. Due to the propagation of the division property, we get balanced bits at \(N^{29}_0\) (from Table 3). We take the xorsum of the partially decrypted ciphertexts and check for which key nibble values the xorsum at \(N^{29}_0\) becomes 0. We take those as possible key choice for the nibbles \(\{0, 4, 8, 12\}\) in \(R^{30}\) and proceed in the next step. Upto this point we get \(2^{12}\) key choices for the nibbles as we have 4 bit filters at \(R^{29}\). In the next step we take all the key values of the nibbles \(N^{29}_j\) for \(j \in \{0, \cdots , 3\}\) and partially decrypt one more round. In the output of \(R^{28}\) we check whether the xorsum of the decrypted ciphertexts becomes 0 or not. As 16 bits at \(R^{28}\) becomes balanced hence the key choices of the nibbles \(N^{30}_j\) for \(j\in \{0, 4, 8, 12\}\) and \(N^{29}_j\) for \(j \in \{0, \cdots , 3\}\) reduces to \(2^{12}\). Thus total 20 bit subkey of the last two rounds can be recovered using this attack. The algorithm for the last round key recovery of this attack is given in Appendix C.1 while the pictorial view of the full attack is given in Fig. 6 (red-colored bits are the balanced bits).

Fig. 6.
figure 6

Partial key recovery attack on PRESENT

Complexity of the Attack: The data complexity for this case is \(2^{16}\) as we have used 16 plaintext-ciphertext pairs to recover the partial subkeys. In this case also instead of decrypting the whole ciphertext at once we partially decrypt the 0-th \(\mathcal{Q}\mathcal{R}\) group i.e. the nibbles \(N^{30}_j\) for \(j\in \{0, 4, 8, 12\}\) in the last round and \(N^{29}_j\) for \(j\in \{0, 1, 2, 3\}\) in the second last. In the last round we have 16 bit keys and 4 bit filters for this \(\mathcal{Q}\mathcal{R}\) group. Hence the number of keys that passes through the filter is \(2^{12}\). For each of the key in the last round we take all possible values in the second last round and check the zero-sum. Hence the complexity for this group is \(2^{12}\times 2^{16} = 2^{28}\) sbox inversions and the number of keys that passes through the filer is \(2^{12}\) for 0-th \(\mathcal{Q}\mathcal{R}\) group.

1.1 C.1 Last Round Key Recovery of PRESENT

figure j

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Kundu, A.K., Ghosh, S., Saha, D., Rahman, M. (2023). Divide and Rule: DiFA  - Division Property Based Fault Attacks on PRESENT and GIFT. In: Tibouchi, M., Wang, X. (eds) Applied Cryptography and Network Security. ACNS 2023. Lecture Notes in Computer Science, vol 13905. Springer, Cham. https://doi.org/10.1007/978-3-031-33488-7_4

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