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I am attempting to find the cartesian product and append specific criteria.
I have four pools of 25 people each. Each person has a score and a price. Each person in each pool looks as such.
[0] => array(
"name" => "jacob",
"price" => 15,
"score" => 100
),
[1] => array(
"name" => "daniel",
"price" => 22,
"score" => 200
)
I want to find the best combination of people, with one person being picked from each pool. However, there is a ceiling price where no grouping can exceed a certain price.
I have been messing with cartesians and permutation functions and cannot seem to figure out how to do this. The only way I know how to code it is to have nested foreach loops, but that is incredibly taxing.
This code below, as you can see, is incredibly inefficient. Especially if the pools increase!
foreach($poolA as $vA) {
foreach($poolb as $vB) {
foreach($poolC as $vC) {
foreach($poolD as $vD) {
// calculate total price and check if valid
// calculate total score and check if greatest
// if so, add to $greatest array
}
}
}
}
I also thought I could find a way to calculate the total price/score ratio and use that to my advantage, but I don't know what I'm missing.
As pointed out by Barmar, sorting the people in each pool allows you to halt the loops early when the total price exceeds the limit and hence reduces the number of cases you need to check. However, the asymptotic complexity for applying this improvement is still O(n4) (where n is the number of people in a pool).
I will outline an alternative approach with better asymptotic complexity as follow:
Construct a pool X that contains all pairs of people with one from pool A and the other from pool B.
Construct a pool Y that contains all pairs of people with one from pool C and the other from pool D.
Sort the pairs in pool X by total price. Then for any pairs with the same price, retain the one with the highest score and discard the remaining pairs.
Sort the pairs in pool Y by total price. Then for any pairs with the same price, retain the one with the highest score and discard the remaining pairs.
Do a loop with two pointers to check over all possible combinations that satisfy the price constraint, where the head pointer starts at the first item in pool X, and the tail pointer starts at the last item in pool Y. Sample code is given below to illustrate how this loop works:
==========================================================================
$head = 0;
$tail = sizeof($poolY) - 1;
while ($head < sizeof($poolX) && $tail >= 0) {
$total_price = $poolX[$head].price + $poolY[$tail].price;
// Your logic goes here...
if ($total_price > $price_limit) {
$tail--;
} else if ($total_price < $price_limit) {
$head++;
} else {
$head++;
$tail--;
}
}
for ($i = $head; $i < sizeof($poolX); $i++) {
// Your logic goes here...
}
for ($i = $tail; $i >= 0; $i--) {
// Your logic goes here...
}
==========================================================================
The complexity of steps 1 and 2 are O(n2), and the complexity of steps 3 and 4 can be done in O(n2 log(n)) using balanced binary tree. And step 5 is essentially a linear scan over n2 items, so the complexity is also O(n2). Therefore the overall complexity of this approach is O(n2 log(n)).
A couple of things to note about your approach here. Speaking strictly from a mathematics perspective, you're calculating way more permutations than is actually necessary to arrive at a definitive answer.
In combinatorics, there are two important questions to ask in order to arrive at the exact number of permutations necessary to yield all possible combinations.
Does order matter? (for your case, it does not)
Is repetition allowed? (for your case, it is not necessary to repeat)
Since the answer to both of these question is no, you need only a fraction of the iterations you're currently doing with your nested loop. Currently you are doing, pow(25, 4) permutations, which is 390625. You only actually need n! / r! (n-r)! or gmp_fact(25) / (gmp_fact(4) * gmp_fact(25 - 4)) which is only 12650 total permutations needed.
Here's a simple example of a function that produces combinations without repetition (and where order does not matter), using a generator in PHP (taken from this SO answer).
function comb($m, $a) {
if (!$m) {
yield [];
return;
}
if (!$a) {
return;
}
$h = $a[0];
$t = array_slice($a, 1);
foreach(comb($m - 1, $t) as $c)
yield array_merge([$h], $c);
foreach(comb($m, $t) as $c)
yield $c;
}
$a = range(1,25); // 25 people in each pool
$n = 4; // 4 pools
foreach(comb($n, $a) as $i => $c) {
echo $i, ": ", array_sum($c), "\n";
}
It would be pretty easy to modify the generator function to check whether the sum of prices meets/exceeds the desired threshhold and only return valid results from there (i.e. abandoning early where needed).
The reason repetition and order are not important here for your use case, is because it doesn't matter whether you add $price1 + $price2 or $price2 + $price1, the result will undoubtedly be the same in both permutations. So you only need to add up each unique set once to ascertain all possible sums.
Similar to chiwangs solutions, you may eliminate up front every group member, where another group member in that group exists, with same or higher score for a lower price.
