Operations with big real numbers in Fortran - algorithm

Question

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

Answer 1

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