AES解密算法的问题……

1920 · 发表于 2010-4-30 20:54:36

马上注册，结交更多好友，享用更多功能，让你轻松玩转社区。

您需要登录才可以下载或查看，没有账号？注册

x

请教一下实现AES解密算法中的逆列混合变化和逆的行移位如何用Verilog代码实现呢，在网上没找到，如果哪位前辈知道如何设计，恳请赐教……

kueikuei · 发表于 2010-5-4 14:37:27

我猜你應該是在 modulus 上面有問題
今天假設 A* B 的結果是C
而你還要針對 C 去做 modulus

1920 · 发表于 2010-5-12 23:03:29

module 都能写错……

悲伤的丢车 · 发表于 2010-5-14 03:09:34

没啥说的啦飘过 ……都麻木了

hllyh · 发表于 2010-5-14 12:14:54

http://www.eetop.cn/bbs/thread-239812-1-1.html
有AES算法源码。

xiaotanglang · 发表于 2010-5-17 16:22:31

楼主，有需求可以讨论，本人有刚写好的，比较实用，比网上疯传的那个好。哈哈
QQ：809174805

huangchangsandy · 发表于 2010-5-24 10:58:44

1# 1920

去中国期刊网下载相关硕士论文，所有问题都有的。

alexander21th · 发表于 2011-6-6 12:23:51

AES是个比较成熟的算法，去开源上看看估计应该是有的。

warmheard · 发表于 2011-6-7 10:54:52

逆列混合其实就是有限域乘法问题，而且是变量与固定数乘法，分析一下有限域乘法怎么做，自然就知道invmixcol怎么实现了。

Finite Field Multiplication
An example is shown as below.
{57}•{83} = {57}•{01} ^ {57}•{02} ^ {57}•{80}
{57}•{01} = {57}
{57}•{02} = {AE} //(Shift 1 bit to left, and XOR{1b} if the MSB before shifting is 1)
{57}•{04} = {AE}•{02} = {1}{01011100}
{57}•{08} = {47}•{02} = {8E}
{57}•{10} = {8E }•{02} = {1}{00011100}
{57}•{20} ={07}•{02} = {0E}
{57}•{40} ={0E}•{02} = {1C}
{57}•{80} ={1C }•{02} = {38}
{57}•{83} = {57} ^ {AE} ^{38} = {C1}

Another way to calculate {57}•{83} which is suitable for hardware implementation.
{57}•{83}
= {57}•{01}^ {57}•{02} ^ {57}•{80}
= {01010111} ^ {10101110} ^ {0101011}{10000000}
= {0101011}{01111001}
Step1: {0101011}{01111001}
= {01011}{01111001} ^ {00011}{01100000}
= {01000}{00011001}
Step2: {01000}{00011001}
= {000}{00011001}^{000}{11011000}
= {11000001} = {C1}

warmheard · 发表于 2011-6-7 10:55:41

逆列混合其实就是有限域乘法问题，而且是变量与固定数乘法，分析一下有限域乘法怎么做，自然就知道invmixcol怎么实现了。

Finite Field Multiplication
An example is shown as below.
{57}•{83} = {57}•{01} ^ {57}•{02} ^ {57}•{80}
{57}•{01} = {57}
{57}•{02} = {AE} //(Shift 1 bit to left, and XOR{1b} if the MSB before shifting is 1)
{57}•{04} = {AE}•{02} = {1}{01011100}
{57}•{08} = {47}•{02} = {8E}
{57}•{10} = {8E }•{02} = {1}{00011100}
{57}•{20} ={07}•{02} = {0E}
{57}•{40} ={0E}•{02} = {1C}
{57}•{80} ={1C }•{02} = {38}
{57}•{83} = {57} ^ {AE} ^{38} = {C1}

Another way to calculate {57}•{83} which is suitable for hardware implementation.
{57}•{83}
= {57}•{01}^ {57}•{02} ^ {57}•{80}
= {01010111} ^ {10101110} ^ {0101011}{10000000}
= {0101011}{01111001}
Step1: {0101011}{01111001}
= {01011}{01111001} ^ {00011}{01100000}
= {01000}{00011001}
Step2: {01000}{00011001}
= {000}{00011001}^{000}{11011000}
= {11000001} = {C1}

[求助] AES解密算法的问题……

马上注册，结交更多好友，享用更多功能，让你轻松玩转社区。

相关帖子

站长推荐 /2