Maybe you can eliminate many members in each group with this approach.
You may then either use this technique, to build two pairs and repeat the filtering (eliminate pairs, where anothr pair exists, with higher score for the same or lower costs) and then combine the pairs the same way, or add a member step by step (one pair, a triple, a quartett).
If there exists some member, who exceed the allowed sum price on their own, they can be eliminated up front.
If you order the 4 groups by score descending, and you find a solution abcd, where the sum price is legal, you found the optimal solution for a given set of abc.
The reponses here helped me figure out the best way for me to do this.
I haven't optimized the function yet, but essentially I looped through each results two at a time to find the combined salaries / scores for each combination in the two pools.
I stored the combined salary -> score combination in a new array, and if the salary already existed, I'd compare scores and remove the lower one.
$results = array();
foreach($poolA as $A) {
foreach($poolB as $B) {
$total_salary = $A['Salary'] + $B['Salary'];
$total_score = $A['Score'] + $B['Score'];
$pids = array($A['pid'], $B['pid']);
if(isset($results[$total_salary]) {
if($total_score > $results[$total_salary]['Score']) {
$results[$total_salary]['Score'] => $total_score;
$results[$total_salary]['pid'] => $pids;
} else {
$results[$total_salary]['Score'] = $total_score;
$results[$total_salary]['pid'] = $pids;
}
}
}
After this loop, I have another one that is identical, except my foreach loops are between $results and $poolC.
foreach($results as $R) {
foreach($poolC as $C) {
and finally, I do it one last time for $poolD.
I am working on optimizing the code by putting all four foreach loops into one.
Thank you everyone for your help, I was able to loop through 9 lists with 25+ people in each and find the best result in an incredibly quick processing time!
I have a set of items. I need to randomly pick one. The problem is that they each have a weight of 1-10. A weight of 2 means that the item is twice as likely to be picked than a weight of 1. A weight of 3 is three times as likely.
I currently fill an array with each item. If the weight is 3, I put three copies of the item in the array. Then, I pick a random item.
My method is fast, but uses a lot of memory. I am trying to think of a faster method, but nothing comes to mind. Anyone have a trick for this problem?
EDIT: My Code...
Apparently, I wasn't clear. I do not want to use (or improve) my code. This is what I did.
//Given an array $a where $a[0] is an item name and $a[1] is the weight from 1 to 100.
$b = array();
foreach($a as $t)
$b = array_merge($b, array_fill(0,$t[1],$t));
$item = $b[array_rand($b)];
This required me to check every item in $a and uses max_weight/2*size of $a memory for the array. I wanted a COMPLETELY DIFFERENT algorithm.
Further, I asked this question in the middle of the night using a phone. Typing code on a phone is nearly impossible because those silly virtual keyboards simply suck. It auto-corrects everything, ruining any code I type.
An yet further, I woke up this morning with an entirely new algorithm that uses virtual no extra memory at all and does not require checking every item in the array. I posted it as an answer below.
This ones your huckleberry.
$arr = array(
array("val" => "one", "weight" => 1),
array("val" => "two", "weight" => 2),
array("val" => "three", "weight" => 3),
array("val" => "four", "weight" => 4)
);
$weight_sum = 0;
foreach($arr as $val)
{
$weight_sum += $val['weight'];
}
$r = rand(1, $weight_sum);
print "random value is $r\n";
for($i = 0; $i < count($arr); $i++)
{
if($r <= $arr[$i]['weight'])
{
print "$r <= {$arr[$i]['weight']}, this is our match\n";
print $arr[$i]['val'] . "\n";
break;
}
else
{
print "$r > {$arr[$i]['weight']}, subtracting weight\n";
$r -= $arr[$i]['weight'];
print "new \$r is $r\n";
}
}
No need to generate arrays containing an item for every weight, no need to fill an array with n elements for a weight of n. Just generate a random number between 1 and total weight, then loop through the array until you find a weight less than your random number. If it isn't less than the number, subtract that weight from the random and continue.
