I am having problems with a do while implementation for a sine taylor series. Editing the do loop to do bb = 1, 10 , 2 gives an expected result well within the margin of error, however when running the desired implementation of the do loop (do while(abs(sineseries) - accuracy > 0), will always give an answer equal to 1. So I have narrowed the possibilities down to the do while loop implementation being faulty.
program taylor
implicit none
real :: x
real :: sineseries, nfactsine
real, parameter :: accuracy = 1.e-10
integer :: signum, bb
nfactsine = 1
signum = 1
write(*,*) "Write your input value"
read(*,*) x
sineseries = 0
do while(abs(sineseries) - accuracy > 0)
sineseries = sineseries + (signum*x**bb)/nfactsine
nfactsine = nfactsine*(bb+1)*(bb+2)
signum = -signum
end do
write(*,*) sineseries, sin(x)
end program taylor
The two types of loops are not doing the same thing.
In the loop
do bb=1, 10, 2
...
end do
you have loop control with variable bb. This variable takes the values 1, 3, ..., 9 at iterations as the loop proceeds.
The do while does not have this control: you must replicate the increment of bb manually:
bb=1
do while (...)
...
bb=bb+2
end do
As Pierre de Buyl commented, you also have an error in the termination condition for the indefinite iteration count. The condition initially evaluates as false, so the loop body isn't executed even once.
Related
Im trying to print prime numbers till 10000. (display the first five for testing)
This is my program
program primes
implicit none
integer :: array(1229)
integer :: i, ind
logical :: is_prime
ind = 1
do i = 2, 10000, 1
if (is_prime(i) .eqv. .true.) then
array(ind) = i
ind = ind + 1
end if
end do
print *, array(1)
print *, array(2)
print *, array(3)
print *, array(4)
print *, array(5)
end program primes
function is_prime(n) result(ispr)
implicit none
integer :: c, i
integer, intent(in) :: n
logical :: ispr
c = 0
do i = 2, n
if (mod(i,2) == 0) then
c = c + 1
end if
end do
ispr = (c == 0)
end function is_prime
I don't know why but this is the output
9175178
6417360
5374044
6750309
7536745
Why does this happen and how to correct?
is_prime should be(n is the only divider of n besides 1 <=> c == 1)
function is_prime(n) result(ispr)
implicit none
integer :: c, i
integer, intent(in) :: n
logical :: ispr
c = 0
do i = 2, n
if (mod(n,i) == 0) then
c = c + 1
end if
end do
ispr = (c == 1)
end function is_prime
Could be optimezed by leaving the loop when c == 1 and i < n(after adding 1 to c)...
See on online fortran compiler
version with exit loop
While I am not familiar with modern Fortran, it looks to me as if function is_prime(n) result(ispr) is not working.
In the do loop in that function, you want a loop that tests thus:
is n divisible by 2?
is n divisible by 3?
is n divisible by 4?
is n divisible by 5?
and so on.
But, what it is actually doing is asking these:
is 2 divisible by 2?
is 3 divisible by 2?
is 4 divisible by 2?
is 5 divisible by 2?
and so on.
As a result, your counter will always have a non-zero value, and your function will always return false.
But, that's not the only problem. From your results, it looks like your Fortran implementation does not automatically initialize variables. Suppose I have statements like the following:
integer :: b
print *,b
What will be the result?
Remember, the names of variables represent locations in the computer's memory. If a variable is not initialized, it's value will be what was in the memory location before your program started to run. This value will not be related to your program.
I have 2 suggestions to fix the 2nd problem:
Prior to do i = 2, 10000, 1, have another loop that sets each value in array.
Set a values of each array (i) inside your do i = 2, 10000, 1 loop. One way to do this is to set one value when (is_prime(i) .eqv. .true.) is true and a different value when it is false.
I'm trying to write some Fortran 90 code to sum up the first 1234 multiples of 3 and 5 (including multiples of both). Here is my code so far:
program sum
implicit none
integer :: x
integer :: y = 5
integer :: z = 3
integer :: n
if (mod(x,y) == 0 .or. mod(x,z) ==0) then
print *, x
n = x
n = x + x
end if
end program sum
However, this code does not print anything to the terminal.
Your code tests the value of x in the if condition:
if (mod(x,y) == 0 .or. mod(x,z) ==0
but the value of x is not set at all. Therefore the result of the program is completely undefined. You need to create some kind of loop. Better two loops.
The most naive approach is to loop from 1 and test all numbers with the above if condition and stop when you have found the desired number of multiples.
