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I wrote a fortran program to code this algorithm (https://en.wikipedia.org/wiki/Reservoir_sampling#Algorithm_A-ExpJ). It works on my computer. But after I asked these two questions (Intrinsic Rand, what is the interval [0,1] or ]0,1] or [0,1[ and How far can we trust calculus with infinity?), I think I could have a problem with log(random()) because call random_number(Xw); Xw = log(Xw) is used.
Indeed, random_number(Xw) could return 0 and log(0)=-infinity.
Therefore, I plan to modify this line as follow call random_number(Xw); Xw = log(1-Xw) to change the random value interval from [0,1[ to ]0,1].
Is it a good idea or is there a best solution ?
While mathematically it is true that if X is uniformly distributed on (the real interval) [0,1) then 1-X is uniformly distributed on (0,1], this does not particularly help you.
As noted in the description of the algorithm to which you link, the underlying assumption is that the base uniform distribution is over the interval (0,1). This is not the same as (0,1].
You can use rejection sampling to generate X uniformly over (0,1) from random_number() (which is [0,1)): throw away all zero occurrences.
Not a good idea. If you want your algorithm to be stable, you need to define bounds. your log function represents a priority, it can likely be as low as you want, but it must be a number. You can bind it to numerical precision:
program t
use iso_fortran_env
implicit none
real(real64), parameter :: SAFE = exp(-0.5d0*huge(0.0_real64))
print *, log(randoms_in_range(100,SAFE,1.0_real64))
contains
elemental real(real64) function in_range(f,low,hi) result(x)
real(real64), intent(in) :: f ! in: [0:1]
real(real64), intent(in) :: low,hi
real(real64) :: frac
frac = max(min(f,1.0_real64),0.0_real64)
x = low+frac*(hi-low)
end function in_range
real(real64) function random_in_range(low,hi) result(x)
real(real64), intent(in) :: low,hi
call random_number(x) ! [0,1]
x = in_range(x,low,hi) ! [low,hi]
end function random_in_range
function randoms_in_range(n,low,hi) result(x)
integer , intent(in) :: n
real(real64), intent(in) :: low,hi
real(real64) :: x(n)
call random_number(x) ! [0,1]
x = in_range(x,low,hi) ! [low,hi]
end function randoms_in_range
end program
I want to perform a Matrix-Vector product in fortran using the SGEMV subroutine from BLAS.
I have a code that is similar to this:
program test
integer, parameter :: DP = selected_real_kind(15)
real(kind=DP), dimension (3,3) :: A
real(kind=DP), dimension (3) :: XP,YP
call sgemv(A,XP,YP)
A is a 3x3 Matrix, XP and YP are Vectors.
In the included module one can see the following code:
PURE SUBROUTINE SGEMV_F95(A,X,Y,ALPHA,BETA,TRANS)
! Fortran77 call:
! SGEMV(TRANS,M,N,ALPHA,A,LDA,X,INCX,BETA,Y,INCY)
USE F95_PRECISION, ONLY: WP => SP
REAL(WP), INTENT(IN), OPTIONAL :: ALPHA
REAL(WP), INTENT(IN), OPTIONAL :: BETA
CHARACTER(LEN=1), INTENT(IN), OPTIONAL :: TRANS
REAL(WP), INTENT(IN) :: A(:,:)
REAL(WP), INTENT(IN) :: X(:)
REAL(WP), INTENT(INOUT) :: Y(:)
END SUBROUTINE SGEMV_F95
I understand that the some of the parameters are optional, so where am i wrong in the method call?
When you look at BLAS or LAPACK routines then you should always have a look at the first letter:
S: single precision
D: double precision
C: single precision complex
Z: double precision complex
You defined your matrix A as well as the vectors XP and YP as a double precision number using the statement:
integer, parameter :: DP = selected_real_kind(15)
So for this, you need to use dgemv or define your precision as single precision.
There is also a difference between calling dgemv and dgemv_f95. dgemv_f95 is part of Intel MKL and not really a common naming. For portability reasons, I would not use that notation but stick to the classic dgemv which is also part of Intel MKL.
DGEMV performs one of the matrix-vector operations
y := alpha*A*x + beta*y, or y := alpha*A**T*x + beta*y,
where alpha and beta are scalars, x and y are vectors and A is an
m by n matrix.
