Neural Aided Statistical Attack for Cryptanalysis Yi Chen, Yantian Shen, Hongbo Yu

Abstract—In CRYPTO’19, Gohr proposed the first key recov- exist in the prepended differential. Up to date, there are no ery attack based on deep learning, which opens the direction of solutions for these two problems. neural aided cryptanalysis. Up to date, neural aided cryptanalysis In this paper, we have explored neural aided cryptanalysis still faces two problems: (1) its attack complexity estimations rely purely on practical experiments. There is no theoretical and made contributions as follows. framework for estimating theoretical complexity. (2) it can not • We figure out the reason why Gohr’s attack does not allow work when there are no enough neutral bits that exist in the the adversary to estimate the theoretical complexities. prepended differential. To our best knowledge, we are the first In [9], each key guess corresponds to a key rank score that to solve these two problems. In this paper, our work contains is directly determined by the output of ND. A key guess three aspects: (1) we propose a Neural Aided Statistical Attack (NASA) that is a generic neural aided cryptanalysis technique. is returned as a candidate when its key rank score exceeds NASA allows us to estimate the theoretical complexities without a threshold. Since the output of ND is unpredictable, performing practical experiments. Moreover, NASA does not rely the threshold is set without any theoretical basis. Then on any special properties including the neutral bit. (2) we propose it is impossible to estimate the attack success rate and three methods for reducing the attack complexities of NASA. One theoretical complexities unless practical experiments are of the methods is based on a newly proposed concept named Informative Bit that reveals an important phenomenon related executed. to neural distinguishers. (3) We prove the superiorities of NASA • We propose a Neural Aided Statistical Attack (NASA), based on applications to round reduced DES and Speck32/64. which supports theoretical complexity estimation and Our work arguably raises a new direction for neural aided does not rely on neutral bits. The theoretical basis of cryptanalysis. NASA is as follows. By analyzing the key recovery Index Terms—Neural Aided Statistical Attack, Deep Learning, process, we find that there are only four different sce- Cryptanalysis, Informative Bit, DES, Speck narios. A statistic designed in this article obeys a normal distribution in each scenario. Based on this statistic, the key recovery is transformed into the distinguishing I.INTRODUCTION between two normal distributions, which tells the required Deep learning has received much expectation in the cryp- data complexity of NASA. tography community since the last century. Rivest in [1] • We propose three methods to reduce the attack complex- reviewed various connections between machine learning and ities of NASA. The first one is reducing the key space cryptography. Some possible directions of research in crypt- by building ND on partial ciphertext bits. The initial analytic applications of machine learning were also suggested. ND proposed by Gohr takes the complete ciphertext Greydanus proved that a simplified version of Enigma can be pair as input, which forces the adversary to guess all simulated by recurrent neural networks [2]. the key bits simultaneously. Adopting the new ND, the Although deep learning has shown its superiorities in var- adversary guesses partial key bits at a time. The second ious fields such as computer vision [3], natural language one is a highly selective Bayesian key search algorithm. It processing [4], and smart medical [5], its application in the allows the adversary to search key guesses that are most field of conventional cryptanalysis has been stagnant. A few likely to be the right key instead of traversing all the key valuable applications are only concentrated in the side-channel guesses. The third one is reducing the data complexity by analysis [6]–[8]. exploiting neutral bits. When there are available neutral In CRYPTO’19, Gohr proposed a deep learning-based dis- bits, the data complexity of NASA can also be reduced. tinguisher [9] that is also called neural distinguisher (ND). • We perform experiments on DES and Speck32/64 to By placing a differential before ND, Gohr developed a key fully analyze NASA. First, experiments on DES prove recovery attack on 11-round Speck32/64. Gohr’s attack shows that NASA can attack a cipher covering more rounds considerable advantages in terms of attack complexities over than Gohr’s attack when there are no enough neutral the traditional differential attack, which opens the direction of bits. Second, the experiment on Speck32/64 shows that neural aided cryptanalysis. NASA can achieve comparable performance with Gohr’s However, Gohr’s attack faces two problems. First, it can not attack when the proposed three optimization methods are be applied to the theoretical security analysis of a cipher. More available. exactly, we do not know what the required data complexity is to attack a specific cipher. We can estimate the attack Organization Section III presents the neural aided statisti- complexities and success rate empirically only if the attack is cal distinguisher. Section IV summarizes NASA and provides finished within an acceptable running time. Second, when we the correctness verification. The three optimization methods add a differential before ND for attacking a cipher covering are introduced in Section V, VI, VII. Applications to DES more rounds, it requires that enough neutral bits [10] must and Speck32/64 are presented in Section VIII, IX. 2

II.RELATED WORK Algorithm 1 Basic version of Gohr’s key recovery attack Require: k neutral bits that exist in ∆P → ∆S; Let (P0,P1) denote a plaintext pair with difference ∆P . The An ND built over ∆S; corresponding intermediate states, ciphertexts are (S0,S1), A key rank score threshold, c1; (C0,C1). A maximum number of iterations. Ensure: A possible key candidate. A. Neutral Bit 1: repeat ∆P → ∆S E 1 1 1 1 Consider a differential . Let denote the 2: Random generate a plaintext pair (P0 ,P1 )|P0 ⊕ P1 = encryption function covering the differential. For any plaintext ∆P ; pair (P0,P1) conforming to the differential, if the following 3: Create a plaintext structure consisting of 2k plaintext condition always holds pairs by k neutral bits; i i E(P ⊕ ej) ⊕ E(P ⊕ ej) = ∆S, ej = 1 j, 4: Collect corresponding ciphertext pairs, (C0,C1), i ∈ 0 1 {1, ··· , 2k}; the j-th bit is called a neutral bit [10]. 5: for each key guess kg do k Based on k neutral bits {j1, ··· , jk} and a plaintext pair 6: Partially decrypt 2 ciphertext pairs with kg; (P0,P1)|P0 ⊕P1 = ∆P , we can generate a plaintext structure 7: Feed decrypted ciphertext pairs to ND and collect k consisting of 2 plaintext pairs. Once (P0,P1) satisﬁes the the outputs; k differential, the remaining 2 − 1 plaintext pairs also conform 8: Calculate the key rank score vkg based on collected to the differential. outputs; 9: if vkg > c1 then B. Neural Distinguisher 10: stop the key search and return kg as the key candidate; The target of ND [9] is to distinguish two classes of 11: end if ciphertext pairs 12: end for 1, if S0 ⊕ S1 = ∆S 13: until a key candidate is returned or the maximum number Y (C0,C1) = , (1) 0, if S0 ⊕ S1 6= ∆S of iterations is reached. where Y = 1 or Y = 0 is the label of (C0,C1). If the difference between S0 and S1 is the target difference ∆S, the pair (C0,C1) is regarded as a positive sample drawn from The rank score is likely to exceed c1 only when the plaintext the target distribution. Or (C0,C1) is regarded as a negative structure passes the prepended differential and kg is the right sample that comes from a uniform distribution. key. If the plaintext structure does not pass the differential or N the key guess is wrong, the rank score should be very low. A neural network is trained over 2 positive samples and N Thus, the right key can be identiﬁed by comparing the rank 2 negative samples. The neural network can be used as an ND if the distinguishing accuracy over a testing database is score with a threshold. When the performance of ND is weak, higher than 0.5. 2k needs to be large. Then more neutral bits are required. Let NDh denotes a neural distinguisher against the cipher The application of Gohr’s attack is limited by two aspects: reduced to h rounds. Given a sample (C0,C1), ND will output • The adversary can not estimate the required attack com- a score Z which is used as the posterior probability plexities and success rate theoretically. Since the output P r(Y = 1 |(C0,C1)) = Z = ND(C0,C1), 0 6 Z 6 1 Z of ND is unpredictable, the threshold c1 is set without (2) any clear theoretical basis. Then it is unknown how many When Z > 0.5, the predicted label of (C0,C1) is 1 [9]. plaintext pairs a plaintext structure should contain. As a In [9], a Residual Network (ResNet) is adopted by Gohr. result, the success rate under an attack setting is also The training pipeline can refer to [9]. unknown. • If the number of neutral bits is not large enough, Gohr’s C. Gohr’s Key Recovery Attack attack does not work. Algorithm 1 summarizes the core idea of the basic version (unaccelerated version) of Gohr’s key recovery attack [9]. Decrypting 2k ciphertext pairs drawn from the target or D. Distinguishing between Two Normal Distributions uniform distribution with a key guess kg, the adversary uses Here, we present the details of the distinguishing between the following formula two normal distributions. 2k Consider two normal distributions: N (µ , σ ), and X Zi r r v = log (3) kg 2 1 − Z N (µw, σw). A sample s is sampled from either N (µr, σr) i=1 i or N (µw, σw). We have to decide if this sample is from to combine the scores Zi of individual decrypted ciphertext N (µr, σr) or N (µw, σw). The decision is made by comparing pairs into a rank score for kg. When the rank score vkg exceeds the value s to some threshold t. Without loss of generality, a threshold c1, kg is regarded as a key candidate. assume that µr > µw. If s > t, the decision is s ∈ N (µr, σr). 3

If s < t, the decision is s ∈ N (µw, σw). Then there are error probabilities of two types:

βr = P r {N (µw, σw) |s ∈ N (µr, σr)} , (4) βw = P r {N (µr, σr) |s ∈ N (µw, σw)} .

