Дипломная работа: Восьмиэлементные ассоциативные кольца
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 2 0 2 2 0 2 0
0 2 0 2 2 0 2 0
0 2 0 2 2 0 2 0
0 2 0 2 2 0 2 0
0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 1
0 0 0 0 2 2 2 2
0 0 0 0 3 3 3 3
0 0 0 0 4 4 4 4
0 0 0 0 5 5 5 5
0 0 0 0 6 6 6 6
0 0 0 0 7 7 7 7
0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 1
0 0 0 0 2 2 2 2
0 0 0 0 3 3 3 3
0 1 2 3 4 5 6 7
0 1 2 3 5 6 7 4
0 1 2 3 6 7 4 5
0 1 2 3 7 4 5 6
0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 1
0 0 0 0 2 2 2 2
0 0 0 0 3 3 3 3
0 1 2 3 5 6 7 4
0 1 2 3 6 7 4 5
0 1 2 3 7 4 5 6
0 1 2 3 4 5 6 7
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 2 0 2 0 2 0 2
0 2 0 2 2 0 2 0
0 2 0 2 0 2 0 2
0 2 0 2 2 0 2 0
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 2 0 2 1 3 1 3
0 2 0 2 3 1 3 1
0 2 0 2 1 3 1 3
0 2 0 2 3 1 3 1
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2
0 2 0 2 2 0 2 0
0 2 0 2 0 2 0 2
0 2 0 2 2 0 2 0
0 2 0 2 0 2 0 2
0 0 0 0 0 0 0 0
0 1 2 3 0 1 2 3
0 2 0 2 0 2 0 2
0 3 2 1 0 3 2 1
0 0 0 0 0 0 0 0
0 1 2 3 0 1 2 3
0 2 0 2 0 2 0 2
0 3 2 1 0 3 2 1
0 0 0 0 0 0 0 0
0 1 2 3 0 1 2 3
0 2 0 2 0 2 0 2
0 3 2 1 0 3 2 1
0 0 0 0 4 4 4 4
0 1 2 3 4 5 6 7
0 2 0 2 4 6 4 6
0 3 2 1 4 7 6 5
0 0 0 0 0 0 0 0
0 1 2 3 0 1 2 3
0 2 0 2 0 2 0 2
0 3 2 1 0 3 2 1
0 4 0 4 0 4 0 4
0 5 2 7 0 5 2 7
0 6 0 6 0 6 0 6
0 7 2 5 0 7 2 5
0 0 0 0 0 0 0 0
0 1 2 3 1 2 3 0
0 2 0 2 2 0 2 0
0 3 2 1 3 2 1 0
0 1 2 3 1 2 3 0
0 2 0 2 2 0 2 0
0 3 2 1 3 2 1 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 2 3 1 2 3 0
0 2 0 2 2 0 2 0
0 3 2 1 3 2 1 0
0 1 2 3 4 5 6 7
0 2 0 2 5 7 5 7
0 3 2 1 6 5 4 7
0 0 0 0 7 7 7 7
0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7
0 2 0 2 0 2 0 2
0 3 2 1 4 7 6 5
0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7
0 2 0 2 0 2 0 2
0 3 2 1 4 7 6 5
0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7
0 2 0 2 0 2 0 2
0 3 2 1 4 7 6 5
0 4 0 4 0 4 0 4
0 5 2 7 4 1 6 3
0 6 0 6 0 6 0 6
0 7 2 5 4 3 6 1
0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7
0 2 0 2 0 2 0 2
0 3 2 1 4 7 6 5
0 4 0 4 2 6 2 6
0 5 2 7 6 3 4 1
0 6 0 6 2 4 2 4
0 7 2 5 6 1 4 3
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
0 0 0 0 4 4 4 4
0 2 0 2 4 6 4 6
0 0 0 0 4 4 4 4
0 2 0 2 4 6 4 6
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
0 2 0 2 2 0 2 0
0 0 0 0 2 2 2 2
0 2 0 2 2 0 2 0
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 2 0 2 1 3 1 3
0 0 0 0 2 2 2 2
0 2 0 2 3 1 3 1
0 1 2 3 4 5 6 7
0 3 2 1 5 4 7 6
0 1 2 3 6 7 4 5
0 3 2 1 7 6 5 4
0 0 0 0 0 0 0 0
0 2 0 2 2 0 2 0
0 0 0 0 0 0 0 0
