Best Approach to Getting Adjustment from Raw Number - php

I would like advice on the best/most efficient way to get an adjusted score for a team based on a raw score. Let me explain.
The teams are contract bridge teams and the raw score for the winner is from 0 (tie) to any number, but greater than 100 would be rare. The raw score is called IMPS and the adjusted score is called VPs, but that is just for clarity.
The adjustment table is:
IMPs VPs
1 thru 2 16 to 14
3 thru 4 17 to 13
5 thru 7 18 to 12
8 thru 11 19 to 11
12 thru 15 20 to 10
16 thru 20 21 to 9
21 thru 26 22 to 8
27 thru 33 23 to 7
34 thru 41 24 to 6
42 thru 50 25 to 5
51 thru 60 26 to 4
61 thru 71 27 to 3
72 thru 83 8 to 2
84 thru 95 29 to 1
96+ 30 to 0
Here is my code. It works fine:
PHP
$teamArawScore = 20; //Actual result will come from form input
if ($teamArawScore >95 )
{
$teamAadjScore = 30;
$teamBadjScore = 0;
}
else
{
$adjustmentArray = array
(15,
16,16,
17,17,
18,18,18,
19,19,19,19,
20,20,20,20,
21,21,21,21,21,
22,22,22,22,22,22,
23,23,23,23,23,23,23,
24,24,24,24,24,24,24,24,
25,25,25,25,25,25,25,25,25,
26,26,26,26,26,26,26,26,26,26,
27,27,27,27,27,27,27,27,27,27,27,
28,28,29,29,28,28,28,28,28,28,28,28,
29,29,29,29,29,29,29,29,29,29,29,29);
$teamAadjScore = $adjustmentArray[$teamArawScore];
$teamBadjScore = 30 - $teamAadjScore;
}
echo "TeamA won by $teamArawScore so it won $teamAadjScore VPs and TeamB won $teamBadjScore VPs.";
My approach just seems inefficient. Since the array is small, I doubt there is any performance issues, but I would like to do the conversion as efficiently as possible.
What do you suggest?

Since there's no consistency in the table you described you will always be populating a lookup array containing the actual logic. You could also structure this differently, with a shorter array containing the sequential amount of points leading to a given score etc., but in the end those would all result in having to loop over the array to see where you end up - swapping memory for CPU cycles.
Effectively, you need a lookup table anyway because there appears to be no algorithm that can reliably map the contents of the lookup table, and your implementation is now O(1) in big-O notation. As such it can by definition not be made more efficient.
For reference on lookup tables (emphasis added in quote):
In computer science, a lookup table is an array that replaces runtime
computation with a simpler array indexing operation. The savings in
terms of processing time can be significant, since retrieving a value
from memory is often faster than undergoing an 'expensive' computation
or input/output operation. The tables may be precalculated and
stored in static program storage, calculated (or "pre-fetched") as
part of a program's initialization phase (memoization), or even stored
in hardware in application-specific platforms.
There's nothing 'bad practice' about using them. Back in the days when CPUs didn't have floating point units on board we'd have entire sine and sqrt tables embedded in programs to circumvent computationally expensive calculations at runtime.

I got a little something for you.Even though I'm not sure it would actualy optimize the execution speed.
Looping through an array, knowing the base value being 16 if we have a positive score (higher than one) from the A team.
This is not optimized at all (mainly because of the $a == 0 condition), but here it is:
<?php
$teamArawScore = 30;
$a = $teamArawScore;
$teamAfinalScore = 16;
$scoreArray = array(3,5,8,12,16,21,27,34,42,51,61,72,84,10000000000);
$count = 0;
foreach($scoreArray as $elem)
{
if($a < $elem)
{
$teamAfinalScore += $count;
break;
}
$count++;
}
if($a ==0)
{
$teamAfinalScore = 15;
$teamBfinalScore = 30 - $teamAfinalScore;
}
echo "Team A: ".$teamAfinalScore. "<br />Team B:".$teamBfinalScore;
?>