If you want to know how to call the function, I suggest to have a look here, but it should, in the end, look something like this:
call DGEMV('N',3,3,ALPHA,A,3,XP,1,BETA,YP,1)
The precisions are incompatible. You are calling sgemv which takes single precision arguments but you are passing double precision arrays and vectors.
Perhaps the trans parameter is required?
trans: Must be 'N', 'C', or 'T'.
(As per the note at the bottom of Developer Reference for Intel® Math Kernel Library - Fortran.)
I have a problem with QR decomposition method. I use dgeqrf subroutine for decomposition but there is no error in the compiler but it gives a problem after that. I haven't found where is the mistake.
Another question is, A=Q*R=> if A matrix has zero, Can decomposition be zero or lose the rank.
program decomposition
!CONTAINS
!subroutine Qrdecomposition(A_mat, R)
real,dimension(2,2) :: A_mat !real,dimension(2,2),intent(inout)
:: A_mat
real,dimension(2,2) :: R !real,dimension(2,2),intent(out)
:: R
real,dimension(2,2) :: A
integer :: M,N,LDA,LWORK,INFO
real,allocatable, dimension(:,:) :: TAU
real,allocatable, dimension(:,:) :: WORK
external dgeqrf
M=2
N=2
LDA=2
LWORK=2
INFO=0
A_mat(1,1)=4
A_mat(1,2)=1
A_mat(2,1)=3
A_mat(2,2)=1
A=A_mat
call dgeqrf(M,N,A,TAU,WORK,LWORK,INFO)
R=A
print *,R,WORK,LWORK
!end subroutine Qrdecomposition
end program decomposition
I see three mistakes in your code:
1) You forgot the LDA argument to dgeqrf,
2) TAU and WORK must be explicitly allocated,
3) All arrays should be declared with double precision to be consistent with dgeqrf interface:
program decomposition
!CONTAINS
!subroutine Qrdecomposition(A_mat, R)
! Note: using '8' for the kind parameter is not the best style but I'm doing it here for brevity.
real(8),dimension(2,2) :: A_mat !real,dimension(2,2),intent(inout)
real(8),dimension(2,2) :: R !real,dimension(2,2),intent(out)
real(8),dimension(2,2) :: A
integer :: M,N,LDA,LWORK,INFO
real(8),allocatable, dimension(:,:) :: TAU
real(8),allocatable, dimension(:,:) :: WORK
external dgeqrf
M=2
N=2
LDA=2
LWORK=2
INFO=0
A_mat(1,1)=4
A_mat(1,2)=1
A_mat(2,1)=3
A_mat(2,2)=1
A=A_mat
allocate(tau(M,N), work(M,N))
call dgeqrf(M,N,A,LDA,TAU,WORK,LWORK,INFO)
R=A
print *,R,WORK,LWORK
!end subroutine Qrdecomposition
end program decomposition
In certain situations, Fortran does perform automatic allocation of arrays, but it should generally not be counted on and it is not the case here.
EDIT Point 3 was pointed out by roygvib, see below.
I have an error with this code and I don't understand why
-"The module or main program array 'u' at (1) must have constant shape."
-Moreover, how can I do this code with a choice of parameters, I mean [U]=vector(N) where I can chose N and it returns me U.
program vector
!declaration
implicit none
integer :: n
integer, parameter :: N=10
real, dimension(N,1) :: U
do n=1,N
U(1,N)=n
end do
print*,U
end program vector
First up, Fortran is caseINsensitive, so n and N are the same thing, and you can't declare two different variables/parameters n and N.
Then you declare U to have shape (N, 1), but seem to use it in the form (1, N).
As for how to auto-generate something like U, you could use something like this:
function vector(n) result(v)
integer, intent(in) :: n
integer :: v(n)
integer :: i
v = [ (i, i=1, n) ]
return
end function vector
One more thing:
You declare U with dimension(1, N) which creates a 2D array with one dimension having length 1. I'm wondering whether you wanted to create a 1D array with range from 1 to N, for which the declaration would need to be dimension(1:N) (or, since Fortran assumes indices start at 1, just dimension(N)).
Addressing the questions in your comment:
The purpose of intent(in) tells the compiler that n is only read, not written to, in this function. Considering that you want to use n as the size of array v, you want that.
