N-queens problem: Difference between revisions
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===Alternate Fortran 77 solution===
<lang fortran>
C This one enumerates permutations, skipping impossible "branches"
C See the 2nd program for Scheme on the "Permutations" page for the main idea
program qtest
integer n
do 10 n=1,16
call queens(n)
10 continue
end
subroutine queens(n)
implicit none
integer a,i,j,m,n,s,z
logical chk
dimension a(n),s(n)
do 10 i=1,n
a(i)=i
s(i)=0
10 continue
m=0
i=1
20 if(i.gt.n) then
m=m+1
go to 60
end if
j=i
30 z=a(i)
a(i)=a(j)
a(j)=z
if(chk(n,a,i)) then
s(i)=j
i=i+1
go to 20
end if
j=j+1
if(j.le.n) go to 30
50 j=j-1
if(j.eq.i) go to 60
z=a(i)
a(i)=a(j)
a(j)=z
go to 50
60 i=i-1
if(i.eq.0) go to 70
j=s(i)
j=j+1
if(j.le.n) go to 30
go to 50
70 print *,n,m
end
C TODO : replace the following loop with a single test on diagonals already visited
C This would need keeping track of "up" and "down" diagonals in vectors of size 2*n-1
function chk(n,a,i)
implicit none
logical chk
integer n,a,i,j
dimension a(n)
chk=.true.
j=1
10 if(j.eq.i) return
if(iabs(a(i)-a(j)).eq.iabs(i-j)) go to 20
j=j+1
go to 10
20 chk=.false.
end</lang>
=={{header|Haskell}}==
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