<?php
$teamArawScore = 1000; //Actual result will come from form input
if ($teamArawScore >95 )
{
$teamAadjScore = 30;
$teamBadjScore = 0;
}
else
{
$adjustmentArray = array('1'=>16,'3'=>17,'5'=>18,'8'=>19,'12'=>20,'16'=>21,'21'=>22,'27'=>23,'34'=>24,'42'=>50,'51'=>60,'61'=>27,'72'=>83,'84'=>95);
$base_score=array(1,3,5,8,12,16,21,27,34,42,51,61,72,84);
$count=count($base_score);
$adjustment_value=$adjustmentArray['1'];
for($i=1; $i<$count-1; $i++){
if($teamArawScore < $base_score[$i+1]){
$adjustment_value=$adjustmentArray[$base_score[$i]];
break;
}
else{
$adjustment_value=$adjustmentArray[$base_score[$i]]; // for values greater than 84
}
}
$teamAadjScore = $adjustment_value;
$teamBadjScore = 30 - $teamAadjScore;
}
echo "TeamA won by $teamArawScore so it won $teamAadjScore VPs and TeamB won $teamBadjScore VPs." ;

Related

Controlling likelyhood of randomly generated numbers

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;
}

Finding out what numbers have been added to come up with a sum

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

php game, formula to calculate a level based on exp

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
) ;

PHP: find two or more numbers from a list of numbers that add up towards a given amount

