How to exit from nested Fortran loops? - fortran

Question

I'm trying to write a program (in Fortran 95) that finds the minimal decomposition of natural numbers up to N into a sum of at most 4 positive integers.
I've been trying to add and remove statements for a while to make it stop at only the minimal decomposition but I'm not getting anywhere. How do I make the program stop as soon as it's found the minimal decomposition?
PROGRAM SummeQuadrat
IMPLICIT NONE
real:: start,finish
integer:: a,b,c,d,g,x,y
write(*,*) "Max n"
read (*,*) y
call cpu_time(start)
do x=1,y,1
do a=0,x,1
do b=a,x-a,1
do c=b,x-b,1
do d=c,x-c,1
if (a**2+b**2+c**2+d**2 .eq. x) then
write(*,*) "x=",x,d,c,b,a
end if
end do
end do
end do
end do
end do
call cpu_time(finish)
write(*,*)finish-start
end program SummeQuadrat

Answer 1

As I explained in the comments, I am not sure you are asking only how to break out of the loops or for more.
You can jump out of any loop using the EXIT statement. To exit from a loop which is not the innermost loop you are currently in you use a labeled loop and use the label in the EXIT statement to exit that particular loop.
outer: do x = 1, y
do a = 0, x
do b = a, x-a
do c = b, x-b
do d = c, x-c
if (a**2+b**2+c**2+d**2 == x) then
write(*,*) "x=",x,d,c,b,a
if (minimal(a,b,c,d)) exit outer
end if
end do
end do
end do
end do
end do outer

Answer 2

Old thread, but it's kind of a fun problem so I thought I might post my own interpretation.
First off, if we cheat a little and peek at the solution it can be seen that all 4 squares are only needed when x=4**k*(8*m+7). Thus we can search cheaply for 1 or 2 square solutions and on failure decide by the above criterion whether to search for a 3- or 4-square solution.
Then when we structure our loops, count down from the largest a such that a**2 <= x, then the largest b <= a such that a**2+b**2 <= x and so on. This takes the problem from O(x**4) down to O(x**1.5) so it can go much quicker.
For output format, by judicious use of the colon format we can write a single format that prints out results in perhaps a more readable fashion.
! squares.f90 -- Prints out minimal decomposition x into squares
! for 1 <= x <= y (user input)
program squares
use ISO_FORTRAN_ENV, only: REAL64
implicit none
! Need this constant so we can take the square root of an
! integer.
real(REAL64), parameter :: half = 0.5_REAL64
real start, finish
integer a,b,c
integer amax,bmax,cmax,dmax
integer amin,bmin,cmin
integer x,y
! Format for printing out decomposition into squares
character(40) :: fmt = '(i0," = ",i0"**2":3(" + ",i0,"**2":))'
integer nzero
! Get uper bound from user
write(*,'(a)',advance='no') 'Please enter the max N:> '
read(*,*) y
call cpu_time(start)
! Loop over requested range
outer: do x = 1, y
amax = sqrt(x+half)
! Check for perfect square
if(amax**2 == x) then
write(*,fmt) x,amax
cycle outer
end if
! Check for sum of 2 squares
amin = sqrt(x/2+half)
try2: do a = amax, amin, -1
bmax = sqrt(x-a**2+half)
if(bmax > a) exit try2
if(a**2+bmax**2 == x) then
write(*,fmt) x,a,bmax
cycle outer
end if
end do try2
! If trailz(x) is even, then x = 4**k*z, where z is odd
! If further z = 8*m+7, then 4 squares are required, otherwise
! only 3 should suffice.
nzero = trailz(x)
if(iand(nzero,1)==0 .AND. ibits(x,nzero,3)==7) then
amin = sqrt(x/4+half)
do a = amax, amin, -1
bmax = sqrt(x-a**2+half)
bmin = sqrt((x-a**2)/3+half)
do b = min(bmax,a), bmin, -1
cmax = sqrt(x-a**2-b**2+half)
cmin = sqrt((x-a**2-b**2)/2+half)
do c = min(cmax,b), cmin, -1
dmax = sqrt(x-a**2-b**2-c**2+half)
if(a**2+b**2+c**2+dmax**2 == x) then
write(*,fmt) x,a,b,c,dmax
cycle outer
end if
end do
end do
end do
else
amin = sqrt(x/3+half)
do a = amax, amin, -1
bmax = sqrt(x-a**2+half)
bmin = sqrt((x-a**2)/2+half)
do b = min(bmax,a), bmin, -1
cmax = sqrt(x-a**2-b**2+half)
if(a**2+b**2+cmax**2 == x) then
write(*,fmt) x,a,b,cmax
cycle outer
end if
end do
end do
end if
! We should have a solution by now. If not, print out
! an error message and abort.
write(*,'(*(g0))') 'Failure at x = ',x
stop
end do outer
call cpu_time(finish)
write(*,'(*(g0))') 'CPU time = ',finish-start
end program squares