Side-Channel Attack: timing attack

1. Side-Channel Attack: timing attack Hiroki Morimoto

2. Overview • Review of traditional attacks • Side-Channel Attacks • Timing Attack • Several ways to compromise RSA • Countermeasures • Conclusion • References

3. Review • Basic Attacks: • exploiting security holes and weakness in the systems/algorithms • choosing inadequate parameters • brute force • social engineering • more … • There is 3 types of cryptanalysis: • Ciphertext-only Attack • Known plaintext Attack • Chosen plaintext Attack

4. What is Side-Channel Attack • Side-Channel Attack don’t belong to the traditional attack • This attack is based on experiments and statistics not mathematical theories • Two types of the side-channel attack • Passive Attacks: • Observe the target such as computer or cell-phone • Gain the “additional” information leaked from the physical implementations/devices caused by any operation • i.e. timing information, power consumptions, electromagnetic leaks, voices/sounds • Active Attacks: • Add “additional” inputs • Change the environment or target itself to let abnormal operations or change the program flow • i.e. add voltage, clock gritching, or tempest virus

6. Goal • Finds information such as: • Algorithm/operation • Cryptographic key • Partial state information • Plaintext/cyphertext • more …

7. Advantages • The Attackers can implement: • With information easy to obtain • With available non-expensive hardware • From remote place • Often quicker than the regular attack • Compared to brute force and dictionaly attacks • From few seconds to few hours • Without damaging regular operations and physical devices • Without notifying the victims

8. Examples of Side-Channel Attacks • Timing Attack • Power Monitoring Attack • Fault Analysis • Magnetic Emanation Attack • Light Emission Attack • Sound Attack (Includes wire-tapping and eavesdropping)‏

9. Timing Attack • Timing attack is an example of an attack that exploits the implementation of an algoritm rather than the algorithm itself • Measure the time it takes for a certain unit to perform an operation • Keep the input, output, and consumed time • Check the correlation between time measurements of guess key or input and empirical result (often statistically)‏

10. Background • Operation takes slightly different amounts of time to process different input because of: • Bypass operations such as branching or conditional statements • RAM cache hit • Processor instruction such as multiplication and division • Others … • Usually consumed time depends on input data, crypt keys, and modulo in cryptosystems

11. Usages • Timing attack is often used to compromise public-key cryptosystem such as RSA • For example, most of smart-card uses RSA. Therefore, inappropriate usage of it revels its secret key easily • Sometimes, the key is tamper-proof • Timing attacks reveal key length, key values, plaintext, etc…

12. RSA review • Multiple prime RSA key generating algorithm 1. Select two primes: p and q 2. Calculate n = p * q 3. Calculate φ(n) = (p-1) * (q-1)‏ 4. Choose e where gcd(e,φ(n)) = 1 5. Calculate d = e-1(mod φ(n))6. Public Key = (e,n) and Private key = (d)‏ • Encryption: c = me mod nDecryption: m = cd mod n

13. Modular Exponentiation • The way of attacks depend on the details of modular exponentiation • For efficiency, modular exponentiation is done via: • Simple multiplication • Repeated squaring • Chinese Remainder Theorem (CRT)‏ • Montgomery multiplication • Sliding window • Karatsuba multiplication

14. Simple Multiplication • The simplest case, the modular exponentiation is done by multiplying the number as many as the values of exponent such as 2^13 = 2 * 2 * 2 * 2 * 2 * 2* ….. • Therefore, the execution time is direct proportional to the exponent value (key value)‏

15. Attacking Scenario: simple multiplication • An attacker eavesdrops the decryption operation where he gets a plaintext and its computation time (the decryption key is 13 which is hidden from the attacker)‏ • He guesses the key is 12. He decrypts with the guess key and it returns small computation time • Then, he guesses the key is 14 and retuned computation time is greater than empirical data • Now, he knows the key is between 12 and 14

16. Repeated Squaring • The most common and fast algorithm • The number of loops is proportional to its key bit length • Kotcher found a possible attack

17. Algorithm • In each step, the number is squared and mod by n • If the current bit is 1, then a modular multiplication is executed • If the current bit is 0, goto the next step

18. Pseudo-Code // Compute c = md (mod n)‏ // where, in binary, d = (d0,d1,d2,…,dnum) with d0 = 1 s = m for i = 1 to num s = s2 (mod n)‏ if di == 1 then s = s  m (mod n)‏ end if next i return s

19. Example • For example: • 520 = 95367431640625 = 25 mod 35 • With repeated squaring • d = 20 = 10100 base 2, m = 5, and n = 35 • Initialize s = 5^1 (d0 == 1)‏ • s = (5 * 5) mod 35 and d1 == 0  s = 25 • s = (25 * 25) mod 35 and d2 == 1 so that (30 * 5) mode 35  s = 10 • s = (10 * 10) mod 35 and d3 == 0  s = 30 • s = (30 * 30) mod 35 and d3 == 0  s = 25 • No huge numbers and it’s efficient • In this example, 5 steps vs 20 multiplications

20. Attacking Scenario: repeated squaring • This attack also measures the correlation between guessed and empirical time measurements • Because the 2nd consuming time depends on the 1st data (s) and second bit of the key, and so forth. In other word, the high-order bits affect a result more than the lower-bits. • Thus the attacker begins the top of the bit, then continues to next bit and so on • The more bits the attacker already knows, the stronger the signal, thus easier to detect (error-correction property)

21. Attacking Scenario: repeated squaring • First, the attacker wants to know the first bit of the secret key where he has a target plaintext and knows its consumed time • He decrypts the plaintext with 1111 • Next he decrypts the plaintext with 0111 • Then he creates two graphs for each pair of consumed times • Then he finds the strong correlation for 0111 especially at the last step. Thus the first bit may be 0. • He continues this procedure to the next bit and so on • He can efficiently recover low-order bits when enough high-order bits are known because of error correlation property

22. Chinese Reminder Theorem • Modular Reduction is done by subtracting multiples of the modules which also takes most of the computation time • Given m = cd (mod n) where n = pq • With CRT, first compute cd modulo p, and them cd modulo q. After that “glue” them together • Two modular reductions of size n1/2 • As opposed to one reduction of size n • CRT provides significant speedup by a factor of 4 • (comment) several researchers claim above two statements. However, I don’t think so !

