For an online game (MMORPG) I want to create characters (players) with random strength values. The stronger the characters are, the less should exist of this sort.
Example:
12,000 strength 1 players
10,500 strength 2 players
8,500 strength 3 players
6,000 strength 4 players
3,000 strength 5 players
Actually, I need floating, progressive strength values from 1.1 to 9.9 but for this example it was easier to explain it with integer strengths.
Do you have an idea how I could code this in PHP? Of course, I would need mt_rand() to generate random numbers. But how can I achieve this pyramid structure?
What function is it? Root function, exponential function, power function or logarithm function?
Thanks in advance!
It should look like this in a graph:
Pyramid graph http://img7.imageshack.us/img7/107/pyramidy.jpg
You can simulate a distribution such as the one you described using a logarithmic function. The following will return a random strength value between 1.1 and 9.9:
function getRandomStrength()
{
$rand = mt_rand() / mt_getrandmax();
return round(pow(M_E, ($rand - 1.033) / -0.45), 1);
}
Distribution over 1000 runs (where S is the strength value floored):
S | Count
--+------
1 - 290
2 - 174
3 - 141
4 - 101
5 - 84
6 - 67
7 - 55
8 - 50
9 - 38
Note:
This answer was updated to include a strength value of 1.1 (which wasn't included before because of the rounding) and to fix the name of the mt_getrandmax() function
The simplest way to do this would be to provide 'bands' for where a random number should go. In your example, you have 15 players so you could have:
rand < 1/15, highest strength
1/15 < rand < 3/15, second highest
3/15 < rand < 6/15, third highest
6/15 < rand < 10/15, fourth highest
10/15 < rand < 15/15, lowest strength
You could also parameterise such a function with a 'max' number of each band that you allow and when the band is filled, it is subsumed into the next lowest existing band (apart from the bottom band, which would be subsumed into the next highest) to ensure only a certain number of each with a random distribution.
Edit adding from my comments:
To get a floating range pyramid structure the best function would most likely be a logarithm. The formula:
11 - log10(rand)
would work (with log10 being a logarithm with base 10) as this would give ranges like:
1 < rand < 10 = 9 < strength < 10
10 < rand < 100 = 8 < strength < 9
100 < rand < 1000 = 7 < strength < 8
etc.
but rand would need to range from 1 to 10^10 which would require a lot of randomness (more than most random generators can manage). To get a random number in this sort of range you could multiply some together. 3 random numbers could manage it:
11 - log10(rand1 * rand2 * rand3)
with rand1 having range 1-10000 and rand2 and rand3 having range 1-1000. This would skew the distribution away from a proper pyramid slightly though (more likely to have numbers in the centre I believe) so it may not be suitable.
workmad3 has the start of it down, I think, but there's a catch - you need to track your bucket sizes and whether or not they're full. A random number generator won't guarantee that. You'll need to assign your bucket values (strenghs) and sizes (number of people), and let your random generator tell you which bucket to drop the player into - if that one is full, 'overflow' to the next lower.
As to assigning the bucket sizes for a given strength value, that's the tricky bit (and I think what you're really working at). The characteristics of your desired distribution are critical. If you want a linear drop (which the pyramid shape hints at), a line of the form
strength = max_strength - m(number_characters)
would work. Varying the value of m would change the speed at which the line drops off, and will basically limit your max number of total characters. If you're looking for a more sophisticated way for the strength values to drop off, you could use a parabolic or hyperbolic curve - these are a bit more complex, but give you very different characteristics.
something like this
<?php
$rand = rand(1,10);
switch ($rand) {
case 1:
echo "band 1";
break;
case 2:
case 3:
echo "band 2";
break;
case 4:
case 5:
case 6:
echo "band 3";
break;
default:
echo "band 4";
break;
}
?>
Band 1 being the strongest, band 4 being the weakest.
Ofcourse this is basic, you would want to refactor this to use loops instead of hardcoded switches, but you get the idea :)
It's probably easiest to use percentages in this case.
From your examples would approximately be (converted to an array for ease of use later):
$strength[1] = .3; // start with a key of 1
$strength[2] = .26;
$strength[3] = .21;
$strength[4] = .15;
$strength[5] = .08;
That way, you can generate a random number using mt_rand() and divide by the maximum possible value to get a number between 0 and 1:
$rand = mt_rand() / mt_getrandmax(); // rand is some random value between 0 and 1
Then you can use a foreach statement to isolate each case:
$comparisonPercentage = 1;
$selectedLevel = count($strength); // covers the case where mt_rand() returns 0
foreach($strength as $level => $currentPercentage)
{
$comparisonPercentage -= $currentPercentage;
if ($rand > $comparisonPercentage)
{
$selectedLevel = $level;
break;
}
}
// $selectedLevel contains the level you need...
