![Outline
Introduction
Continued Fractions
On the use of continued fractions for stream cipher
Questions
Preliminaries
It is not possible to find an irrational number α simply on the
basis of knowledge of the partial quotients [am+1, . . . , am+n].
Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph](https://image.slidesharecdn.com/ontheuseofcontinuedfractionforstreamciphersver1-150601171017-lva1-app6892/85/On-the-use-of-continued-fraction-for-stream-ciphers-9-320.jpg)
![Outline
Introduction
Continued Fractions
On the use of continued fractions for stream cipher
Questions
Preliminaries
It is not possible to find an irrational number α simply on the
basis of knowledge of the partial quotients [am+1, . . . , am+n].
The knowledge of a = [am+1, . . . , am+n] does not allow to
know any other partial quotients of continued fraction
expansion.
Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph](https://image.slidesharecdn.com/ontheuseofcontinuedfractionforstreamciphersver1-150601171017-lva1-app6892/85/On-the-use-of-continued-fraction-for-stream-ciphers-10-320.jpg)
![Outline
Introduction
Continued Fractions
On the use of continued fractions for stream cipher
Questions
Preliminaries
It is not possible to find an irrational number α simply on the
basis of knowledge of the partial quotients [am+1, . . . , am+n].
The knowledge of a = [am+1, . . . , am+n] does not allow to
know any other partial quotients of continued fraction
expansion.
r
log(A) is transcendental.
Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph](https://image.slidesharecdn.com/ontheuseofcontinuedfractionforstreamciphersver1-150601171017-lva1-app6892/85/On-the-use-of-continued-fraction-for-stream-ciphers-11-320.jpg)
![Outline
Introduction
Continued Fractions
On the use of continued fractions for stream cipher
Questions
Continued fraction cipher
Khinchin’s Attack
Applications
Example of Khinchin’s Attack on π
The first partial quotients of π are :
[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2...]
17
i=1
ai
1/17
≈ 2.6929721 . . .
Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph](https://image.slidesharecdn.com/ontheuseofcontinuedfractionforstreamciphersver1-150601171017-lva1-app6892/85/On-the-use-of-continued-fraction-for-stream-ciphers-19-320.jpg)
![Outline
Introduction
Continued Fractions
On the use of continued fractions for stream cipher
Questions
Continued fraction cipher
Khinchin’s Attack
Applications
Example of Khinchin’s Attack on π
The first partial quotients of π are :
[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2...]
17
i=1
ai
1/17
≈ 2.6929721 . . .
let’s suppose that the plaintext is 11111111111111111.
keystream : 0111 1111 0001 100100100 .....0010 0010
plaintext : 0001 0001 0001 0001 .....0001 0001
cipher : 0110 1110 0000 100100101 ....0011 0011
Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph](https://image.slidesharecdn.com/ontheuseofcontinuedfractionforstreamciphersver1-150601171017-lva1-app6892/85/On-the-use-of-continued-fraction-for-stream-ciphers-20-320.jpg)
![Outline
Introduction
Continued Fractions
On the use of continued fractions for stream cipher
Questions
Continued fraction cipher
Khinchin’s Attack
Applications
Example of Khinchin’s Attack on π
The first partial quotients of π are :
[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2...]
17
i=1
ai
1/17
≈ 2.6929721 . . .
let’s suppose that the plaintext is 11111111111111111.
keystream : 0111 1111 0001 100100100 .....0010 0010
plaintext : 0001 0001 0001 0001 .....0001 0001
cipher : 0110 1110 0000 100100101 ....0011 0011
In base 10, the cipher will be: 6 14 1 293 1 1 1 3 1 2 1 15 3 1
1 3 3.
Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph](https://image.slidesharecdn.com/ontheuseofcontinuedfractionforstreamciphersver1-150601171017-lva1-app6892/85/On-the-use-of-continued-fraction-for-stream-ciphers-21-320.jpg)
On the use of continued fraction for stream ciphers
- 1. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Presentation: On the use of continued fractions for stream cipher Amadou Moctar Kane KSecurity amadou1@gmail.com May 4, 2015 Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 2. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions 1 Introduction 2 Continued Fractions 3 On the use of continued fractions for stream cipher Continued fraction cipher Khinchin’s Attack Applications 4 Questions Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 3. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Goals After Diffie-Hellman: Fermat’s little theorem, Linearization XL, graph theory. . . Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 4. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Goals After Diffie-Hellman: Fermat’s little theorem, Linearization XL, graph theory. . . Continued Fraction Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 5. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Goals After Diffie-Hellman: Fermat’s little theorem, Linearization XL, graph theory. . . Continued Fraction How to use? Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 6. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Goals After Diffie-Hellman: Fermat’s little theorem, Linearization XL, graph theory. . . Continued Fraction How to use? Quadratic irrational? Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 7. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Goals After Diffie-Hellman: Fermat’s little theorem, Linearization XL, graph theory. . . Continued Fraction How to use? Quadratic irrational? Γ? Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 8. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued Fractions An expression of the form α := a0 + b0 a1 + b1 a2 + b2 ... is called a generalized continued fraction. Typically, the numbers a1, . . . , b1, . . . may be real or complex, and the expansion may be finite or infinite. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 9. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Preliminaries It is not possible to find an irrational number α simply on the basis of knowledge of the partial quotients [am+1, . . . , am+n]. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 10. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Preliminaries It is not possible to find an irrational number α simply on the basis of knowledge of the partial quotients [am+1, . . . , am+n]. The knowledge of a = [am+1, . . . , am+n] does not allow to know any other partial quotients of continued fraction expansion. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 11. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Preliminaries It is not possible to find an irrational number α simply on the basis of knowledge of the partial quotients [am+1, . . . , am+n]. The knowledge of a = [am+1, . . . , am+n] does not allow to know any other partial quotients of continued fraction expansion. r log(A) is transcendental. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 12. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Stream Ciphers First Algorithm:Stream Cipher Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 13. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Stream Ciphers One time pad. random key ⊕ plaintext Unbreakable system. Easy to implement. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 14. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Stream Ciphers One time pad. random key ⊕ plaintext Unbreakable system. Easy to implement. Stream Ciphers. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 15. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Continued fraction cipher We suppose that z ∈R N, and m is the secret message. Table: Continued fraction cipher. Alice Bob computes t ≡ ze mod n t =⇒ computes z ≡ td mod n. Computes X = e log(z) Computes X = e log(z) Computes the CFE of X Computes the CFE of X. Concatenates some PQ’s Concatenates some PQ’s. Produces the keystream k1 Produces the keystream k1. Computes m1 := m ⊕ k1 m1 =⇒ receives m1. Computes m := m1 ⊕ k1 Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 16. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Efficiency analysis Table: Comparison with Blum-Blum-Shub. Number of bits producted Computing time in seconds BBS 150000 2.358 Our algorithm 150000 0.007 We worked with an irrational X ∈ Γ, and the number of digits of the partial numerator (bi ’s) was around 5000. For BBS, n had 949 digits, the results are listed below. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 17. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Khinchin Aleksandr Khinchin proved in 1935 that for almost all real numbers x, the infinitely many partial quotients ai of the continued fraction expansion of x have an astonishing property: their geometric mean is a constant, known as Khinchin’s constant, which is independent of the value of x. That is, for x = a1 + 1 a2 + 1 ... lim n→∞ n i=1 ai 1/n = K ≈ 2, 6854520010 . . . where K is Khinchin’s constant. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 18. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Khinchin’s Attack The attacker Eve needs the cipher only to find a part of the message in these following steps: Eve eavesdrops a long cipher text Tn, splits it in bytes and computes K1 = lim n→∞ n i=1 di 1/n . where di is the integer corresponding to the byte i. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 19. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Example of Khinchin’s Attack on π The first partial quotients of π are : [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2...] 17 i=1 ai 1/17 ≈ 2.6929721 . . . Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 20. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Example of Khinchin’s Attack on π The first partial quotients of π are : [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2...] 17 i=1 ai 1/17 ≈ 2.6929721 . . . let’s suppose that the plaintext is 11111111111111111. keystream : 0111 1111 0001 100100100 .....0010 0010 plaintext : 0001 0001 0001 0001 .....0001 0001 cipher : 0110 1110 0000 100100101 ....0011 0011 Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 21. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Example of Khinchin’s Attack on π The first partial quotients of π are : [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2...] 17 i=1 ai 1/17 ≈ 2.6929721 . . . let’s suppose that the plaintext is 11111111111111111. keystream : 0111 1111 0001 100100100 .....0010 0010 plaintext : 0001 0001 0001 0001 .....0001 0001 cipher : 0110 1110 0000 100100101 ....0011 0011 In base 10, the cipher will be: 6 14 1 293 1 1 1 3 1 2 1 15 3 1 1 3 3. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 22. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Khinchin’s Attack Eve computes the geometric mean of the cipher: (6∗14∗1∗293∗1∗1∗1∗3∗1∗2∗1∗15∗3∗1∗1∗3∗3)(1/17) = 2.867 Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 23. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Khinchin’s Attack Eve computes the geometric mean of the cipher: (6∗14∗1∗293∗1∗1∗1∗3∗1∗2∗1∗15∗3∗1∗1∗3∗3)(1/17) = 2.867 Eve Makes a conclusion, for example there are a lot of zeros in the plain text. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 24. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Khinchin’s Attack Eve computes the geometric mean of the cipher: (6∗14∗1∗293∗1∗1∗1∗3∗1∗2∗1∗15∗3∗1∗1∗3∗3)(1/17) = 2.867 Eve Makes a conclusion, for example there are a lot of zeros in the plain text. She modifies the cipher and computes the geometric mean of the new cipher K2 = (6 ∗ 14 ∗ 1 ∗ 292 ∗ · · · ∗ 2)(1/17) = 2.595 Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 25. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Khinchin’s Attack Eve computes the geometric mean of the cipher: (6∗14∗1∗293∗1∗1∗1∗3∗1∗2∗1∗15∗3∗1∗1∗3∗3)(1/17) = 2.867 Eve Makes a conclusion, for example there are a lot of zeros in the plain text. She modifies the cipher and computes the geometric mean of the new cipher K2 = (6 ∗ 14 ∗ 1 ∗ 292 ∗ · · · ∗ 2)(1/17) = 2.595 . . . Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 26. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Applications Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 27. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Applications Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 28. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Conclusion 1 Goal 1: I tried to find new techniques using continued fraction in cryptography. 2 Goal 2: I was interested in finding new methods of cryptanalysis. 3 Goal 3: I tried to create a renewal of interest around continued fractions. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 29. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Conclusion 1 Goal 1: I tried to find new techniques using continued fraction in cryptography. Result: I designed a new pseudo random generator statistically tested. 2 Goal 2: I was interested in finding new methods of cryptanalysis. 3 Goal 3: I tried to create a renewal of interest around continued fractions. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 30. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Conclusion 1 Goal 1: I tried to find new techniques using continued fraction in cryptography. Result: I designed a new pseudo random generator statistically tested. 2 Goal 2: I was interested in finding new methods of cryptanalysis. Result: I designed a weak version which can be attacked by the Khinchin constant. 3 Goal 3: I tried to create a renewal of interest around continued fractions. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 31. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions Continued fraction cipher Khinchin’s Attack Applications Conclusion 1 Goal 1: I tried to find new techniques using continued fraction in cryptography. Result: I designed a new pseudo random generator statistically tested. 2 Goal 2: I was interested in finding new methods of cryptanalysis. Result: I designed a weak version which can be attacked by the Khinchin constant. 3 Goal 3: I tried to create a renewal of interest around continued fractions. Result: I introduced the works of Khinchin, Kuzmin, Levy, and Lochs in cryptology. Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph
- 32. Outline Introduction Continued Fractions On the use of continued fractions for stream cipher Questions For your attention Thank you! Amadou Moctar Kane Presentation: On the use of continued fractions for stream ciph