0 2 0 2 2 0 2 0
0 2 0 2 0 2 0 2
0 0 0 0 2 2 2 2
0 2 0 2 0 2 0 2
0 0 0 0 2 2 2 2
0 0 0 0 0 0 0 0
0 2 0 2 2 0 2 0
0 0 0 0 0 0 0 0
0 2 0 2 2 0 2 0
0 2 0 2 5 7 5 7
0 0 0 0 7 7 7 7
0 2 0 2 5 7 5 7
0 0 0 0 7 7 7 7
0 0 0 0 0 0 0 0
0 2 0 2 3 1 3 1
0 0 0 0 2 2 2 2
0 2 0 2 1 3 1 3
0 3 2 1 5 4 7 6
0 1 2 3 4 5 6 7
0 3 2 1 7 6 5 4
0 1 2 3 6 7 4 5
0 0 0 0 0 0 0 0
0 4 0 4 0 4 0 4
0 0 0 0 0 0 0 0
0 4 0 4 0 4 0 4
0 0 0 0 0 0 0 0
0 4 0 4 0 4 0 4
0 0 0 0 0 0 0 0
0 4 0 4 0 4 0 4
0 0 0 0 0 0 0 0
0 4 0 4 2 6 2 6
0 0 0 0 0 0 0 0
0 4 0 4 2 6 2 6
0 2 0 2 0 2 0 2
0 6 0 6 2 4 2 4
0 2 0 2 0 2 0 2
0 6 0 6 2 4 2 4
0 0 0 0 0 0 0 0
0 4 0 4 6 2 6 2
0 0 0 0 0 0 0 0
0 4 0 4 6 2 6 2
0 6 0 6 6 0 6 0
0 2 0 2 0 2 0 2
0 6 0 6 6 0 6 0
0 2 0 2 0 2 0 2
§3.
Кольца, образованные аддитивной группой 
Для нахождения колец с данной группой по сложению использовалась программа на языке Pascal (Приложение 2). Принцип действия данной программы, аналогичен принципу, описанному в предыдущем параграфе. Добавляется только пункт по нахождению всех базисов данной аддитивной группы.
Всего колец с аддитивной
группой 
будет 355. Выпишем
некоторые из них:
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 0 1 0 1 1 1
0 0 0 1 0 1 1 1
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 3 0 3 3 3
0 0 0 0 0 0 0 0
0 0 0 3 0 3 3 3
0 0 0 3 0 3 3 3
0 0 0 3 0 3 3 3
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 1 0 1 0 1 1
0 0 0 0 0 0 0 0
0 0 1 0 1 0 1 1
0 0 1 0 1 0 1 1
0 0 1 0 1 0 1 1
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 1 1 1 1 0 0
0 0 0 0 0 0 0 0
0 0 1 1 1 1 0 0
0 0 1 1 1 1 0 0
0 0 1 1 1 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 2 3 2 3 6 6
0 0 0 0 0 0 0 0
0 0 2 3 2 3 6 6
0 0 2 3 2 3 6 6
0 0 2 3 2 3 6 6
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 2 5 2 5 7 7
0 0 0 0 0 0 0 0
0 0 2 5 2 5 7 7
0 0 2 5 2 5 7 7
0 0 2 5 2 5 7 7
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 1 3 0 5 5 3
0 0 0 0 0 0 0 0
0 1 1 3 0 5 5 3
0 1 1 3 0 5 5 3
0 1 1 3 0 5 5 3
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 1 7 0 6 6 7
0 0 0 0 0 0 0 0
0 1 1 7 0 6 6 7
0 1 1 7 0 6 6 7
0 1 1 7 0 6 6 7
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7
0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7
0 1 2 3 4 5 6 7
0 1 2 3 4 5 6 7
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 4 0 3 4 7 3 7
0 0 0 0 0 0 0 0
0 4 0 3 4 7 3 7
0 4 0 3 4 7 3 7
0 4 0 3 4 7 3 7
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 4 0 6 4 5 6 5
0 0 0 0 0 0 0 0
0 4 0 6 4 5 6 5
0 4 0 6 4 5 6 5
0 4 0 6 4 5 6 5
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 0 1 0 1 1 1