I am trying to create a little php script that can make my life a bit easier.
Basically, I am going to have 21 text fields on a page where I am going to input 20 different numbers. In the last field I will enter a number let's call it the TOTAL AMOUNT. All I want the script to do is to point out which numbers from the 20 fields added up will come up to TOTAL AMOUNT.
Example:
field1 = 25.23
field2 = 34.45
field3 = 56.67
field4 = 63.54
field5 = 87.54
....
field20 = 4.2
Total Amount = 81.90
Output: field1 + fields3 = 81.90
Some of the fields might have 0 as value because sometimes I only need to enter 5-15 fields and the maximum will be 20.
If someone can help me out with the php code for this, will be greatly appreciated.
If you look at oezis algorithm one drawback is immediately clear: It spends very much time summing up numbers which are already known not to work. (For example if 1 + 2 is already too big, it doesn't make any sense to try 1 + 2 + 3, 1 + 2 + 3 + 4, 1 + 2 + 3 + 4 + 5, ..., too.)
Thus I have written an improved version. It does not use bit magic, it makes everything manual. A drawback is, that it requires the input values to be sorted (use rsort). But that shouldn't be a big problem ;)
function array_sum_parts($vals, $sum){
$solutions = array();
$pos = array(0 => count($vals) - 1);
$lastPosIndex = 0;
$currentPos = $pos[0];
$currentSum = 0;
while (true) {
$currentSum += $vals[$currentPos];
if ($currentSum < $sum && $currentPos != 0) {
$pos[++$lastPosIndex] = --$currentPos;
} else {
if ($currentSum == $sum) {
$solutions[] = array_slice($pos, 0, $lastPosIndex + 1);
}
if ($lastPosIndex == 0) {
break;
}
$currentSum -= $vals[$currentPos] + $vals[1 + $currentPos = --$pos[--$lastPosIndex]];
}
}
return $solutions;
}
A modified version of oezis testing program (see end) outputs:
possibilities: 540
took: 3.0897309780121
So it took only 3.1 seconds to execute, whereas oezis code executed 65 seconds on my machine (yes, my machine is very slow). That's more than 20 times faster!
Furthermore you may notice, that my code found 540 instead of 338 possibilities. This is because I adjusted the testing program to use integers instead of floats. Direct floating point comparison is rarely the right thing to do, this is a great example why: You sometimes get 59.959999999999 instead of 59.96 and thus the match will not be counted. So, if I run oezis code with integers it finds 540 possibilities, too ;)
Testing program:
// Inputs
$n = array();
$n[0] = 6.56;
$n[1] = 8.99;
$n[2] = 1.45;
$n[3] = 4.83;
$n[4] = 8.16;
$n[5] = 2.53;
$n[6] = 0.28;
$n[7] = 9.37;
$n[8] = 0.34;
$n[9] = 5.82;
$n[10] = 8.24;
$n[11] = 4.35;
$n[12] = 9.67;
$n[13] = 1.69;
$n[14] = 5.64;
$n[15] = 0.27;
$n[16] = 2.73;
$n[17] = 1.63;
$n[18] = 4.07;
$n[19] = 9.04;
$n[20] = 6.32;
// Convert to Integers
foreach ($n as &$num) {
$num *= 100;
}
$sum = 57.96 * 100;
// Sort from High to Low
rsort($n);
// Measure time
$start = microtime(true);
echo 'possibilities: ', count($result = array_sum_parts($n, $sum)), '<br />';
echo 'took: ', microtime(true) - $start;
// Check that the result is correct
foreach ($result as $element) {
$s = 0;
foreach ($element as $i) {
$s += $n[$i];
}
if ($s != $sum) echo '<br />FAIL!';
}
var_dump($result);
sorry for adding a new answer, but this is a complete new solution to solve all problems of life, universe and everything...:
function array_sum_parts($n,$t,$all=false){
$count_n = count($n); // how much fields are in that array?
$count = pow(2,$count_n); // we need to do 2^fields calculations to test all possibilities
# now i want to look at every number from 1 to $count, where the number is representing
# the array and add up all array-elements which are at positions where my actual number
# has a 1-bit
# EXAMPLE:
# $i = 1 in binary mode 1 = 01 i'll use ony the first array-element
# $i = 10 in binary mode 10 = 1010 ill use the secont and the fourth array-element
# and so on... the number of 1-bits is the amount of numbers used in that try
for($i=1;$i<=$count;$i++){ // start calculating all possibilities
$total=0; // sum of this try
$anzahl=0; // counter for 1-bits in this try
$k = $i; // store $i to another variable which can be changed during the loop
for($j=0;$j<$count_n;$j++){ // loop trough array-elemnts
$total+=($k%2)*$n[$j]; // add up if the corresponding bit of $i is 1
$anzahl+=($k%2); // add up the number of 1-bits
$k=$k>>1; //bit-shift to the left for looking at the next bit in the next loop
}
if($total==$t){
$loesung[$i] = $anzahl; // if sum of this try is the sum we are looking for, save this to an array (whith the number of 1-bits for sorting)
if(!$all){
break; // if we're not looking for all solutions, make a break because the first one was found
}
}
}
asort($loesung); // sort all solutions by the amount of numbers used
// formating the solutions to getting back the original array-keys (which shoud be the return-value)
foreach($loesung as $val=>$anzahl){
$bit = strrev(decbin($val));
$total=0;
$ret_this = array();
for($j=0;$j<=strlen($bit);$j++){