If you do it this way, you only have to change the $strength array if you need to fiddle with the percentages.
generate a random number between 0 and 40000, if its between 0 and 12000, assign strength 1, between 12000 and 22500 assign 2 etc.
Edit: for progressive values between 0 and 10 use the square root of a random number between 0 and 100, then substract if from 10
rand -> strengh
0-1 -> 9.9 -> 9 (1%)
2-4 -> 9 -> 8 (2%)
...
81 - 100 -> 1 - 0 (19%)
For results between 1.1 and 9.9 the formula would be in pseudocode)
strength = 10 - sqrt(rand(1..79))
Related
My task:
Generate random numbers between 1 and 20, to 1 decimal place.
However my issue as simple as mt_rand. I want most of the numbers generated to be lower around 0.5 - 4.5 with the occasional number being between 4.5-10 and very rarely say once every 12-20 hours being between 10-20.
I've been using the following but have no idea where to go from. I am a very basic self-taught programmer.
$min = 1;
$max = 20;
$suisse_interest = mt_rand ($min*10, $max*10) / 10
Maybe if I briefly explain why I want this it may help..
I own an online game and want to add 3 "banks" with each bank generating different interests each hour. Most of the time I want it low, but sometimes higher and very rarely very high (15-20%).
With the above code the random number goes too high to often.
Any help with this is greatly appreciated!
You need an exponential calculation. If you use a function similar to the following function, the probability for low numbers increases. Of course you need to adapt the numbers a bit to provide an output suiting your needs.
$i = 0;
while($i<30) {
$i++;
$rand = mt_rand(0, 7000) / 100; // 0.0 ~ 70.0
// This is the most important line:
$output = round( 20*(pow(0.95,$rand)) , 1);
echo "$output ";
}
Sample output:
1.8 4.3 2.6 5.5 3.7 15.5 1.6 0.6 0.6 1.6 5.8
1.3 6.1 3.2 0.8 1.7 14.7 7.9 1.3 10.3 5.5 12.6
1.5 8.4 1.5 0.9 13.3 5.8 7.5 1.7
As you see, mostly smaller number are printed.
The probability to get 20 is around 1.4% in my code whereas the probability to get a number smaller than 5 is around 78%
Try this.The probability to 1.0~4.5 is around 96%, 4.5~10.0 is around 2%, and 10.0~20.0 is around 2%.
<?php
// 1.0~4.5 96%
// 4.5~10.0 2%
// 10.0~20.0 2%
function fun() {
$num = mt_rand(1, 100);
if ($num > 0 && $num <= 96) {
$return = mt_rand(10, 45) / 10; // 96%
} else if ($num > 96 && $num <= 98) {
$return = mt_rand(45, 100) / 10; // 2%
} else {
$return = mt_rand(100, 200) / 10; // 2%
}
return sprintf("%01.1f",$return);
}
echo fun();
?>
This is not a PHP-specific problem.
What you need is a non-linear probability law, that you can then implement in PHP.
If you want something centered around an average value, the ideal would be a gaussian aka normal distribution, but computing it requires various complicated tricks, most of them being optimized for rapid generation at the cost of increasing complexity.
If you generate only a few values each hour, performance will not be an issue.
A reasonable approximation would be to sum 3 or 4 random variables, taking advantage of the central limit theorem.
Summing random values between 0 and twice your middle rate will create an approximation of a gaussian centered around your middle value.
You can then clamp values inferior to the middle point if you don't want low rates. The net result would be 50% chances of getting middle rate and a steadily decreasing chance to get up to twice that value.
An increasing number of sums will "narrow" the curve, making it less likely to get a high value.
for instance:
define ("INTEREST_MEAN", 10);
define ("INTEREST_SPREAD", 5);
function random_interest ()
{
$res = 0;
for ($i = 0 ; $i != INTEREST_SPREAD ; $i++) $res += mt_rand(0, 2*INTEREST_MEAN);
$res /= INTEREST_SPREAD; // normalize the sum to get a mean-centered result
$res = max ($res, INTEREST_MEAN); // clamp lower values
}
If I wanted a random number between one and three I could do $n = mt_rand(1,3).
