IIT KHARAGPURDestroying Fault Invariant with Randomization

-A Countermeasure for AESagainst Differential Fault Attacks

Authors:Harshal Tupsamudre, Shikha Bisht, Debdeep Mukhopadhyay

(IIT KHARAGPUR)

CHES 2014

South Korea, Busan

September 24, 2014

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 1 / 48

Outline

1 Preliminaries

2 LatinCrypt 2012 Infection Countermeasure

3 FDTC 2013 Attack

4 A Major Loop Hole in LatinCrypt 2012 Countermeasure

5 Piret and Quisquater’s Attack on Infection CountermeasureAttack Without Random Dummy RoundsComplexity AnalysisAttack in Presence of Random Dummy Rounds

6 Improved Countermeasure

7 Summary & Conclusion

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Preliminaries

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 3 / 48

AES128

Plaintext ⊕ Key

Ciphertext

Round 1

Round i

Round 10

SubByte

ShiftRow

MixColumn

Add Round Key

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 4 / 48

AES128: Round Function

I0 I4 I8 I12I1 I5 I9 I13I2 I6 I10 I14I3 I7 I11 I15

—S—

S[I0] S[I4] S[I8] S[I12]S[I1] S[I5] S[I9] S[I13]S[I2] S[I6] S[I10] S[I14]S[I3] S[I7] S[I11] S[I15]

—SR—

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—MC—

2 3 1 11 2 3 11 1 2 33 1 1 2

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—Add key—

I′0 ⊕ k0 I

′4 ⊕ k4 I

′8 ⊕ k8 I

′12 ⊕ k12

I′1 ⊕ k1 I

′5 ⊕ k5 I

′9 ⊕ k9 I

′13 ⊕ k13

I′2 ⊕ k2 I

′6 ⊕ k6 I

′10 ⊕ k10 I

′14 ⊕ k14

I′3 ⊕ k3 I

′7 ⊕ k7 I

′11 ⊕ k11 I

′15 ⊕ k15

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 5 / 48

AES128: Round Function

I0 I4 I8 I12I1 I5 I9 I13I2 I6 I10 I14I3 I7 I11 I15

—S—

S[I0] S[I4] S[I8] S[I12]S[I1] S[I5] S[I9] S[I13]S[I2] S[I6] S[I10] S[I14]S[I3] S[I7] S[I11] S[I15]

—SR—

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—MC—

2 3 1 11 2 3 11 1 2 33 1 1 2

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—Add key—

I′0 ⊕ k0 I

′4 ⊕ k4 I

′8 ⊕ k8 I

′12 ⊕ k12

I′1 ⊕ k1 I

′5 ⊕ k5 I

′9 ⊕ k9 I

′13 ⊕ k13

I′2 ⊕ k2 I

′6 ⊕ k6 I

′10 ⊕ k10 I

′14 ⊕ k14

I′3 ⊕ k3 I

′7 ⊕ k7 I

′11 ⊕ k11 I

′15 ⊕ k15

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AES128: Round Function

I0 I4 I8 I12I1 I5 I9 I13I2 I6 I10 I14I3 I7 I11 I15

—S—

S[I0] S[I4] S[I8] S[I12]S[I1] S[I5] S[I9] S[I13]S[I2] S[I6] S[I10] S[I14]S[I3] S[I7] S[I11] S[I15]

—SR—

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—MC—

2 3 1 11 2 3 11 1 2 33 1 1 2

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—Add key—

I′0 ⊕ k0 I

′4 ⊕ k4 I

′8 ⊕ k8 I

′12 ⊕ k12

I′1 ⊕ k1 I

′5 ⊕ k5 I

′9 ⊕ k9 I

′13 ⊕ k13

I′2 ⊕ k2 I

′6 ⊕ k6 I

′10 ⊕ k10 I

′14 ⊕ k14

I′3 ⊕ k3 I

′7 ⊕ k7 I

′11 ⊕ k11 I

′15 ⊕ k15

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 5 / 48

AES128: Round Function

I0 I4 I8 I12I1 I5 I9 I13I2 I6 I10 I14I3 I7 I11 I15

—S—

S[I0] S[I4] S[I8] S[I12]S[I1] S[I5] S[I9] S[I13]S[I2] S[I6] S[I10] S[I14]S[I3] S[I7] S[I11] S[I15]

—SR—

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—MC—

2 3 1 11 2 3 11 1 2 33 1 1 2

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—Add key—

I′0 ⊕ k0 I

′4 ⊕ k4 I

′8 ⊕ k8 I

′12 ⊕ k12

I′1 ⊕ k1 I

′5 ⊕ k5 I

′9 ⊕ k9 I

′13 ⊕ k13

I′2 ⊕ k2 I

′6 ⊕ k6 I

′10 ⊕ k10 I

′14 ⊕ k14

I′3 ⊕ k3 I

′7 ⊕ k7 I

′11 ⊕ k11 I

′15 ⊕ k15

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 5 / 48

AES128: Round Function

I0 I4 I8 I12I1 I5 I9 I13I2 I6 I10 I14I3 I7 I11 I15

—S—

S[I0] S[I4] S[I8] S[I12]S[I1] S[I5] S[I9] S[I13]S[I2] S[I6] S[I10] S[I14]S[I3] S[I7] S[I11] S[I15]

—SR—

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—MC—

2 3 1 11 2 3 11 1 2 33 1 1 2

S[I0] S[I4] S[I8] S[I12]S[I5] S[I9] S[I13] S[I1]S[I10] S[I14] S[I2] S[I6]S[I15] S[I3] S[I7] S[I11]

—Add key—

I′0 ⊕ k0 I

′4 ⊕ k4 I

′8 ⊕ k8 I

′12 ⊕ k12

I′1 ⊕ k1 I

′5 ⊕ k5 I

′9 ⊕ k9 I

′13 ⊕ k13

I′2 ⊕ k2 I

′6 ⊕ k6 I

′10 ⊕ k10 I

′14 ⊕ k14

I′3 ⊕ k3 I

′7 ⊕ k7 I

′11 ⊕ k11 I

′15 ⊕ k15

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 5 / 48

Fault Attack

Cipher Process

Plaintext Plaintext

Cipher Process

Fault

Correct Ciphertext

Output Analysis

⊕ Faulty Ciphertext

Only one fault sufficient to retrieve the entire secret key of AES.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 6 / 48

Fault Attack

Cipher Process

Plaintext Plaintext

Cipher Process

Fault

Correct Ciphertext

Output Analysis

⊕ Faulty Ciphertext

Only one fault sufficient to retrieve the entire secret key of AES.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 6 / 48

Fault Attack

1 Fault models to model the strength of adversary1 Bit flip Fault Model : Affects a bit of the intermediate result2 Constant Byte Fault Model : Requires control over fault value and

position3 Random Byte Fault Model : No control over fault value and position

2 Attacks that require both the correct and faulty ciphertext are knownas differential fault attacks

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 7 / 48

Countermeasures Against Fault Attacks

Detection Countermeasure

PT PT*

CT*CT

PT

CT

PT

CT

CT = CT* ? PT = PT* ?