Sample output:
# php wr.php
random value is 8
8 > 1, subtracting weight
new $r is 7
7 > 2, subtracting weight
new $r is 5
5 > 3, subtracting weight
new $r is 2
2 <= 4, this is our match
four
This should also support fractional weights.
modified version to use array keyed by weight, rather than by item
$arr2 = array(
);
for($i = 0; $i <= 500000; $i++)
{
$weight = rand(1, 10);
$num = rand(1, 1000);
$arr2[$weight][] = $num;
}
$start = microtime(true);
$weight_sum = 0;
foreach($arr2 as $weight => $vals) {
$weight_sum += $weight * count($vals);
}
print "weighted sum is $weight_sum\n";
$r = rand(1, $weight_sum);
print "random value is $r\n";
$found = false;
$elem = null;
foreach($arr2 as $weight => $vals)
{
if($found) break;
for($j = 0; $j < count($vals); $j ++)
{
if($r < $weight)
{
$elem = $vals[$j];
$found = true;
break;
}
else
{
$r -= $weight;
}
}
}
$end = microtime(true);
print "random element is: $elem\n";
print "total time is " . ($end - $start) . "\n";
With sample output:
# php wr2.php
weighted sum is 2751550
random value is 345713
random element is: 681
total time is 0.017189025878906
measurement is hardly scientific - and fluctuates depending on where in the array the element falls (obviously) but it seems fast enough for huge datasets.
This way requires two random calculations but they should be faster and require about 1/4 of the memory but with some reduced accuracy if weights have disproportionate counts. (See Update for increased accuracy at the cost of some memory and processing)
Store a multidimensional array where each item is stored in the an array based on its weight:
$array[$weight][] = $item;
// example: Item with a weight of 5 would be $array[5][] = 'Item'
Generate a new array with the weights (1-10) appearing n times for n weight:
foreach($array as $n=>$null) {
for ($i=1;$i<=$n;$i++) {
$weights[] = $n;
}
}
The above array would be something like: [ 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ... ]
First calculation: Get a random weight from the weighted array we just created
$weight = $weights[mt_rand(0, count($weights)-1)];
Second calculation: Get a random key from that weight array
$value = $array[$weight][mt_rand(0, count($array[$weight])-1)];
Why this works: You solve the weighted issue by using the weighted array of integers we created. Then you select randomly from that weighted group.
Update: Because of the possibility of disproportionate counts of items per weight, you could add another loop and array for the counts to increase accuracy.
foreach($array as $n=>$null) {
$counts[$n] = count($array[$n]);
}
foreach($array as $n=>$null) {
// Calculate proportionate weight (number of items in this weight opposed to minimum counted weight)
$proportion = $n * ($counts[$n] / min($counts));
for ($i=1; $i<=$proportion; $i++) {
$weights[] = $n;
}
}
What this does is if you have 2000 10's and 100 1's, it'll add 200 10's (20 * 10, 20 because it has 20x the count, and 10 because it is weighted 10) instead of 10 10's to make it proportionate to how many are in there opposed the minimum weight count. So to be accurate, instead of adding one for EVERY possible key, you are just being proportionate based on the MINIMUM count of weights.
I greatly appreciate the answers above. Please consider this answer, which does not require checking every item in the original array.
// Given $a as an array of items
// where $a[0] is the item name and $a[1] is the item weight.
// It is known that weights are integers from 1 to 100.
for($i=0; $i<sizeof($a); $i++) // Safeguard described below
{
$item = $a[array_rand($a)];
if(rand(1,100)<=$item[1]) break;
}
This algorithm only requires storage for two variables ($i and $item) as $a was already created before the algorithm kicked in. It does not require a massive array of duplicate items or an array of intervals.
In a best-case scenario, this algorithm will touch one item in the original array and be done. In a worst-case scenario, it will touch n items in an array of n items (not necessarily every item in the array as some may be touched more than once).
If there was no safeguard, this could run forever. The safeguard is there to stop the algorithm if it simply never picks an item. When the safeguard is triggered, the last item touched is the one selected. However, in millions of tests using random data sets of 100,000 items with random weights of 1 to 10 (changing rand(1,100) to rand(1,10) in my code), the safeguard was never hit.
I made histograms comparing the frequency of items selected among my original algorithm, the ones from answers above, and the one in this answer. The differences in frequencies are trivial - easy to attribute to variances in the random numbers.
EDIT... It is apparent to me that my algorithm may be combined with the algorithm pala_ posted, removing the need for a safeguard.
In pala_'s algorithm, a list is required, which I call an interval list. To simplify, you begin with a random_weight that is rather high. You step down the list of items and subtract the weight of each one until your random_weight falls to zero (or less). Then, the item you ended on is your item to return. There are variations on this interval algorithm that I've tested and pala_'s is a very good one. But, I wanted to avoid making a list. I wanted to use only the given weighted list and never touch all the items. The following algorithm merges my use of random jumping with pala_'s interval list. Instead of a list, I randomly jump around the list. I am guaranteed to get to zero eventually, so no safeguard is needed.
// Given $a as the weighted array (described above)
$weight = rand(1,100); // The bigger this is, the slower the algorithm runs.
while($weight>0)
{
$item = $a[array_rand($a)];
$weight-= $item[1];
}
// $item is the random item you want.
