【现代密码学】大作业-RSA大礼包RSA加密体制破译

RSA加密体制破译

题目来源

首届（2016 ）全国高校密码数学挑战赛赛题三

问题描述

RSA密码算法是使用最为广泛的公钥密码体制。该体制简单且易于实现，只需要选择5个参数即可（两个素数????和????、模数????=????????、加密指数????和解密指数????。设????为待加密消息RSA体制破译相当于已知$m^{e}modN$能否还原????的数论问题。目前模数规模为1024比特的RSA算法一般情况下是安全的，但是如果参数选取不当，同样存在被破译的可能。

有人制作了一个RSA加解密软件采用的RSA体制的参数特点描述见密码背景部分）。已知该软件发送某个明文的所有参数和加密过程的全部数据（加密案例文件详见附件3-1。Alice使用该软件发送了一个通关密语，且所有加密数据已经被截获，请问能否仅从加密数据恢复该通关密语及RSA体制参数？如能请给出原文和参数，如不能请给出已恢复部分并说明剩余部分不能恢复的理由？

1. 原始明文

This is a test of my RSA system.

2. Frame0

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

3. Frame1

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

4. Frame2

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

5. Frame3

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

6. 过程及参数

符号说明： n 模数、p 素数、q素数、e加密指数、d 解密指数、m 明文分片、c 密文分片、“0X”十六进制数据表示

明文： "This is a test of my RSA system."将其分割为4个8字符长度消息(注意：空格也是一个字符)

This is 	该8字符对应的ASCII为		54	68	69	73	20	69	73	20	将其视为64比特整数为==>	0X5468697320697320
a test o	该8字符对应的ASCII为		61	20	74	65	73	74	20	6F	将其视为64比特整数为==>	0X612074657374206F
f my RSA	该8字符对应的ASCII为		66	20	6D	79	20	52	53	41	将其视为64比特整数为==>	0X66206D7920525341
 system.	该8字符对应的ASCII为		20	73	79	73	74	65	6D	2E	将其视为64比特整数为==>	0X2073797374656D2E

选择前缀为0xFFFFFFFFFFFFFFFF,再添加通信序号和若干个0，最终填充后的4条消息依次为

0xFFFFFFFFFFFFFFFF0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000005468697320697320
0xFFFFFFFFFFFFFFFF000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000612074657374206F
0xFFFFFFFFFFFFFFFF00000002000000000000000000000000000000000000000000000000000000000000000000000000000000000000000066206D7920525341
0xFFFFFFFFFFFFFFFF0000000300000000000000000000000000000000000000000000000000000000000000000000000000000000000000002073797374656D2E

第0个明文分片参数及加密结果

p= 0XC60C5F1B997ED8A5E340023F33D2E269CFB423A3CF66B46D3F686747403A92B1265CB12B9A4E0135B890254F31A2C3F96A0427B39A36DEFDEEB85C57A80A9641
q= 0XD684DA331AB6157DA338B6D7B08AB4C1B72C29BB7F9EF445466056DFDBF29809C4D4A2435986A40DE688AFE7CC5A5C519F7C63CB486E44D523B0E1EF21C22199
n= 0XA5F51EB02EA9C0CC9B96926A08A761FE3E7CDB6E5B348DBEAEC761DBCFCDB15A6C76F8EE08196008AE60E396D7E228C6DAFC3CC1127F16EC87576B89C151F20F99098621FD46872BC92CDA8C915B758E5C0CACB994F55B8705B938126E08589E2502A7B9019C9A62E82392E8449E00CFC7DA17B8CDE92F9516CE9A2009F42DD9
e= 0X10001
d= 0X4C5340AAECBB1BB5BE74F09F9D9D45BF4583ECF38334D75FF44834A4809CEC4D57071C9374DC1EC3BF574634B0D30DC7EF1D04E0131EAA2F5C4B8364D6A95676C23F9DADAAB4523A6F5B22EC5904650BF558B3FDF39E3B13EA4771FB1D297DA03C8E1E82F4759B31A9492C56E4D1C690A66ECEC430849A17C027D1A7480F1E01
m= 0XFFFFFFFFFFFFFFFF0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000005468697320697320
c= 0X9726C82FED1E6CD58DE825528AE5634653C9921CAE02AFF7325F20D6E7085B7C8E3DC78D7518A78A8BC7D07E2E837083324579510851827794AE3D1FB9BAB360B1413A8F171A83804CEA73DFBC1248139BB27EB7D5BAD724AD8B08F51888B90562AF950725ACDD698F817AE62746CEA09479A191A6552B0116830355C68D0F61