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Infection Countermeasure

Cipher Round

Redundant Round

Diffusion

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LatinCrypt 2012 InfectionCountermeasure

LatinCrypt 2012 Infection CountermeasureSNLF operates on a byte and SNLF(0) = 0

SNLF

Redundan

tCipher

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LatinCrypt 2012 Infection CountermeasureSNLF operates on a byte and SNLF(0) = 0

SNLF

Redundan

tCipher

Redundan

tCipher

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LatinCrypt 2012 Infection CountermeasureDummy rounds occur randomly

SNLF

Redundan

tCipher

Redundan

tCipher

Dummy

Dummy

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LatinCrypt 2012 Infection CountermeasureRoundFunction(β, k0) = β

Dummy

β

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LatinCrypt 2012 Infection CountermeasureRoundFunction(β, k0) = β

Redundan

tDummy

Redundan

t

β

Dummy

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LatinCrypt 2012 Infection CountermeasureRoundFunction(β, k0) = β

Redundan

tDummy

Cipher

Redundan

tCipher

β

Dummy

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FDTC 2013 Attack

FDTC 2013 Attack

Fault f in I 101 , i .e., first byte of the second row in the input of 10th

cipher round of AES128

Countermeasure infects the faulty computation twice

I After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 12 / 48

FDTC 2013 Attack

Fault f in I 101 , i .e., first byte of the second row in the input of 10th

cipher round of AES128

Countermeasure infects the faulty computation twice

I After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 12 / 48

FDTC 2013 Attack

Fault f in I 101 , i .e., first byte of the second row in the input of 10th

cipher round of AES128

Countermeasure infects the faulty computation twice

I After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 12 / 48

FDTC 2013 Attack

Fault f in I 101 , i .e., first byte of the second row in the input of 10th

cipher round of AES128

Countermeasure infects the faulty computation twiceI After the execution of 10th cipher round

I After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 12 / 48

FDTC 2013 Attack

Fault f in I 101 , i .e., first byte of the second row in the input of 10th

cipher round of AES128

Countermeasure infects the faulty computation twiceI After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 12 / 48

FDTC 2013 Attack

f

Step 6

R0 R1

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FDTC 2013 Attack

εf

Step 6 Step 7

R0 R1

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FDTC 2013 Attack

ε

ε SNLF[ε]

f

Step 6 Step 7 Step 11

Step 10 SNLF[ε]

R0 R1

β R2

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FDTC 2013 Attack

∆1

∆2

∆3

∆4

SNLF[ε]

ε

ε SNLF[ε]

f

Step 6 Step 7 Step 11

Step 10 Step 14

β R2

R0 R1

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FDTC 2013 Attack

∆1

∆2

∆3

∆4

∆1

∆2

∆3

∆4

ε

ε SNLF[ε]

β R2

f

C C*

Step 6 Step 7 Step 11

Step 10 Step 14SNLF[ε]

Step 14

R0 R1

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 13 / 48

FDTC 2013 Attack: Infection Caused by the 10th CipherRound

1 The difference between correct (R1) and faulty computation (R0) is:0 0 0 0f 0 0 00 0 0 00 0 0 0

—S—

0 0 0 0ε 0 0 00 0 0 00 0 0 0

—SR—

0 0 0 00 0 0 ε0 0 0 00 0 0 0

2 After Infection Step, the difference is:

R0 ⊕ R1 =

0 0 0 00 0 0 ε⊕ SNLF [ε]0 0 0 00 0 0 0

where ε = S [I 101 ⊕ f ]⊕ S [I 101 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 14 / 48

FDTC 2013 Attack: Infection Caused by the 10th CipherRound

1 The difference between correct (R1) and faulty computation (R0) is:0 0 0 0f 0 0 00 0 0 00 0 0 0

—S—

0 0 0 0ε 0 0 00 0 0 00 0 0 0

—SR—

0 0 0 00 0 0 ε0 0 0 00 0 0 0

2 After Infection Step, the difference is:

R0 ⊕ R1 =

0 0 0 00 0 0 ε⊕ SNLF [ε]0 0 0 00 0 0 0

where ε = S [I 101 ⊕ f ]⊕ S [I 101 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 14 / 48

FDTC 2013 Attack: Infection Caused by the 10th CipherRound

1 The difference between correct (R1) and faulty computation (R0) is:0 0 0 0f 0 0 00 0 0 00 0 0 0

—S—

0 0 0 0ε 0 0 00 0 0 00 0 0 0

—SR—

0 0 0 00 0 0 ε0 0 0 00 0 0 0

2 After Infection Step, the difference is:

R0 ⊕ R1 =

0 0 0 00 0 0 ε⊕ SNLF [ε]0 0 0 00 0 0 0

where ε = S [I 101 ⊕ f ]⊕ S [I 101 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 14 / 48

FDTC 2013 Attack: Infection Caused by the 10th CipherRound

1 The difference between correct (R1) and faulty computation (R0) is:0 0 0 0f 0 0 00 0 0 00 0 0 0

—S—

0 0 0 0ε 0 0 00 0 0 00 0 0 0

—SR—

0 0 0 00 0 0 ε0 0 0 00 0 0 0

2 After Infection Step, the difference is:

R0 ⊕ R1 =

0 0 0 00 0 0 ε⊕ SNLF [ε]0 0 0 00 0 0 0

where ε = S [I 101 ⊕ f ]⊕ S [I 101 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 14 / 48

FDTC 2013 Attack: Infection Caused by the CompulsoryDummy Round

3 The differential of R2 and β is:

R2 ⊕ β =

0 0 0 00 0 0 SNLF [ε]0 0 0 00 0 0 0

4 When R2 = β, RoundFunction(R2, k0)⊕ β = 0

5 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

6 ∴ RoundFunction(R2, k0)⊕ β =

0 0 ∆1 00 0 ∆2 00 0 ∆3 00 0 ∆4 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 15 / 48

FDTC 2013 Attack: Infection Caused by the CompulsoryDummy Round

3 The differential of R2 and β is:

R2 ⊕ β =

0 0 0 00 0 0 SNLF [ε]0 0 0 00 0 0 0

4 When R2 = β, RoundFunction(R2, k

0)⊕ β = 0

5 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

6 ∴ RoundFunction(R2, k0)⊕ β =

0 0 ∆1 00 0 ∆2 00 0 ∆3 00 0 ∆4 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 15 / 48

FDTC 2013 Attack: Infection Caused by the CompulsoryDummy Round

3 The differential of R2 and β is:

R2 ⊕ β =

0 0 0 00 0 0 SNLF [ε]0 0 0 00 0 0 0

4 When R2 = β, RoundFunction(R2, k

0)⊕ β = 0

5 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

6 ∴ RoundFunction(R2, k0)⊕ β =

0 0 ∆1 00 0 ∆2 00 0 ∆3 00 0 ∆4 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 15 / 48

FDTC 2013 Attack: Infection Caused by the CompulsoryDummy Round

3 The differential of R2 and β is:

R2 ⊕ β =

0 0 0 00 0 0 SNLF [ε]0 0 0 00 0 0 0

4 When R2 = β, RoundFunction(R2, k

0)⊕ β = 0

5 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

6 ∴ RoundFunction(R2, k0)⊕ β =

0 0 ∆1 00 0 ∆2 00 0 ∆3 00 0 ∆4 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 15 / 48