When a sample s is sampled from N (µr, σr), the probability that the decision is s ∈ N (µw, σw) is βr. Here a condition is given on µr, µw, σr, σw such that the error probabilities are βr and βw. The proof can refer to related research [11], [12]. Fig. 1. Four situations of decrypting a ciphertext pair with a key guess. Proposition 1: For the test to have error probabilities of at most βr and βw, the parameters of the normal distribution N (µr, σr) and N (µw, σw) with µr 6= µw have to be such • Decrypting a positive sample with wrong keys: the that ciphertext pair satisfies the differential but the key guess z1−β × σr + z1−β × σw r w = 1 (5) is wrong. |µr − µw| • Decrypting a negative sample with the right key: the where z1−βr and z1−βw are the quantiles of the standard ciphertext pair does not satisfy the differential but the key normal distribution. guess is right. Proposition 1 is the theoretical basis of our work in this • Decrypting a negative sample with wrong keys: the article. ciphertext pair does not satisfy the differential and the key guess is wrong. III.NEURAL AIDED STATISTICAL DISTINGUISHER Given an ND, we denote the probability of Z > c2 as A. A Chosen Plaintext Statistical Distinguisher p1, p2, p3, p4 for the four situations respectively. Then the Consider a cipher E with a block size of m, and a distributions of the statistic (formula 6) in these four situations p0 m are differential ∆P −→ ∆S|∆P, ∆S ∈ F2 where p0 is the p transition probability. Build an ND over ∆S. Randomly T1 ∼ N (µ1, σ1), µ1 = N1 × p1, σ1 = N1 × p1(1 − p1) N ∆P p generate plaintext pairs with a difference , and col- T2 ∼ N (µ2, σ2), µ2 = N2 × p2, σ2 = N2 × p2(1 − p2) lect corresponding ciphertext pairs. The adversary needs to p T3 ∼ N (µ3, σ3), µ3 = N3 × p3, σ3 = N3 × p3(1 − p3) distinguish between this cipher and a random permutation. p T4 ∼ N (µ4, σ4), µ4 = N4 × p4, σ4 = N4 × p4(1 − p4) The concrete process is as follows. For each ciphertext pair (7) Ci ,Ci , i ∈ {1, ··· ,N}, the adversary feeds it into ND 0 1 if N1, N2, N3, N4 are high enough. N (µi, σi) is a normal and obtains its output Z . Setting a threshold value c , the i 2 distribution with mean µi and standard deviation σi, i ∈ adversary calculates the statistic T {1, 2, 3, 4}. An empirical condition is N X 1, if Zi > c2 Ni × pi > 5,Ni × (1 − pi) > 5, i ∈ {1, 2, 3, 4} (8) T = φ (Zi), φ (Zi) = . (6) 0, if Zi 6 c2 i=1 Now come back to the key recovery attack. If the probability −m When p0 > 2 holds, it’s expected that the value of the of the differential ∆P → ∆S is p0 and N ciphertext pairs are statistic T for the cipher is higher than that for a random collected randomly, then permutation. In a key recovery setting, the right key will N = N = N × p ,N = N = N × (1 − p ) . (9) result in the statistic T being among the highest values for 1 2 0 3 4 0 all candidate keys if N is large enough. In the sequel, we give Besides, the distributions of the statistic (formula 6) under the this a theoretical analysis. right key and wrong keys are both a mixture of two normal distributions. B. Distribution of the Statistic under Right and Wrong keys First, we regard a ciphertext pair as a point in a high- 1) Right key guess: This case contains two situations dimensional space. For a given threshold c2, it’s equivalent to in which corresponding distributions are N (µ1, σ1) and creating a stable classification hyperplane in this space using N (µ3, σ3). Since a mixture of two independent normal dis- an ND. Thus the classification over a random ciphertext pair tributions is still a normal distribution, the distribution of the is modeled as a Bernoulli experiment. It provides us with a statistic (formula 6) under the right key guess is: theoretical analysis framework. T = T + T ∼ N (µ , σ ) (10) According to the key recovery process, there are four r 1 3 r r possible situations when we decrypt a ciphertext pair with a µr = N × (p0p1 + (1 − p0) p3) (11) key guess as shown in Fig. 1: σ = pN × p × p (1 − p ) + N (1 − p ) p (1 − p ) • Decrypting a positive sample with the right key: the r 0 1 1 0 3 3 ciphertext pair satisfies the differential and the key guess (12) is right. 4

2) Wrong key guess: This case also contains two situa- D. Estimation of p1, pn tions in which corresponding distributions are N (µ , σ ) and 2 2 Consider an ND against a cipher, the values of p1, pn are N (µ4, σ4). Then the distribution of the statistic (formula 6) estimated as: under wrong key guesses is: 1) Randomly generate M positive / negative samples and decrypt them for 1 round with the right / wrong subkeys. T = T + T ∼ N (µ , σ ) (13) w 2 4 w w 2) Feed partially decrypted samples into ND. 3) Calculate the final ratio of Z > c2. µw = N × (p0p2 + (1 − p0) p4) (14) The ratio is the statistical expectation of p1 or pn. A large M p can make the statistical expectation accurate enough. σw = N × p0 × p2 (1 − p2) + N (1 − p0) p4 (1 − p4) (15) Negative samples in the high-dimensional space approxi- E. Further Analysis and Estimation of p2 mately obey uniform distribution, thus p3 = p4 holds theo- When we decrypt a positive sample with a wrong key guess retically and experiments also verify it. Since the accuracy (Fig. 1(2)), the final value of p2 is related to the Hamming of NDs is higher than 0.5, p1 > p2 also holds with a high distance between the wrong key guess and the right key. Such probability. When we set c2 = 0.5, we can ensure p1 > p2. a phenomenon is based on Property 1 and Property 2. Thus µr > µw also holds. Property 1: Decrypt a ciphertext for one round with two Since the distributions of Tr,Tw are different, the key recovery attack can be performed based on Proposition 1. different subkeys, 1 Ch−1 = DecOneRound(Ch, kg1) 2 Ch−1 = DecOneRound(Ch, kg2). C. Data Complexity of the Statistical Distinguisher If kg1 and kg2 are only different at a few bits (e.g. just 1 bit Based on Proposition 1, one obtains the condition: 1 2 or 2 bits), the Hamming distance between Ch−1 and Ch−1 z σ + z σ will be very small in high probability. 1−βr r 1−βw w = 1 (16) µr − µw Property 2: Consider a neural network F (·) solving a binary classification problem. If two input samples I1, I2 where the values of µr, σr, µw, σw refer to formula 11, 12, 14, 15 respectively. In a key recovery setting, are very close to each other in the input space, two outputs F (I1),F (I2) of the neural network may satisfy F (I1) ≈ 1 − βr is the minimum probability that the right key survives, F (I2) with a high probability. βw is the maximum probability that wrong keys survive. Since we can not know the real classification hyperplane Although the distance metric in the input space of neural learned by the ND, p1, p2, p3, and p4 are estimated ex- networks is complex and unknown, the Hamming distance is perimentally. Then the estimated values of p3 and p4 will still a good alternative. Thus it is expected that p2 is related to be slightly different even they should be theoretically equal. the Hamming distance between the right key and wrong key When the probability p0 of the differential transition is very guesses. low, the slight distinction p3 − p4 may dominate µr − µw, Suppose we decrypt a positive sample (Ch+x,0,Ch+x,1) which is wrong. Thus we neglect the minor difference and with x subkey guesses simultaneously replace p3, p4 with pn. Then the condition (formula 16) is simplified Ch+j−1,0/1 = DecOneRound(Ch+j,0/1, kgh+j), j ∈ [1, x] √ where kg is the key guess of the (h + j)-th round. z1−βr × X1 + z1−βw × X2 h+j N = , (17) (Ch,0,Ch,1) is fed into an NDh for estimating p2. (p1 − p2) × p0 When the last x − 1 key guesses kgh+j, j ∈ [2, x] are all where right, (Ch+1,0,Ch+1,1) is a positive sample. The probability p Z > c p kg , j ∈ {2, ··· , x} X1 = p0a1 + (1 − p0)a3, (18) of 2 is 2. If h+j are not all right, then (Ch+1,0,Ch+1,1) is not a positive sample anymore. The p resulted probability of Z > c2 is closer to pn. X2 = p0a2 + (1 − p0)a3, (19) Since x key guesses have different influences on the probability of Z > c2, we consider x Hamming distances for a1 = p1(1−p1), a2 = p2(1−p2), a3 = pn(1−pn). (20) estimating p2. Let dj denotes the Hamming distance between The decision threshold t is: the right key and key guess in the (h + j)-th round, and