第1个明文分片参数及加密结果

p= 0XC60C5F1B997ED8A5E340023F33D2E269CFB423A3CF66B46D3F686747403A92B1265CB12B9A4E0135B890254F31A2C3F96A0427B39A36DEFDEEB85C57A80A9641
q= 0XD684DA331AB6157DA338B6D7B08AB4C1B72C29BB7F9EF445466056DFDBF29809C4D4A2435986A40DE688AFE7CC5A5C519F7C63CB486E44D523B0E1EF21C22199
n= 0XA5F51EB02EA9C0CC9B96926A08A761FE3E7CDB6E5B348DBEAEC761DBCFCDB15A6C76F8EE08196008AE60E396D7E228C6DAFC3CC1127F16EC87576B89C151F20F99098621FD46872BC92CDA8C915B758E5C0CACB994F55B8705B938126E08589E2502A7B9019C9A62E82392E8449E00CFC7DA17B8CDE92F9516CE9A2009F42DD9
e= 0X10001
d= 0X4C5340AAECBB1BB5BE74F09F9D9D45BF4583ECF38334D75FF44834A4809CEC4D57071C9374DC1EC3BF574634B0D30DC7EF1D04E0131EAA2F5C4B8364D6A95676C23F9DADAAB4523A6F5B22EC5904650BF558B3FDF39E3B13EA4771FB1D297DA03C8E1E82F4759B31A9492C56E4D1C690A66ECEC430849A17C027D1A7480F1E01
m= 0XFFFFFFFFFFFFFFFF000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000612074657374206F
c= 0X789018FDB800AD59A54D27A77F280515BB3BFAB1CD75CB0A255A116D4A44849459FD887FF091F87C0B3E6305F019700E4E4CB3646D1DF276DFB87C4F64245F77377508EC6A796236F8ABB125023D3F4B898F55E3342D0A852193AF890990EA82F12FC85917BF132F2A58C449648D6E934B24E80307AB092DB18110D77BBA0F8E

第2个明文分片参数及加密结果

p= 0XD502B3D96C648A9393966CDD37188D37576AA221290C861B347ED7A57640993F7ED2D16992B42AA3CA66936D3268DE47EB3A61B1495C982BF54EC0350B907C4F3CA272F9ED04EEB355367DFDA1B89357130A25411DAC4E3B8A1EECC594E0435F0E7298897B54D6C334062C8D8508AC67CEDAECD1A5FCA84BF2EE5D
q= 0XE00258CB6F
n= 0XBA645145D9DE58B0FFA6FC4624A2815092D2A2DC405E7A2515F985727D3C52F479A4D04694568CA9B08391BE79BD122808CF6034AB7251088687BFF5916A4F4723FE1372DCF9B069CAB269A9F8F47CB50078D3279B9452C9B3B65A07B49C793783EDB8EB8D8F1A220D9EFED33147483103A2551A96932738255493F13B511953
e= 0X10001
d= 0X524EB244F1CE59966C273B91AC40B620CFB55BA2030E871F01147E11844888B6224C5D4DE14551DFDB93C984DAD94A4359643B247ED6CC7DE774A15440D525E26FE9CF4328DBEF2AAA8E402922596F1C23B8F117C018870777434C93B68F1028295DFA6E69FA8E00FFC4EFEF747C348EDBC99C529B7C3B649813647FF90A8261
m= 0XFFFFFFFFFFFFFFFF00000002000000000000000000000000000000000000000000000000000000000000000000000000000000000000000066206D7920525341
c= 0X534810A0D1B2F6FB257DC3BBDA30AA76157B89038E52D05EE1E5DB06C2D79FAE84892950EF5FD8ADC4F241C3741AD7C97002902C8CA4D96574F28EDCEF3BEF15303335FA8D250102B4EE77E3B405E30F6B81E92403A8881285B65F29668E05B9CD6AC44FC1CD193CF4A5811A2649BE0EDEFBA91FA7143266286C5EC6EE8077D6

第3个明文分片参数及加密结果

p= 0XD502B3D96C648A9393966CDD37188D37576AA221290C861B347ED7A57640993F7ED2D16992B42AA3CA66936D3268DE47EB3A61B1495C982BF54EC0350B907C4F3CA272F9ED04EEB355367DFDA1B89357130A25411DAC4E3B8A1EECC594E0435F0E7298897B54D6C334062C8D8508AC67CEDAECD1A5FCA84BF2EE5D
q= 0XE00258CB6F
n= 0XBA645145D9DE58B0FFA6FC4624A2815092D2A2DC405E7A2515F985727D3C52F479A4D04694568CA9B08391BE79BD122808CF6034AB7251088687BFF5916A4F4723FE1372DCF9B069CAB269A9F8F47CB50078D3279B9452C9B3B65A07B49C793783EDB8EB8D8F1A220D9EFED33147483103A2551A96932738255493F13B511953
e= 0X10001
d= 0X524EB244F1CE59966C273B91AC40B620CFB55BA2030E871F01147E11844888B6224C5D4DE14551DFDB93C984DAD94A4359643B247ED6CC7DE774A15440D525E26FE9CF4328DBEF2AAA8E402922596F1C23B8F117C018870777434C93B68F1028295DFA6E69FA8E00FFC4EFEF747C348EDBC99C529B7C3B649813647FF90A8261
m= 0XFFFFFFFFFFFFFFFF0000000300000000000000000000000000000000000000000000000000000000000000000000000000000000000000002073797374656D2E
c= 0X8805A937DABF25FE760F9F398C7D9D5955EF7468FEC89119F8DD69874FB009AB2C424BD6A8E85401C4CD130B48D0490586DFBD81C8154EDCEFC3AFC4F80338432197EB059AB54CF109B231416FB65E2F9BE4F01D455E25486D8E155A5874E8A910E8C65F73ACD953D316B35A148D5AC5834D86F66AD415EBA38AD3908B32780A