There is a 33% chance that $n = 1, a 33% chance it's 2, and a 33% chance that it's 3.
What if I want to make it more difficult to get a 3 than a 1?
Say I want a 50% chance that a 1 is drawn, a 30% chance that a 2 is drawn and a 20% chance that a 3 is drawn?
I need a scalable solution as the possible range will vary between 1-3 and 1-100, but in general I'd like the lower numbers to be drawn more often than the higher ones.
How can I accomplish this?
There is a simple explanation of how you can use standard uniform random variable to produce random variable with a distribution similar to the one you want:
https://math.stackexchange.com/a/241543
This is maths.
In your example the just chose a random number between 0 and 99.
Values returned between 0 to 49 - call it 1
Values returned between 50 - 69 - Call it 2
Values returned between 70 - 99 - Call it 3
Simple if statement will do this or populate an array for the distribution required
Assuming a 1 - 10 scale, you can use a simple if statement and have the numbers represent percentages. And just have each if statement set $n to a specific. Only downfall, it isn't universal.
$dummy = mt_rand(1,10);
// represents 50%
if ($dummy <= 5) {
$n = 1;
}
// represents 40%
if ($dummy >= 6 && $dummy <= 9) {
$n = 2;
} else {
// represents 10%
$n = 3;
}
Sorry for the title. I wasn't sure how to ask this question.
I have a form on a website that asks a question. The answers are in check box form. Each answer is saved into my database with a 'score', the values look like this:
Allergy 1
Cardiology 2
Chest Disease 4
Dermatology 8
Emergency Room 16
Ambulance Trips 32
Gastroenterology 64
General Medicine 128
Gynecology 256
Hematology 512
Neurology 1024
Obstetrics 2048
Opthamology 4096
Orthopedics 8192
Physical Therapy 16384
Plastic Surgery 32768
Podiatry 65536
Proctology 131072
Psychiatry 262144
Surgery Performed 524288
Thoracic Surgery 1048576
Urology 2097152
Outside X-Rays 4194304
Diagnostic Tests (outside) 8388608
As you can see, the score is the previous value times two. When a user fills out the form, the answer is saved in the database as one value - all the answers added together.
For example, a user selected the values: Allergy, General Medicine, Hematology, Obstetrics. In the database, the answer for this question is saved as 2689.
Is there a way to figure out what answers have been selected by only having the answer to the question?
For example, I would query my database and pull the 2689 value, and I need to determine what answers were checked.
edit: I was hoping to reverse engineer the answer in PHP.
Yes, this is a common pattern called bit masking. Use your language's binary AND operator on the value corresponding to a given answer and the value submitted from the form to see if the given answer was one of the selected choices. For example, if the answer submitted and saved is 2689 as in your example, you can check whether "chest disease" was one of the selected choices by seeing if 2689 & 4 is nonzero. (& should be substituted with whatever the binary AND operator is in your language of choice.)
Note that this only works as long as all the values corresponding to individual choices are powers of 2. In general, the question posed in your title, about finding out what numbers from a given set have been added to come up with a given sum, is an instance of something called the knapsack problem and is only known to be solvable by checking every possible combination, which is very inefficient. (NP-complete, specifically)
You can find the values by ANDing with powers of 2.
20 = 1
21 = 2
22 = 4
23 = 8
...
223 = 8388608
You can find out the value of 2n using binary shifting like this: 1 << n
php like code:
$item[] = {"Allergy", "Cardiology", ..., "Diagnostic Tests (outside)"};
$answer = 2689;
for ( $power = 0; $power < count($item); $power++ ) {
if ( 1 << $power & $answer ) {
echo $item[$power] . "\n";
}
}
Edit: made it more php friendly
Yes, there is. Note that each k'th "score" is of the form 2^(k - 1), which corresponds to a bitstring with only the k'th bit set. If you know which bits are set, you can reconstruct the sum.
