摘要
对分组密码算法差分能量攻击的样本数量选取问题进行研究。通过建立差分能量信号的信噪比模型,推导出样本数量的数学表达式为N>(8σ2+ε2(αn+n-d))/d2ε2,根据σ和ε计算得到攻击所需样本数量为8000。分别用5000组和8000组随机明文对高级加密标准算法进行差分能量攻击,结果证明,当样本数量为8000时可以得到正确密钥,效果结果优于5000组明文的情况。
Aiming at the problem of sample amount to differential power attack of block cipher,by establishing the SNR model of differential power signal,this paper proposes the expression of the sample amount： N（8σ2＋ε2（αn＋n-d））/d2ε2.After measuring the parameters σ and ε,the numerical value is got,which is about 8 000.Using the 5 000 samples and 8 000 samples separately to finish the Differential Power Attack（DPA） to Advanced Encryption Standard（AES）,and gets the right key when the samples’ amount is 8 000.The result is better than when it is 5 000,so the expression proposed is reasonable.
出处
《计算机工程》
CAS
CSCD
2012年第24期128-132,共5页
Computer Engineering
关键词
分组密码算法
高级加密标准
差分能量攻击
样本数量
block cipher algorithm
Advanced Encryption Standard（AES）
Differential PowerAttack（DPA）
sample amount
