Question

我正在尝试使用jacobi迭代解决ax = b，我的串行代码工作正常，但MPI版本甚至不会运行。有谁可以帮助我？

串行

program jacobis

implicit none

integer, parameter :: n=10
integer :: i,j,k,ni,s,seed
double precision :: tol,t1,t2,sig
double precision, dimension(0:n-1,0:n-1) :: A
double precision, dimension(0:n-1) :: B, x, xb, buff

ni=1000

seed=time()
call srand(seed)

do i=0, n-1
  do j=0, n-1
    A(i,j)=rand(0)
    B(i)=rand(0)
  end do
end do

do i = 0, n-1
 A(i,i) = sum(A(i,:)) + 1
enddo

!do i=0,n-1
 !A(i,i)=4
!end do  

print *, "a", A
print *, "b", B

x=B
call cpu_time(t1)
do k=1,ni
 xb=x
 do i=0,n-1
    s=0
    do j=0,n-1
    if (j/=i) then
         s=s+A(i,j)*xb(j)
        endif
    end do
    x(i)=(B(i)-s)/A(i,i) 

   sig=(x(i)-xb(i))*(x(i)-xb(i))
   tol=tol+sig
   tol=sqrt(tol)
 end do


 print *, "x", x

 !print *, "tol=", tol

 print *, "iter =",k

 if (tol<1.000001) EXIT
 if (k==(ni-1)) then
    print *, "Numero Maximo de Iteracoes" 
    EXIT
 endif
end do

 call cpu_time(t2)
 print *, "t=",t2-t1


end

MPI版

program jacobis

use mpi
implicit none

integer, parameter :: n=2
integer :: i_local,i_global,j,k,ni,s,m
double precision :: tol,t,t2,sig
double precision, dimension(:,:), ALLOCATABLE :: A_local
double precision, dimension(:), ALLOCATABLE :: B_local, x_local, x_temp1,x_temp2,x_old,x_new, buff
INTEGER, DIMENSION (MPI_STATUS_SIZE) :: STATUS
integer :: rank,procs,tag,ierror


CALL MPI_INIT(ierror)
CALL MPI_COMM_RANK(MPI_COMM_WORLD,rank,ierror)
CALL MPI_COMM_SIZE(MPI_COMM_WORLD,procs,ierror)

ni=100
m=n/procs

ALLOCATE (A_local(0:n-1,0:n-1))
ALLOCATE (B_local(0:m-1))
ALLOCATE (x_temp1(0:m-1))
ALLOCATE (x_temp2(0:m-1))

A_local=0
B_local=2

do i_global=0,n-1
 A_local(i_global,i_global)=2
end do  

CALL MPI_ALLGATHER(B_local, m, MPI_DOUBLE, x_temp1, m, MPI_DOUBLE, MPI_COMM_WORLD,ierror)

x_new=x_temp1
x_old=x_temp2

print *, "a", A_local
print *, "b", B_local


t=mpi_wtime()
do k=1,ni
 x_old=x_new
 do i_local=0,m-1
    i_global=i_local+rank*m 
    !x_local(i_local)=b_local(i_local)
    s=0
    do j=0,n-1
    if (j/=i_local) then
         s=s+A_local(i_local,j)*x_old(j)
        endif
    end do
    x_local(i_local)=(B_local(i_local)-s)/A_local(i_local,i_global) 

 end do
 CALL MPI_ALLGATHER(x_local,m, MPI_DOUBLE, x_new, m, MPI_DOUBLE, MPI_COMM_WORLD,ierror)

 do i_global=0,n-1
   sig=(x_new(i_global)-x_old(i_global))*(x_new(i_global)-x_old(i_global))
   tol=tol+sig
   tol=sqrt(tol)
 end do

 print *, "x", x_local

 print *, "tol=", tol

 print *, "iter =",k

 if (tol<1.000001) EXIT
 if (k==(ni-1)) then
    print *, "Numero Maximo de Iteracoes" 
    EXIT
 endif
end do

 t2=mpi_wtime()-t;
 print *, "t=",t2

CALL MPI_FINALIZE(ierror)
end

谁能指出我做错了什么？这是指数问题吗？请我今天真的需要解决这个问题，否则我将不及格。我花了无数个小时才完成这项工作。

好的你是对的！现在我有一个分段错误，但找不到它！用新版本取代了代码

Answer 1

您的程序有几个我可以看到的问题。您包含的错误消息表示此调用中未分配的接收缓冲区：

CALL MPI_ALLGATHER(B_local, m, MPI_DOUBLE, x_temp1, m, MPI_DOUBLE, MPI_COMM_WORLD)