FDTC 2013 Attack: Final Difference

7 Infection caused by compulsory dummy round does not affect ε.

C ⊕ C ∗ =

0 0 ∆1 00 0 ∆2 ε⊕ SNLF [ε]0 0 ∆3 00 0 ∆4 0

8 Infection SNLF[ε] caused by 10th cipher round is ineffective.

9 Attacker uses the value of ε = S [I 101 ⊕ f ]⊕ S [I 10] to make hypotheseson I 101 and key byte k1113 .

10 Repeat this process with two more pairs of faulty and correctciphertexts, using constant byte fault model.

11 The attack targets last three rows of the 10th round input.

12 Recover remaining 4 bytes of top row using brute force search.

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FDTC 2013 Attack: Final Difference

7 Infection caused by compulsory dummy round does not affect ε.

C ⊕ C ∗ =

0 0 ∆1 00 0 ∆2 ε⊕ SNLF [ε]0 0 ∆3 00 0 ∆4 0

8 Infection SNLF[ε] caused by 10th cipher round is ineffective.

9 Attacker uses the value of ε = S [I 101 ⊕ f ]⊕ S [I 10] to make hypotheseson I 101 and key byte k1113 .

10 Repeat this process with two more pairs of faulty and correctciphertexts, using constant byte fault model.

11 The attack targets last three rows of the 10th round input.

12 Recover remaining 4 bytes of top row using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 16 / 48

FDTC 2013 Attack: Final Difference

7 Infection caused by compulsory dummy round does not affect ε.

C ⊕ C ∗ =

0 0 ∆1 00 0 ∆2 ε⊕ SNLF [ε]0 0 ∆3 00 0 ∆4 0

8 Infection SNLF[ε] caused by 10th cipher round is ineffective.

9 Attacker uses the value of ε = S [I 101 ⊕ f ]⊕ S [I 10] to make hypotheseson I 101 and key byte k1113 .

10 Repeat this process with two more pairs of faulty and correctciphertexts, using constant byte fault model.

11 The attack targets last three rows of the 10th round input.

12 Recover remaining 4 bytes of top row using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 16 / 48

FDTC 2013 Attack: Final Difference

7 Infection caused by compulsory dummy round does not affect ε.

C ⊕ C ∗ =

0 0 ∆1 00 0 ∆2 ε⊕ SNLF [ε]0 0 ∆3 00 0 ∆4 0

8 Infection SNLF[ε] caused by 10th cipher round is ineffective.

9 Attacker uses the value of ε = S [I 101 ⊕ f ]⊕ S [I 10] to make hypotheseson I 101 and key byte k1113 .

10 Repeat this process with two more pairs of faulty and correctciphertexts, using constant byte fault model.

11 The attack targets last three rows of the 10th round input.

12 Recover remaining 4 bytes of top row using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 16 / 48

FDTC 2013 Attack: Final Difference

7 Infection caused by compulsory dummy round does not affect ε.

C ⊕ C ∗ =

0 0 ∆1 00 0 ∆2 ε⊕ SNLF [ε]0 0 ∆3 00 0 ∆4 0

8 Infection SNLF[ε] caused by 10th cipher round is ineffective.

9 Attacker uses the value of ε = S [I 101 ⊕ f ]⊕ S [I 10] to make hypotheseson I 101 and key byte k1113 .

10 Repeat this process with two more pairs of faulty and correctciphertexts, using constant byte fault model.

11 The attack targets last three rows of the 10th round input.

12 Recover remaining 4 bytes of top row using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 16 / 48

FDTC 2013 Attack: Final Difference

7 Infection caused by compulsory dummy round does not affect ε.

C ⊕ C ∗ =

0 0 ∆1 00 0 ∆2 ε⊕ SNLF [ε]0 0 ∆3 00 0 ∆4 0

8 Infection SNLF[ε] caused by 10th cipher round is ineffective.

9 Attacker uses the value of ε = S [I 101 ⊕ f ]⊕ S [I 10] to make hypotheseson I 101 and key byte k1113 .

10 Repeat this process with two more pairs of faulty and correctciphertexts, using constant byte fault model.

11 The attack targets last three rows of the 10th round input.

12 Recover remaining 4 bytes of top row using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 16 / 48

Flaws Exploited by FDTC 2013 attack

1 The last cipher round is always the penultimate round: The attackercan verify target round using side channel.

2 A fault in last three rows of 10th round =⇒Infection caused by compulsory dummy round does not affect theerroneous byte.

Remark

What happens if the infection caused by compulsory dummy round affectsthe erroneous byte of 10th round??

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 17 / 48

Flaws Exploited by FDTC 2013 attack

1 The last cipher round is always the penultimate round: The attackercan verify target round using side channel.

2 A fault in last three rows of 10th round =⇒Infection caused by compulsory dummy round does not affect theerroneous byte.

Remark

What happens if the infection caused by compulsory dummy round affectsthe erroneous byte of 10th round??

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 17 / 48

Further Loop Holes in LatinCrypt 2012Countermeasure

Extending FDTC 2013 Attack to the Top Row

Fault f in I 100 , i .e., first byte of the top row in the input of 10th

cipher round

Countermeasure infects the faulty computation twice

I After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 18 / 48

Extending FDTC 2013 Attack to the Top Row

Fault f in I 100 , i .e., first byte of the top row in the input of 10th

cipher round

Countermeasure infects the faulty computation twice

I After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 18 / 48

Extending FDTC 2013 Attack to the Top Row

Fault f in I 100 , i .e., first byte of the top row in the input of 10th

cipher round

Countermeasure infects the faulty computation twice

I After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 18 / 48

Extending FDTC 2013 Attack to the Top Row

Fault f in I 100 , i .e., first byte of the top row in the input of 10th

cipher round

Countermeasure infects the faulty computation twiceI After the execution of 10th cipher round

I After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 18 / 48

Extending FDTC 2013 Attack to the Top Row

Fault f in I 100 , i .e., first byte of the top row in the input of 10th

cipher round

Countermeasure infects the faulty computation twiceI After the execution of 10th cipher roundI After the execution of compulsory dummy round

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Extending FDTC 2013 Attack to the Top Row

f

Step 6

R0 R1

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Extending FDTC 2013 Attack to the Top Row

ε

Step 6 Step 7

R0 R1

f

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 19 / 48

Extending FDTC 2013 Attack to the Top Row

Step 6 Step 7 Step 11

Step 10

SNLF[ε]

R0 R1

β R2

f εε SNLF[ε]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 19 / 48

Extending FDTC 2013 Attack to the Top Row

Step 6 Step 7 Step 11

Step 10

β R2

R0 R1

f ε

SNLF[ε]

Step 14

α1 ∆1

α2 ∆2

α3 ∆3

α4 ∆4

ε SNLF[ε]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 19 / 48

Extending FDTC 2013 Attack to the Top Row

α2

α3

α4

C C*

Step 6 Step 7

Step 14

Step 14

R0 R1

f ε

α1 ∆1

α2 ∆2

α3 ∆3

α4 ∆4

Step 10

β R2

SNLF[ε]