if($bit[$j]=='1'){
$ret_this[] = $j;
}
}
$ret[]=$ret_this;
}
return $ret;
}
// Inputs
$n[0]=6.56;
$n[1]=8.99;
$n[2]=1.45;
$n[3]=4.83;
$n[4]=8.16;
$n[5]=2.53;
$n[6]=0.28;
$n[7]=9.37;
$n[8]=0.34;
$n[9]=5.82;
$n[10]=8.24;
$n[11]=4.35;
$n[12]=9.67;
$n[13]=1.69;
$n[14]=5.64;
$n[15]=0.27;
$n[16]=2.73;
$n[17]=1.63;
$n[18]=4.07;
$n[19]=9.04;
$n[20]=6.32;
// Output
$t=57.96;
var_dump(array_sum_parts($n,$t)); //returns one possible solution (fuc*** fast)
var_dump(array_sum_parts($n,$t,true)); // returns all possible solution (relatively fast when you think of all the needet calculations)
if you don't use the third parameter, it returns the best (whith the least amount numbers used) solution as array (whith keys of the input-array) - if you set the third parameter to true, ALL solutions are returned (for testing, i used the same numbers as zaf in his post - there are 338 solutions in this case, found in ~10sec on my machine).
EDIT:
if you get all, you get the results ordered by which is "best" - whithout this, you only get the first found solution (which isn't necessarily the best).
EDIT2:
to forfil the desire of some explanation, i commented the essential parts of the code . if anyone needs more explanation, please ask
1. Check and eliminate fields values more than 21st field
2. Check highest of the remaining, Add smallest,
3. if its greater than 21st eliminate highest (iterate this process)
4. If lower: Highest + second Lowest, if equal show result.
5. if higher go to step 7
6. if lower go to step 4
7. if its lower than add second lowest, go to step 3.
8. if its equal show result
This is efficient and will take less execution time.
Following method will give you an answer... almost all of the time. Increase the iterations variable to your taste.
<?php
// Inputs
$n[1]=8.99;
$n[2]=1.45;
$n[3]=4.83;
$n[4]=8.16;
$n[5]=2.53;
$n[6]=0.28;
$n[7]=9.37;
$n[8]=0.34;
$n[9]=5.82;
$n[10]=8.24;
$n[11]=4.35;
$n[12]=9.67;
$n[13]=1.69;
$n[14]=5.64;
$n[15]=0.27;
$n[16]=2.73;
$n[17]=1.63;
$n[18]=4.07;
$n[19]=9.04;
$n[20]=6.32;
// Output
$t=57.96;
// Let's try to do this a million times randomly
// Relax, thats less than a blink
$iterations=1000000;
while($iterations-->0){
$z=array_rand($n, mt_rand(2,20));
$total=0;
foreach($z as $x) $total+=$n[$x];
if($total==$t)break;
}
// If we did less than a million times we have an answer
if($iterations>0){
$total=0;
foreach($z as $x){
$total+=$n[$x];
print("[$x] + ". $n[$x] . " = $total<br/>");
}
}
?>
One solution:
[1] + 8.99 = 8.99
[4] + 8.16 = 17.15
[5] + 2.53 = 19.68
[6] + 0.28 = 19.96
[8] + 0.34 = 20.3
[10] + 8.24 = 28.54
[11] + 4.35 = 32.89
[13] + 1.69 = 34.58
[14] + 5.64 = 40.22
[15] + 0.27 = 40.49
[16] + 2.73 = 43.22
[17] + 1.63 = 44.85
[18] + 4.07 = 48.92
[19] + 9.04 = 57.96
A probably inefficient but simple solution with backtracking
function subset_sums($a, $val, $i = 0) {
$r = array();
while($i < count($a)) {
$v = $a[$i];
if($v == $val)
$r[] = $v;
if($v < $val)
foreach(subset_sums($a, $val - $v, $i + 1) as $s)
$r[] = "$v $s";
$i++;
}
return $r;
}
example
$ns = array(1, 2, 6, 7, 11, 5, 8, 9, 3);
print_r(subset_sums($ns, 11));
result
Array
(
[0] => 1 2 5 3
[1] => 1 2 8
[2] => 1 7 3
[3] => 2 6 3
[4] => 2 9
[5] => 6 5
[6] => 11
[7] => 8 3
)
i don't think the answer isn't as easy as nik mentioned. let's ay you have the following numbers:
1 2 3 6 8
looking for an amount of 10
niks solution would do this (if i understand it right):
1*8 = 9 = too low
adding next lowest (2) = 11 = too high
now he would delete the high number and start again taking the new highest
1*6 = 7 = too low
adding next lowest (2) = 9 = too low
adding next lowest (3) = 12 = too high
... and so on, where the perfect answer would simply
be 8+2 = 10... i think the only solution is trying every possible combination of
numbers and stop if the amaunt you are looking for is found (or realy calculate all, if there are different solutions and save which one has used least numbers).
EDIT: realy calculating all possible combiations of 21 numbers will end up in realy, realy, realy much calculations - so there must be any "intelligent" solution for adding numbers in a special order (lik that one in niks post - with some improvements, maybe that will bring us to a reliable solution)
Without knowing if this is a homework assignment or not, I can give you some pseudo code as a hint for a possible solution, note the solution is not very efficient, more of a demonstration.
Hint:
Compare each field value to all field value and at each iteration check if their sum is equal to TOTAL_AMOUNT.
Pseudo code:
for i through field 1-20
for j through field 1-20
if value of i + value of j == total_amount
return i and j
Update:
What you seem to be having is the Subset sum problem, given within the Wiki link is pseudo code for the algorithm which might help point you in the right direction.

Generate random player strengths in a pyramid structure (PHP)

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))

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