接收缓冲区数组x_temp1需要在此上下文中使用之前进行分配。

修复此问题只会让您获得更多信息，并且您将获得信息量较少的分段错误。在MPI实现中查找MPI_AllGather的正确用法会很有用。大多数MPI例程最后都有一个整数错误状态参数：

MPI_ALLGATHER(SENDBUF, SENDCOUNT, SENDTYPE, RECVBUF, RECVCOUNT,
        RECVTYPE, COMM, IERROR)
    <type>    SENDBUF (*), RECVBUF (*)
    INTEGER    SENDCOUNT, SENDTYPE, RECVCOUNT, RECVTYPE, COMM,
    INTEGER    IERROR

这可以帮助您完成作业。确保分配您使用的所有allocatable数组，并为MPI实现和编译器手册使用适当的文档。

Answer 2

我已经解决了这个问题，现在它正确地计算了迭代次数，使用相同的矩阵证明了串行程序。这是一个分配和索引问题。感谢前面的回答，非常有帮助。

program jacobis

use mpi
implicit none

integer, parameter :: n=1000
integer :: i_local,i_global,j,k,ni,s,m,seed
double precision :: tol,t,t2,sig
double precision, dimension(:,:), ALLOCATABLE :: A_local
double precision, dimension(:), ALLOCATABLE :: B_local, x_local, x_temp1,x_old,x_new, buff
INTEGER, DIMENSION (MPI_STATUS_SIZE) :: STATUS
integer :: rank,procs,tag,ierror


CALL MPI_INIT(ierror)
CALL MPI_COMM_RANK(MPI_COMM_WORLD,rank,ierror)
CALL MPI_COMM_SIZE(MPI_COMM_WORLD,procs,ierror)

ni=1000
m=n/procs

ALLOCATE (A_local(0:n-1,0:n-1))
ALLOCATE (B_local(0:n-1))
ALLOCATE (x_local(0:n-1))
ALLOCATE (x_temp1(0:n-1))
ALLOCATE (x_new(0:n-1))

!A_local=23
!B_local=47

seed=time()
call srand(seed)

do k=0, n-1
 do j=0, n-1
    A_local(k,j)=rand(0)
    B_local(k)=rand(0)
 end do
end do

do i_global = 0, m-1
 A_local(i_global,i_global) = sum(A_local(i_global,:)) + n
enddo

CALL MPI_ALLGATHER(B_local, m, MPI_DOUBLE, x_temp1, m, MPI_DOUBLE, MPI_COMM_WORLD,ierror)

x_new=x_temp1

print *, "a", A_local
print *, "b", B_local


t=mpi_wtime()
do k=1,ni
 x_old=x_new
 do i_local=0,m-1
    i_global=i_local+rank*m 
    !x_local(i_local)=b_local(i_local)
    s=0
    do j=0,n-1
    if (j/=i_local) then
         s=s+A_local(i_local,j)*x_old(j)
        endif
    end do
    x_local(i_local)=(B_local(i_local)-s)/A_local(i_local,i_global) 
 end do
 CALL MPI_ALLGATHER(x_local,m, MPI_DOUBLE, x_new, m, MPI_DOUBLE, MPI_COMM_WORLD,ierror)
 do j=0,n-1
   sig=(x_new(j)-x_old(j))*(x_new(j)-x_old(j))
   tol=tol+sig
   tol=sqrt(tol)
 end do

 print *, "x", x_local

 print *, "tol=", tol

 print *, "iter =",k

 if (tol<1.01) EXIT
 if (k==(ni-1)) then
    print *, "Numero Maximo de Iteracoes" 
    EXIT
 endif
end do

 t2=mpi_wtime()-t;
 print *, "t=",t2

CALL MPI_FINALIZE(ierror)
end

Answer 3

您的程序存在严重问题，您可能会收到错误的结果。变量s声明为整数，而它被赋予非整数值。将其重新声明为双精度以获得正确的结果。（发布给那些复制此代码的人）

Fortran和MPI中的Jacobi迭代

3 个答案: