密碼編碼學(xué)與網(wǎng)絡(luò)安全講義-全文預(yù)覽
【正文】 stic Signature Scheme) ? new, remended RSASSAPKCS1v1_5 ? obsolete ? RSASSAPSS EMSAPSS＋ RSASP1/RSAVP1 ? RSASSAPKCS1v1_5 EMSAPKCS1v1_5＋ RSASP1/RSAVP1 B RSASSAPSS ? RSASSAPSSSIGN (K=(n,d), M) – EM = EMSAPSSENCODE (M, modBits – 1) – m = OS2IP (EM) – s = RSASP1 (K, m) – S = I2OSP (s, k) ? RSASSAPSSVERIFY ((n, e), M, S) B RSASSAPKCS1v1_5 ? RSASSAPKCS1V1_5SIGN (K＝ (n, d), M) – EM = EMSAPKCS1V1_5ENCODE (M, k) – m = OS2IP (EM) – s = RSASP1 (K, m) – S = I2OSP (s, k) ? RSASSAPKCS1V1_5VERIFY((n, e), M, S) B Encoding methods for signatures with appendix ? EM – EMSAPSS – EMSAPKCS1v1_5 ? EMSAPSS – EMSAPSSENCODE (M, emBits) – EMSAPSSVERIFY (M, EM, emBits) ? EMSAPKCS1v1_5 – EMSAPKCS1v1_5ENCODE (M, emLen) B EMSAPSS encoding operation bc pa dd in g2 m a s k ed DB M M pa dd in g1 m H a s h s a lt s a lt H ’ M G F M ’ = D B = EM = a H a s h H a s h B EMSAPKCS1v1_5 ? EMSAPKCS1v1_5ENCODE (M, emLen) – Option: Hash of hLenbyte – emLen at least tLen + 11 – H = Hash (M) – T = HashID＋ H ( ) ? DigestInfo ::= SEQUENCE { digestAlgorithm AlgorithmIdentifier, digest OCTET STRING } – PS: emLentLen3(8+B) octets with value 0xff – EM = 0x00 || 0x01 || PS || 0x00 || T B 對 HASH的 (in EMSAPKCS1v1_5) ? MD2 ref layman’s guide 30 20 30 0c 06 08 2a 86 48 86 f7 0d 02 02 05 00 04 10 || H ? MD5 30 20 30 0c 06 08 2a 86 48 86 f7 0d 02 05 05 00 04 10 || H ? SHA1 30 21 30 09 06 05 2b 0e 03 02 1a 05 00 04 14 || H ? SHA256 30 31 30 0d 06 09 60 86 48 01 65 03 04 02 01 05 00 04 20 || H ? SHA384 30 41 30 0d 06 09 60 86 48 01 65 03 04 02 02 05 00 04 30 || H ? SHA512 30 51 30 0d 06 09 60 86 48 01 65 03 04 02 03 05 00 04 40 || H B Reading B RSA簽名示例 用 OpenSSL函數(shù)簽名的例子 ? 讀入或產(chǎn)生 RSA的 key ? 簽署一個(gè)報(bào)文 ? 驗(yàn)證之 demo_sign,v B RSA四種操作總結(jié) ? 公鑰加密 RSA_public_encrypt() ? 私鑰解密 RSA_private_decrypt() ? 私鑰加密 RSA_private_encrypt() ? 公鑰解密 RSA_public_decrypt() ? 簽名 RSA_sign() ? (私鑰加密 ) ? 驗(yàn)證 RSA_verify() ? (公鑰解密 ) B { 回顧 ElGamal加密體制 } ? 準(zhǔn)備 –素?cái)?shù) p， Zp*中本原元 g，公開參數(shù) –私鑰 a，公鑰 b=ga mod p ? 加密 –對明文 1=m=p1，選隨機(jī)數(shù) k –密文 (c1, c2) c1=gk mod p, c2=mbk mod p ? 解密 – m＝ c2 (c1a)1＝ mbk ((gk)a)1 ＝ m(ga)k (gka) ＝ m mod p B ElGamal簽名方案 ? Zp滿足離散對數(shù)問題難解， α是生成元 設(shè) P ＝ Zp* ， A＝ Zp* Zp1 K ＝ {(p,α, a,β ), β =αa (mod p) } 私鑰是 a ? 簽名時(shí)，取秘密隨機(jī)數(shù) k∈ Zp1*， 定義
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