Taking 2689 as an example, we first need to write it out in binary:
2689 = 101010000001b
Counting from the right, we see that the first, eighth, tenth and twelfth bits are set, so (as you can verify)
2689 = 2^0 + 2^7 + 2^9 + 2^11
= 1 + 128 + 512 + 2048
The actual implementation of this can be done efficiently using bitwise operations. By taking the AND of the value and each of the "scores" in turn, then checking whether that gives a non-zero value, we can check which scores went into the sum.
this will do exactly what you wanted :
<?php
Print bindecValues("2689");
function bindecValues($decimal, $reverse=false, $inverse=false) {
$bin = decbin($decimal);
if ($inverse) {
$bin = str_replace("0", "x", $bin);
$bin = str_replace("1", "0", $bin);
$bin = str_replace("x", "1", $bin);
}
$total = strlen($bin);
$stock = array();
for ($i = 0; $i < $total; $i++) {
if ($bin{$i} != 0) {
$bin_2 = str_pad($bin{$i}, $total - $i, 0);
array_push($stock, bindec($bin_2));
}
}
$reverse ? rsort($stock):sort($stock);
return implode(", ", $stock);
}
?>
Happy coding
Remember that integers are stored in binary - so each of these flags (Allergy = 1) etc. will correspond to a single bit being true or false in the binary representation of the sum.
For example, 2689 in binary is 0000 1010 1000 0001 which, if you think of it as an array of bits, where the least significant bit (right most in that array) is the least significant flag (allergy) then we can easily see that the first (allergy), eighth (gen. medicine), tenth (hematology) and twelfth (obs) slots of the array are marked with a 1 for true.
The largest value in your array of flags is 24th bit in a 32 bit integer. You could define up to 8 more flags in this system before having to use a larger integer.
Since all your numbers seem to be powers of two, you just need to store the input value in a long enough integer to hold it, then bit mask.
if( value & 1 ) then 1 was part of the selection
if( value & 2 ) then 2 was part of the selection
if( value & 3 ) then 3 was part of the selection
and so on
Im making a browser based PHP game and in my database for the players it has a record of that players total EXP or experience.
What i need is a formula to translate that exp into a level or rank, out of 100.
So they start off at level 1, and when they hit say, 50 exp, go to level 2, then when they hit maybe 125/150, level 2.
Basically a formula that steadily makes each level longer (more exp)
Can anyone help? I'm not very good at maths :P
Many formulas may suit your needs, depending on how fast you want the required exp to go up.
In fact, you really should make this configurable (or at least easily changed in one central location), so that you can balance the game later. In most games these (and other) formulas are determined only after playtesting and trying out several options.
Here's one formula: First level-up happens at 50 exp; second at 150exp; third at 300 exp; fourth at 500 exp; etc. In other words, first you have to gather 50 exp, then 100 exp, then 150exp, etc. It's an Arithmetic Progression.
For levelup X then you need 25*X*(1+X) exp.
Added: To get it the other way round you just use basic math. Like this:
y=25*X*(1+X)
0=25*X*X+25*X-y
That's a standard Quadratic equation, and you can solve for X with:
X = (-25Âħsqrt(625+100y))/50
Now, since we want both X and Y to be greater than 0, we can drop one of the answers and are left with:
X = (sqrt(625+100y)-25)/50
So, for example, if we have 300 exp, we see that:
(sqrt(625+100*300)-25)/50 = (sqrt(30625)-25)/50 = (175-25)/50 = 150/50 = 3
Now, this is the 3rd levelup, so that means level 4.
If you wanted the following:
Level 1 # 0 points
Level 2 # 50 points
Level 3 # 150 points
Level 4 # 300 points
Level 5 # 500 points etc.
An equation relating experience (X) with level (L) is:
X = 25 * L * L - 25 * L
To calculate the level for a given experience use the quadratic equation to get:
L = (25 + sqrt(25 * 25 - 4 * 25 * (-X) ))/ (2 * 25)
This simplifies to:
L = (25 + sqrt(625 + 100 * X)) / 50
Then round down using the floor function to get your final formula:
L = floor(25 + sqrt(625 + 100 * X)) / 50
Where L is the level, and X is the experience points
It really depends on how you want the exp to scale for each level.
Let's say
LvL1 : 50 Xp
Lvl2: LvL1*2=100Xp
LvL3: LvL2*2=200Xp
Lvl4: LvL3*2=400Xp
This means you have a geometric progression
The Xp required to complete level n would be
`XPn=base*Q^(n-1)`
In my example base is the inital 50 xp and Q is 2 (ratio).
Provided a player starts at lvl1 with no xp:
when he dings lvl2 he would have 50 total Xp
at lvl3 150xp
at lvl4 350xp
and so forth
The total xp a player has when he gets a new level up would be:
base*(Q^n-1)/(Q-1)
In your case you already know how much xp the player has. For a ratio of 2 the formula gets simpler:
base * (2^n-1)=total xp at level n
to find out the level for a given xp amount all you need to do is apply a simple formula
$playerLevel=floor(log($playerXp/50+1,2));
But with a geometric progression it will get harder and harder and harder for players to level.