p2|d1,··· ,dx denotes the probability of Z > c2. Algorithm 2 is proposed for the estimation of p . t = µr − z1−βr σr = µw + z1−βw σw. (21) 2|d1,··· ,dx Veriﬁcation. Gohr provided ND5, ND6, ND7, ND8 The data complexity N is directly calculated when βr and against Speck32/64 [9], which are built over a plaintext βw are set. The impacts of p0, p1, p2, pn on N are about difference (0x0040, 0). We have performed tests on these four −2 −2 7 O(p0 ), O((p1 − p2) ), O(pn) respectively. The proof is distinguishers. Let M = 10 , Table I and Table II show the presented in Appendix. estimation results of p2|d1 and p2|d1,d2 respectively. 5

TABLE I

THE ESTIMATION OF p2|d1 OF 4 NEURALDISTINGUISHERSAGAINSTROUNDREDUCED SPECK32/64. FOR ND5, ND6, ND7, c2 = 0.55.FOR ND8, c2 = 0.5. p2|d1=0 = p1.

d1 0 1 2 3 4 5 6 7 8 ∼ 16 ND5 p2|d1 0.8889 0.5151 0.3213 0.2168 0.1556 0.1189 0.0956 0.08 6 0.0694 d1 0 1 2 3 4 5 6 7 8 ∼ 16 ND6 p2|d1 0.6785 0.4429 0.3135 0.2384 0.1947 0.1684 0.1518 0.1408 6 0.1334 d1 0 1 2 3 4 5 6 7 8 ∼ 16 ND7 p2|d1 0.4183 0.3369 0.2884 0.2607 0.2442 0.234 0.2276 0.2236 6 0.2211 d1 0 1 2 3 4 5 6 7 8 ∼ 16 ND8 p2|d1 0.5184 0.5056 0.4993 0.4957 0.4939 0.4927 0.4925 0.4918 6 0.4917

TABLE II

THE ESTIMATION OF p2|d1,d2 OF ND7 AGAINST SPECK32/64. c2 = 0.55. THECOLUMNSCORRESPONDTO d2. THEROWSCORRESPONDTO d1. ALL RESULTS ONLY RETAIN TWO DECIMAL PLACES. THESAMEVALUEISREPLACEDBYANUPPERCASELETTER. Y = 0.21, E = 0.22, J = 0.23, U = 0.25, AND V = 0.26. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0 0.42 V E E Y Y Y Y Y Y Y Y Y Y Y Y Y 1 0.33 U E Y Y Y Y Y Y Y Y Y Y Y Y Y E 2 0.29 J E Y Y Y Y Y Y Y Y Y Y Y Y Y Y 3 V J E Y Y Y Y Y Y Y Y Y Y Y Y Y Y 4 J J E Y Y Y Y Y Y Y Y Y Y Y Y Y Y 5 E E E Y Y Y Y Y Y Y Y Y Y Y Y Y Y 6 E Y E Y Y Y Y Y Y Y Y Y Y Y Y Y Y 7 E E Y Y Y Y Y Y Y Y Y Y Y Y Y Y E 8 E Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y 9 ∼ 16 6 Y

Algorithm 2 Estimation of p2|d ,··· ,d 1 x when two subkeys are guessed simultaneously, p2|d1,d2 will Require: a cipher with a subkey size of m; decrease sharply even if the key guess of the last round is i i i i M random plaintext pairs, (P0,P1),P0 ⊕ P1 = ∆P, i ∈ wrong at only 1 bit. {1, ··· ,M}; Thus, the choice of p2 depends on the target of the key an NDh built over ∆P . recovery attack. If we think the attack is successful as long M random master keys, MKi, i ∈ {1, ··· ,M}; as the Hamming distance between the key guess and the right The threshold c2. key is not larger than a threshold d, the value of p2 should be

Ensure: p2|d1,··· ,dx . i i p2 = max p2|d1,··· ,dx |d1 + ··· + dx > d (22) 1: Encrypt each plaintext pair (P0,P1) with a master key MKi for h + x rounds; This choice is based on the following truth. By setting a proper i i 2: Save resulting ciphertext pair (C0,C1); threshold c2 such as c2 0.5, we ensure i > 3: Save the (h + j)-th subkey skh+j, j ∈ {1, ··· , x}; p 0.5, if d1 + ··· + dx > d (23) 4: for d1 = 0 to m, ··· , dx = 0 to m do 2|d1,··· ,dx 6 5: for i = 1 to M do p i According to formula 17, the higher 2 is, the higher the 6: Randomly draw x key guesses kgj, j ∈ {1, ··· , x} required data complexity is. The decision threshold also in- i i where the Hamming distance between kgj and skh+j creases when p2 increases. Thus we only need to focus on the is dj; highest data complexity required for ﬁltering wrong keys. i 7: i i Decrypt (C0,C1) with kgj, j ∈ {1, ··· , x} for x Take ND7 as an example. Let d = 2, it means that the rounds; attack is successful if the recovered key is different from the 8: ND Feed the decrypted ciphertext pair into h and right key at most 2 bits. Then p2 = p2|3 = 0.2607 or p2 = save the output as Z ; i|d1,··· ,dx p2|0,1 = p2|3,0 = 0.26. 9: end for 10: Count the number of Z > c , and denote it as i|d1,··· ,dx 2 IV. NEURAL AIDED STATISTICAL ATTACK Td1,··· ,dx ; Td ,··· ,d A. Key Recovery Attack Model 11: 1 x Save p2|d1,··· ,dx = M . 12: end for This neural aided statistical distinguisher is used to deter- mine whether a key guess may be the right key. This is done by the Statistical Test as shown in Algorithm 3. Algorithm 4 summarizes the Neural Aided StatisticalAttack (NASA) based The test results have veriﬁed the analysis of p2. Besides, on the statistical distinguisher. 6

Algorithm 3 Statistical test for a key guess TABLE III Require: An ND; A key guess, kg; TWO OPTIONS OF THE PREPENDED DIFFERENTIAL OF SPECK32/64. nr IS THE NUMBER OF ENCRYPTION ROUNDS COVERED BY THE DIFFERENTIAL. A posterior probability threshold, c2; The decision