23. Algorithm • To compute Cd (mod N) where N = pq • First pre-computes: • dp = d (mod (p  1)) • dq = d (mod (q  1))‏ • Second, pre-find a and b such that • a = 1 (mod p) and a = 0 (mod q) • b = 0 (mod p) and b = 1 (mod q)‏ • Now computes: • Solution is:

24. Example • Suppose N = 33, p = 11, q = 3 and d = 7 • Pre-compute • dp = 7 (mod 10) = 7 • dq = 7 (mod 2) = 1 • Pre-find, a = 12 and b = 22 • Suppose decrypt C = 5 • Cp = 5 (mod 11) = 5 and Cq = 5 (mod 3) = 2 • xp = 57 = 3 (mod 11), xq = 21 = 2 (mod 3)‏ • Solution: 57 = 312 + 222 = 14 (mod 33)‏ • Regular Operation: Cd = 57 (mod 33) = 14

25. Limitation: • Factors p and q of N must be known • Only for private key operations

26. Attacking Scenario: CRT • The attacker doesn’t have to know anything • As we mentioned before, the CRT operates first computes cd modulo p, and then cd modulo q • First guess cd and measure the consumed time for first (or second) operation. • If the p is smaller than cd, takes no time. • If the p is larger than cd, it must subtract p at least once • Then extract the p (or q)‏

27. Attacking Scenario: CRT • The attacker wants to know decryption key (d)‏ • First, he tries to extract the value p so that he runs the program with cd = 1, 3, 5, 7, 11 …. and measures the consumed times • The consumed times are constant from 1 to 5, but increase after 7. Thus,pmight be 7 • Then he does the same operation to find q • Now, he knows q and p • Thus, he can calculate n = p * q and φ(n) = (p-1) * (q-1)‏ • Because e is public so that d = e-1(mod φ(n))‏

28. Countermeasures • How To prevent or make difficult to do timing attack • Reduce or eliminate coherence between the execution time and parameters such as input data, modulo, and keys OR • Add noises because the number of samples needed to obtain enough information are proportional to the noises

29. Examples of Countermeasures • Constant Time Calculation • Random Time Calculation • RSA Blinding • Avoid Conditional Operation • Time Equalization of Multiplication and Exponentiation

30. Constant Time Calculation • In this strategy, the time it takes to do any operation must be independent from input and key (constant and equal at every time)‏ • Thus, every operation takes the slowest operational time by waiting • However, this strategy raises the execution time dramatically (corresponding to the worst case)‏

31. Random Time Calculation • In this strategy, the time it takes to do any operation changes every operation at each time • It is done by waiting a random time before going to the next execution • However, this strategy also raises the execution time and its random variance must be large and completely random

32. RSA Blinding • The idea is same as the random time calculation; time it takes to do any operation changes every operation at each time • However, randomized time is done via multiplying the random seed before the operation and multiplying the inverse of the seed after the operation. In other word, it changes m (plaintext) or c (ciphertext)‏ • This strategy adds slight execution time

33. Algorithm and Example • Algorithm: • Generate random r • First multiply re: m” = rec (mod N)‏ • Then decrypt: m’ = m” d (mod N)‏ • Finally, multiply by r1 (mod N): m = r1m’ = r1(rec)d = r1rcd = cd (mod N)‏ • Example: c = 3, r = 2, e = 3, d = 7, and N = 33 • m” = 23 * 3 (mod 33) = 24 • m’ = 247 (mod 33) = 18 • m = ½ * 18 = 9 • Regular Operation: m = 37 (mod 33) = 9

34. Avoid Branch and Conditional Operation • Conditional Statement often depends on input or key • As we mentioned before, branch and condition statements (i.e. if statement) changes the consuming time • So that eliminates any branch and conditional statement to equalize the computational time • Also the calculation must be performed via elementary operations (such as AND, OR, and XOR)‏

35. Time Equalization of Multiplication and Exponentiation • Time taken by multiplication and exponentiation (especially squaring) are different • Therefore, when one need to equalize them by performing both operations when one of the operations required and discards unnecessary result • So, the attacker will not be able to learn when and how many multiplications and exponentiations are made • This strangely also adds overhead

36. Conclusion • Side-Channel Attack is a real threat with wide range ofpossibility and a large impact • Side-Channel Attack is not a traditional cryptanalysis • Side-Channel Attack is easy, quick, inexpensive, and few risk to be notified by victims • When one design algorithm or system such as cryptosystem, one must think about additional output leaked from the devices, too.

37. References • Bar-El Hagai “Introduction to Side Channel Attack” • Kocher Paul. “Timing Attacks On Implementation of DH, RSA, DSS and Other Systems” • Haas Job. “Side Channel Analysis and Embedded Systems Impact and Coutner measure” • Endrodi, Csilla, “Side-Channel Attack of RSA” • Cid Carlos. “Cryptanalysts of RSA: A Survey”