Crypto: How the Code Rebels Beat the Government Saving by Steven Levy

By Steven Levy

If you've ever made a safe buy together with your bank card over the web, then you definately have visible cryptography, or "crypto," in motion. From Steven Levy-the writer who made "hackers" a family word-comes this account of a revolution that's already affecting each citizen within the twenty-first century. Crypto tells the interior tale of the way a gaggle of "crypto rebels"-nerds and visionaries became freedom fighters-teamed up with company pursuits to overcome colossal Brother and make sure our privateness on the net. Levy's heritage of 1 of the main arguable and demanding themes of the electronic age reads just like the top futuristic fiction.

"Gripping and illuminating." (The Wall road Journal)

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Additional resources for Crypto: How the Code Rebels Beat the Government Saving Privacy in the Digital Age

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7. Define a “near-collision” to be a pair of elements from two sequences as in Floyd’s cycle-finding algorithm which differ by at most 2. Adapt Floyd’s algorithm to this case. 8. Redo Problem 5 to find a “near-collision” as defined in Problem 7. Computer-Assisted Problems 9. Use Floyd’s cycle-finding algorithm to find two different locations with the same value in the sequence x0 = 3 and f (xi ) = 2xi2 + 7 in a congruence system modulo 111. 10. html. This is a well-known site on integer sequences.

16 Let q = 2663 and a = 2 be public, where a is a primitive root modulo 2663. Alice chooses xA = 1085; Bob chooses xB = 1701. Alice computes 21085 ≡ 252 (mod 2663); • Bob computes 21701 ≡ 1524 (mod 2663). • Both people publish these values. DISCRETE LOGARITHMS AND DIFFIE – HELLMAN KEY EXCHANGE 31 • Alice takes Bob’s 1524 and uses her exponent to get 15241085 ≡ 2103 (mod 2663). • Bob takes Alice’s 252 and uses his exponent to get 2521701 ≡ 2103 (mod 2663). They can now both use k = 2103 as a shared secret key.

Therefore, a q−1 2 ≡ −1 (mod q). Returning to y = ax (mod q), raise both sides to the power (q − 1)/2 to get: y q−1 2 ≡ a q−1 2 x ≡ (−1)x (mod q). We know y and q and so can compute: if y if y q−1 2 q−1 2 ≡ 1 (mod q), then x must be even; ≡ −1 (mod q), then x must be odd. 32 CONGRUENCE EQUATIONS Since an attacker can easily compute this power of y, she knows the parity of the secret value, thus cutting her work to find it in half. There are several algorithms available for computing discrete logarithms, usually assuming once again that certain conditions hold.

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