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Digital Signature

Digital Signature. Integrity Authentication Unforgeable Non-repudiation. Conventional signature VS. Digital signature A signature is physically part of the document being signed. A digital signature is not attached physically to the message that is signed.

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Digital Signature

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  1. Digital Signature Integrity Authentication Unforgeable Non-repudiation

  2. Conventional signature VS.Digital signature • A signature is physically part of the document being signed. A digital signature is not attached physically to the message that is signed. • A signature is verified by comparing it to authentic signatures. Digital signatures can be verified using a publicly known verification algorithm. • A copy of a signed paper document can usually be distinguished from an original. A “copy” of a signed digital message is identical to the original.

  3. 傳統(對稱性)加密法之數位簽章需可信賴之第三者之助方得竟其功.{參考下圖}傳統(對稱性)加密法之數位簽章需可信賴之第三者之助方得竟其功.{參考下圖} M=Dka(C) C’=Ekb(M) TTP 3. C’ 2. C 1. C=Eka(M) M=Dkb(C’) Alice 與 Bob 沒有約定session key,兩人與TTP 分別約定 session keys ka及 kb, Bob 收到密文后直接轉給 TTP,而 TTP 用與 Alice 約定之session key 解得明文,再用與Bob約定之 session key 加密后,送給Bob. TTP被完全信任?

  4. RSA數位簽章 1. 本法乃為了確定送方身份而存在. 2. 所謂簽章,就RSA而言只不過送方先用唯獨自己知道的私鑰(private key) 加一次密之謂. Alice Bob Alice送出 C = (MdA mod mA)eBmod mB 簽章 加密 Bob 解得M = (CdB mod mB)eAmod mA

  5. Probabilistic encryption systems: 機率式公開金匙加密系統: PKC中，如Knapsack，RSA等均是確定式的(明密文一對一固定方式)，所以Intruder可以從截收到的密文C=E(M)中，針對數個最有可能的明文，如M’，進行加密而得C’=E(M’)。然後再將C與C’比對，若C=C’則可知M=M’。質言之,確定式的保密系統對Intruder而言，其破密資訊是可以累積的。 機率式加密系統對每一個明文使用不同的亂數(當然是在加密時)，因此一個明文可以對應到很多不同的密文。而解密時卻是解得唯一的明文。 • Rabin Public Key cryptosystem(1979) Rabin的方法是一個密文C可以對應到四個明文M(之中只有一個是正確的)。因此，在加密時必須加入一些有意義且易於分辨的訊息於明文中，使得解密時能夠明確地還原出原來的明文(參見次頁) 。

  6. 方法簡介: 選定n=p*q; 其中p與q是大質數。令明文為M，密文為C，公開加密金匙為 (b,n)，秘密解密金匙為(p,q)。 [加密程序]: C = M * (M + b) mod n， 其中b是亂數。 [解密程序]: 根據上式可知M2 + M*b - C = 0 mod n. 故明文可由下述四者之一算出: M = -b/2 (b/2)2+C mod p M = -b/2 (b/2)2+C mod q

  7. ElGamal 數位簽章 ElGamal 採機率式;即同一明文簽兩次會有相異簽署文 大質數 P, mod P 之原根g , 簽署者 B 任選之整數(私鑰)x , 1<x<P-1 B 算出之公鑰 y = g x mod P{用原根g 以免被解 離散對數} 系統已知 簽署作業 [ 此法之簽體(即明文 M)不能似 RSA簽章法之可直接還原] 對明文M, 1MP-1, B選一整數 k, gcd(k, P-1) = 1, B 計算簽署文 (r, s)滿足: r = g k mod P, s = k -1 (M - x r) mod P-1(或 M = x r + k s mod P-1). B 將 (r, s)送給 A. 驗證 驗證者 A verifies 下式是否為真? g M = yrr s mod P  gM = gx r g k s =yr r s mod P