I'm trying to use an if statement in a do loop which is supposed to generate prime numbers. For that I used modulo to sort out the numbers. After it found a prime number I want it to go a step further and add 1 so that the next prime number can be found and added to the array pzahl. My problem is that the loop seems to ignore that it should go a step further with plauf after it found a prime number so that it just keeps going till infinity... I tried to rearrange the contents of the loop and if statement but it's just not working. Here is the code:
PROGRAM Primzahlen
IMPLICIT NONE
INTEGER :: start, plauf, n, a
INTEGER, ALLOCATABLE, DIMENSION(:) :: pzahlen !array into which the prime numbers should be added
INTEGER :: input
INTEGER, DIMENSION(:), ALLOCATABLE :: alle
PRINT *, "How many prime numbers should be listed"
READ (*,*) input
ALLOCATE (pzahlen(input))
pzahlen(1) = 1
start = 2
plauf = 1
loop1: DO
ALLOCATE(alle(start))
loop2: DO n = 1,start
alle(n)= MODULO(start,n)
END DO loop2
IF (minval(alle) /= 0) THEN ! This is what it seems to ignore.
plauf= plauf + 1
pzahlen(plauf) = start
PRINT *, plauf
END IF
start = start + 1
IF (plauf == eingabe) then
EXIT
END IF
PRINT *, alle
DEALLOCATE(alle)
END DO loop1
PRINT *, "prime numbers:" , pzahlen(1:input)
END PROGRAM Primzahlen
I use the gfortran compiler and write it in Emacs if that helps to know.
It's not ignoring it, it executes correctly:
loop2: DO n = 1,start
alle(n)= MODULO(start,n)
END DO loop2
It doesn't matter what start is, alle(1) will always be zero, as every integer is evenly divisible by 1. That means that minval(alle) will also always be zero, which means that the condition minval(alle) /= 0 is never true, and the statement will never execute.
Added: The last value, alle(start), will also be zero, as every number is evenly divisible by itself.
Suppose you have a file.dat of the form:
1
1
1
2
2
3
3
3
3
...
I want to count how many equal numbers there are and save them iteratively in a string. For instance:
m = 3 (times 1),
m = 2 (times 2),
m = 4 (times 3).
I put here my code:
program sele
implicit none
integer::j,k,s,n,l,r,m
real*8,allocatable::ID(:)
real*8:: j_r8,i_r8
open(10,file='data.dat')
n=0
DO
READ(10,*,END=100)
n=n+1
END DO
100 continue
rewind(10)
allocate(ID(n))
s=0
do s=1, n
read(10,*) ID(s)
end do
do r=1,n-1
if (ID(r)-ID(r+1) .EQ. 0) then
m = m + 1
print*, m
end if
end do
end program
The last do is the condition I'd like to expand, with something like:
if (condition is true) then
save an index of the number of equal digits
use this to do some operations:
do i = 1, number of equal digits
if (condition is not true) then
restart with the other digits.
If the values you want to read are integer values in a given limited range (for instance from 1 to 100), then the simplest way is the following :
program sele
implicit none
integer, parameter :: vmin=1
integer, parameter :: vmax=100
integer :: list(vmin:vmax)
integer :: value,i
open(10,file='data.dat')
list=0
do
read(10,*,end=10) value
if(value < vmin .OR. value > vmax) then
write(*,*) 'invalid value ',value
stop
endif
list(value)=list(value)+1
enddo
10 continue
do i=vmin,vmax
if(list(i) > 0) then
write(*,*) list(i),' times ',i
endif
enddo
end program
Which gives on your example :
3 times 1
2 times 2
4 times 3
It is possible to improve easily that program to manage variable vmin and vmax (the array list must then be declared allocatable and allocated at the right size).
If the range is too large, then a simple array is not accurate anymore and the right algorithm becomes more complicated : it must avoid to store unused values.
I'm trying to write a program to find the mean, median, mode of an integer array but am having some complications in finding the mode. The following is the code that I've written so far.
First, the program will prompt user to enter a value for the number of integers that will be entered followed by request to enter that number of integers. The integers are then sorted in ascending order and the mean and median are found.
The problem I am having is when I try to get the mode. I am able to count the number of occurrence of a repetitive value. By finding the value with highest occurrence, we'll be able to find Mode. But I am unsure how to do this. Is there any intrinsic function in Fortran to calculate number of occurrence of input values and the value with highest occurrence?