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 0 1 0 1 1 1
0 0 0 1 0 1 1 1
0 0 0 1 0 1 1 1
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 1 0 1 0 1 1
0 0 0 1 0 1 1 1
0 0 1 0 1 0 1 1
0 0 1 1 1 1 0 0
0 0 1 1 1 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 1 1 1 1 0 0
0 0 0 1 0 1 1 1
0 0 1 1 1 1 0 0
0 0 1 0 1 0 1 1
0 0 1 0 1 0 1 1
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 1
0 0 1 2 1 2 4 4
0 0 0 1 0 1 1 1
0 0 1 2 1 2 4 4
0 0 1 4 1 4 2 2
0 0 1 4 1 4 2 2
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 2 0 2 2 2
0 0 0 3 0 3 3 3
0 0 0 2 0 2 2 2
0 0 0 3 0 3 3 3
0 0 0 6 0 6 6 6
0 0 0 6 0 6 6 6
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 2 0 2 2 2
0 0 0 5 0 5 5 5
0 0 0 2 0 2 2 2
0 0 0 5 0 5 5 5
0 0 0 7 0 7 7 7
0 0 0 7 0 7 7 7
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 2 0 2 2 2
0 0 2 3 2 3 6 6
0 0 0 2 0 2 2 2
0 0 2 3 2 3 6 6
0 0 2 6 2 6 3 3
0 0 2 6 2 6 3 3
Библиографический список.
1. Кострикин А.И. Введение в алгебру. М.: Наука, 1997.
2. Скорняков Л.А. Элементы алгебры. М.: Наука, 1986.
Приложение
1. Текст программы
поиска колец, имеющих аддитивную группу 
.
Программа 1. Формирование мультипликативных полугрупп:
program form;
uses crt;
type mas=array [0..7,0..7] of integer;
dmas=array [1..7,1..7] of integer;
var a:mas;
b:dmas;
f:text;
procedure sc(var a:mas);
var i,j:integer;
begin
assign(f,'a:\mat.txt');
reset(f);
for i:=0 to 7 do begin
for j:=0 to 7 do read(f,a[i,j]);
readln(f);
end;
close(f);
end;
procedure zap1(k1,k2,k3,k4:integer;a:mas; var b:dmas);
var t2,t3,t5,t6,t7:integer;
begin
b[1,1]:=k1; b[1,4]:=k2; b[4,1]:=k3; b[4,4]:=k4;
b[1,2]:=a[b[1,1],b[1,1]]; b[1,3]:=a[b[1,1],b[1,2]];
b[1,5]:=a[b[1,1],b[1,4]]; b[1,6]:=a[b[1,1],b[1,5]];
b[1,7]:=a[b[1,1],b[1,6]];
for t2:=1 to 7 do b[2,t2]:=a[b[1,t2],b[1,t2]];
for t3:=1 to 7 do b[3,t3]:=a[b[1,t3],b[2,t3]];
b[4,2]:=a[b[4,1],b[4,1]]; b[4,3]:=a[b[4,1],b[4,2]];
b[4,5]:=a[b[4,1],b[4,4]]; b[4,6]:=a[b[4,1],b[4,5]];
b[4,7]:=a[b[4,1],b[4,6]];
for t5:=1 to 7 do b[5,t5]:=a[b[1,t5],b[4,t5]];
for t6:=1 to 7 do b[6,t6]:=a[b[1,t6],b[5,t6]];
for t7:=1 to 7 do b[7,t7]:=a[b[1,t7],b[6,t7]];
end;
procedure zap(a:mas);
var f1:text;
i,j,k1,k2,k3,k4,r:integer;
begin
assign(f1,'a:\a2.txt');
rewrite(f1);
r:=0;
for k1:=0 to 7 do
for k2:=0 to 7 do
for k3:=0 to 7 do
for k4:=0 to 7 do begin
zap1(k1,k2,k3,k4,a,b);
for i:=1 to 7 do begin
for j:=1 to 7 do write(f1,b[i,j]);
writeln(f1);
end;
writeln(f1); inc(r);
end;
close(f1);
writeln(r);
end;
begin
clrscr;
sc(a);
zap(a);
writeln('ok');
readln;
end.
Программа 2. Проверка на ассоциативность:
program assoc;
uses crt;
type mas=array [0..7,0..7] of integer;
var a,b:mas;
i,j,k1,k2,k3,n,i1,i2,r,q:integer;
f,f1:text;
begin
clrscr;
q:=0;
assign(f,'a:\w03.txt');
reset(f);
assign(f1,'a:\aw03.txt');
rewrite(f1);
n:=1; r:=0;
while n<=2048 do begin
for i:=0 to 7 do begin
for j:=0 to 7 do read(f,a[i,j]);
readln(f);
end;
for k1:=1 to 7 do
for k2:=1 to 7 do
for k3:=1 to 7 do
if a[k1,a[k2,k3]]<>a[a[k1,k2],k3] then q:=1;
if q<>1 then begin
for i1:=0 to 7 do begin
for i2:=0 to 7 do write(f1,a[i1,i2],' ');
writeln(f1);
end;
inc(r);
writeln(f1);
end;
readln(f);inc(n);q:=0;
end;
close(f);
close(f1);
writeln(r);
writeln('ok');
readln;
end.
Программа 3. Нахождение изоморфных колец:
program izomorf;
uses crt;
const p=1;
type mas=array [0..7,0..7] of integer;
mas1=array [1..8,1..8] of integer;
var a,b:mas;
y:mas1;
i,j,n,i1,i2,r,w,h,g,d:integer;
f,f1,f2:text;
label 1;
procedure alfa(a:mas; var b:mas);
var k1,k2,j1,j2,z1,z2,z3,z4,q3,w3:integer;
begin
for k1:=0 to 7 do begin
b[0,k1]:=0;
b[1,k1]:=a[7,k1];
b[2,k1]:=a[2,k1];
b[3,k1]:=a[5,k1];
b[4,k1]:=a[4,k1];
b[5,k1]:=a[3,k1];
b[6,k1]:=a[6,k1];
b[7,k1]:=a[1,k1];
end;
for k2:=0 to 7 do begin
z1:=b[k2,3];z2:=b[k2,5];
b[k2,5]:=z1;b[k2,3]:=z2;
z3:=b[k2,1];z4:=b[k2,7];
b[k2,7]:=z3;b[k2,1]:=z4;
end;
for j1:=0 to 7 do
for j2:=0 to 7 do begin
if b[j1,j2]=1 then b[j1,j2]:=-1;
if b[j1,j2]=7 then b[j1,j2]:=-7;
end;
for j1:=0 to 7 do
for j2:=0 to 7 do begin
if b[j1,j2]=-1 then b[j1,j2]:=7;
if b[j1,j2]=-7 then b[j1,j2]:=1;
end;
for j1:=0 to 7 do
for j2:=0 to 7 do begin
if b[j1,j2]=5 then b[j1,j2]:=-5;
if b[j1,j2]=3 then b[j1,j2]:=-3;
end;
for j1:=0 to 7 do
for j2:=0 to 7 do begin
if b[j1,j2]=-5 then b[j1,j2]:=3;
if b[j1,j2]=-3 then b[j1,j2]:=5;
end;
for q3:=0 to 7 do begin
for w3:=0 to 7 do write(b[q3,w3],' ');
writeln;
end;
end;
procedure prov(b:mas);
var q,i3,j3,t,m,x,i4,j4:integer;
a1:mas;
begin
t:=1;
reset(f);
rewrite(f1);
for m:=1 to 9 do readln(f);
while t<=p do begin
for i3:=0 to 7 do begin
for j3:=0 to 7 do read(f,a1[i3,j3]);
readln(f);
end;
q:=0;
for i3:=0 to 7 do
for j3:=0 to 7 do begin
if a1[i3,j3]=b[i3,j3] then inc(q); end;
if q<>64 then
{for x:=1 to 9 do readln(f) else}
begin
for i4:=0 to 7 do begin
for j4:=0 to 7 do write(f1,a1[i4,j4],' ');
writeln(f1);
end;
writeln(f1);
end;
readln(f);
inc(t);