  8. 安全性分析 1. 若可由 y = g x mod P 中之 y 及 g 推算出 x ,則本法危矣! (Discrete Logarithm Problem.) 2. Intruder A’ 欲偽造 (r, s)以滿足 g M = yr r s mod P同樣面臨解 DLP. 3. Intruder A’ 若獲得明文 M及其簽署 (r , s), 則因為 M = x r + k s mod P-1, 可知 想算出 x 及k 兩未知數難矣. 但若 B 利用相同的 k簽署 M1與 M2而得 簽署文 (r , s1) 及 (r , s2)  r = gk mod P, 則 聯立 M1 = x r + k s1 mod P-1 M2 = x r + k s2 mod P-1 而求解 x 與 k 易矣 ! 4.intruder 可偽造 M之合法簽署 (r, s) ;此intruder 偽造簽署文 (r, s) 如下: intruder 任選兩亂數 u and w, 1 < u, w < P-1 and gcd(w, P-1)=1, 計算 r = g u y -w mod P ………(1) s = r w -1 mod P-1 ………(2) 及 M = u s mod P-1 ………(3) {注意:M無法由intruder事先自定,它是被算出的因變數} 得 yr r s = yr (gu y -w)s = yr gus y -ws = yr gus y -r = gus = gMmod P. 但明文M無法由intruder事先自定;即M可能是無意義的數字爾 !故此種偽簽嚴格來看是nonsense.不過所有DS均有此攻擊法,而解決之道即引進單向赫序函數 ! {

  9. 多數數位簽章法均有類似的攻擊法存在,若能使用one-way hash function 合併之,則可高枕無憂矣. 請參照後續之介紹. Why One-way: (1) hash fun 對任意長度的明文,產生固定長度的密文. (2) hash fun對任意明文M, 藉 HW/SW 易求得 h(M). 防攻擊 : (3) 對任意hash fun 值 x, 計算上不可能藉 x=h(M)求M. (4) 對任一明文M1,要找另一明文M2使得 h(M1 )=h(M2) 在計算上不可行. (5) 要找任一對 (M1 ,M2)明文對,使得h(M1 )=h(M2) 在計算上不可行.  Unfortunately, although there are many functions that are believed to be one-way, there currently do not exist functions that can be proved to be one-way.  Zm is defined to be the set {0, 1, . . . , m-1}, equipped with two operators, + and *. Addition and multiplication in Zm work exactly like real addition and multiplication, except that the results are reduced modulo m.  相信f(x) = xb mod n, gcd(x, n)=1 是one-way function.  For a (finite) multiplicative group G, define the order of an element gG to be the smallest positive integer m such that gm = 1.

  10. Digital Signature Algorithm (DSA) 1991年，美國National Institute of Standards and Technology (NIST)公佈DSA為國家數位簽章標準(DSS)。DSA公佈後，引發以下幾點重大爭議(原則上，業界及學界仍是接受此一標準)： (1) DSA不能用來做加密或金鑰分配之用，只能用來做數位簽章。 (2) DSA是由美國國家安全局(NSA)所發展出來的，不像RSA或 ElGamal數位簽章法是由學界人士所設計出來的，普遍上仍是存在 是否存有暗門(trapdoor)的疑慮。 (3) DSA的計算速度比RSA要來得慢。簽署時間與RSA大約相同，但 驗證簽章的時間要比RSA慢上約10至40倍。 (4) RSA雖然不是政府頒布的一項標準(牽涉到專利的問題)，但是全世 界的使用者早已將之視為一項重要的數位簽章標準來使用。

  11. Schnorr's DS {CRYPTO’89} 系統已知 1. 大質數 P 及 q 滿足 q|P-1, q 2160及 P  2512 2. g  Zp且滿足 gq = 1 mod P, g  1 3. h 為 one-way hash fun. 4. Signer Bob 之私鑰 x , 1 < x < q 5. Bob 之公鑰 y = gx mod P. 1. Bob 任選一整數 k , 1 < k < q, 並求出r = g k mod P 2. 求出 e = h(r, M) 3. 求出 s = k - x e mod q, (e, s) 即 Bob 對 M 之簽署文. 簽署作業 驗證者(Verifier) Alice 求出: 1. r’ = gs ye mod P 2. 檢查 h(r’ , M) = e 驗證 ?  r’ = gsye = gk-xegxe = gk = r (mod P) h(r’ , M) = h(r , M) = e

