利用交错矩阵构造带仲裁的认证码  

Using alternation matrices to construct cartesian authentication codes with arbitration

作　　者：孔德宝[1] 

机构地区：[1]呼伦贝尔学院数学系,内蒙古海拉尔021008

出　　处：《辽宁师范大学学报（自然科学版）》2010年第1期26-30,共5页Journal of Liaoning Normal University:Natural Science Edition

摘　　要：设Fq是q元有限域,q是素数的幂.令信源集S为Fq上所有的n×n交错矩阵的合同标准型,编码规则集ET和解码规则集ER为Fq上所有的n×n非奇异矩阵,信息集为Fq上所有的n×n奇异的交错矩阵,构造映射f:S×ET→Mg:M×ER→S∪{欺诈}(K′(ν,n),P)→PK′(ν,n)Pt,(A,X)→{K′(ν,n)如果XKAKXt=K′(v,n),秩A=2ν欺诈,其他其中K=[In-1000].证明了该六元组(S,ET,ER,M;f,g)是一个带仲裁的Cartesian认证码,并计算了该认证码的参数.进而,当收方与发方的编码规则按照等概率均匀分布选取时,计算出该码敌方模仿攻击成功的概率PI,敌方替换攻击成功的概率PS,发方模仿攻击成功的概率PT,收方模仿攻击成功的概率PR0,收方替换攻击成功的概率PR1.Abstract.Let Fq be the finite field with q elements, where q is a power of a prime. Suppose the set of source states S is a cogredient normal form of all the n × n alternate matrices over Fq ,the set of encoding rules Er and decoding rules ER are all of the n × n nonsingular matrices over Fq, and the set of messages M is all of the n × n singular alternate matrices over Fq. Construct the maps f：s×Er→Mg：M×Ei→S∪{reject）（K＇（v,R）,P）→P＇（v,R）P^t,（A,X）→{K＇（v,R）if XKAKX^t=K＇（v,n）,rank A=2v reject, otherwise.where K=（In=1 0 0 0）. In this paper, we prove that the sets （S,Er,ER ,M;f,g） is a Cartesian authen tication code and the associated parameters are calculated. When the encoding rules obey a umIorm probability distribution,we calculate PI, Ps, PT,PR0 and PR1 , which denote the largest probabilities of a successful impersonation attacks by the opponent, a successful substitution attacks by the opponent, a successful impersonation attacks by the transmitter,a successful impersonation attacks by the receiver and a successful substitution attacks by the receiver, respectively.

关 键 词：带仲裁的认证码 交错矩阵 有限域 

分 类 号：O157.4[理学—数学]