threshold, t; ID ∆P → ∆S p0 nr i i i i i −2 N ciphertext pairs (C0,C1) encrypted from (P0,P1),P0⊕ 1 (0x2800, 0x10) → (0x0040, 0) 2 1 i −6 P1 = ∆P, i ∈ [1,N]. 2 (0x211, 0xa04) → (0x0040, 0) 2 2 1: Decrypt N ciphertext pairs with kg; 2: Feed decrypted ciphertext pairs into ND, and collect the outputs Z , i ∈ [1,N]; i 3) Check the following observation indexes: 3: Calculate the statistic T in formula 6; a) The ratio that the right key (d = 0) passes the 4: if T t then 1 > statistical test. 5: Return kg as a key candidate. b) The average number of surviving keys in 100 trails. 6: end if c) The ratio that the number of surviving keys is smaller than the expected upper bound. Algorithm 4 Neural Aided Statistical Attack Table IV summarizes the settings related to four attacks. Require: The attacked cipher; p 0 Table I shows the estimations of p2|d1 related to ND5, ND7, The differential with a probability of p0, ∆P −→ ∆S; ND8. The value of p2 should be p2|d1=3 in four attacks. Two maximum error probabilities, βr, βw; A posterior probability threshold, c2. 1) Attack 1: recover sk10 of 10-round Speck32/64: In Ensure: All possible key candidates. the first attack setting, we get N = 3309 ≈ 211.69 ( see 1: Train an ND over ∆S; formula 17). The decision threshold is t = 818. The right key 2: Estimate p1, pn, p2 using ND (Section III-D, Algo- (d1 = 0) should survive with a 1 − βr = 99.5% probability at rithm 2); least. Wrong keys (d1 > 3) should survive with a βw = 0.3% 3: Calculate the data complexity N and the decision thresh- probability at most. The number of surviving keys should not old t (Section III-C); exceed 137 + (216 − 137) × 0.003 = 333.197. i i i i 4: Randomly generate N plaintext pairs (P0,P1),P0 ⊕P1 = After performing this attack 100 times, we find: ∆P, i ∈ {1, ··· ,N}; • The right key (d = 0) has survived in all the 100 5: Collect corresponding N ciphertext pairs, (Ci ,Ci ), i ∈ 1 0 1 experiments. {1, ··· ,N}; • The average number of surviving keys is 124.21 that is 6: for each key guess kg do smaller than 333.197. 7: Perform the statistical test (Algorithm 3); • The number of surviving keys is smaller than 333.197 8: end for in 97 experiments. 9: Test surviving key candidates against a necessary number of plaintext-ciphertext pairs according to the unicity distance for the attacked cipher. Based on the Hamming distance between the right key and key guess, the whole subkey space is divided into 17 subspaces. We further calculate the average ratio that keys in B. Verification of the Key Recovery Attack Model each subspace survive the attack. Table V shows the average surviving ratios of 17 key subspaces. In order to verify NASA, four practical attacks on round reduced Speck32/64 are performed. For Speck32/64 reduced 2) Attack 2: recover sk10 of 10-round Speck32/64: In the 12.34 to h rounds, our target is to recover the last subkey skh. It’s second attack setting, N = 5274 ≈ 2 and t = 1325. The expected that returned key guesses are different from the right number of surviving keys should not exceed 137 + (216 − key at most d = 2 bits. 137) × 2−16 ≈ 137.998. NASA should work as long as the adopted ND has a After performing this attack 100 times, we find: distinguishing accuracy higher than 0.5. Besides, the data • d = 0 p0 The right key ( 1 ) has survived in 99 experiments. complexity should be correctly estimated once ∆P −→ ∆S, • The average number of surviving keys is 63.54 that is ND, d, βr, and βw are provided. Thus, different settings about smaller than 137.998. these factors are considered. • The number of surviving keys is smaller than 137.998 Three distinguishers ND5, ND7, ND8 provided by in 98 experiments. Gohr [9] are adopted. Table III shows two different differentials of Speck32/64 adopted in the verification. Since Table VI shows the average surviving ratios of 17 subspaces. no key addition happens in Speck before the first nonlinear operation, these two differentials can be extended to a 2/3- 3) Attack 3: recover sk10 of 10-round Speck32/64: ND8 is round differential respectively. a very weak distinguisher. Its distinguishing accuracy is only The verification plan consists of three steps: about 0.518. In the third attack setting, N = 25680 ≈ 214.65 1) Set the value of βr, βw. Calculate the data complexity N. and t = 13064. The number of surviving keys should not 2) Perform NASA 100 times with N samples. exceed 137 + (216 − 137) × 2−16 ≈ 137.998. 7

TABLE IV SETTINGSOFTHEFOURATTACKSAGAINSTROUNDREDUCED SPECK32/64. DID IS THE DIFFERENTIAL TRANSITION’S ID IN TABLE 4.

Attack ID Attack rounds ND DID p0 c2 p1 d p2 pn βr βw −2 1 10 ND7 1 2 0.55 0.4183 2 0.2607 0.2162 0.005 0.003 −2 −16 2 10 ND7 1 2 0.55 0.4183 2 0.2607 0.2162 0.005 2 −16 3 10 ND8 - 1 0.5 0.5184 2 0.4957 0.4914 0.001 2 −6 −16 4 9 ND5 2 2 0.55 0.8889 2 0.2168 0.0384 0.005 2

TABLE V TABLE VIII AVERAGE SURVIVING RATIOS (SR) OF KEY GUESSES IN ATTACK 1. AVERAGE SURVIVING RATIOS (SR) OF KEY GUESSES IN ATTACK 4. d 0 1 2 3 4 5 ∼ 16 d1 0 1 2 3 4 5 ∼ 16 1 SR 1 0.2781 0.0562 0.0082 0.0007 0.0001 SR 1 0.4706 0.1786 0.0591 0.017 6 0.0041 6

TABLE VI AVERAGE SURVIVING RATIOS (SR) OF KEY GUESSES IN ATTACK 2. distance metric for the estimation of p2. The correctness of NASA is also well veriﬁed. d1 0 1 2 3 4 5 ∼ 16 SR 0.99 0.3788 0.1245 0.0343 0.008 6 0.0014 V. REDUCETHE KEY SPACE So far we still need to guess all the bits of the subkey After performing this attack 100 times, we ﬁnd: simultaneously, since ND takes the complete ciphertext pairs (C0,C1) as input. When the subkey has a large size, this is a • The right key (d1 = 0) has survived in all the 100 serious bottleneck. experiments. • The average number of surviving keys is 77.47 that is smaller than 137.998. A. An Intuitive Method for Reducing the Key Space • The number of surviving keys is smaller than 137.998 An intuitive method for reducing the key space is building in 85 experiments. ND on partial ciphertext bits

Table VII shows the average surviving ratios of 17 sub- Ci = Ci[L − 1]|| · · · ||Ci[0], i ∈ [0, 1] (24) spaces. Γ = {x1, x2, ··· , xk}, x1 > ··· > xk, k <= L (25)

TABLE VII ϕ(Ci,Γ ) = Ci[x1]||Ci[x2]||Ci[xk], i ∈ [0, 1] (26) AVERAGE SURVIVING RATIOS (SR) OF KEY GUESSES IN ATTACK 3. d1 0 1 2 3 4 5 ∼ 16 1, if S0 ⊕ S1 = ∆S Y (ϕ(C0,Γ ), ϕ(C1,Γ )) = (27) SR 1 0.4444 0.1459 0.0406 0.0095 6 0.0019 0, if S0 ⊕ S1 6= ∆S

P r(Y = 1 |(ϕ(C0,Γ ), ϕ(C1,Γ ))) = ND(ϕ(C0,Γ ), ϕ(C1,Γ )) (28) 4) Attack 4: recover sk9 of 9-round Speck32/64: In the where C [0] is the least significant bit of the ciphertext C , Γ 13.957 i i fourth attack setting, N = 15905 ≈ 2 and t = 758. The is the subscript set of selected ciphertext bits. 16 number of surviving keys should not exceed 137 + (2 − Such a method significantly reduces the key space to be −16 137) × 2 ≈ 137.998. searched. But which ciphertext bits should we select for After performing this attack 100 times, we find: building ND? Can we develop a generic and efficient • The right key (d1 = 0) has survived in all the 100 framework for guiding this selection? In order to better experiments. introduce our work for solving these problems, three new • The average number of surviving keys is 18.41 that is concepts are proposed first. smaller than 137.998. Definition 1: An informative bit is the ciphertext bit that is • The number of surviving keys is smaller than 137.998 in helpful to distinguish between the cipher and a pseudo-random 100 experiments. permutation. Table VIII shows the average surviving ratios of 17 subspaces. Definition 2: For a cipher reduced to h rounds, the neural It’s clear that these four attacks have achieved the most distinguisher trained on the complete ciphertexts (C0,C1) is t important two targets of NASA. Although the ratio that keys denoted as the teacher distinguisher NDh, the neural distin- of each subset survive is not strictly consistent with the guisher trained on partial ciphertext bits (ϕ(C0,Γ ), ϕ(C1,Γ )) s theoretical value, the total number of surviving keys is within is denoted as the student distinguisher NDh. The teacher the upper bound. This shows the Hamming distance is a good distinguisher is viewed as a special student distinguisher. 8