With result(v) I tell the compiler that I want to use the name v to refer to the result of the function, not the default (which is the function name). I do this to avoid confusion.
integer :: v(n) is the same as integer, dimension(n) :: v
I wrote a Fortran code that calculates the ith-permutation of a given list {1,2,3,...,n}, without computing all the others, that are n! I needed that in order to find the ith-path of the TSP (Travelling salesman problem).
When n! is big, the code gives me some error and I tested that the ith-permutation found is not the exact value. For n=10, there are not problems at all, but for n=20, the code crashes or wrong values are found. I think this is due to errors that Fortran makes operating with big numbers (sums of big numbers).
I use Visual Fortran Ultimate 2013. In attached you find the subroutine I use for my goal. WeightAdjMatRete is the distance matrix between each pair of knots of the network.
! Fattoriale
RECURSIVE FUNCTION factorial(n) RESULT(n_factorial)
IMPLICIT NONE
REAL, INTENT(IN) :: n
REAL :: n_factorial
IF(n>0) THEN
n_factorial=n*factorial(n-1)
ELSE
n_factorial=1.
ENDIF
ENDFUNCTION factorial
! ith-permutazione di una lista
SUBROUTINE ith_permutazione(lista_iniziale,n,i,ith_permutation)
IMPLICIT NONE
INTEGER :: k,n
REAL :: j,f
REAL, INTENT(IN) :: i
INTEGER, DIMENSION(1:n), INTENT(IN) :: lista_iniziale
INTEGER, DIMENSION(1:n) :: lista_lavoro
INTEGER, DIMENSION(1:n), INTENT(OUT) :: ith_permutation
lista_lavoro=lista_iniziale
j=i
DO k=1,n
f=factorial(REAL(n-k))
ith_permutation(k)=lista_lavoro(FLOOR(j/f)+1)
lista_lavoro=PACK(lista_lavoro,MASK=lista_lavoro/=ith_permutation(k))
j=MOD(j,f)
ENDDO
ENDSUBROUTINE ith_permutazione
! Funzione modulo, adattata
PURE FUNCTION mood(k,modulo) RESULT(ris)
IMPLICIT NONE
INTEGER, INTENT(IN) :: k,modulo
INTEGER :: ris
IF(MOD(k,modulo)/=0) THEN
ris=MOD(k,modulo)
ELSE
ris=modulo
ENDIF
ENDFUNCTION mood
! Funzione quoziente, adattata
PURE FUNCTION quoziente(a,p) RESULT(ris)
IMPLICIT NONE
INTEGER, INTENT(IN) :: a,p
INTEGER :: ris
IF(MOD(a,p)/=0) THEN
ris=(a/p)+1
ELSE
ris=a/p
ENDIF
ENDFUNCTION quoziente
! Vettori contenenti tutti i payoff percepiti dagli agenti allo state vector attuale e quelli ad ogni sua singola permutazione
SUBROUTINE tuttipayoff(n,m,nodi,nodi_rete,sigma,bvector,MatVecSomma,VecPos,lista_iniziale,ith_permutation,lunghezze_percorso,WeightAdjMatRete,array_perceived_payoff_old,array_perceived_payoff_neg)
IMPLICIT NONE
INTEGER, INTENT(IN) :: n,m,nodi,nodi_rete
INTEGER, DIMENSION(1:nodi), INTENT(IN) :: sigma
INTEGER, DIMENSION(1:nodi), INTENT(OUT) :: bvector
REAL, DIMENSION(1:m,1:n), INTENT(OUT) :: MatVecSomma
REAL, DIMENSION(1:m), INTENT(OUT) :: VecPos
INTEGER, DIMENSION(1:nodi_rete), INTENT(IN) :: lista_iniziale
INTEGER, DIMENSION(1:nodi_rete), INTENT(OUT) :: ith_permutation
REAL, DIMENSION(1:nodi_rete), INTENT(OUT) :: lunghezze_percorso
REAL, DIMENSION(1:nodi_rete,1:nodi_rete), INTENT(IN) :: WeightAdjMatRete
REAL, DIMENSION(1:nodi), INTENT(OUT) :: array_perceived_payoff_old,array_perceived_payoff_neg
INTEGER :: i,j,k
bvector=sigma
FORALL(i=1:nodi,bvector(i)==-1)
bvector(i)=0
ENDFORALL
FORALL(i=1:m,j=1:n)
MatVecSomma(i,j)=bvector(m*(j-1)+i)*(2.**REAL(n-j))
ENDFORALL
FORALL(i=1:m)
VecPos(i)=1.+SUM(MatVecSomma(i,:))
ENDFORALL
DO k=1,nodi
IF(VecPos(mood(k,m))<=factorial(REAL(nodi_rete))) THEN
CALL ith_permutazione(lista_iniziale,nodi_rete,VecPos(mood(k,m))-1.,ith_permutation)
FORALL(i=1:(nodi_rete-1))
lunghezze_percorso(i)=WeightAdjMatRete(ith_permutation(i),ith_permutation(i+1))
ENDFORALL
lunghezze_percorso(nodi_rete)=WeightAdjMatRete(ith_permutation(nodi_rete),ith_permutation(1))
array_perceived_payoff_old(k)=(1./SUM(lunghezze_percorso))
ELSE
array_perceived_payoff_old(k)=0.
