I am converting f77 code to f90 code, and part of the code needs to sum over elements of a 3d matrix. In f77 this was accomplished by using 3 loops (over outer,middle,inner indices). I decided to use the f90 intrinsic sum (3 times) to accomplish this, and much to my surprise the answers differ. I am using the ifort compiler, have debugging, check-bounds, no optimization all turned on
Here is the f77-style code
r1 = 0.0
do k=1,nz
do j=1,ny
do i=1,nx
r1 = r1 + foo(i,j,k)
end do
end do
end do
and here is the f90 code
r = SUM(SUM(SUM(foo, DIM=3), DIM=2), DIM=1)
I have tried all sorts of variations, such as swapping the order of the loops for the f77 code, or creating temporary 2D matrices and 1D arrays to "reduce" the dimensions while using SUM, but the explicit f77 style loops always give different answers from the f90+ SUM function.
I'd appreciate any suggestions that help understand the discrepancy.
By the way this is using one serial processor.
Edited 12:13 pm to show complete example
! ifort -check bounds -extend-source 132 -g -traceback -debug inline-debug-info -mkl -o verify verify.f90
! ./verify
program verify
implicit none
integer :: nx,ny,nz
parameter(nx=131,ny=131,nz=131)
integer :: i,j,k
real :: foo(nx,ny,nz)
real :: r0,r1,r2
real :: s0,s1,s2
real :: r2Dfooxy(nx,ny),r1Dfoox(nx)
call random_seed
call random_number(foo)
r0 = 0.0
do k=1,nz
do j=1,ny
do i=1,nx
r0 = r0 + foo(i,j,k)
end do
end do
end do
r1 = 0.0
do i=1,nx
do j=1,ny
do k=1,nz
r1 = r1 + foo(i,j,k)
end do
end do
end do
r2 = 0.0
do j=1,ny
do i=1,nx
do k=1,nz
r2 = r2 + foo(i,j,k)
end do
end do
end do
!*************************
s0 = 0.0
s0 = SUM(SUM(SUM(foo, DIM=3), DIM=2), DIM=1)
s1 = 0.0
r2Dfooxy = SUM(foo, DIM = 3)
r1Dfoox = SUM(r2Dfooxy, DIM = 2)
s1 = SUM(r1Dfoox)
s2 = SUM(foo)
!*************************
print *,'nx,ny,nz = ',nx,ny,nz
print *,'size(foo) = ',size(foo)
write(*,'(A,4(ES15.8))') 'r0,r1,r2 = ',r0,r1,r2
write(*,'(A,3(ES15.8))') 'r0-r1,r0-r2,r1-r2 = ',r0-r1,r0-r2,r1-r2
write(*,'(A,4(ES15.8))') 's0,s1,s2 = ',s0,s1,s2
write(*,'(A,3(ES15.8))') 's0-s1,s0-s2,s1-s2 = ',s0-s1,s0-s2,s1-s2
write(*,'(A,3(ES15.8))') 'r0-s1,r1-s1,r2-s1 = ',r0-s1,r1-s1,r2-s1
stop
end
!**********************************************
sample output
nx,ny,nz = 131 131 131
size(foo) = 2248091
r0,r1,r2 = 1.12398225E+06 1.12399525E+06 1.12397238E+06
r0-r1,r0-r2,r1-r2 = -1.30000000E+01 9.87500000E+00 2.28750000E+01
s0,s1,s2 = 1.12397975E+06 1.12397975E+06 1.12398225E+06
s0-s1,s0-s2,s1-s2 = 0.00000000E+00-2.50000000E+00-2.50000000E+00
r0-s1,r1-s1,r2-s1 = 2.50000000E+00 1.55000000E+01-7.37500000E+00
First, welcome to StackOverflow. Please take the tour! There is a reason we expect a Minimal, Complete, and Verifiable example because we look at your code and can only guess at what might be the case and that is not too helpful for the community.
I hope the following suggestions helps you figure out what is going on.
Use the size() function and print what Fortran thinks are the sizes of the dimensions as well as printing nx, ny, and nz. As far as we know, the array is declared bigger than nx, ny, and nz and these variables are set according to the data set. Fortran does not necessarily initialize arrays to zero depending on whether it is a static or allocatable array.
You can also try specifying array extents in the sum function:
r = Sum(foo(1:nx,1:ny,1:nz))
If done like this, at least we know that the sum function is working on the exact same slice of foo that the loops loop over.
If this is the case, you will get the wrong answer even though there is nothing 'wrong' with the code. This is why it is particularly important to give that Minimal, Complete, and Verifiable example.
I can see the differences now. These are typical rounding errors from adding small numbers to a large sum. The processor is allowed to use any order of the summation it wants. There is no "right" order. You cannot really say that the original loops make the "correct" answer and the others do not.
What you can do is to use double precision. In extreme circumstances there are tricks like the Kahan summation but one rarely needs that.