B. Identify Informative Bits by Bit Sensitivity Test Algorithm 5 Bit Sensitivity Test Require: a cipher with a block size of m; It’s clear that student distinguishers should be built on an NDt against this cipher; informative bits. However, it’s hard to identify informative bits a test dataset consisting of M positive samples and M according to Definition 1. Thus we propose an approximate 2 2 negative samples. definition of the informative bit. Ensure: An array sen that saves the bit sensitivity of m Definition 3: For an NDt, if the distinguishing accuracy is ciphertext bits. t greatly affected by the j-th bit of C0 or C1, the j-th ciphertext 1: Test the distinguishing accuracy of ND on the test bit is an informative bit. dataset. Save it to sen[m]; 2: for j = 0 to m − 1 do An NDt works since it has learned knowledge from ci- 3: for i = 1 to M do phertext bits. According to Definition 1, only informative 4: Generate a random mask η ∈ {0, 1}; bits provide knowledge. Thus the ciphertext bit that has a i,new i t 5: C0 = C0 ⊕ (η j); significant influence on the distinguishing accuracy of ND i,new t 6: Feed the new sample (C ,Ci ) to ND ; must be an informative bit. 0 1 7: end for Definition 3 can not ensure each informative bit that obeys 8: Count the current accuracy cp; the Definition 1 is identified successfully. But we only care 9: sen[j] = sen[m] − cp; about informative bits that are captured by an NDt. This 10: end for approximate definition helps develop a simple but effective framework for identifying informative bits. 0.22

The proposed framework is named Bit Sensitivity Test 0.2 (BST). Its core idea is to test whether the distinguishing 0.18 accuracy of an NDt drops after we remove some knowledge 0.16 0.14 related to the speciﬁc bit. 0.12 t Gohr in [9] has proved that NDh, h ∈ {5, 6, 7, 8} against 0.1 Speck32/64 captures the knowledge about the ciphertext differ- 0.08 ence and some unknown features. Consider the j-th ciphertext 0.06 0.04 bit. We remove the knowledge about the j-th ciphertext bit 0.02

0 difference by 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 sen-0 sen-1 sen-01 C = C ⊕ (η j) or C = C ⊕ (η j) (29) 0 0 1 1 t 6 Fig. 2. Results of BST of ND5 against Speck32/64, M = 10 . sen0 is the η 0 1 results of performing C0 ⊕ (η j), sen1 is the results of performing C1 ⊕ where is a random mask that could be or . (η j), sen is the results of performing two operations simultaneously, t 0,1 We have performed an extreme test on NDh, h ∈ j ∈ {0, ··· , 31}. {5, 6, 7, 8} against Speck32/64. If we XOR each bit of C0 t or C1 with a random mask, ND , h ∈ {5, 6, 7, 8} can not h 0.16 distinguish positive samples and negative samples anymore. 0.14 These tests imply that knowledge about unknown features is also removed by one of the two operations presented in 0.12 formula 29. 0.1 After the knowledge related to a ciphertext bit is removed, 0.08 the accuracy decrease of NDt is named Bit Sensitivity, which 0.06 is used to identify informative bits. Algorithm 5 summarizes 0.04 the BST. 0.02

0 Examples and analysis. We have applied the BST to 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 t sen-0 sen-1 sen-01 NDh, h ∈ {5, 6, 7} against Speck32/64. The results of the BST under three scenarios are shown in Fig. 2, Fig. 3, and Fig. 3. Results of BST of NDt against Speck32/64, M = 106. Fig. 4 respectively. 6 We observe that sen0 ≈ sen1. This proves that C0 ⊕ (η j) is equivalent to C1 ⊕ (η j). Besides, we know Reverse verification about identified informative bits. • If sen0[j] > 0, the j-th ciphertext bit is an informative To further verify Definition 3, a reverse verification about bit. identified informative bits is performed. First, select some • If sen0,1[j] > 0, the j-th ciphertext bit provides some informative bits. Second, train an NDs on selected informative useful unknown features. Since the knowledge about the bits and observe the distinguishing accuracy. bit difference is not removed, then only useful unknown t Taking ND against Speck32/64 as an example, we have features can lead to a decrease in the accuracy. 7 performed the reverse verification based on results in Fig. 4. • If sen [j] ≈ sen [j], the j-th ciphertext bit difference 0 0,1 Table IX shows the distinguishing accuracies under two set- has little influence on NDt . h tings. For Speck32/64, the j-th and (j +16)-th bit are directly 9

0.07 Then there are m + 1 resulting distributions of the statistic

0.06 (formula 6)

0.05 Td1 ∼ N (µd1 , σd1 ), (30)

0.04 µd1 = N × (p0 × p2|d1 + (1 − p0)pn), 0.03 q σ = N × p × p 1 − p + N (1 − p ) p (1 − p ). 0.02 d1 0 2|d1 2|d1 0 n n

0.01 The parameters of these m + 1 distributions are obtained

0 off-line. These distributions are used as prior knowledge to 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 sen-0 sen-1 sen-01 improve the key search strategy of NASA.

Fig. 4. Results of BST of NDt against Speck32/64, M = 106. 7 B. Bayesian Key Search Strategy Algorithm 6 summarizes the newly proposed Bayesian key search algorithm, which is the second technique for reducing related to the same subkey bit. Thus the 8-th and 1st ciphertext the attack complexities of NASA. bits are also considered. Algorithm 6 Bayesian Key Search Algorithm TABLE IX i i ACCURACIESOFNEURALDISTINGUISHERSTRAINEDONSELECTED Require: Ciphertext pairs, (C0,C1), i ∈ {1, ··· ,N}; CIPHERTEXT BITS A neural distinguisher, ND; Γ Accuracy Prior knowledge µd1 and σd1 , d1 ∈ {0, ··· , m}; {30 ∼ 23, 14 ∼ 7} 0.5414 The number of key guess candidates to be generated {30 ∼ 23, 21 ∼ 17, 14 ∼ 7, 5 ∼ 1} 0.6065 {31 ∼ 0} 0.6067 within each iteration, ncand; t The number of iterations, niter. The accuracy of ND7 is 0.6067. When all the identified s Ensure: The list L of tuples of recommended keys and informative bits are considered, the resulted ND7 obtains a corresponding statistics. distinguishing accuracy of 0.6065, which is almost the same as 1: K = {kg1, ··· , kgcand} ← choose ncand values at 0.6067. Such an experiment shows that Definition 3 can help random without replacement from the set of all subkey identify all the ciphertext bits that have a significant influence candidates. on teacher distinguishers. 2: L ← {} Once informative bits are identified by the Bit Sensitiv- 3: for t = 1 to niter do ity Test, the whole key space can be divided into several 4: for each kg ∈ K do subspaces. In each subspace, NASA is performed to recover 5: for i = 1 to N do specific key bits. i i 6: Decrypt C0,C1 with kg. This informative-bit-based method is the first generic tech- 7: Feed partially decrypted ciphertext pair into ND. nique for reducing the attack complexities of NASA. 8: Collect the output Zi,kg of ND. 9: end for VI.SELECTIVE KEY SEARCH 10: Compute the statistic T kg (formula 6). kg In the basic NASA as shown in Algorithm 4, each possible 11: L ← L||(kg, T ). 12: end for key guess kg is tested. Inspired by the analysis of p2 in m Section III-E, we develop a highly selective key search strategy 13: for sk ∈ {0, ··· , 2 − 1} do P kg 2 2 for further reducing the attack complexity. 14: λsk = kg∈K T − µhd(kg⊕sk) /σhd(kg⊕sk). Specifically, we do not need to traverse all the key guesses. /* hd(kg ⊕ sk) is the Hamming distance between kg Some key guesses that are most likely to be the right key are and sk */ recommended based on the keys that have been tested. 15: end for 16: K ← argsortsk(λ)[0 : ncand − 1]. /* Pick ncand key guesses with the ncand smallest score A. Distribution of the Statistic Under Different Keys to form the new set of key guess candidates K */ In this section, we first discuss the distribution of the statistic 17: end for (formula 6) under different keys again. 18: Return L Here we still take Fig. 1 as the example. Suppose that the size of the key guess kg is m. According to the analysis in Section III-E, we know there are the following m + 1 VII.REDUCETHE DATA COMPLEXITY p probabilities Consider the prepended differential ∆P −→0 ∆S. As we have presented in Section III-C, the impact of p0 on the data p2|d1 = P r{z > c2|S0 ⊕ S1 = ∆S}, 0 6 d1 6 m, −2 complexity is about O(p0 ). where d1 is the Hamming distance between kg and the right Neutral bits seldom exist in a long differential character- key. istic. But there usually are numerous neutral bits in a short 10 differential characteristic. This section shows how to reduce B. Further Improvement Based On Early Stopping the data complexity of NASA by neutral bits. 1 In Algorithm 7, q plaintext structures are generated. But only one plaintext structure P is expected to satisfy the differential ∆P → ∆B. A. Improved Neural Aided Statistical Attack This plaintext structure is called valid plaintext structure We still take the key recovery attack with 1-round decryp- while other plaintext structures are called invalid plaintext tion as an example. Fig. 5 shows the scheme of the improved structures. If a valid plaintext structure is identified once it NASA. arises, Algorithm 7 can be early stopped at step 4. We propose an identification method that does not change the process of Algorithm 7. At a high level, the identification method is as follows: 1) Generate a plaintext structure P consisting of M plaintext pairs. 2) Filter key guesses based on the statistic T (formula 6) and a decision threshold tM . 3) If the number of surviving key guesses exceeds a threshold tP , P is a valid plaintext structure. By setting proper parameters tM , the number of surviving key guesses exceeds tP only when P is a valid plaintext structure. Next, we present an theoretical analysis of M, tM , tP . For Fig. 5. The scheme of the improved neural aided statistical attack. The convenience, we rewrite the statistic T (formula 6) as statistical distinguisher only covers the second differential ∆B → ∆S. M Neutral bits that exist in ∆P → ∆B are exploited. X 1, if Zi > c2 T = φ (Zi), φ (Zi) = . (31) 0, if Zi c2 i=1 6 Its core idea is dividing the initial long differential into two The four situations shown in Fig. 1 also exist in the improved q p short ones: ∆P −→ ∆B, and ∆B −→ ∆S where p0 = q × p. NASA (see Fig. 5). The following notations introduced in The statistical distinguisher only covers the second differential Section III are adopted again: ∆B → ∆S. Neutral bits that exist in the first part ∆P → • p2|d1 : the probability P r(Z > c2|S0 ⊕ S1 = ∆S) when ∆B are exploited. Algorithm 7 summarizes the details of the the Hamming distance between the key guess and the improved NASA. right key is d1 ∈ [0, m]. • pn : the probability P r(Z > c2|S0 ⊕ S1 6= ∆S). Algorithm 7 Improved neural aided statistical attack Require: The attacked cipher; 1) Distribution of the statistic under valid plaintext struc- q p The prepended differential, ∆P −→ ∆B −→ ∆S; tures: When P is a valid plaintext structure that satisfies Neutral bits that exist in ∆P → ∆B; ∆P → ∆B, there are M ×p positive samples and M ×(1−p) Two maximum error probabilities, βr, βw; negative samples.

