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lizi_n · Feb 7th, 2008, 5:09 PM

I need to randomly generate an equal number of +,- 1's. I have used [call random_number] and made every number >0.5=1 and any number <0.5=-1. It's in a loop from 1..n but I can't work out how to make sure that there are exactly the same amount of each. Please help!
Lizi

Sane · Feb 7th, 2008, 6:09 PM

You want an equal amount, but you want it random? A little strange, but okay. I'd suggest the following.

If you want there to be 10 numbers, then, make an array of 5 ones and 5 negative ones.

{+1, +1, +1, +1, +1, -1, -1, -1, -1, -1}

Loop through each index. Each time, generate a random number between (0 - 9). Use that number to determine which position the current element should be swapped with. Swapping 10 elements 10 times, will effectively produce a random array of perfectly uniform distribution.

(Toggle Plain Text)

0th index
Random Number: 9
Swap 9th element with 0th element
{-1, +1, +1, +1, +1, -1, -1, -1, -1, +1}

1st index
Random Number: 2
Swap 2nd element with 1st element
{-1, +1, +1, +1, +1, -1, -1, -1, -1, +1}

... etc ...

9th index
Random Number: 5
Swap 9th element with 5th element
{+1, +1, -1, -1, +1, +1, -1, -1, +1, -1}

In Pseudocode:

(Toggle Plain Text)

n = 10

Make array X [Size n]
X[0 .. n/2-1] = {1}
X[n/2 .. n-1] = {-1} 

for i : 0 ... n-1
    j = random number (0, n-1)
    swap(X[i], X[j])

output X

lizi_n · Feb 11th, 2008, 3:09 PM

Thank you that's awesome. Could you just tell me what I write in order to swap the elements. At the moment I think I'm just replacing them by typing

x(i)=x(j)
x(j)=x(i)

so I end up with an uneven amount of +1 and -1.
Thank you!

lizi_n · Feb 11th, 2008, 3:18 PM

This is what I've got so far if it helps.

(Toggle Plain Text)

integer:: n
parameter(n=10)

integer:: i,t,h
real:: s
integer, dimension(n) :: x

do i=1,n/2
x(i)=1
end do

do i=n/2+1,n
x(i)=-1
end do

print '(10i2)',x

do i=1,n

call random_number(s)
s=n*s

do h=0,n-1
if(s.gt.h.and.s.lt.h+1) t=h+1
end do


x(i)=x(t)
x(t)=x(i)

print*,x(i)


end do

Sane · Feb 11th, 2008, 3:30 PM

(Toggle Plain Text)

x(i)=x(j)
x(j)=x(i)

That is close, but examine what happens when you swap two elements. The value of the first element is overwritten, by the time you assign it a new value.

(Toggle Plain Text)

E.G. Let x(i) be 5, and x(j) be 6.

x(i) becomes x(j) = 6
x(j) becomes x(i) = 6 as well

The solution to this is to use a temporary variable, let's say temp, to remember the first value, before the swapping occurs.

(Toggle Plain Text)

temp = x(i)
x(i) = x(j)
x(j) = temp

Feb 7th, 2008, 5:09 PM	#1
lizi_n Newbie Join Date: Feb 2008 Posts: 3 Rep Power: 0	random numbers fortran 90 I need to randomly generate an equal number of +,- 1's. I have used [call random_number] and made every number >0.5=1 and any number <0.5=-1. It's in a loop from 1..n but I can't work out how to make sure that there are exactly the same amount of each. Please help! Lizi