  12. Observations 1.公鑰 y = gx mod P is precomputable, 故此法特別適用於 IC 卡上. 2. 此法之簽署文較ElGamal 為短, 因為e 的長度由 h函數決定, s 的長度小於 |q |. 3. Schnorr’s DS 中之g ( > 2160)非 ZP中之原根(其序為P-1), 故安全不若 ElGamal’s DS. ElGamal-type Signatures ElGamal簽章法乃以DLP為基,其systematic design 詳情可見 於“第八屆全國資訊安全會議論文集”之韓亮教授大作 pp. 1-16. 一般而言, 除了{ r, s, M, k, x } 五參數外,尚有one-way hash fun h 亦存在於簽章環境中,而簽署之對象是 h(M), 只是論文中已簡化成 M矣. 利用此五參數之排列組合,即可最少組合至 18 種簽章之多,當然此中有優劣之別. 另, since DSA is a special form of the original ElGamal scheme, which utilizes another parameter q, where q is a prime factor of P-1, all 18 ElGamal-type variants can be easily converted into DSA-type variants. 次頁簡介六種 DSA-type variants, 而其后再介紹 18種 ElGamal-type variants.

  13. DSA-type Variants

  14. (一) rr’ = gs ym ? (mod p) 左 = ( g k)r’ = g k r’ 右 = gs ym = gs (gx)m = gs+xm = g k r’ = 左

  15. ElGamal-type variants 前提如 DSA-type form, 然而 此處   g , {r, s} is the digital signature of message m, k and x are secrets of the signer. In ElGamal original scheme, there are five parameters {r, s, m, k, x} in the signature signing equation. Although additional generalization and variants can generate more than 13,000 variants, but these 18 variants are the most efficient variants since each of them permutes these five parameters directly. 1. mx = rk + s mod p-1 ym = r rs mod p 2. mx = sk + r mod p-1 ym = r sr mod p 3. rx = mk + s mod p-1 yr = r ms mod p 驗證第一種: mx = rk + s mod p-1 ym = r rs mod p 左 = ym = (gx)m = gxm 右 = r rs = (gk)r gs = gkr+s = gxm = 左

  16. 4. rx = sk + m mod p-1 yr = r ms mod p 5. sx = rk + m mod p-1 ys = r rm mod p 6. sx = mk + r mod p-1 ys = r mr mod p 7. rmx = k + s mod p-1 yr m = r s mod p 8. x = mrk + s mod p-1 y = r mrs mod p 9. sx = k + mr mod p-1 ys = r mr mod p 10. x = sk + rm mod p-1 y = r sr m mod p 11. rmx = sk + 1 mod p-1 yr m = r s mod p 12. sx = rmk + 1 mod p-1 ys = r mr mod p 13. (r+m)x = k + s mod p-1 yr+m = r s mod p 14. x = (m+r)k + s mod p-1 y = r m+rs mod p 15. sx = k + (m+r) mod p-1 ys = r m+r mod p 16. x = sk + (r+m) mod p-1 y = r sm+r mod p 17. (r+m)x = sk + 1 mod p-1 yr+m = r s mod p 18. sx = (r+m)k + 1 mod p-1 ys = r m+r mod p

  17. ( D. Chaum & H. van Antwerpen at CRYPTO’89) undeniable signature Known facts: a large prime P, and a primitive element g, are made public, and used by a group of signers. Alice has a private key, x, and a public key , gx. To sign a message, Alice computesz = Mx. That’s all she needs to do. Verification is a litter more complicated: (B 要 A 印證 z 是她簽署的) (1) Bob chooses two random numbers, a and b, both less then P, and sends Alice: c = z a(gx)b (mod P) (2) Alice computesx -1 (mod P-1),and sends Bob: d = cx -1 (mod P) (3) Bob confirms that d  Magb (mod P) (此式子可馬上導證;讀者自己試試看) If it is, he accepts the signature as genuine. 1. 此簽章法無法將還原明文M. 2. 除了signer Alice 外,無人知道其私鑰 x , 故不能否認簽署之事實. 3. 下一方法是D.Chaum 在EUROCRYP90所提之zero-knowledge undeniable signature .