A posterior probability threshold, c2. When d1 is not set clearly, we denote p2|d1 as pV . The Ensure: All possible key candidates. distribution of the statistic(formula 31) is 1: Train an ND over ∆S; TV ∼ N (µV , σV ), (32) 2: Estimate p1, pn, p2 using ND (Section III-D, Algo- rithm 2); µV = M[p × pV + (1 − p)pn], (33) 3: Calculate the data complexity N1 based on p, p1, pn, p2 p σV = M × p × pV (1 − pV ) + M(1 − p)pn(1 − pn). (Section III-C); 1 (34) 4: for j from 1 to q do Select a speciﬁc d1, we have pV = p2|d1 . Let KV denote 5: Based on ∆P and neutral bits, randomly generate a the set of key guesses with a Hamming distance d1 from the plaintext structure P consisting of N1 plaintext pairs. right key. Then only kg ∈ KV makes the above TV hold. 6: Perform the basic NASA based on P (Algorithm 4). 7: end for 2) Distribution of the statistic under invalid plaintext struc- 8: Test surviving key candidates against a necessary num- tures: When P is an invalid plaintext structure, all the M ber of plaintext-ciphertext pairs according to the unicity samples are negative samples. distance for the attacked cipher. The distribution of the statistic(formula 31) is

TI ∼ N (µI , σI ) (35) Now, only the impact of p on the total data complexity p is O(p−2). The impact of q on the total data complexity is µI = M × pn, σI = M × pn(1 − pn) (36) O(q−1). Thus, the total impact of the prepended differential Let K denote the set of all possible key guesses. Then any −1 −2 −2 −2 −2 is O(q p ) instead of O(p0 ) = O(q p ). The data kg ∈ K makes the above TI hold. complexity is reduced by a factor of about q−1. 11

3) Distinguishing between TV and TI : Since TV and Algorithm 8 Identify valid plaintext structures TI are two different normal distributions, the technique in Require: a plaintext structure P with a size of Section II-D is used to distinguish these two distributions. M(formula 40); According to Proposition 1, the condition for distinguishing an ND trained over ∆S; TV and TI is the posterior probability threshold c2; z × σ + z × σ a decision threshold tM for ﬁltering subkey guesses; 1−βV V 1−βI I = 1 (37) a decision threshold tP for identifying valid plaintext µV − µI structures. where Ensure: the classiﬁcation of P. βV = P r {N (µI , σI ) |s ∈ N (µV , σV )} , (38) 1: Collect the M ciphertext pairs corresponding to P; βI = P r {N (µV , σV ) |s ∈ N (µI , σI )} , 2: Initialize a counter cp ← 0; and s stands for a sample. 3: for each possible subkey guess kg do Here, we present an explanation about the two error prob- 4: Decrypt M ciphertext pairs with kg; 5: Feed partially decrypted ciphertext pairs into ND; abilities βI , βV . When P is an invalid plaintext structure, the maximum probability that key guesses kg ∈ K survive 6: Save the outputs of ND, Zi, i ∈ [1,M]; 7: Count the number of Zi > c, and denote it as TM ; the attack is βI . When P is a valid plaintext structure, the 8: if TM > tM then minimum probability that key guesses kg ∈ KV survive the 9: cp ← cp + 1; attack is 1 − βV . By simplifying formula 37, the data complexity M is 10: end if 11: end for a = p (1 − p ), a = p (1 − p ), V V V n n n (39) 12: if cp > tP then √ p √ 13: Return 1 (P is a valid plaintext structure). z1−βV × p × aV + (1 − p)an + z1−βI × an M = . 14: else (pV − pn) × p (40) 15: Return 0 (P is an invalid plaintext structure). 16: end if And the decision threshold tM is

tM = µV − z1−βV σV = µI + z1−βI σI , (41) If pV increases, the numerator will decrease and the denomi- where z1−βV and z1−βI are the quantiles of the standard normal distribution. nator will increase. Then M will decrease.

When the Hamming distance d1 increases, p2|d1 will de- 4) Identify valid plaintext structures by counting surviving crease in high probability. But the number of subkey guesses keys: When P is a valid plaintext structure, The lower bound in the subspace may increase sharply when d1 increases. As of the number of surviving subkeys is |KV | × (1 − βV ). When long as the condition (formula 42) holds, we advise pV = p2|d1 P is an invalid plaintext structure, The upper bound of the where d1 should be as small as possible. number of surviving subkeys is |K| × βI . By setting two proper error probabilities βV , βI , we ensure VIII.APPLICATION TO DES the following condition always holds This section proves that NASA surpasses Gohr’s attack |KR| × (1 − βV ) >> |K| × βI . (42) when enough neutral bits do not exist in the prepended long differential. Let t satisfy the following condition S DES [13] is a block cipher with a block size of 64 bits. The |KR| × (1 − βV ) > tS >> |K| × βI . (43) structure of DES is the classical Feistel structure as shown in f Then valid plaintext structures is identified by by comparing Fig. 6. Its round function is composed of eight different S-boxs. More details refer to [13], please. We perform key the number of surviving subkey guesses with tS. Algorithm 8 summarizes the concrete identification process. recovery attacks on round reduced DES. Since the identification is also based on the same statistic as the key recovery, Algorithm 8 and Algorithm 7 are able to be performed simultaneously. The necessary condition is that the size of a plaintext structure P should exceed N1 and M.

5) Further analysis about pV .: The data complexity M is related to pV . And pV is related to the Hamming distance d1. When pV increases, M (Equation 40) decreases since √ √ z × pp × a + (1 − p)a + z × a M = 1−βV V n 1−βI n (pV − pn) × p q √ (1−p)an z1−βI × an z1−β × p(1 − pV ) + + Fig. 6. A round of encryption of DES. V pV pV = √ pn . ( pV − √ ) × p pV 12

A. Prepended Differentials E. Time Complexity Two optimal 2-round iterative differentials found in [14] are Since the output of an S-box only contains 4 bits, P r= 1 we build a look-up table off-line for saving the tuple 0x19600000/0 −−−−−→234 0x19600000/0, s ((C0,C1), ND (C0,C1)). Then the time complexity is not 1 (44) 5 P r= 234 s 0x1B600000/0 −−−−−→ 0x1B600000/0. related to ND5 anymore. In other words, the time complexity of the basic NASA is N × 26 = 246.814. Based on these 2-round differentials, we can get the iterative By building NDs over outputs of different S-box and differential of any even-numbered round. These iterative dif- 5 guessing partial key bits, the complete master key is recovered. ferentials are used as the prepended differential ∆P → ∆S for attacking round reduced DES. According to the definition of the neutral bit, We measure IX.APPLICATION TO SPECK32/64 the neutrality of each ciphertext bit experimentally. We find When there are enough neutral bits in the prepended differ- p0 that 18 neutral bits {33, ··· , 50} exist in the above 2-round ential ∆P −→ ∆S, the attack complexities of Gohr’s attack iterative differentials. As for 4-round iterative differentials, no are much lower than the differential attack. For Gohr’s attack, −1 neutral bits exist anymore. the impact of p0 on the data complexity is only O(p0 ). For basic NASA, the impact of p0 on the data complexity B. Build Neural Distinguishers Against DES −2 is O(p0 ). Let ∆S = 0x19600000/0, we build teacher distinguishers In this section, we prove that NASA can achieve comparable against DES up to 5 rounds. The distinguishing accuracy of performance as Gohr’s attack when the three optimisation the 5-round teacher distinguisher is 0.58. techniques introduced in Section V, VI, VII are available. The Bit Sensitivity Test identifies 12 informative bits Speck32/64 is one variant of the Speck family that is {39, 50, 56, 61, 59, 37, 43, 49, 42, 32, 53, 52}. The 4 bits designed by the NSA Research Directorate [15]. Fig. 7 shows {39, 50, 56, 61} are related to the fifth S-box S5, and the round function of Speck. {59, 37, 43, 49} are related to the eighth S-box S8. In order to introduce the next experiment more clearly, s we focus on the student distinguisher ND5 built over the bits {39, 50, 56, 61} for now. Let the posterior probability threshold be c2 = 0.5, we get p1 = 0.6032 and pn = 0.4883.

Table X shows the estimation of p2|d1 .

TABLE X s THE ESTIMATION OF p2|d1 OF ND5 AGAINSTROUNDREDUCED DES. s c2 = 0.5. ND5 IS BUILT OVER 4 BITS {39, 50, 56, 61}.

d1 0 1 2 3 4 5 6

p2|d1 0.6034 0.505 0.507 0.509 0.5101 0.504 0.496

Fig. 7. The round function of Speck32/64. SKi+1 : the round key. C. Gohr’s Attack on DES Li+1||Ri+1 : the state of the (i + 1)-th round. For Speck32/64, α = 7, β = −1 By placing the 2-round differential 0x19600000/0 −−−−→234 2. s 0x19600000/0 before the student distinguisher ND5, with the help of the 18 neutral bits {33, ··· , 50}, 8-round DES is broken by Gohr’s attack. The 6 key bits related to S5 of A. Prepended Differential the last round are recovered. To attack 11-round Speck32/64, Gohr adopted the following Since no neutral bits exist in the 4-round iterative differen- −2 2-round prepended differential tial 0x19600000/0 −−−−→234 0x19600000/0, 10-round DES can p =2−6 not be attacked by Gohr’s attack. ∆P = (0x211, 0x204) −−−−−→0 ∆S = (0x40, 0x0).

D. Neural Aided Statistical Attack on DES There are only 3 neutral bits {20, 21, 22} that exist in ∆P → ∆S. Besides, there are 2 high probabilistic neutral Consider the basic NASA (Algorithm 4). The 4-round 234−2 bits {14, 15} whose neutrality exceeds 0.95. differential 0x19600000/0 −−−−→ 0x19600000/0 is the Dividing this differential into two 1-round differentials prepended differential. The target is also recovering the 6 key s q=2−4 bits related to S5 of the last round by ND5. ∆P = (0x211, 0x204) −−−−→ ∆B = (0x2800, 0x10) −6 Let p = 0.5101, β = 0.005, and β = 2 . Since p = p=2−2 2 r w 0 ∆B = (0x2800, 0x10) −−−−→ ∆S = (0x40, 0), 234−2, the required data complexity is N = 240.814 chosen- plaintext pairs. Thus, 10-round DES is broken. we measure the neutrality of each ciphertext bit. There are 8 If we adopt the improved NASA, the data complexity is neutral bits {0, 11, 14, 15, 20, 21, 22, 26, } that exist in ∆P → smaller. Thus, when there are no enough neutral bits, NASA ∆B. Moreover, there are 8 high probabilistic neutral bits surpasses Gohr’s attack. {1, 3, 4, 5, 23, 24, 27, 28} whose neutrality exceeds 0.95. 13