Addition of a small number to a large sum is imprecise and especially so in single precision. You still have four significant digits in your result.
One typically does not use the DIM= argument, that is used in certain special circumstances.
If you want to sum all elements of foo, use just
s0 = SUM(foo)
That is enough.
What
s0 = SUM(SUM(SUM(foo, DIM=3), DIM=2), DIM=1)
does is that it will make a temporary 2D arrays with each element be the sum of the respective row in the z dimension, then a 1D array with each element the sum over the last dimension of the 2D array and then finally the sum of that 1D array. If it is done well, the final result will be the same, but it well eat a lot of CPU cycles.
The sum intrinsic function returns a processor-dependant approximation to the sum of the elements of the array argument. This is not the same thing as adding sequentially all elements.
It is simple to find an array x where
summation = x(1) + x(2) + x(3)
(performed strictly left to right) is not the best approximation for the sum treating the values as "mathematical reals" rather than floating point numbers.
As a concrete example to look at the nature of the approximation with ifort, we can look at the following program. We need to enable optimizations here to see effects; the importance of order of summation is apparent even with optimizations disabled (with -O0 or -debug).
implicit none
integer i
real x(50)
real total
x = [1.,(EPSILON(0.)/2, i=1, SIZE(x)-1)]
total = 0
do i=1, SIZE(x)
total = total+x(i)
print '(4F17.14)', total, SUM(x(:i)), SUM(DBLE(x(:i))), REAL(SUM(DBLE(x(:i))))
end do
end program
If adding up in strict order we get 1., seeing that anything smaller in magnitude than epsilon(0.) doesn't affect the sum.
You can experiment with the size of the array and order of its elements, the scaling of the small numbers and the ifort floating point compilation options (such as -fp-model strict, -mieee-fp, -pc32). You can also try to find an example like the above using double precision instead of default real.
I'm trying to generate a random integral and assign it to the variable.
import random
import time
Op = lambda: random.randint(1300, 19000)
op = "https://duckduckgo.com/html?q="
variable = int(Op())
grow = 0
while x < 3:
print(Op())
grow = grow + 1
time.sleep(1)
In here everything works fine, function "print" prints different result every time with 3 attempts.
However when I want to format this code like this:
Op = lambda: random.randint(1300, 19000)
op = "https://duckduckgo.com/html?q="
Op1 = int(Op())
pop = str("{}{}").format(op, Op1)
grow = 0
while grow < 3:
print(pop)
grow = grow + 1
time.sleep(1)
Then the function print gives me the same number three times.
For example:
>>>https://duckduckgo.com/html?q=44543
>>>https://duckduckgo.com/html?q=44543
>>>https://duckduckgo.com/html?q=44543
And I would like to get three random numbers. For example:
>>>https://duckduckgo.com/html?q=44325
>>>https://duckduckgo.com/html?q=57323
>>>https://duckduckgo.com/html?q=35691
I was trying to use %s - %d formatting but the result is the same.
Because you never changes the value of 'pop'.
In you first example you are creating instance of Op in every iteration but in second example you created instance once outside the loop and print the same value.
Try this:
Op = lambda: random.randint(1300, 19000)
op = "https://duckduckgo.com/html?q="
grow = 0
while grow < 3:
pop = str("{}{}").format(op, int(Op()))
print(pop)
grow = grow + 1
time.sleep(1)
Lambda functions are by definition anonymous. If you need to "remember" a lambda's procedure, just use def statement. But actually you don't even need this:
import random
import time
url_base = "https://duckduckgo.com/html?q={}"
grow = 0
while grow < 3:
print(url_base.format(random.randint(1300, 19000))
grow = grow + 1
time.sleep(1)
Your main problem is that you are trying to assign fixed values to variables and expect them to behave like procedures.
You need to apply randomness at every iteration. Instead you calculate a random number once and plug it in to every loop.
I want to terminate solving when it finds one solution, How can I do that?
Now it returns something like below:
* 62: obj = -8.419980000e+05 inf = 0.000e+00 (0)
OPTIMAL LP SOLUTION FOUND
Integer optimization begins...
+ 62: mip = not found yet >= -inf (1; 0)
+ 149: >>>>> -1.370260000e+05 >= -7.939630000e+05 479.4% (26; 0)
+ 1390: >>>>> -1.375090000e+05 >= -4.261680000e+05 209.9% (264; 27)
+ 28323: mip = -1.375090000e+05 >= -1.921510000e+05 39.7% (2232; 1534)
+ 52571: mip = -1.375090000e+05 >= -1.781890000e+05 29.6% (2983; 3596)
I think there is no nice way doing this.
These kind of advanced usages are usually done with more direct access to the solver (opposed to wrappers like this one; i'm assuming you are still using cvxopt like in your other questions).
