CHAPTERTENSecret SharingA secret sharing scheme is a means for n parties to carry shares or parts s i of a message s,called the secret, such that the complete set s 1 , . . . s n of the parts determines the message.The secret sharing scheme is said to be perfect if no proper subset of shares leaks anyinformation regarding the secret.Two party secret sharing. Let s be a secret, encoding as an integer in Z/mZ. Lets 1 ∈ Z/mZ be generated at random by a trusted party. Then the two shares are definedto be s 1 and s − s 1 . The secret is recovered as s = s 1 + s 2 .Multiple party secret sharing. Let s ∈ Z/mZ be a secret to be shared among nparties. Generate the first n − 1 shares s 1 , . . . , s n−1 at random and setThe secret is recovered as s = ∑ ni=1 s i.∑n−1s n = s − .A (t, n) threshold secret sharing scheme is a method for n parties to carry shares s i of amessage s such that any t of the them to reconstruct the message, but so that no t − 1 ofthem can easy do so. The threshold scheme is perfect if knowledge of t − 1 or fewer sharesprovides no information regarding s.Shamir’s (t, n)-threshold scheme. A scheme of Shamir provide an elegant constructionof a perfect (t, n)-threshold scheme using a classical algorithm called Lagrange interpolation.First we introduce Lagrange interpolation as a theorem.i=1Theorem 10.1 (Lagrange interpolation) Given t distinct points (x i , y i ) of the form(x i , f(x i )), where f(x) is a polynomial of degree less that t, then f(x) is determined byf(x) =t∑i=1y i∏1≤j≤ti≠jx − x jx i − x j. (10.1)83
Shamir’s scheme is defined for a secret s ∈ Z/pZ with p prime, by setting a 0 = s, andchoosing a 1 , . . . , a t−1 at random in Z/pZ. The trusted party computes f(i), where∑t−1f(x) = a k x k ,k=0for all 1 ≤ i ≤ n. The shares (i, f(i)) are distributed to the n distinct parties. Since thesecret is the constant term s = a 0 = f(0), the secret is reovered from any t shares (i, f(i)),for I ⊂ {1, . . . , n} bys = ∑ i∈Ic i f(i), where each c i = ∏ j∈Ij≠iij − i .Properties. Shamir’s secret sharing scheme is (1) perfect — no information is leaked bythe shares, (2) ideal — every share is of the same size p as the secret, and (3) involves nounproven hypotheses. In comparison, most public key cryptosystems rely on certain wellknownproblems (integer factorization, discrete logarithm problems) to be hard in order toguarantee security.Proof of Lagrange interpolation theorem. Let g(x) be the right hand side of (10.1).For each x i in we verify directly that f(x i ) = g(x i ), so that f(x)−g(x) is divisible by x−x i .It follows thatt∏(x − x i ) ∣ (f(x) − g(x)), (10.2)i=1but since deg(f(x)−g(x)) ≤ t, the only polynomial of this degree satisfying equation (10.2)is f(x) − g(x) = 0.Example. Shamir secret sharing with p = 31. Let the threshold be t = 3, and the secretbe 7 ∈ Z/31Z. We choose elements at random a 1 = 19 and a 2 = 21 in Z/31Z, and setf(x) = 7 + 19x + 21x 2 . As the trusted pary, we can now generate as many shares as welike,(1, f(1)) = (1, 16) (5, f(5)) = (5, 7)(2, f(2)) = (2, 5) (6, f(6)) = (6, 9)(3, f(3)) = (3, 5) (7, f(7)) = (7, 22)(4, f(4)) = (4, 16) (8, f(8)) = (8, 15)which are distributed to the holders of the share recipients, and the original polynomialf(x) is destroyed. The secret can be recovered from the formulaf(x) =t∑ ∏y ii=11≤i≤ti≠jx − x jx i − x j=〉 f(0) =t∑ ∏y ii=11≤i≤ti≠jx jx j − x i84 Chapter 10. Secret Sharing
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Author (David R. Kohel) /Title (Cry
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CONTENTS1 Introduction to Cryptogra
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PrefaceWhen embarking on a project
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information. We introduce here some
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ut strings in A ∗ map injectively
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CHAPTERTWOClassical Cryptography2.1
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LV MJ CW XP QO IG EZ NB YH UA DS RK
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As a special case, consider 2-chara
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Note that if d k = 1, then we omit
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ExercisesSubstitution ciphersExerci
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Ciphertext-only AttackThe cryptanal
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of size n, suppose that p i is the
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Note that ZKZ and KZA are substring
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Checking possible keys, the partial
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sage: X = pt.frequency_distribution
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sage: a5 := a^n5sage: c5 := c^n5sag
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The application of this function E
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5. (∗) How many elements a of G h
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1. The value f(0) of the polynomial