Feb 7th, 2008, 6:09 PM	#2
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,886 Rep Power: 5	Re: random numbers fortran 90 You want an equal amount, but you want it random? A little strange, but okay. I'd suggest the following. If you want there to be 10 numbers, then, make an array of 5 ones and 5 negative ones. `{+1, +1, +1, +1, +1, -1, -1, -1, -1, -1}` Loop through each index. Each time, generate a random number between (0 - 9). Use that number to determine which position the current element should be swapped with. Swapping 10 elements 10 times, will effectively produce a random array of perfectly uniform distribution. (Toggle Plain Text) 0th index Random Number: 9 Swap 9th element with 0th element {-1, +1, +1, +1, +1, -1, -1, -1, -1, +1} 1st index Random Number: 2 Swap 2nd element with 1st element {-1, +1, +1, +1, +1, -1, -1, -1, -1, +1} ... etc ... 9th index Random Number: 5 Swap 9th element with 5th element {+1, +1, -1, -1, +1, +1, -1, -1, +1, -1} 0th index Random Number: 9 Swap 9th element with 0th element {-1, +1, +1, +1, +1, -1, -1, -1, -1, +1} 1st index Random Number: 2 Swap 2nd element with 1st element {-1, +1, +1, +1, +1, -1, -1, -1, -1, +1} ... etc ... 9th index Random Number: 5 Swap 9th element with 5th element {+1, +1, -1, -1, +1, +1, -1, -1, +1, -1} In Pseudocode: (Toggle Plain Text) n = 10 Make array X [Size n] X[0 .. n/2-1] = {1} X[n/2 .. n-1] = {-1} for i : 0 ... n-1 j = random number (0, n-1) swap(X[i], X[j]) output X n = 10 Make array X [Size n] X[0 .. n/2-1] = {1} X[n/2 .. n-1] = {-1} for i : 0 ... n-1 j = random number (0, n-1) swap(X[i], X[j]) output X

Feb 11th, 2008, 3:09 PM	#3
lizi_n Newbie Join Date: Feb 2008 Posts: 3 Rep Power: 0	Re: random numbers fortran 90 Thank you that's awesome. Could you just tell me what I write in order to swap the elements. At the moment I think I'm just replacing them by typing x(i)=x(j) x(j)=x(i) so I end up with an uneven amount of +1 and -1. Thank you!

Feb 11th, 2008, 3:18 PM	#4
lizi_n Newbie Join Date: Feb 2008 Posts: 3 Rep Power: 0	Re: random numbers fortran 90 This is what I've got so far if it helps. (Toggle Plain Text) integer:: n parameter(n=10) integer:: i,t,h real:: s integer, dimension(n) :: x do i=1,n/2 x(i)=1 end do do i=n/2+1,n x(i)=-1 end do print '(10i2)',x do i=1,n call random_number(s) s=ns do h=0,n-1 if(s.gt.h.and.s.lt.h+1) t=h+1 end do x(i)=x(t) x(t)=x(i) print,x(i) end do integer:: n parameter(n=10) integer:: i,t,h real:: s integer, dimension(n) :: x do i=1,n/2 x(i)=1 end do do i=n/2+1,n x(i)=-1 end do print '(10i2)',x do i=1,n call random_number(s) s=ns do h=0,n-1 if(s.gt.h.and.s.lt.h+1) t=h+1 end do x(i)=x(t) x(t)=x(i) print,x(i) end do

Feb 11th, 2008, 3:30 PM	#5
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,886 Rep Power: 5	Re: random numbers fortran 90 (Toggle Plain Text) x(i)=x(j) x(j)=x(i) x(i)=x(j) x(j)=x(i) That is close, but examine what happens when you swap two elements. The value of the first element is overwritten, by the time you assign it a new value. (Toggle Plain Text) E.G. Let x(i) be 5, and x(j) be 6. x(i) becomes x(j) = 6 x(j) becomes x(i) = 6 as well E.G. Let x(i) be 5, and x(j) be 6. x(i) becomes x(j) = 6 x(j) becomes x(i) = 6 as well The solution to this is to use a temporary variable, let's say `temp`, to remember the first value, before the swapping occurs. (Toggle Plain Text) temp = x(i) x(i) = x(j) x(j) = temp temp = x(i) x(i) = x(j) x(j) = temp

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Programming Forums > Application Development > Other Programming Languages

random numbers fortran 90