Step 11

ε SNLF[ε]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 19 / 48

Extending FDTC 2013 Attack to the Top Row

1 The differential between correct (R1) and faulty computation (R0) is:f 0 0 00 0 0 00 0 0 00 0 0 0

—S—

ε 0 0 00 0 0 00 0 0 00 0 0 0

—SR—

ε 0 0 00 0 0 00 0 0 00 0 0 0

2 After Infection Step, the differential is:

R0 ⊕ R1 =

ε⊕ SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

where ε = S [I 100 ⊕ f ]⊕ S [I 100 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 20 / 48

Extending FDTC 2013 Attack to the Top Row

1 The differential between correct (R1) and faulty computation (R0) is:f 0 0 00 0 0 00 0 0 00 0 0 0

—S—

ε 0 0 00 0 0 00 0 0 00 0 0 0

—SR—

ε 0 0 00 0 0 00 0 0 00 0 0 0

2 After Infection Step, the differential is:

R0 ⊕ R1 =

ε⊕ SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

where ε = S [I 100 ⊕ f ]⊕ S [I 100 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 20 / 48

Extending FDTC 2013 Attack to the Top Row

1 The differential between correct (R1) and faulty computation (R0) is:f 0 0 00 0 0 00 0 0 00 0 0 0

—S—

ε 0 0 00 0 0 00 0 0 00 0 0 0

—SR—

ε 0 0 00 0 0 00 0 0 00 0 0 0

2 After Infection Step, the differential is:

R0 ⊕ R1 =

ε⊕ SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

where ε = S [I 100 ⊕ f ]⊕ S [I 100 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 20 / 48

Extending FDTC 2013 Attack to the Top Row

1 The differential between correct (R1) and faulty computation (R0) is:f 0 0 00 0 0 00 0 0 00 0 0 0

—S—

ε 0 0 00 0 0 00 0 0 00 0 0 0

—SR—

ε 0 0 00 0 0 00 0 0 00 0 0 0

2 After Infection Step, the differential is:

R0 ⊕ R1 =

ε⊕ SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

where ε = S [I 100 ⊕ f ]⊕ S [I 100 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 20 / 48

Extending FDTC 2013 Attack to the Top Row

3 The differential of R2 and β is:

R2 ⊕ β =

SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

4 RoundFunction(R2, k0)⊕ β =

α1 0 0 0α2 0 0 0α3 0 0 0α4 0 0 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 21 / 48

Extending FDTC 2013 Attack to the Top Row

3 The differential of R2 and β is:

R2 ⊕ β =

SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

4 RoundFunction(R2, k

0)⊕ β =α1 0 0 0α2 0 0 0α3 0 0 0α4 0 0 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 21 / 48

Extending FDTC 2013 Attack to the Top Row

5 Infection caused by compulsory dummy round affects ε.

C ⊕ C ∗ =

α1 ⊕ ε⊕ SNLF [ε] 0 0 0

α2 0 0 0α3 0 0 0α4 0 0 0

6 Attack of FDTC 2013 will not work.

7 α1 has to be unmasked.

We show that αi are interrelated and infection caused by compulsorydummy round is ineffective.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 22 / 48

Extending FDTC 2013 Attack to the Top Row

5 Infection caused by compulsory dummy round affects ε.

C ⊕ C ∗ =

α1 ⊕ ε⊕ SNLF [ε] 0 0 0

α2 0 0 0α3 0 0 0α4 0 0 0

6 Attack of FDTC 2013 will not work.

7 α1 has to be unmasked.

We show that αi are interrelated and infection caused by compulsorydummy round is ineffective.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 22 / 48

Extending FDTC 2013 Attack to the Top Row

5 Infection caused by compulsory dummy round affects ε.

C ⊕ C ∗ =

α1 ⊕ ε⊕ SNLF [ε] 0 0 0

α2 0 0 0α3 0 0 0α4 0 0 0

6 Attack of FDTC 2013 will not work.

7 α1 has to be unmasked.

We show that αi are interrelated and infection caused by compulsorydummy round is ineffective.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 22 / 48

Extending FDTC 2013 Attack to the Top Row

5 Infection caused by compulsory dummy round affects ε.

C ⊕ C ∗ =

α1 ⊕ ε⊕ SNLF [ε] 0 0 0

α2 0 0 0α3 0 0 0α4 0 0 0

6 Attack of FDTC 2013 will not work.

7 α1 has to be unmasked.

We show that αi are interrelated and infection caused by compulsorydummy round is ineffective.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 22 / 48

A Major Flaw in the Infection Scheme

Since RoundFunction(β, k0) = β we can write:

RoundFunction(R2, k0)⊕ β = RoundFunction(R2, k

0)⊕ RoundFunction(β, k0)

= MC (SR(S(R2)))⊕ k0 ⊕MC (SR(S(β)))⊕ k0

= MC (SR(S(R2)))⊕MC (SR(S(β)))

= MC (SR(S(R2)⊕ S(β)))

1 When R2 = β, RoundFunction(R2, k0)⊕ β = 0

2 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 23 / 48

A Major Flaw in the Infection Scheme

Since RoundFunction(β, k0) = β we can write:

RoundFunction(R2, k0)⊕ β = RoundFunction(R2, k

0)⊕ RoundFunction(β, k0)

= MC (SR(S(R2)))⊕ k0 ⊕MC (SR(S(β)))⊕ k0

= MC (SR(S(R2)))⊕MC (SR(S(β)))

= MC (SR(S(R2)⊕ S(β)))

1 When R2 = β, RoundFunction(R2, k0)⊕ β = 0

2 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 23 / 48

A Major Flaw in the Infection Scheme

Since RoundFunction(β, k0) = β we can write:

RoundFunction(R2, k0)⊕ β = RoundFunction(R2, k

0)⊕ RoundFunction(β, k0)

= MC (SR(S(R2)))⊕ k0 ⊕MC (SR(S(β)))⊕ k0

= MC (SR(S(R2)))⊕MC (SR(S(β)))

= MC (SR(S(R2)⊕ S(β)))

1 When R2 = β, RoundFunction(R2, k0)⊕ β = 0

2 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 23 / 48

A Major Flaw in the Infection Scheme

Since RoundFunction(β, k0) = β we can write:

RoundFunction(R2, k0)⊕ β = RoundFunction(R2, k

0)⊕ RoundFunction(β, k0)

= MC (SR(S(R2)))⊕ k0 ⊕MC (SR(S(β)))⊕ k0

= MC (SR(S(R2)))⊕MC (SR(S(β)))

= MC (SR(S(R2)⊕ S(β)))

1 When R2 = β, RoundFunction(R2, k0)⊕ β = 0

2 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 23 / 48

A Major Flaw in the Infection Scheme

Since RoundFunction(β, k0) = β we can write:

RoundFunction(R2, k0)⊕ β = RoundFunction(R2, k

0)⊕ RoundFunction(β, k0)

= MC (SR(S(R2)))⊕ k0 ⊕MC (SR(S(β)))⊕ k0

= MC (SR(S(R2)))⊕MC (SR(S(β)))

= MC (SR(S(R2)⊕ S(β)))

1 When R2 = β, RoundFunction(R2, k0)⊕ β = 0

2 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 23 / 48

A Major Flaw in the Infection Scheme

Since RoundFunction(β, k0) = β we can write:

RoundFunction(R2, k0)⊕ β = RoundFunction(R2, k

0)⊕ RoundFunction(β, k0)

= MC (SR(S(R2)))⊕ k0 ⊕MC (SR(S(β)))⊕ k0

= MC (SR(S(R2)))⊕MC (SR(S(β)))

= MC (SR(S(R2)⊕ S(β)))

1 When R2 = β, RoundFunction(R2, k0)⊕ β = 0

2 When R2 6= β, RoundFunction(R2, k0)⊕ β 6= 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 23 / 48

Infection Removal of Compulsory Dummy Round

3 The differential of R2 and β is:

R2 ⊕ β =

SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

4 RoundFunction(R2, k0)⊕ β = MC (SR(S(R2)⊕ S(β)))

SNLF [ε] 0 0 00 0 0 00 0 0 00 0 0 0

–S & SR–

y 0 0 00 0 0 00 0 0 00 0 0 0

–MC–

2y 0 0 01y 0 0 01y 0 0 03y 0 0 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 24 / 48

Infection Removal of Compulsory Dummy Round

3 The differential of R2 and β is:

R2 ⊕ β =

SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

4 RoundFunction(R2, k

0)⊕ β = MC (SR(S(R2)⊕ S(β)))SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

–S & SR–

y 0 0 00 0 0 00 0 0 00 0 0 0

–MC–

2y 0 0 01y 0 0 01y 0 0 03y 0 0 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 24 / 48

Infection Removal of Compulsory Dummy Round

3 The differential of R2 and β is:

R2 ⊕ β =

SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

4 RoundFunction(R2, k

0)⊕ β = MC (SR(S(R2)⊕ S(β)))SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

–S & SR–

y 0 0 00 0 0 00 0 0 00 0 0 0

–MC–

2y 0 0 01y 0 0 01y 0 0 03y 0 0 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 24 / 48

Infection Removal of Compulsory Dummy Round

3 The differential of R2 and β is:

R2 ⊕ β =

SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

4 RoundFunction(R2, k

0)⊕ β = MC (SR(S(R2)⊕ S(β)))SNLF [ε] 0 0 0

0 0 0 00 0 0 00 0 0 0

–S & SR–

y 0 0 00 0 0 00 0 0 00 0 0 0

–MC–

2y 0 0 01y 0 0 01y 0 0 03y 0 0 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 24 / 48

Infection Removal of Compulsory Dummy Round

5 Therefore we can write the difference between correct and faultycomputation as:

C ⊕ C ∗ =

2y ⊕ ε⊕ SNLF [ε] 0 0 0

1y 0 0 01y 0 0 03y 0 0 0

6 y can be deduced from the above matrix.

7 2y can be unmasked.

8 And the attack of FDTC 2013 can be mounted.

9 Now, this attack can target any 12 bytes of 10th round input.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 25 / 48

Infection Removal of Compulsory Dummy Round

5 Therefore we can write the difference between correct and faultycomputation as:

C ⊕ C ∗ =

2y ⊕ ε⊕ SNLF [ε] 0 0 0

1y 0 0 01y 0 0 03y 0 0 0

6 y can be deduced from the above matrix.

7 2y can be unmasked.

8 And the attack of FDTC 2013 can be mounted.

9 Now, this attack can target any 12 bytes of 10th round input.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 25 / 48

Infection Removal of Compulsory Dummy Round

5 Therefore we can write the difference between correct and faultycomputation as:

C ⊕ C ∗ =

2y ⊕ ε⊕ SNLF [ε] 0 0 0

1y 0 0 01y 0 0 03y 0 0 0

6 y can be deduced from the above matrix.

7 2y can be unmasked.

8 And the attack of FDTC 2013 can be mounted.

9 Now, this attack can target any 12 bytes of 10th round input.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 25 / 48

Infection Removal of Compulsory Dummy Round

5 Therefore we can write the difference between correct and faultycomputation as:

C ⊕ C ∗ =

2y ⊕ ε⊕ SNLF [ε] 0 0 0

1y 0 0 01y 0 0 03y 0 0 0

6 y can be deduced from the above matrix.

7 2y can be unmasked.

8 And the attack of FDTC 2013 can be mounted.

9 Now, this attack can target any 12 bytes of 10th round input.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 25 / 48

Infection Removal of Compulsory Dummy Round

5 Therefore we can write the difference between correct and faultycomputation as:

C ⊕ C ∗ =

2y ⊕ ε⊕ SNLF [ε] 0 0 0

1y 0 0 01y 0 0 03y 0 0 0

6 y can be deduced from the above matrix.

7 2y can be unmasked.

8 And the attack of FDTC 2013 can be mounted.

9 Now, this attack can target any 12 bytes of 10th round input.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 25 / 48

FDTC 2013 Attack Extended to the Top Row

1y

1y

3y

C C*

Step 6 Step 7

Step 14

Step 14

R0 R1

f ε

2y ∆1

1y ∆2

1y ∆3

3y ∆4

Step 10

β R2

SNLF[ε]

Step 11

ε SNLF[ε]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 26 / 48

Piret and Quisquater’s Attack

Relaxing the Restrictions of FDTC 2013 Attack

1 The attack assumes constant byte fault model which requiresprecise control over fault position and value.

2 The attack can retrieve only last 3 rows of k11 using 12*3 = 36faults.

3 The top row of k11 has to be recoverd using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 27 / 48

Relaxing the Restrictions of FDTC 2013 Attack

1 The attack assumes constant byte fault model which requiresprecise control over fault position and value.

2 The attack can retrieve only last 3 rows of k11 using 12*3 = 36faults.

3 The top row of k11 has to be recoverd using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 27 / 48

Relaxing the Restrictions of FDTC 2013 Attack

1 The attack assumes constant byte fault model which requiresprecise control over fault position and value.

2 The attack can retrieve only last 3 rows of k11 using 12*3 = 36faults.

3 The top row of k11 has to be recoverd using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 27 / 48

Relaxing the Restrictions of FDTC 2013 Attack

1 The attack assumes constant byte fault model which requiresprecise control over fault position and value.

2 The attack can retrieve only last 3 rows of k11 using 12*3 = 36faults.

3 The top row of k11 has to be recoverd using brute force search.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 27 / 48

Piret and Quisquater’s Attack in absence of RandomDummy Rounds

The attack targets the penultimate round of AES, e.g, in case ofAES128, input of 9th round is the target.

Fault f in I 90 , i .e., first byte of the top row in the input of 9th cipherround

Countermeasure infects faulty computation thrice

I After the execution of 9th cipher roundI After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 28 / 48

Piret and Quisquater’s Attack in absence of RandomDummy Rounds

The attack targets the penultimate round of AES, e.g, in case ofAES128, input of 9th round is the target.

Fault f in I 90 , i .e., first byte of the top row in the input of 9th cipherround

Countermeasure infects faulty computation thrice

I After the execution of 9th cipher roundI After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 28 / 48

Piret and Quisquater’s Attack in absence of RandomDummy Rounds

The attack targets the penultimate round of AES, e.g, in case ofAES128, input of 9th round is the target.

Fault f in I 90 , i .e., first byte of the top row in the input of 9th cipherround

Countermeasure infects faulty computation thrice

I After the execution of 9th cipher roundI After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 28 / 48

Piret and Quisquater’s Attack in absence of RandomDummy Rounds

The attack targets the penultimate round of AES, e.g, in case ofAES128, input of 9th round is the target.

Fault f in I 90 , i .e., first byte of the top row in the input of 9th cipherround

Countermeasure infects faulty computation thriceI After the execution of 9th cipher round

I After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 28 / 48

Piret and Quisquater’s Attack in absence of RandomDummy Rounds

The attack targets the penultimate round of AES, e.g, in case ofAES128, input of 9th round is the target.

Fault f in I 90 , i .e., first byte of the top row in the input of 9th cipherround

Countermeasure infects faulty computation thriceI After the execution of 9th cipher roundI After the execution of 10th cipher round

I After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 28 / 48

Piret and Quisquater’s Attack in absence of RandomDummy Rounds

The attack targets the penultimate round of AES, e.g, in case ofAES128, input of 9th round is the target.

Fault f in I 90 , i .e., first byte of the top row in the input of 9th cipherround

Countermeasure infects faulty computation thriceI After the execution of 9th cipher roundI After the execution of 10th cipher roundI After the execution of compulsory dummy round

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 28 / 48

Differential after 9th round

1 Without Countermeasure

R0 ⊕ R1 =

2f ′ 0 0 0f ′ 0 0 0f ′ 0 0 0

3f ′ 0 0 0

2 With Countermeasure

R0 ⊕ R1 =

2f ′ ⊕ SNLF [2f ′] 0 0 0f ′ ⊕ SNLF [f ′] 0 0 0f ′ ⊕ SNLF [f ′] 0 0 0

3f ′ ⊕ SNLF [3f ′] 0 0 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 29 / 48

Differential after 9th round

1 Without Countermeasure

R0 ⊕ R1 =

2f ′ 0 0 0f ′ 0 0 0f ′ 0 0 0

3f ′ 0 0 0

2 With Countermeasure

R0 ⊕ R1 =

2f ′ ⊕ SNLF [2f ′] 0 0 0f ′ ⊕ SNLF [f ′] 0 0 0f ′ ⊕ SNLF [f ′] 0 0 0

3f ′ ⊕ SNLF [3f ′] 0 0 0

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 29 / 48

Differential after 10th round

1 Without Countermeasure

R0 ⊕ R1 =

S[I 100 ] ⊕ S[I 100 ⊕ P0] 0 0 0

0 0 0 S[I 101 ] ⊕ S[I 101 ⊕ P1]

0 0 S[I 102 ] ⊕ S[I 102 ⊕ P2] 0

0 S[I 103 ] ⊕ S[I 103 ⊕ P3] 0 0

2 With Countermeasure

R0 ⊕ R1 =

z0 ⊕ SNLF [z0] 0 0 0

0 0 0 z1 ⊕ SNLF [z1]0 0 z2 ⊕ SNLF [z2] 00 z3 ⊕ SNLF [z3] 0 0

where zi = S [I 10i ] ⊕ S [I 10i ⊕ Pi ⊕ SNLF [Pi ]], i ∈ {0, . . . , 3}.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 30 / 48

Differential after 10th round

1 Without Countermeasure

R0 ⊕ R1 =

S[I 100 ] ⊕ S[I 100 ⊕ P0] 0 0 0

0 0 0 S[I 101 ] ⊕ S[I 101 ⊕ P1]

0 0 S[I 102 ] ⊕ S[I 102 ⊕ P2] 0

0 S[I 103 ] ⊕ S[I 103 ⊕ P3] 0 0

2 With Countermeasure

R0 ⊕ R1 =

z0 ⊕ SNLF [z0] 0 0 0

0 0 0 z1 ⊕ SNLF [z1]0 0 z2 ⊕ SNLF [z2] 00 z3 ⊕ SNLF [z3] 0 0

where zi = S [I 10i ] ⊕ S [I 10i ⊕ Pi ⊕ SNLF [Pi ]], i ∈ {0, . . . , 3}.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 30 / 48

Equations for the keys

1 Without Countermeasure

2 · f ′ = S−1[T0 ⊕ k110 ]⊕ S−1[T ∗0 ⊕ k11

0 ]

1 · f ′ = S−1[T13 ⊕ k1113 ]⊕ S−1[T ∗13 ⊕ k11

13 ]

1 · f ′ = S−1[T10 ⊕ k1110 ]⊕ S−1[T ∗10 ⊕ k11

10 ]

3 · f ′ = S−1[T7 ⊕ k117 ]⊕ S−1[T ∗7 ⊕ k11

7 ]

where T and T ∗ is correct and faulty ciphertext resp.

2 With Countermeasure

2 · f ′ ⊕ SNLF [2 · f ′] = S−1[T0 ⊕ k110 ]⊕ S−1[T ∗0 ⊕ k110 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T13 ⊕ k1113 ]⊕ S−1[T ∗13 ⊕ k1113 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T10 ⊕ k1110 ]⊕ S−1[T ∗10 ⊕ k1110 ]

3 · f ′ ⊕ SNLF [3 · f ′] = S−1[T7 ⊕ k117 ]⊕ S−1[T ∗7 ⊕ k117 ]

where T and T ∗ is correct and faulty ciphertext resp.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 31 / 48

Equations for the keys

1 Without Countermeasure

2 · f ′ = S−1[T0 ⊕ k110 ]⊕ S−1[T ∗0 ⊕ k11

0 ]

1 · f ′ = S−1[T13 ⊕ k1113 ]⊕ S−1[T ∗13 ⊕ k11

13 ]

1 · f ′ = S−1[T10 ⊕ k1110 ]⊕ S−1[T ∗10 ⊕ k11

10 ]

3 · f ′ = S−1[T7 ⊕ k117 ]⊕ S−1[T ∗7 ⊕ k11

7 ]

where T and T ∗ is correct and faulty ciphertext resp.

2 With Countermeasure

2 · f ′ ⊕ SNLF [2 · f ′] = S−1[T0 ⊕ k110 ]⊕ S−1[T ∗0 ⊕ k110 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T13 ⊕ k1113 ]⊕ S−1[T ∗13 ⊕ k1113 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T10 ⊕ k1110 ]⊕ S−1[T ∗10 ⊕ k1110 ]

3 · f ′ ⊕ SNLF [3 · f ′] = S−1[T7 ⊕ k117 ]⊕ S−1[T ∗7 ⊕ k117 ]

where T and T ∗ is correct and faulty ciphertext resp.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 31 / 48

Infection of Compulsory dummy round

1 Due to the presence of compulsory dummy round, the differencebetween the final faulty and correct ciphertext:

T ⊕ T ∗ =

m0 ⊕ cdr0 cdr4 cdr8 cdr12cdr1 cdr5 cdr9 m1 ⊕ cdr13cdr2 cdr6 m2 ⊕ cdr10 cdr14cdr3 m3 ⊕ cdr7 cdr11 cdr15

mj = zj ⊕ SNLF [zj ], j ∈ {0, . . . , 3}.