PROGRAM STATISTICS
!Created by : Rethnaraj Rambabu
IMPLICIT NONE
REAL, DIMENSION(:), ALLOCATABLE:: VAL
REAL TEMP, MEDIAN
REAL EVEN, MEAN, SUM, FMODE
INTEGER N, I,J
WRITE(*,*)' WHAT IS THE VALUE FOR N? '
READ(*,*) N
ALLOCATE(VAL(N))
WRITE(*,*) 'ENTER THE NUMBERS'
OPEN(1,FILE='FILE.TXT')
READ(1,*)(VAL(I),I=1,N)
CLOSE(1)
WRITE(*,*) VAL
!/---FOR SORTING----/!
DO I=1,N-1
DO J=1,N-1
IF(VAL(J) > VAL(J+1)) THEN
TEMP=VAL(J)
VAL(J)=VAL(J+1)
VAL(J+1)=TEMP
END IF
END DO
END DO
WRITE(*,*) VAL
!/-----MEDIAN----/!
IF ((N/2*2) /= N) THEN
MEDIAN=VAL((N+1)/2)
ELSE IF ((N/2*2) == N) THEN
EVEN= (VAL(N/2)+VAL((N+2)/2))
MEDIAN=EVEN/2
END IF
WRITE(*,*)'MEDIAN=', MEDIAN
!/----MEAN----/
SUM=0
DO I=1,N
SUM=SUM+VAL(I)
END DO
MEAN=SUM/N
WRITE(*,*)'MEAN=', MEAN
!/------MODE----/
FMODE=1
DO I=1,N-1
IF (VAL(I) == VAL(I+1)) THEN
FMODE=FMODE+1
END IF
END DO
WRITE(*,*)FMODE
END PROGRAM
The FILE.TXT contains
10 8 1 9 8 9 9 7 5 9 3 5 6
But, how to do that? Or is there any intrinsic function in Fortran to calculate number of occurrence of input values and the value with highest occurrence.
No, there is not. You'll have to calculate the mode by hand.
The following code should work (on a sorted array):
FMODE = VAL(1)
COUNT = 1
CURRENTCOUNT = 1
DO I = 2, N
! We are going through the loop looking for values == VAL(I-1)...
IF (VAL(I) == VAL(I-1)) THEN
! We spotted another VAL(I-1), so increment the count.
CURRENTCOUNT = CURRENTCOUNT + 1
ELSE
! There are no more VAL(I-1)
IF (CURRENTCOUNT > COUNT) THEN
! There were more elements of value VAL(I-1) than of value FMODE
COUNT = CURRENTCOUNT
FMODE = VAL(I-1)
END IF
! Next we are looking for values == VAL(I), so far we have spotted one...
CURRENTCOUNT = 1
END
END DO
IF (CURRENTCOUNT > COUNT) THEN
! This means there are more elements of value VAL(N) than of value FMODE.
FMODE = VAL(N)
END IF
Explanation:
We keep the best-so-far mode in the FMODE variable, and the count of the FMODE in the COUNT variable. As we step through the array we count the number of hits that are equal to what we are looking at now, in the CURRENTCOUNT variable.
If the next item we look at is equal to the previous, we simply increment the CURRENTCOUNT. If it's different, then we need to reset the CURRENTCOUNT, because we will now count the number of duplications of the next element.
Before we reset the CURRENTCOUNT we check if it's bigger than the previous best result, and if it is, we overwrite the previous best result (the FMODE and COUNT variables) with the new best results (whatever is at VAL(I) and CURRENTCOUNT), before we continue.
This reset doesn't happen at the end of the loop, so I inserted another check at the end in case the most frequent element happens to be the final element of the loop. In that case we overwrite FMODE, like we would have done in the loop.
It is a bit lengthy, you could probably get rid of the optional argument, but there is an example provided here. They use the quick sort algorithm as implemented here.
Alternatively, you could use
integer function mode(arr) result(m)
implicit none
integer, dimension(:), intent(in) :: arr
! Local variables
integer, dimension(:), allocatable :: counts
integer :: i, astat
character(len=128) :: error_str
! Initialise array to count occurrences of each value.
allocate(counts(minval(arr):maxval(arr)), stat=astat, errmsg=error_str)
if (astat/=0) then
print'("Allocation of counts array failed.")'
print*, error_str
end if
counts = 0
! Loop over inputted array, counting occurrence of each value.
do i=1,size(arr)
counts(arr(i)) = counts(arr(i)) + 1
end do
! Finally, find the mode
m = minloc(abs(counts - maxval(counts)),1)
end function mode
This doesn't require any sorting.