Send x% of the users to A, the rest to B

[DreamT]

Hi,
br> I have a task where i'm supposed to send a certain percentage of the users to one location, and the rest to another location.

I have an idea where I want to use the session key and modulus, but I don't know how to implement it. Some pseudocode:

# 1. Convert every character to a number using ord?

# 2. ???

# 3. use modulus 100 on the converted string to get 1 or 0

if (($converted_string % 100) < $percentage)
{
    $location = "new location";
}
else
{
    $location = "old location";
}

As you can see I'm quite clueless on this, but it's more of a mathematical problem than a code problem. I know that I need to do some calculations on the string before I can use modulus on it, but how?

[BrowserUk]

This one-liner shows that if you use rand() < 0.5 to decide between locations -- even if the rand calls are in different invocations of perl -- the distribution will be fairly split:

perl -E"{++$t;$r=`perl -E\"say rand()\"`;$r<.5?++$a:++$b; printf qq[\r
+%.3f : %.3f], $a*100/$t, $b*100/$t; redo}"
49.350 : 50.650

[Perleone]

Cool snippet! Interesting to see that (at least on my Debian box) it takes a few dozen iterations to actually reach middle ground:

perl -Mstrict -Mwarnings -E 'my $WIDTH = 50;
my ( $try, $i, $j ) = (0) x 3;
for ( 1 .. 40 ) {
    ++$try;
    my $rand = `perl -E "say rand()"`;
    $rand < .5 ? ++$i : ++$j;
    printf qq[%${WIDTH}s : %s\n],
      "#" x ( $i * $WIDTH / $try ),
      "#" x ( $j * $WIDTH / $try );
}'

[BrowserUk]

Interesting to see that ... it takes a few dozen iterations to actually reach middle ground:

That is the nature of randomness. If it was exactly 50:50, it wouldn't be random.

But the more trials you run, the closer it will tend. This is more evident if you run this version with various different resets:

C:\test>perl 
-E"{++$t;$r=`perl -E\"say rand()\"`;$r<.5?++$a:++$b; printf qq[\r%.3f 
+: %.3f], $a*100/$t, $b*100/$t; $t>$ARGV[0] and say and $t=$a=$b=0; re
+do}" 10
45.455 : 54.545
27.273 : 72.727
36.364 : 63.636
54.545 : 45.455
72.727 : 27.273
63.636 : 36.364
36.364 : 63.636
45.455 : 54.545
27.273 : 72.727
36.364 : 63.636
72.727 : 27.273
45.455 : 54.545
54.545 : 45.455
63.636 : 36.364

C:\test>perl 
-E"{++$t;$r=`perl -E\"say rand()\"`;$r<.5?++$a:++$b; printf qq[\r%.3f 
+: %.3f], $a*100/$t, $b*100/$t; $t>$ARGV[0] and say and $t=$a=$b=0; re
+do}" 100
52.475 : 47.525
52.475 : 47.525
52.475 : 47.525
46.535 : 53.465
45.545 : 54.455
51.485 : 48.515
41.584 : 58.416
46.535 : 53.465
42.574 : 57.426
40.594 : 59.406

C:\test>perl 
-E"{++$t;$r=`perl -E\"say rand()\"`;$r<.5?++$a:++$b; printf qq[\r%.3f 
+: %.3f], $a*100/$t, $b*100/$t; $t>$ARGV[0] and say and $t=$a=$b=0; re
+do}" 1000
49.750 : 50.250
51.349 : 48.651
48.651 : 51.349
53.047 : 46.953

[choroba]

Cannot you just decide based on rand 100 < $percentage ?

[DreamT]

It would work - But I need to remember the choice per user. As I think of it, I could of course store the selection for the current session(user) instead:)

[Laurent_R]

If you have only two locations, you probably don't need to convert the full session ID into an integer, but just a small part or it. I recently had to build a kind of hashing function for splitting my processing of a very large data chunk into 10 sub-processes. One of the mandatory IDs of my data was a telephone number. I simply used the last digit of the phone number to decide to which process to allocate a given record and obtained an excellent load balancing. The advantage is that the simpler the splitting function is, the less performance overhead you get.

[talexb]

If a user can make multiple requests, and you always want to send a user to the same place, then hashing some user attribute and picking a destination based on that should work. For example, you could hash their username and look at the LSB of the result to decide where to send them.

[Anonymous Monk]

Are you simply trying to achieve a load-balancer? That's a device.