2 Using the relation:RoundFunction(R2, k

0)⊕ β = MC (SR(S(R2)⊕ S(β))) we have:

T ⊕ T ∗ =

m0 ⊕ g1(F1,F2) 1F3 h1(F4,F5,F6) 3F7

g2(F1,F2) 1F3 h2(F4,F5,F6) m1 ⊕ 2F7

g3(F1,F2) 3F3 m2 ⊕ h3(F4,F5,F6) 1F7

g4(F1,F2) m3 ⊕ 2F3 h4(F4,F5,F6) 1F7

Fi , i ∈ {1, . . . , 7} is infection caused by compulsory dummy round andgj and hj , j ∈ {1, . . . , 4} are linear functions.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 32 / 48

Infection of Compulsory dummy round

1 Due to the presence of compulsory dummy round, the differencebetween the final faulty and correct ciphertext:

T ⊕ T ∗ =

m0 ⊕ cdr0 cdr4 cdr8 cdr12cdr1 cdr5 cdr9 m1 ⊕ cdr13cdr2 cdr6 m2 ⊕ cdr10 cdr14cdr3 m3 ⊕ cdr7 cdr11 cdr15

mj = zj ⊕ SNLF [zj ], j ∈ {0, . . . , 3}.

2 Using the relation:RoundFunction(R2, k

0)⊕ β = MC (SR(S(R2)⊕ S(β))) we have:

T ⊕ T ∗ =

m0 ⊕ g1(F1,F2) 1F3 h1(F4,F5,F6) 3F7

g2(F1,F2) 1F3 h2(F4,F5,F6) m1 ⊕ 2F7

g3(F1,F2) 3F3 m2 ⊕ h3(F4,F5,F6) 1F7

g4(F1,F2) m3 ⊕ 2F3 h4(F4,F5,F6) 1F7

Fi , i ∈ {1, . . . , 7} is infection caused by compulsory dummy round andgj and hj , j ∈ {1, . . . , 4} are linear functions.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 32 / 48

P&Q’s Attack on LatinCrypt 2012 Countermeasure:Infection Removal

1 After removing infection caused by compulsory dummy round weobtain:

T ⊕ T ∗ =

m0 0 0 00 0 0 m1

0 0 m2 00 m3 0 0

where mj = zj ⊕ SNLF [zj ], j ∈ {0, . . . , 3}.

2 We can deduce zj(two possibilities) from mj which gives 24

possibilities for T ∗.3 Now, we can make hypotheses on 4 bytes of last round key k11.

2 · f ′ ⊕ SNLF [2 · f ′] = S−1[T0 ⊕ k110 ]⊕ S−1[T ∗0 ⊕ k110 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T13 ⊕ k1113 ]⊕ S−1[T ∗13 ⊕ k1113 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T10 ⊕ k1110 ]⊕ S−1[T ∗10 ⊕ k1110 ]

3 · f ′ ⊕ SNLF [3 · f ′] = S−1[T7 ⊕ k117 ]⊕ S−1[T ∗7 ⊕ k117 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 33 / 48

P&Q’s Attack on LatinCrypt 2012 Countermeasure:Infection Removal

1 After removing infection caused by compulsory dummy round weobtain:

T ⊕ T ∗ =

m0 0 0 00 0 0 m1

0 0 m2 00 m3 0 0

where mj = zj ⊕ SNLF [zj ], j ∈ {0, . . . , 3}.

2 We can deduce zj(two possibilities) from mj which gives 24

possibilities for T ∗.

3 Now, we can make hypotheses on 4 bytes of last round key k11.

2 · f ′ ⊕ SNLF [2 · f ′] = S−1[T0 ⊕ k110 ]⊕ S−1[T ∗0 ⊕ k110 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T13 ⊕ k1113 ]⊕ S−1[T ∗13 ⊕ k1113 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T10 ⊕ k1110 ]⊕ S−1[T ∗10 ⊕ k1110 ]

3 · f ′ ⊕ SNLF [3 · f ′] = S−1[T7 ⊕ k117 ]⊕ S−1[T ∗7 ⊕ k117 ]

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 33 / 48

P&Q’s Attack on LatinCrypt 2012 Countermeasure:Infection Removal

1 After removing infection caused by compulsory dummy round weobtain:

T ⊕ T ∗ =

m0 0 0 00 0 0 m1

0 0 m2 00 m3 0 0

where mj = zj ⊕ SNLF [zj ], j ∈ {0, . . . , 3}.

2 We can deduce zj(two possibilities) from mj which gives 24

possibilities for T ∗.3 Now, we can make hypotheses on 4 bytes of last round key k11.

2 · f ′ ⊕ SNLF [2 · f ′] = S−1[T0 ⊕ k110 ]⊕ S−1[T ∗0 ⊕ k110 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T13 ⊕ k1113 ]⊕ S−1[T ∗13 ⊕ k1113 ]

1 · f ′ ⊕ SNLF [1 · f ′] = S−1[T10 ⊕ k1110 ]⊕ S−1[T ∗10 ⊕ k1110 ]

3 · f ′ ⊕ SNLF [3 · f ′] = S−1[T7 ⊕ k117 ]⊕ S−1[T ∗7 ⊕ k117 ]CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 33 / 48

Complexity Analysis

1 24 values of T ∗ gives 24 ∗ 1036 candidate values for 4 bytes of k11.

2 Repeating the attack with another pair of faulty and correctciphertext gives atmost 2 candidate values.

3 Total 8 faulty ciphertexts required to retrieve all 16 bytes of k11.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 34 / 48

Complexity Analysis

1 24 values of T ∗ gives 24 ∗ 1036 candidate values for 4 bytes of k11.

2 Repeating the attack with another pair of faulty and correctciphertext gives atmost 2 candidate values.

3 Total 8 faulty ciphertexts required to retrieve all 16 bytes of k11.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 34 / 48

Complexity Analysis

1 24 values of T ∗ gives 24 ∗ 1036 candidate values for 4 bytes of k11.

2 Repeating the attack with another pair of faulty and correctciphertext gives atmost 2 candidate values.

3 Total 8 faulty ciphertexts required to retrieve all 16 bytes of k11.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 34 / 48

Attack in Presence of Random Dummy Rounds

Compulsory dummy round

10th Cipher Round

10th Redundant Round

9th Cipher Round

FavourableCase

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 35 / 48

Attack in Presence of Random Dummy Rounds

Compulsory dummy round

10th Cipher Round

10th Redundant Round

9th Cipher Round

Case 2

10th Red Round Random Dummy

FavourableCase

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 35 / 48

Attack in Presence of Random Dummy Rounds

Compulsory dummy round

10th Cipher Round

10th Redundant Round

9th Cipher Round

Case 2

10th Red Round Random Dummy

Random Dummy 10th Red Round

Case 3

FavourableCase

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 35 / 48

Attack in Presence of Random Dummy Rounds

Compulsory dummy round

10th Cipher Round

10th Redundant Round

9th Cipher Round

Case 2

10th Red Round Random Dummy

Random Dummy 10th Red Round

Case 3

Random Dummy Random Dummy

Case 4

FavourableCase

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 35 / 48

Attack in Presence of Random Dummy Rounds

1 Number of random dummy rounds : d

2 Total number of rounds : 22 + d + 1

3 Target round of fault injection : (22 + d − 2)th RoundFunction.

4 (22 + d)th RoundFunction: 10th cipher round.