t B. Build Neural Distinguishers Against Speck32/64 • stage 3: Recover sk11 by ND7 (Algorithm 4). s t t • stage 4: Recover (sk11, sk10[7 ∼ 0]) by ND (Algo- Two teacher distinguishers ND6, ND7 built by Gohr over 6 ∆S = (0x40, 0) are adopted in this section. Table I shows the rithm 4). t t t • stage 5: Recover (sk11, sk10) by ND6 (Algorithm 4). estimation of p2|d1 of ND6, ND7 when c2 = 0.55. NDt , NDt −8 Based on the bit sensitivity test results of 6 7 as In stage 1, we set pV = p2|d1=1, βV = 0.1, βI = 2 . It shown in Fig. 3 and Fig. 4 (see Section V), we build 2 student means that the number of surviving subkey guess kg11[7 ∼ 0] s s distinguishers ND6, ND7 by setting Γ = {30 ∼ 23, 14 ∼ 7}. should exceed 8(1 − 0.1) = 7.2 when P is a valid plaintext These 16 ciphertext bits are related to the least significant 8 structure. Or the number of surviving subkey should not s s 8 −8 subkey bits. Later, ND6, ND7 are used to recover sk10[7 ∼ 0] exceed 2 × 2 = 1. Based on this setting, each plaintext 15.235 and sk11[7 ∼ 0] respectively. Let c2 = 0.55, Table XI shows structure P should contain M = 2 plaintext pairs. The s s the estimation of p2|d1 of ND6, ND7. decision threshold for filtering subkey guess is tM = 11285. When the number of surviving subkey guess is greater than TABLE XI or equal to tP = 8, P is a valid plaintext structure. s s −8 THE ESTIMATION OF p2|d1 OF ND6, ND7 AGAINST SPECK32/64. In stage 2, we set p2 = p2|d1=3, βr = 0.005, βw = 2 . c2 = 0.55.THESUBSCRIPTSETOFSELECTEDCIPHERTEXTBITSIS Γ = {30 ∼ 23, 14 ∼ 7} Then the number of surviving subkey guess kg11[7 ∼ 0] should not exceed 37. The required data complexity is N = d 0 1 2 3 ∼ 8 NDs 1 214.465 plaintext pairs, and the decision threshold for filtering 6 p 0.5132 0.4057 0.3402 0.3025 2|d1 6 subkey guess is t = 6703. s d1 0 1 2 3 ∼ 8 ND7 kg kg [7 ∼ p2|d 0.3576 0.3230 0.3036 6 0.2940 In stage 3, we filter 11 based on each surviving 11 1 −16 0]. Let p2 = p2|d1=2, βr = 0.005, βw = 2 , the number of surviving subkey guess kg11 should not exceed 137. Besides, Let c2 = 0.55, Table XII summarizes the estimation of 12.373 s s t t we have N = 2 , t = 1333. p1, pn of ND6, ND7, ND6, ND7 against Speck32/64. In stage 4, we filter (kg10[7 ∼ 0], kg11) based on each surviving kg . Let p = p , β = 0.001, β = 2−14, the TABLE XII 11 2 2|d1=1 r w s s t t number of surviving subkey guess kg should not exceed 16. THE ESTIMATION OF p1, pn OF ND6, ND7, ND6, ND7 AGAINST 11 12.36 SPECK32/64. c2 = 0.55. Besides, we have N = 2 , t = 1598. s s t t ND6 ND7 ND6 ND7 In stage 5, we filter (kg10, kg11) based on each surviving p 0.5129 0.3575 0.6787 0.4183 1 (kg10[7 ∼ 0], kg11). Let p2 = p2|d1=1, βr = 0.001, βw = p −16 n 0.2604 0.2865 0.1162 0.2162 2 , the number of surviving subkey guess kg11 should not exceed 16. Besides, we have N = 29.694, t = 180. We use 16 high probabilistic neutral bits {0, 1, 3 ∼ 5, 11, 14, 15, 20 ∼ 24, 26 ∼ 28} that exist in ∆P → ∆B C. Gohr’s Attack on Speck32/64 to generate plaintext structures P consisting of M = 215.235 Based on the 2-round prepended differential ∆P → ∆S t t plaintext pairs with a difference ∆P = (0x211, 0xa04). The and ND7, ND6, Gohr presented a key recovery attack on 11- probability that P is a valid plaintext structure is about 2−4.2. round Speck32/64. By adopting some optimization techniques In stage 1, if no valid plaintext structures occur after 32 for accelerating the basic key (Section II-C), Gohr broken 11- plaintext structures are generated, the attack is stopped and 38 round Speck32/64 with a time complexity of 2 . viewed as a failure. The target is to recover the last two subkeys sk10, sk11. Once one valid plaintext structure P is found, the remaining Gohr counted a key guess as successful if the last subkey was 4 stages are performed based on this structure P. More exactly, guessed correctly and if the second subkey was at hamming 214.465, 212.373, 212.36, 29.694 plaintext pairs are randomly se- distance at most two of the real key sk10. Finally, the success lected from this valid plaintext structure respectively. rate of Gohr’s attack is about 52%. In stage 1 ∼ 5, we do not traverse each possible subkey We have performed this accelerated attack again based on guess. The proposed Bayesian key search strategy (Algo- the code provided by Gohr. By adopting an Intel(R) Core(TM) rithm 6) is applied in each stage, which does not influence i5-7500 CPU and one graphics card (NVIDIA GeForce GTX the above attack setting. 1060(6GB)), we find that the average time consumption of The settings related to the Bayesian key search strategy are performing this attack one time is 45 seconds. as follows. In stage 1 and stage 2, the number of iterations is niter = 3. For each iteration, we search ncand = 32 subkey D. Neural Aided Statistical Attack on Speck32/64 guesses. In stage 3, we set niter = 4, ncand = 32. In stage 4 We also perform the key recovery attack on 11-round and stage 5, we set niter = 32, ncand = 32. Speck32/64 by adopting NASA. At a high level, the attack We count a key guess as successful if the right subkey pair is composed of five stages: (sk10, sk11) survives and if the average number of surviving • stage 1: Identify the valid plaintext structure P that key guesses in stage 5 does not exceed 16. Under these s satisfies ∆P → ∆B by ND7 (Algorithm 8). The subkey conditions, the attack was successful 585 out of 1000 trials. to be searched is sk11[7 ∼ 0]. The average number of generated plaintext structures is 18. s • stage 2: Recover sk11[7 ∼ 0] by ND7 (Algorithm 4). For comparison, under the same hardware environment, the 14 average time consumption of this attack one time is 87 A. The Differential’s Impact on the Data Complexity seconds. According to formula 17, the data complexity N is affected by the probability p0 of the differential as p p E. Time Complexity √ z1−β p0a1 + (1 − p0)a3 + z1−β p0a2 + (1 − p0)a3 N = r w (p − p ) × p Once the valid plaintext structure is found, the key recovery 1 2 0 z pa + (a − a )p + z pa + (a − a )p process is very fast. Compared with the time consumption of ∝ 1−βr 3 1 3 0 1−βw 3 2 3 0 recovering (sk10, sk11), the time consumption of recovering p0 the remaining subkeys is negligible. pa + (a − a )p + a × pa + (a − a )p ∝ 3 1 3 0 4 3 2 3 0 Under the same hardware environment, the average time p0 consumption of NASA is about twice as long as that of Gohr’s z1−βw 38 where a4 = . We further know attack. The time complexity of Gohr’s attack is 2 . Thus, the z1−βr 39 time complexity of NASA is 2 . −2 2 2 2 N ∝ p0 a3 + a4a3 + (a1 − a3 + a4a2 − a4a3)p0 + a5 p where a5 = 2 × a4 × a3 + (a1 − a3)p0 × X.CONCLUSION p a3 + (a2 − a3)p0. Thus the impact of the probability −2 In this article, we propose a Neural Aided Statistical Attack p0 of the differential is O(p0 ). (NASA) and three methods for reducing the complexity of NASA. NASA recovers the right key based on distinguishing B. The Neural Distinguisher’s Impact on the Data Complexity between two different normal distributions. NASA is the Three probabilities p1, p2, pn are related to the neural dis- first deep learning-based cryptanalysis technique that supports tinguisher. Since pn is related to negative samples and p1, p2 theoretical complexity estimation and does not rely on any are related to positive samples, we can discuss pn separately. special properties such as neutral bits. Applications to round 1) The impact of pn: In formula 17, only a3 is related to reduced DES and Speck32/64 also prove that NASA is full of pn. potential. √ z pp a + (1 − p )a + z pp a + (1 − p )a Our work in this article also provides many inspirations N = 1−βr 0 1 0 3 1−βw 0 2 0 3 for neural aided cryptanalysis. First, if we replace the neural (p1 − p2) × p0 p p network with other machine learning models, NASA still ∝ p0a1 + (1 − p0)a3 + a4 × p0a2 + (1 − p0)a3 works. Compared with neural networks, other shallow machine ⇒ N ∝ p × a + a2 × p × a + (1 − p )(1 + a2)a + a learning models such as logistic regression are far faster. 0 1 4 0 2 0 4 3 5 Thus, it is possible to further accelerate NASA by adopting Next, we focus on attack scenarios where the p0 is very other machine learning-based distinguishers. Second, when small. This can make the discussion easier and more concise. we try to reduce the key space, we find that ciphertext bits This simplification is also reasonable. Because when we attack have a different influence on the neural distinguisher. This a cipher for more rounds, the probability of the differential is finding is not only useful for neural aided cryptanalysis. The generally small. traditional differential attack may be improved by exploiting When p0 → 0, a5 → 2 × a4 × a3. Thus the impact of a3 knowledge extracted from neural distinguishers. Third, the on the data complexity is O(a3). Since pn < 1 always holds 2 data complexity for distinguishing two normal distributions is and a3 = pn − pn, the impact of pn on the data complexity very high, which makes the data complexity of basic NASA is also O(pn). is also high. If some new probability distributions are more 2) The impact of p1, p2: In formula 17, a1, a2 are related suitable for simulating the key recovery process, new neural to p1, p2 respectively. The impacts of a1, a2 are adjusted by aided attacks with a lower complexity are able to be developed. p0. Due to this property, we also focus on attack scenarios where p0 → 0. √ z pp a + (1 − p )a + z pp a + (1 − p )a ACKNOWLEDGMENT N = 1−βr 0 1 0 3 1−βw 0 2 0 3 (p1 − p2) × p0 This work is supported by the National Key Research z p(1 − p )a + z p(1 − p )a and Development Program of China (2018YFB0803405, ≈ 1−βr 0 3 1−βw 0 3 ∝ (p − p )−1 (p − p ) × p 1 2 2017YFA0303903). 1 2 0 Thus the impact of p1, p2 on the data complexity is O((p1 − −2 p2) ). APPENDIX

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Yi Chen received his B.E. degree and M.S. degree both from the Department of Electronics and Information Engineering, Huazhong University of Science and Technology(HUST), Wuhan, China, in 2016, 2018 respectively. He is currently pursuing the Ph.D. degree in the school of computer science and technology at Tsinghua University, Beijing, China. His current research interests include cryptanalysis, multimedia encryption and deep learning.

Yantian Shen is currently pursuing the B.E. degree in the department of computer science and technology at Tsinghua University, Beijing, China. His current research interests include cryptanalysis and deep learning.