Some remarks:
I didn't find any solver-parameter supporting this kind of early abort
cbc (which i prefer to glpk; supports this through setMaximumSolutions, but i think there is no wrapper within cvxopt)
What you could try:
set objll or objul docs
objul (default: +DBL_MAX)
Upper limit of the objective function. If the objective function reaches this limit and continues increasing, the solver stops the search. This parameter is used in the dual simplex only.
objll (default: -DBL_MAX)
Lower limit of the objective function. If the objective function reaches this limit and continues decreasing, the solver stops the search. This parameter is used in the dual simplex method only.
So: if you are minimizing, set objll to a huge value (or better: the expected worst-solution value = biggest value)
if you are maximizing, set objul to a tiny value (possibly negative; or better: the expected worst-solution value = smallest value)
I was assigned the following problem:
Make a Fortran program which will be able to read a degree[0-360] checking validity range(not type) and it will be able to calculate and print the cos(x) from the following equation, where x is in radians:
cos(x)=1-x^2/2! + x^4/4!-x^6/6!+x^8/8!-...
As a convergence criteria assume 10^(-5) using the absolute error between two successive repeats (I suppose it means do's).
For the calculation of the ! the greatest possible kind of integer should be used. Finally the total number of repeats should be printed on screen.
So my code is this:
program ex6_pr2
implicit none
!Variables and Constants
integer::i
real*8::fact,fact2 !fact=factorial
real,parameter::pi=3.14159265
double precision::degree,radiants,cosradiants,s,oldcosradiants,difference !degree,radiants=angle
print*,'This program reads and calculates an angle`s co-sinus'
print*,'Please input the degrees of the angle'
read*,degree
do while(degree<0 .or. degree>360) !number range
read*,degree
print*,'Error input degree'
cycle
end do
radiants=(degree*pi/180)
fact=1
fact2=1
s=0
cosradiants=0
!repeat structure
do i=2,200,1
fact=fact*i
fact2=fact2*(i+2)
oldcosradiants=cosradiants
cosradiants=(-(radiants)**i/fact)+(((radiants)**(i+2))/fact2)
difference=cosradiants-oldcosradiants
s=s+cosradiants
if(abs(difference)<1e-5) exit
end do
!Printing results
print*,s+1.
end program
I get right results for angles such as 45 degrees (or pi/4) and wrong for other for example 90 degrees or 180.
I have checked my factorials where I believe the error is hidden (at least for me).
Well I created another code which seems unable to run due to the following error:FUNCTION name,(RESULT of PROJECT2_EX6~FACT),used where not expected,perhaps missing '()'
program project2_ex6
implicit none
integer(kind=3)::degrees,i,sign
integer::n
double precision::x,err_limit,s_old,s
real,parameter::pi=3.14159265359
print*,'This program calculates the cos(x)'
print*,"Enter the angle's degrees"
read*,degrees
do
if(degrees<0.or.degrees>360) then
print*,'Degrees must be between 0-360'
else
x=pi*degrees/180
exit
end if
end do
sign=1
sign=sign*(-1)
err_limit=1e-5
n=0
s=0
s_old=0
do
do i=1,n
end do
s=(((-1.)**n/(fact(2.*n)))*x**(2.*n))*sign
s=s+s_old
n=n+1
if(abs(s-s_old)<1e-5) then
exit
else
s_old=s
cycle
end if
end do
print*,s,i,n
contains
real function fact(i)
double precision::fact
integer::i
if(i>=1) then
fact=i*fact(i-1)
else
fact=1
end if
return
end function
end program
Although it is your homework, I will help you here. The first thing which is wrong is ýour factorial which you need to replace with
fact = 1
do j = 1,i
fact = fact*j
enddo
second it is easier if you let your do loop do the job so run it as
do i=4,200,2
and predefine cosradians outside the do loob with
cosradiants = 1-radiants**2/2
additionally you need to take into account the changing sign which you can do in the loop using
sign = sign*(-1)
and starting it off with sign = 1 before the loop
in the loop its then
cosradiants= cosradiants+sign*radiants**i/fact
If you have included these things it should work (at least with my code it does)
I have a problem while using a do loop in fortran,
REAL W,V,X
DO 50 W = 0.5,5.0,0.5
DO 50 V = 10.0,1000.0,10.0
DO 50 X = 1.0,10,1.0
C=(W*V*X)/1000.0
WRITE(*,*) W,V,X,C
50 CONTINUE
STOP
END
If I gave this it is showing that only integers needs to be used in do loop, is there any way to give integers in do loop or any other way to do it?
Use integers as your looping indices
REAL W,V,X
INTEGER I,J,K
DO 50 I = 1,10
DO 50 J = 1,100
DO 50 K = 1,10
W = 0.5 * I
V = 10.0 * J
X = 1.0 * K
C=(W*V*X)/1000.0
WRITE(*,*) W,V,X,C
50 CONTINUE
STOP
END
You should be able to accomplish the same thing by incrementing a real variable by adding your step value, and using an if then to exit the loop. Clunky, but should work.
The last time I programmed in Fortran, I used punch cards and an IBM-360, so I'm not going to pretend I remember the syntax.