Reputation: 1684
Earlier I wrote a code in Matlab for this sort of lottery function, just to test if it was possible. However, I actually needed it in PHP so I've just rewritten the code and it does seem to work, but as it involves a lot of looping I want to make sure I'm doing it as efficiently as possible.
What the code does:
You can call the function $lotto -> type($users,$difficulty)
and it will return two numbers. Here's the explanation, $users
is the number of users registered on the website, i.e the people who will potentially buy a ticket. $difficulty
is a number between 1 and 10, where 5 is normal, 1 is easy and 10 is hard. Difficulty here means how hard it is to match all numbers on a lottery ticket.
So what are the numbers that the function returns? That would be $n
and $r
. $n
is the amount of numbers there will be on the lottery ticket, and $r
is the amount of numbers you can choose from the lottery ticket. For example, in the UK a national lottery ticket has 49 numbers if which you choose 6. I.e $n = 49
and $r = 6
.
How does the function calculate these two numbers? In the UK national lottery there are 13,983,816 different possible ticket combinations. If I were to run $lotto -> type(13983816,1)
it would return array(49,6)
. Basically it tried to make it so there are as many combinations of tickets as there are registered users.
tl;dr, here's the code:
<?php
class lotto {
public function type($users,$difficulty){
$current_r = $r = 2;
$current_n = 0;
$difficulty = ($difficulty + 5) / 10; // sliding scale from 1 - 10
$last_tickets_sold = 200; // tickets sold in last lotto
$last_users = 100; // how many users there were in the last lotto
$last_factor = $last_tickets_sold / $last_users; // tickets per user
$factor = $last_factor * $difficulty;
$users *= $factor;
while($r <= 10){
$u = 0;
$n = $r;
while($u < $users && $n < 50){
$u = $this -> nCr(++$n,$r);
}
if($r == 2){
$current_n = $n;
} elseif(abs($this -> nCr($n,$r) - $users) < abs($this -> nCr($current_n,$current_r) - $users)){
// this is a better match so update current n and r
$current_r = $r;
$current_n = $n;
}
$r++;
}
return array($current_n,$current_r);
}
private function nCr($n,$r){
return $this -> factorial($n) / (
$this -> factorial($r) * $this -> factorial($n - $r)
);
}
private function factorial($x){
$f = $x;
while(--$x){
$f *= $x;
}
return $f;
}
}
$lotto = new lotto;
print_r($lotto -> type(1000,5));
?>
Upvotes: 0
Views: 1394
Reputation: 10888
The range of factorial() in your algo is [0,50], so why not just precompute this statically?
private static $factorial=array(1);
private static genFactorial($max) {
if( count( self::$factorial ) > $max ) return;
foreach ( range(count(self::$factorial), $max) as $n ) {
self::$factorial[$n] = $i*self::$factorial[$n-1];
}
}
Now add a self::genFactorial(50);
to __construct()
or to type()
and replace references to $this -> factorial($n)
by self::$factorial[$n]
.
This is just a quick code dump; not even compile checked so forgive any typos, etc. but what this does is to replace a function call (which includes a while loop) by an array element fetch.
Upvotes: 0
Reputation: 11181
I did a quick scan and spotted a few places that can be further optimized.
Combination
Your algorithm is a brute force one and can be further optimized
private function nCr($n,$r){
return $this -> factorial($n) / (
$this->factorial($r) * $this->factorial($n - $r)
);
}
to
function nCr($n,$r) {
$top = 1;
$sub = 1;
for($i = $r+1; $i <= $n; $i++)
$top *= $i;
$n -= $r;
for($i = 2; $i <= $n; $i++)
$sub *= $i;
return $top / $sub;
}
Too Much Combination Calculation
Calculate combination is expensive.
$u = 0;
$n = $r;
while($u < $users && $n < 50){
$u = $this -> nCr(++$n,$r);
}
to
$n = $r + 1;
$u = nCr($n, $r);
while ($u < $users && $n < 50) {
$n++;
$u *= $n;
$u /= ($n - $r);
}
Upvotes: 2
Reputation:
Regardless detailed examination of your code, are you sure that your loops does not need continue or break?
Upvotes: 0
Reputation: 3931
An immediate observation is that you have the possibility of a divide by 0 error
$last_factor = $last_tickets_sold / $last_users;
Could be solved by putting a simple if statement around it
$last_factor = ($last_users == 0) ? 0 : $last_tickets_sold / $last_users;
Upvotes: 1