OpenEuphoria

Re: 32-bit random numbers

• Posted by "Igor Kachan" <kinz at peterlink.ru> Jul 12, 2004
• 646 views

Hi Juergen,

This subject seems to be near the end,  for now  

Try please new function, it can give random 
integer atoms in full EU range with step=1.
Distribution is flat, as far as I can see.

include misc.e

type range(integer N)
    return N>=1 and N<=8388608
end type

constant K = 1073741823 -- the max EU integer
-- 1073741823 / 8388608 = 127.99999... -- i.e. N << K

global function rand_int_atom(range N)
       return K * (rand(N) - 1) + rand(K) - 1
end function
 
-- this function gives random integer atoms 
-- in [0.. up to 9007199246352383, by 1073741823] range 
-- with step = 1, flat distribution.


--- Testing -- flat or not flat

sequence RR

RR = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}

atom a01,a02,a03,a04,a05,a06,a07,a08,a09,a10,
     a11,a12,a13,a14,a15,a16,a17,a18,a19,a20
     
 atom SS
      SS = 9007199246352383 / 20
     a01 = 01 * SS
     a02 = 02 * SS
     a03 = 03 * SS
     a04 = 04 * SS
     a05 = 05 * SS
     a06 = 06 * SS
     a07 = 07 * SS
     a08 = 08 * SS
     a09 = 09 * SS
     a10 = 10 * SS
     a11 = 11 * SS
     a12 = 12 * SS
     a13 = 13 * SS
     a14 = 14 * SS
     a15 = 15 * SS
     a16 = 16 * SS
     a17 = 17 * SS
     a18 = 18 * SS
     a19 = 19 * SS
     a20 = 20 * SS
     
atom NN  
     
NN = 100000

atom Z
for i=1 to NN do
 Z=rand_int_atom(8388608)  --<-- max range 1073741823 * 8388608 - 1
    if Z>=  0 and Z< a01 then
  RR[01] += 1
 elsif Z>=a01 and Z< a02 then
  RR[02] += 1
 elsif Z>=a02 and Z< a03 then
  RR[03] += 1
 elsif Z>=a03 and Z< a04 then
  RR[04] += 1
 elsif Z>=a04 and Z< a05 then
  RR[05] += 1
 elsif Z>=a05 and Z< a06 then
  RR[06] += 1
 elsif Z>=a06 and Z< a07 then
  RR[07] += 1
 elsif Z>=a07 and Z< a08 then
  RR[08] += 1
 elsif Z>=a08 and Z< a09 then
  RR[09] += 1
 elsif Z>=a09 and Z< a10 then
  RR[10] += 1
 elsif Z>=a10 and Z< a11 then
  RR[11] += 1
 elsif Z>=a11 and Z< a12 then
  RR[12] += 1
 elsif Z>=a12 and Z< a13 then
  RR[13] += 1
 elsif Z>=a13 and Z< a14 then
  RR[14] += 1
 elsif Z>=a14 and Z< a15 then
  RR[15] += 1
 elsif Z>=a15 and Z< a16 then
  RR[16] += 1
 elsif Z>=a16 and Z< a17 then
  RR[17] += 1
 elsif Z>=a17 and Z< a18 then
  RR[18] += 1
 elsif Z>=a18 and Z< a19 then
  RR[19] += 1
 elsif Z>=a19 and Z< a20 then
  RR[20] += 1
 end if
end for

pretty_print(1, RR / NN, {0,1,1,1})

--- flat, probabilities are almost the same
--- in all intervals.

while get_key()=-1 do end while
 clear_screen()
 
pretty_print(1, RR, {0,1,1,1})

--- or number of cases in each interval is almost
--- the same.

To get the top bound of EU integer atoms  for
this task, I used the following program:

atom k, t
integer K, z, y

K = 1073741823

k=0
z=0
t=0
y=0

while 1 do
 z += 1
 k += K
  if  
    k + 1 = k  then  exit  --  Rob suggested this probe
  end if
 t += K
 y += 1
end while

printf(1, "%d\n", {k}) -- the first bad bound
printf(1, "%d\n", {z}) -- the number of first bad iteration
printf(1, "%d\n", {t}) -- the last good bound
printf(1, "%d\n", {y}) -- the number of last good iteration
printf(1, "%d\n", {t/y}) -- control
printf(1, "%d\n", {1073741823}) -- control

Regards,
Igor Kachan
kinz at peterlink.ru