To display the XP required for next level you can just calculate total XP for next level.
$totalXpNextLevel=50*(pow(2,$playerLevel+1)-1);
$reqXp=$totalXpNextLevel - $playerXp;
Check start of the post:
to get from lvl1 -> lvl2 you need 50 xp
lvl2 ->lvl3 100xp
to get from lvl x to lvl(x+1)
you would need
$totalXprequired=50*pow(2,$playerLevel-1);
Google gave me this:
function experience($L) {
$a=0;
for($x=1; $x<$L; $x++) {
$a += floor($x+300*pow(2, ($x/7)));
}
return floor($a/4);
}
for($L=1;$L<100;$L++) {
echo 'Level '.$L.': '.experience($L).'<br />';
}
It is supposed the be the formula that RuneScape uses, you might me able to modify it to your needs.
Example output:
Level 1: 0
Level 2: 55
Level 3: 116
Level 4: 184
Level 5: 259
Level 6: 343
Level 7: 435
Level 8: 536
Level 9: 649
Level 10: 773
Here is a fast solution I used for a similar problem. You will likely wanna change the math of course, but it will give you the level from a summed xp.
$n = -1;
$L = 0;
while($n < $xp){
$n += pow(($L+1),3)+30*pow(($L+1),2)+30*($L+1)-50;
$L++;
}
echo("Current XP: " .$xp);
echo("Current Level: ".$L);
echo("Next Level: " .$n);
I take it what you're looking for is the amount of experience to decide what level they are on? Such as:
Level 1: 50exp
Level 2: 100exp
Level 3: 150exp ?
if that's the case you could use a loop something like:
$currentExp = x;
$currentLevel;
$i; // initialLevel
for($i=1; $i < 100; $i *= 3)
{
if( ($i*50 > $currentExp) && ($i < ($i+1)*$currentExp)){
$currentLevel = $i/3;
break;
}
}
This is as simple as I can make an algorithm for levels, I haven't tested it so there could be errors.
Let me know if you do use this, cool to think an algorithm I wrote could be in a game!
The original was based upon a base of 50, thus the 25 scattered across the equation.
This is the answer as a real equation. Just supply your multiplier (base) and your in business.
$_level = floor( floor( ($_multipliter/2)
+ sqrt( ($_multipliter^2) + ( ($_multipliter*2) * $_score) )
)
/ $_multipliter
) ;
1st number: 50
2. 30
3. 70
4. 40
5. 11
and other number is 33
I need to calculate which two numbers the last number is between (using php) .. any help?
Iterate over your list and find the following two values:
The largest number that is smaller than your target number.
The smallest number that is larger than your target number.
In pseudo-code:
lowerlimit = Unknown
upperlimit = Unknown
for each n in list:
if (n <= target) and (lowerlimit is Unknown or n > lowerlimit):
lowerlimit = n
if (n >= target) and (upperlimit is Unknown or n < upperlimit):
upperlimit = n
Then lowerlimit and upperlimit are your answer. This algorithm requires O(n) time and O(1) extra space.
If you are going to test the same list with many different target numbers then it could make sense to sort the list first requiring O(n log(n)) time, but then you can find the limits in just O(log(n)) time using a binary search.
I'll not give you the code but give you some guidelines for your homework.
You need to do these steps to solve your problem.
Sort your list of numbers. (I guess you are storing them in an array so sort the array.)
With a for loop search for the place where the element N is bigger than your number and the element N-1 is smaller. That will give you your position.
Oh and to avoid really long loop. use "break" after you find your position.
Sorted List:
11
30
// your 33 is bigger than 30 and smaller than 40, so this is the position you want.
40
50
70
function isBetween($several_numbers, $number)
{
$return_numbers = array();
sort($several_numbers);
$j = 0;
//find the first number in the array that $number is bigger than
while($number > $several_numbers[$j]) $j++;
if ($j == 0 || $j > count($several_numbers) - 1) return array();
$return_numbers[0] = $several_numbers[$j-1];
while($number > $several_numbers[$j]) $j++;
if ($j > count($several_numbers)-1) return array();
$return_numbers[1] = $several_numbers[$j];
return $return_numbers;
}
print_r(isBetween(array(50, 30, 70, 40, 10), 33));
I don't know if I understood correctly but this seems to be it