  18. ZKUS ( D. Chaum at EUROCRYPTO’90) Known facts: a large prime P, and a primitive element g, are made public, and used by a group of signers. Alice has a private key, x, and a public key , gx. To sign a message, Alice computesz = Mx.  To verify a signature: (B 要 A 印證 [with Zero-Knowledge] z 是她簽署的) (1) Bob chooses two random numbers, a and b, both less then P, and sends Alice: c = Magb (mod P)  (2) Alice chooses a random number, q, less than P, and computes and sends to Bob: s 1 = cg q s 2 = (cg q)x (mod P) (3) Bob sends Alice a and b, so that Alice can confirm that Bob did not cheat in step (1). (4) Alice sends Bob q, so that Bob can use Mx and reconstructs s 1 and s 2. If s 1 = cg q s 2 = (gx)b+q za (此式子可馬上導證;讀者自己試試看) If they are, he accepts the signature as genuine.   

  19. 數學式推導 d  Magb (mod P) ? (此式子可馬上導證): d = cx -1 = (za (gx)b) x -1 = za/x (g1)b = Mx(a/x) gb = Magb (mod P) s 2 = (cg q)x = c x g xq= (Magb)xg xq= Maxgbx g xq = g xq+xbMax = (gx)b+q za s 2 = (gx)b+q za ?(此式子可馬上導證)

  20. Nyberg-Rueppel Signature 系統已知 Let P be a large prime number, Q be a prime factor of P-1, and G be a primitive element to mod P.{ all these three are known to public}, Now, the signer holds SK[1, Q-1] as secret key and PK=GSK mod P as public key , plaintext M [1, P-1] and a random number r [1, Q-1] . 簽署作業 SG1 = M * G r mod P SG2 = SK * SG1 + r mod Q 任何人可用下列方式驗證送方身份並且可recover message M : M G -SG2 * PKSG1 * SG1 mod P  G -SK*SG1 -r * GSK * SG1 * M * G r mod P  M * G -SK*SG1 -r + SK * SG1 +r mod P 驗證 ? 1994年 Carmenisch, Piveteau & Stadler(CPS)利用上述精神提出一盲簽章法

  21. D. Chaum (1982) Blind Signature 系統參數: signer 之public key (e, n), private key d, p, q (n=p*q). 簽署程序: A 請 B 簽署一份信息M,然 B 不被告知 M 之內容 (1) A 任選一亂數 r(即 blinding factor), 1< r < n, 且計算 t = M * r e mod n , 給B. (2) B 簽署 t, 即 = td = (M* r e) d mod n, 給A. (3) A 計算(unblind)出M 之簽章為: s =  * r -1 = td * r -1 = M d (mod n). 1.r -1 如何求出? [根據下述探討知 必另有求反元素之方法] If gcd(a,n)=1, then how to compute x s.t. a*x mod n = 1? @只需將 x 設為 x = a(n)-1 mod n即可!{ 不必求 (n) ? } 2.盲簽章應用於 digital money, electronic voting. 3.即使 A 公佈了 M 與 s , B 也無法找出 (M, s) 與 (t,  )之對應關係; 即 signer 無從追蹤明文 與當時盲簽章之相互關係. Think 利用Euclid's algorithm.(所謂法二)

  22. CPS blind-signature 系統參數: signer holds SK[1, Q-1] as secret key and PK=GSK mod P as public key , and a random number r [1, Q-1] . signer requester 1 K = G r mod P, r[1, Q-1] 任選兩亂數(盲因子) ,  [1, Q-1] 算 SG1 = M *G *Kmod P C = SG1 *  -1 mod Q 2 SGC = C * SK + r mod Q 3 算 SG2 = SGC *  +  mod Q (4) 任何人可以下述驗證 (5) ? M  G-SG2* PK SG1 * SG1 mod P = (G -SGC*  - ) * (GSK * SG1) * (M *G *K )mod P = (G -(C*SK + r)*  - ) * (GSK * SG1) * (M *G *G r * )mod P = (G -SG1*SK -r *  - ) * (GSK * SG1) * (M *G *G r * )mod P = M* G -SG1*SK -r *  -  +SG1*SK +r *  +  mod P.

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