How to convert to a parallel program?
이전 댓글 표시
n=41;
h=1/(n-1);
nt=40;
x(1)=0; y(1)=0;
for i=2:n
x(i)=x(i-1)+h;
y(i)=y(i-1)+h;
end
for j=1:n
for i=2:n
p0(i,j)=1;
end
end
tic
tau=0.01;
for jt=1:nt % Vaqt bo`yicha tsikl}
t(jt)=jt*tau;
for j=1:n
for i=1:n
pt(i,j)=exp(-2*x(i)*y(j)*t(jt));
end
end
% k+0.s cloy
for j=2:21 % Tenglamalar koeffitsientini hisoblash
for i=2:n-1
a(i)=1;
c(i)=1;
b(i)=2+2*h*h/tau;
ff=h*h*(-2*(2*t(jt)*t(jt)*(x(i)*x(i)+y(j)*y(j))+x(i)*y(j))*exp(-2*x(i)*y(j)*t(jt)));
d(i)=2*h*h/tau*p0(i,j)+(p0(i,j-1)-2*p0(i,j)+p0(i,j+1))+ff;
end
%Pragonka
aa1(1)=0; bb1(1)=1;
for i=2:n-1
aa1(i)=c(i)/(b(i)-aa1(i-1)*a(i));
bb1(i)=(d(i)+bb1(i-1)*a(i))/(b(i)-aa1(i-1)*a(i));
end
p(n,j)=exp(-2*y(j)*t(jt));
for i=n-1:-1:1
p(i,j)=aa1(i)*p(i+1,j)+bb1(i);
end
end %j
for j=40:-1:22 % Tenglamalar koeffitsientini hisoblash
for i=2:n-1
a(i)=1;
c(i)=1;
b(i)=2+2*h*h/tau;
ff=h*h*(-2*(2*t(jt)*t(jt)*(x(i)*x(i)+y(j)*y(j))+x(i)*y(j))*exp(-2*x(i)*y(j)*t(jt)));
d(i)=2*h*h/tau*p0(i,j)+(p0(i,j-1)-2*p0(i,j)+p0(i,j+1))+ff;
end
%Pragonka
aa1(1)=0; bb1(1)=1;
for i=2:n-1
aa1(i)=c(i)/(b(i)-aa1(i-1)*a(i));
bb1(i)=(d(i)+bb1(i-1)*a(i))/(b(i)-aa1(i-1)*a(i));
end
p(n,j)=exp(-2*y(j)*t(jt));
for i=n-1:-1:1
p(i,j)=aa1(i)*p(i+1,j)+bb1(i);
end
end %j
for i=1:n
p(i,n)=(4.0*p(i,n-1)-p(i,n-2))/3;
p(i,1)=(4.0*p(i,2)-p(i,3))/3;
end
for j=1:n
for i=1:n
p0(i,j)=p(i,j);
end
end
% k+1 sloy
for i=2:21 % Tenglamalar koeffitsientini hisoblash }
for j=2:n-1
a(j)=1;
c(j)=1;
b(j)=2+2*h*h/tau;
ff=h*h*(-2*(2*t(jt)*t(jt)*(x(i)*x(i)+y(j)*y(j))+x(i)*y(j))*exp(-2*x(i)*y(j)*t(jt)));
d(j)=2*h*h/tau*p0(i,j)+(p0(i-1,j)-2*p0(i,j)+p0(i+1,j))+ff;
end
%Pragonka
aa1(1)=0; bb1(1)=1;
for j=2:n-1
aa1(j)=c(j)/(b(j)-aa1(j-1)*a(j));
bb1(j)=(d(j)+bb1(j-1)*a(j))/(b(j)-aa1(j-1)*a(j));
end
p(i,n)=exp(-2*x(i)*t(jt));
for j=n-1:-1:1
p(i,j)=aa1(j)*p(i,j+1)+bb1(j);
end
end %i
for i=40:-1:22 % Tenglamalar koeffitsientini hisoblash }
for j=2:n-1
a(j)=1;
c(j)=1;
b(j)=2+2*h*h/tau;
ff=h*h*(-2*(2*t(jt)*t(jt)*(x(i)*x(i)+y(j)*y(j))+x(i)*y(j))*exp(-2*x(i)*y(j)*t(jt)));
d(j)=2*h*h/tau*p0(i,j)+(p0(i-1,j)-2*p0(i,j)+p0(i+1,j))+ff;
end
%Pragonka
aa1(1)=0; bb1(1)=1;
for j=2:n-1
aa1(j)=c(j)/(b(j)-aa1(j-1)*a(j));
bb1(j)=(d(j)+bb1(j-1)*a(j))/(b(j)-aa1(j-1)*a(j));
end
p(i,n)=exp(-2*x(i)*t(jt));
for j=n-1:-1:1
p(i,j)=aa1(j)*p(i,j+1)+bb1(j);
end
end %i
for j=1:n
p(n,j)=(4.0*p(n-1,j)-p(n-2,j))/3;
p(1,j)=(4.0*p(2,j)-p(3,j))/3;
end
for i=1:n
for j=1:n
p0(i,j)=p(i,j);
end
end
end % jt
toc
for i=1:n
for j=1:n
b(j)=(p(i,j)-pt(i,j));
max=b(j);
end
end
for j=1:n
px(j)=p(21,j);
pxt(j)=pt(21,j);
end
t
p
pt
b
max
댓글 수: 4
Jan
2021년 9월 17일
What does "parallel program" mean? Do you want to vectrorize the code to increase the speed?
We cannot run the code due to the missing variables nt, n, x, y.
The outer loop is useless, because the contents of pt is overwritten in each iteration. Then it is much cheaper to omit the outer loop and run the code for jt=nt only.
Do you pre-allocate pt before the loops? Iteratively growing arrays need an exponentially growing amount of RAM, even if the final output is not such large.
The current version of your code does not need loops:
pt = exp(-2 * x(:) .* y(:).' * (nt * tau));
% ^^^ ^^^ maybe this can be omitted depending on
% what x and y are
Shixnazar Ismailov
2021년 9월 17일
Shixnazar Ismailov
2021년 9월 17일
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