Anonymous user
Dutch national flag problem: Difference between revisions
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=={{header|FORTRAN}}==
Please find the example run along with compilation instructions on a GNU/linux platform in the comments at the beginning of the FORTRAN 2008 program source. The Netherlands program, using equal numbers of colors, solved the problem at three sample sizes. Swaps number 2/3 the total of samples, convincingly demonstrating the O(n) time behavior that's directly provable by inspection. The color strings are chosen for ASCII sort. Feature not used.
Abhor code duplication. I've repeated code anyway to demonstrate FORTRAN pointers, which behave like an alias. A subroutine with traditional arguments including the number of valid elements of the array is appropriate. I'd use one long array instead of 3 arrays and the size intrinsic.
<lang>
!-*- mode: compilation; default-directory: "/tmp/" -*-
!Compilation started at Mon Jun 3
!
!a=./f && make FFLAGS='-O0 -g' $a && OMP_NUM_THREADS=2 $a < unixdict.txt
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! 9999 items, 6666 swaps.
!
!Compilation finished at Mon Jun 3
program Netherlands
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character(len=6), parameter, dimension(3) :: colors = (/'RED ', 'WHITE ', 'blue '/)
integer, dimension(12) :: sort_me
integer, dimension(999), target :: a999
integer, dimension(9999), target :: a9999
integer, dimension(:), pointer :: pi
integer :: i, swaps
data sort_me/4*1,4*2,4*3/
call shuffle(sort_me, 5)
write(6,*)'Original and flag sequences'
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write(6,*) (colors(sort_me(i)), i = 1, size(sort_me))
write(6,*) 12,'items,',swaps,' swaps.'
pi => a999
call shuffle(a999, 333+1)▼
do i=1, size(pi)
call partition3way(a999, 2, swaps)▼
pi(i) = 1 + L(size(pi)/3 .lt. i) + L(2*size(pi)/3 .lt. i)
write(6,*) 999,'items,',swaps,' swaps.'▼
end do
call
pi => a9999
do i=1, size(pi)
pi(i) = 1 + L(size(pi)/3 .lt. i) + L(2*size(pi)/3 .lt. i)
end do
write(6,*) size(pi),'items,',swaps,' swaps.'
contains
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end program Netherlands
</lang>
=={{header|Go}}==
<lang go>package main
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