我编写了一个Fortran代码,该代码可以计算给定列表的ith- 排列{1,2,3,...,n},而无需计算所有其他列表,这是n!我需要的以便找到TSP的ith-路径(旅行商问题)。
{1,2,3,...,n}
n!
当n!大,代码给了我一些错误,我测试的第i个置换发现的是不是确切值。对于n = 10,根本没有问题,但是对于n=20,代码崩溃或找到错误的值。我认为这是由于Fortran进行大数运算(大数加法)而导致的错误。
n=20
我使用Visual Fortran Ultimate2013。在附件中,您找到了我用于目标的子例程。WeightAdjMatRete是网络的每对结之间的距离矩阵。
WeightAdjMatRete
! 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
不要使用浮点数来表示阶乘;阶乘是整数的乘积,因此最好用整数表示。
阶乘增长很快,因此使用实数可能很诱人,因为实数可以表示1.0e + 30之类的巨大数字。但是浮点数仅在其大小上是精确的。它们的尾数仍然有限,它们可能很大,因为它们的指数可能很大。
32位实数可以表示精确的整数,最大约为1600万。之后,只有每个偶数整数最多可以表示3200万,而每个第四个整数最多可以表示6400万。64位整数更好,因为它们可以表示9兆位以内的精确整数。
64位整数可以进一步移动1024倍:它们可以表示2 ^ 63或大约9百万个整数(9e + 18)整数。这足以代表20 !:
20! = 2,432,902,008,176,640,000 2^63 = 9,223,372,036,854,775,808
Fortran允许您根据其应代表的小数位数选择一种整数:
integer, (kind=selected_int_kind(18))
使用它可以对64位整数进行计算。这将使您的阶乘最多为20!。不过,它不会比这更进一步:大多数计算机仅支持最大64位的整数,因此selected_int_kind(19)会给您一个错误。
selected_int_kind(19)
这是程序的64位整数置换部分。注意所有类型转换如何表示地板和天花板消失。
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