5 ∴ The probability of (22 + d − 2)th RoundFunction being a 9th cipher

round: (19+d)!/((19)!·(d)!)(21+d)!/((21)!·(d)!)

6 If d = 20 then the probability that 40th RoundFunction is a 9th cipherround is nearly 0.26.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 36 / 48

Attack in Presence of Random Dummy Rounds

1 Number of random dummy rounds : d

2 Total number of rounds : 22 + d + 1

3 Target round of fault injection : (22 + d − 2)th RoundFunction.

4 (22 + d)th RoundFunction: 10th cipher round.

5 ∴ The probability of (22 + d − 2)th RoundFunction being a 9th cipher

round: (19+d)!/((19)!·(d)!)(21+d)!/((21)!·(d)!)

6 If d = 20 then the probability that 40th RoundFunction is a 9th cipherround is nearly 0.26.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 36 / 48

Attack in Presence of Random Dummy Rounds

1 Number of random dummy rounds : d

2 Total number of rounds : 22 + d + 1

3 Target round of fault injection : (22 + d − 2)th RoundFunction.

4 (22 + d)th RoundFunction: 10th cipher round.

5 ∴ The probability of (22 + d − 2)th RoundFunction being a 9th cipher

round: (19+d)!/((19)!·(d)!)(21+d)!/((21)!·(d)!)

6 If d = 20 then the probability that 40th RoundFunction is a 9th cipherround is nearly 0.26.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 36 / 48

Attack in Presence of Random Dummy Rounds

1 Number of random dummy rounds : d

2 Total number of rounds : 22 + d + 1

3 Target round of fault injection : (22 + d − 2)th RoundFunction.

4 (22 + d)th RoundFunction: 10th cipher round.

5 ∴ The probability of (22 + d − 2)th RoundFunction being a 9th cipher

round: (19+d)!/((19)!·(d)!)(21+d)!/((21)!·(d)!)

6 If d = 20 then the probability that 40th RoundFunction is a 9th cipherround is nearly 0.26.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 36 / 48

Attack in Presence of Random Dummy Rounds

1 Number of random dummy rounds : d

2 Total number of rounds : 22 + d + 1

3 Target round of fault injection : (22 + d − 2)th RoundFunction.

4 (22 + d)th RoundFunction: 10th cipher round.

5 ∴ The probability of (22 + d − 2)th RoundFunction being a 9th cipher

round: (19+d)!/((19)!·(d)!)(21+d)!/((21)!·(d)!)

6 If d = 20 then the probability that 40th RoundFunction is a 9th cipherround is nearly 0.26.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 36 / 48

Attack in Presence of Random Dummy Rounds

1 Number of random dummy rounds : d

2 Total number of rounds : 22 + d + 1

3 Target round of fault injection : (22 + d − 2)th RoundFunction.

4 (22 + d)th RoundFunction: 10th cipher round.

5 ∴ The probability of (22 + d − 2)th RoundFunction being a 9th cipher

round: (19+d)!/((19)!·(d)!)(21+d)!/((21)!·(d)!)

6 If d = 20 then the probability that 40th RoundFunction is a 9th cipherround is nearly 0.26.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 36 / 48

Simulation Results

0 5 10 15 20 25 30

0

100

200

300

400

Number of Random Dummy Rounds: d

Ave

rage

Nu

mb

erof

Fau

lty

En

cryp

tion

s

Figure: Piret & Quisquater’s Attack on Algorithm 1

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 37 / 48

Flaws in LatinCrypt 2012 Countermeasure

1 The last cipher round is always the penultimate round: The attackercan verify target round using side channel.

2 A fault in last three rows of 10th round =⇒Infection caused by compulsory dummy round does not affect theerroneous byte.

3 Countermeasure uses same value to infect erroneous as well asnon-erroneous byte.

4 The effect of infection varies for different rounds.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 38 / 48

Flaws in LatinCrypt 2012 Countermeasure

1 The last cipher round is always the penultimate round: The attackercan verify target round using side channel.

2 A fault in last three rows of 10th round =⇒Infection caused by compulsory dummy round does not affect theerroneous byte.

3 Countermeasure uses same value to infect erroneous as well asnon-erroneous byte.

4 The effect of infection varies for different rounds.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 38 / 48

Flaws in LatinCrypt 2012 Countermeasure

1 The last cipher round is always the penultimate round: The attackercan verify target round using side channel.

2 A fault in last three rows of 10th round =⇒Infection caused by compulsory dummy round does not affect theerroneous byte.

3 Countermeasure uses same value to infect erroneous as well asnon-erroneous byte.

4 The effect of infection varies for different rounds.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 38 / 48

Flaws in LatinCrypt 2012 Countermeasure

1 The last cipher round is always the penultimate round: The attackercan verify target round using side channel.

2 A fault in last three rows of 10th round =⇒Infection caused by compulsory dummy round does not affect theerroneous byte.

3 Countermeasure uses same value to infect erroneous as well asnon-erroneous byte.

4 The effect of infection varies for different rounds.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 38 / 48

Improved Countermeasure

Improved Countermeasure

Redundan

tCipher

BLFN

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 39 / 48

Improved Countermeasure

BLFN

Redundan

tCipher

BLFN

Dummy

β

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 39 / 48

Improved Countermeasure

Redundan

tCipher

BLFN BLFN

OR

γ δ Dummy

β

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 39 / 48

Improved Countermeasure

~x

Redundan

tCipher

BLFN BLFN

OR

γ δ Dummy

β

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 39 / 48

Improved Countermeasure

BLFN

Redundan

tCipher

BLFN

OR~x x

γ δ Dummy

β

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 39 / 48

Improved Countermeasure

~x

Redundan

tCipher

BLFN

Dummy

BLFN

OR

γ δ

OR x

Cipher

β

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 39 / 48

Improved Countermeasure

~x

Redund antCipher

BLFNDum

my

BLFN

OR

γ=0 δ=0

OR x

Cipher

β

1 0

Cipher Matrix

0

Zero Matrix

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 39 / 48

Improved Countermeasure

~x

Redundan

tCipher

BLFN

Dummy

BLFN

OR

γ=1 δ

OR x

Cipher

β

0 1

Zero Matrix

1

Random Matrix

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 39 / 48

Improved Countermeasure

1 Fault injection in any of the cipher, redundant or dummy round =⇒Every byte in the resulting ciphertext is infected with a differentvalue.

2 The resulting infected faulty ciphertext is completely random.

3 More than one random dummy round after the last cipher round.

4 The improved countermeasure protects both SPN ciphers and Feistelciphers.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 40 / 48

Summary & Conclusion

1 The infection mechanism of LatinCrypt 2012 countermeasure isshown to be ineffective.

2 An improved countermeasure is developed, which outputs acompletely random value in case of fault injection so that fault attackis impossible.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 41 / 48

Summary & Conclusion

1 The infection mechanism of LatinCrypt 2012 countermeasure isshown to be ineffective.

2 An improved countermeasure is developed, which outputs acompletely random value in case of fault injection so that fault attackis impossible.

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 41 / 48

Thank You !

CHES 2014 (South Korea, Busan) IIT KHARAGPUR September 24, 2014 42 / 48

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