ENDIF
IF(VecPos(mood(k,m))-SIGN(1,sigma(m*(quoziente(k,m)-1)+mood(k,m)))*2**(n-quoziente(k,m))<=factorial(REAL(nodi_rete))) THEN
CALL ith_permutazione(lista_iniziale,nodi_rete,VecPos(mood(k,m))-SIGN(1,sigma(m*(quoziente(k,m)-1)+mood(k,m)))*2**(n-quoziente(k,m))-1.,ith_permutation)
FORALL(i=1:(nodi_rete-1))
lunghezze_percorso(i)=WeightAdjMatRete(ith_permutation(i),ith_permutation(i+1))
ENDFORALL
lunghezze_percorso(nodi_rete)=WeightAdjMatRete(ith_permutation(nodi_rete),ith_permutation(1))
array_perceived_payoff_neg(k)=(1./SUM(lunghezze_percorso))
ELSE
array_perceived_payoff_neg(k)=0.
ENDIF
ENDDO
ENDSUBROUTINE tuttipayoff
Don't use floating-point numbers to represent factorials; factorials are products of integers and are therefore best represented as integers.
Factorials grow big fast, so it may be tempting to use reals, because reals can represent huge numbers like 1.0e+30. But floating-point numbers are precise only with relation to their magnitude; their mantissa still has a limited size, they can be huge because their exponents may be huge.
A 32-bit real can represent exact integers up to about 16 million. After that, only every even integer can be represented up to 32 million and every fourth integer up to 64 million. 64-bit integers are better, because they can represent exact integers up to 9 quadrillion.
64-bit integers can go 1024 times further: They can represent 2^63 or about 9 quintillion (9e+18) integers. That is enough to represent 20!:
20! = 2,432,902,008,176,640,000
2^63 = 9,223,372,036,854,775,808
Fortran allows you to select a kind of integer based on the decimal places it should be able to represent:
integer, (kind=selected_int_kind(18))
Use this to do your calculations with 64-bit integers. This will give you factorials up to 20!. It won't go further than that, though: Most machines support only integers up to 64 bit, so selected_int_kind(19) will give you an error.
Here's the permutation part of your program with 64-bit integers. Note how all the type conversions ald floors and ceilings disappear.
program permute
implicit none
integer, parameter :: long = selected_int_kind(18)
integer, parameter :: n = 20
integer, dimension(1:n) :: orig
integer, dimension(1:n) :: perm
integer(kind=long) :: k
do k = 1, n
orig(k) = k
end do
do k = 0, 2000000000000000_long, 100000000000000_long
call ith_perm(perm, orig, n, k)
print *, k
print *, perm
print *
end do
end program
function fact(n)
implicit none
integer, parameter :: long = selected_int_kind(18)
integer(kind=long) :: fact
integer, intent(in) :: n
integer :: i
fact = 1
i = n
do while (i > 1)
fact = fact * i
i = i - 1
end do
end function fact
subroutine ith_perm(perm, orig, n, i)
implicit none
integer, parameter :: long = selected_int_kind(18)
integer, intent(in) :: n
integer(kind=long), intent(in) :: i
integer, dimension(1:n), intent(in) :: orig
integer, dimension(1:n), intent(out) :: perm
integer, dimension(1:n) :: work
integer :: k
integer(kind=long) :: f, j
integer(kind=long) :: fact
work = orig
j = i
do k = 1, n
f = fact(n - k)
perm(k) = work(j / f + 1)
work = pack(work, work /= perm(k))
j = mod(j, f)
end do
end subroutine ith_perm