離散代數(shù)系統(tǒng)上機(jī)求幺元逆元

1、精選優(yōu)質(zhì)文檔-傾情為你奉上#include <iostream>#include <iomanip>using namespace std;int judgejiehe(char a4,char b) int i,j,k,l,x,y,flag=1; for(i=0;i<4;i+) for(j=0;j<4;j+) for(k=0;k<4;k+) for(l=0;l<4;l+) if(aij=bl) x=l;/aij=a*b if(ajk=bl) y=l;/ajk=b*c if(aiy!=axk)/aiy=a*(b*c) cout<<&q

2、uot;("<<bi<<"*"<<bj<<")*"<<bk<<"="<<axk<<","<<bi<<"*("<<bj<<"*"<<bk<<")="<<aiy<<",該運(yùn)算不可結(jié)合！"<<endl; flag=0;/不滿足結(jié)合

3、return 2; if(flag) cout<<"nn 運(yùn)算可結(jié)合！<A，*>構(gòu)成半群"<<endl; return 0;int judgeyaoniyuan(char a4,char b) int i,j,flag=1,counter=0; char c; for(i=0;i<4;i+) if(ai0=b0)&&(ai1=b1)&&(ai2=b2)&&(ai3=b3) for(j=0;j<4;j+) if(a0j=b0)&&(a1j=b1)&&

4、(a2j=b2)&(a3j=b3) cout<<" 幺元為"<<aij<<"，故<A，*>是獨(dú)異點(diǎn)"<<endl;flag=0; c = aij; if(flag) cout <<"無幺元！"<<endl; return 0; for(i=0;i<4;i+) for(j=0;j<4;j+) if(aij=c) if(i=j) cout<<" "<<bi<<"自身互為逆

5、元"<<endl; counter+; if(i!=j) cout<<" "<<bi<<"、"<<bj<<"互為逆元"<<endl; counter+; if(counter=4) cout<<" 每個(gè)元素都存在逆元，<A，*>構(gòu)成群"<<endl;int judgerepeat(char a4,char b) int i,j,counter=0,counter1=0,counter2=

6、0,counter3=0,counter4=0; for(i=0;i<4;i+) for(j=0;j<4;j+) if(aij=b0) counter1+; if(aij=b1) counter2+; if(aij=b2) counter3+; if(aij=b3) counter4+; counter = counter1+counter2+counter3+counter4; if(counter>16) cout<<" 該二元運(yùn)算的二維矩陣中某行列中有重復(fù)元素，故該二元運(yùn)算不是群"<<endl; if(counter=16)

7、cout<<" 該二元運(yùn)算的二維矩陣中每行每列沒有重復(fù)元素"<<endl;int main() char a44='a','b','c','e','b','a','e','c','c','e','b','a','e','c','a','b' char b4='a','b

8、9;,'c','e' char c; int i,j,counter=0,count1; cout<<"n設(shè)集合A=a,b,c,e,“*”是A上的二元運(yùn)算(滿足封閉性),運(yùn)算表通過二維矩陣進(jìn)行賦值.n"<<endl; cout <<" * |"<<setw(6)<<"a"<<setw(12)<<"b"<<setw(12)<<"c"<<setw(1

9、2)<<"e"<<endl; cout <<"-"<<endl; cout<<" a"<<setw(6)<<" | " for(i=0;i<4;i+) for(j=0;j<4;j+) cout<<aij<<setw(12); counter+; switch(counter) case 4: cout<<"n b"<<setw(6)<<" | "break; case 8: cout<<"n c"<<setw(6)<<" | "break; case 12: cout<<"n e"<<setw(6)