实例破解

在本次竞赛问题中，我们选取了一个具体加密实例供大家破解，整个算法与加密过程描述如下，截获的加密数据见附件3-2。

1. RSA 密码算法描述如下，包含体制参数选取和加解密过程。
   1）RSA 体制参数选取
   Step1.每个使用者，任意选择两个大素数 ????和 ????，并求出其乘积????=????????。
   Step2.令 ????(????)=(????−1)(????−1)选择整数 ????，使得 GCD(????,????(????))=1，并求出????模 ????(????)的逆元 ????，即????????≡1 mod ????(????)
   Step3.将数对 (????,????)公布为 公钥，????保存为私钥。
   2）加解密过程Bob欲传递明文????给 Alice，则Bob首先由公开途径找出 Alice 的公钥 (????,????)，Bob 计算加密的信息????为????≡???????? mod ????。
   Bob将密文 ????传送给 Alice 。随后 Alice 利用自己的私钥????解密：????????≡(????????)????≡????????????≡???? mod ????
2. Alice 使用的 RSA 密码体制有以下事项需要说明：
   1）模数????=????????规模为1024比特，其中????，????为素数；
   2）素数????由某一随机数发生器生成；
   3）素数????可以随机选择，也可以由2）中的随机数发生器产生；
   4）可以对文本加密，每次加密最多8个明文字符；
   5）明文超过8个字符时，对明文分片，每个分片不超过8个字符；
   6）分片明文填充为512比特消息后再进行加密，填充规则为高位添加64比特标志位，随后加上32比特通信序号，再添加若干个0，最后64比特为明文分片字符对应的ASCII码（注：填充方式参见加密案例，但注意每次通信的标志位可能变化）
   7）分片加密后发送一个加密帧数据，帧数据文件名称为FrameXX，其中XX表示接收序号，该序号不一定等于通信序号；
   8）帧数据的数据格式如下，其中数据都是16进制表示，结构如下1024bit模数N | 1024bit加密指数e | 1024bit密文$m^{e}modN$。
   9）由于Alice初次使用该软件，可能会重复发送某一明文分片。

截获的加密数据

1. Frame0

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

2. Frame1

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

3. Frame2

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

4. Frame3

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

5. Frame4

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

6. Frame5

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7. Frame6

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成绩评判

通过数论方法获得的原始明文及RSA 参数数量，数量多者获胜。

国内外研究进展与现状

RSA的安全性是基于大整数素因子分解的困难性，而大整数因子分解问题是数学上的著名难题。数域筛法是目前RSA攻击的首选算法。在1999年，一台Cray超级电脑用了5个月时间分解了512比特长的密钥。在512比特RSA算法破解10年之后,即2009年12月9日，768比特RSA算法即232数位数字的RSA-768被分解。分解一个768比特RSA密钥所需时间是512位的数千倍，而1024比特所需时间则是768比特的一千多倍，因此在短时间内1024比特仍然是安全的。除此之外，目前对于RSA算法的攻击主要有以下方式:选择密文攻击、公共模数攻击、低加密密指数攻击、低解密指数攻击、定时攻击等等，详细的RSA安全分析参见有关文献。

参考文献与可能用到的软件

1. 陈少真，密码学基础，科学出版社 , 2008-05-30
2. 任伟 , 现代密码学 , 北京邮电大学出版社 , 2011年4月
3. 冯登国 等译，密码学原理与实践（第三版），电子工业出版社
4. 谢建全阳春华RSA 算法中几种可能泄密的参数选择 , 《计算机工程》2006年16 期
5. Don Coppersmith Finding a Small Root of a Univariate Modular Equation EUROCRYPT 1996 L NCS 1070, Springe r 155 16 5
6. GMP package, GNU Multiple Precision Arithmetic Library
7. Magma Computational Algebra System
8. Pari
9. NTL: A Library for doing Number Theory

解答方法：

最后

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