I wish I could select both pala_ and this answer as the correct answers.
I'm not sure if this is "faster", but I think it may be more "balance"d between memory usage and speed.
The thought is to transform your current implementation (500000 items array) into an equal-length array (100000 items), with the lowest "origin" position as key, and origin index as value:
<?php
$set=[["a",3],["b",5]];
$current_implementation=["a","a","a","b","b","b","b","b"];
// 0=>0 means the lowest "position" 0
// points to 0 in the set;
// 3=>1 means the lowest "position" 3
// points to 1 in the set;
$my_implementation=[0=>0,3=>1];
And then randomly picks a number between 0 and highest "origin" position:
// 3 is the lowest position of the last element ("b")
// and 5 the weight of that last element
$my_implemention_pick=mt_rand(0,3+5-1);
Full code:
<?php
function randomPickByWeight(array $set)
{
$low=0;
$high=0;
$candidates=[];
foreach($set as $key=>$item)
{
$candidates[$high]=$key;
$high+=$item["weight"];
}
$pick=mt_rand($low,$high-1);
while(!array_key_exists($pick,$candidates))
{
$pick--;
}
return $set[$candidates[$pick]];
}
$cache=[];
for($i=0;$i<100000;$i++)
{
$cache[]=["item"=>"item {$i}","weight"=>mt_rand(1,10)];
}
$time=time();
for($i=0;$i<100;$i++)
{
print_r(randomPickByWeight($cache));
}
$time=time()-$time;
var_dump($time);
3v4l.org demo
3v4l.org have some time limitation on codes, so the demo didn't finished. On my laptop the above demo finished in 10 seconds (i7-4700 HQ)
ere is my offer in case I've understand you right. I offer you take a look and if there are some question I'll explain.
Some words in advance:
My sample is with only 3 stages of weight - to be clear
- With outer while I'm simulating your main loop - I count only to 100.
- The array must to be init with one set of initial numbers as shown in my sample.
- In every pass of main loop I get only one random value and I'm keeping the weight at all.
<?php
$array=array(
0=>array('item' => 'A', 'weight' => 1),
1=>array('item' => 'B', 'weight' => 2),
2=>array('item' => 'C', 'weight' => 3),
);
$etalon_weights=array(1,2,3);
$current_weights=array(0,0,0);
$ii=0;
while($ii<100){ // Simulates your main loop
// Randomisation cycle
if($current_weights==$etalon_weights){
$current_weights=array(0,0,0);
}
$ft=true;
while($ft){
$curindex=rand(0,(count($array)-1));
$cur=$array[$curindex];
if($current_weights[$cur['weight']-1]<$etalon_weights[$cur['weight']-1]){
echo $cur['item'];
$array[]=$cur;
$current_weights[$cur['weight']-1]++;
$ft=false;
}
}
$ii++;
}
?>
I'll use this input array for my explanation:
$values_and_weights=array(
"one"=>1,
"two"=>8,
"three"=>10,
"four"=>4,
"five"=>3,
"six"=>10
);
The simple version isn't going to work for you because your array is so large. It requires no array modification but may need to iterate the entire array, and that's a deal breaker.
/*$pick=mt_rand(1,array_sum($values_and_weights));
$x=0;
foreach($values_and_weights as $val=>$wgt){
if(($x+=$wgt)>=$pick){
echo "$val";
break;
}
}*/
For your case, re-structuring the array will offer great benefits.
The cost in memory for generating a new array will be increasingly justified as:
array size increases and
number of selections increases.
The new array requires the replacement of "weight" with a "limit" for each value by adding the previous element's weight to the current element's weight.
Then flip the array so that the limits are the array keys and the values are the array values.
The selection logic is: the selected value will have the lowest limit that is >= $pick.
// Declare new array using array_walk one-liner:
array_walk($values_and_weights,function($v,$k)use(&$limits_and_values,&$x){$limits_and_values[$x+=$v]=$k;});
//Alternative declaration method - 4-liner, foreach() loop:
/*$x=0;
foreach($values_and_weights as $val=>$wgt){
$limits_and_values[$x+=$wgt]=$val;
}*/
var_export($limits_and_values);
$limits_and_values looks like this:
array (
1 => 'one',
9 => 'two',
19 => 'three',
23 => 'four',
26 => 'five',
36 => 'six',
)
Now to generate the random $pick and select the value:
// $x (from walk/loop) is the same as writing: end($limits_and_values); $x=key($limits_and_values);
$pick=mt_rand(1,$x); // pull random integer between 1 and highest limit/key
while(!isset($limits_and_values[$pick])){++$pick;} // smallest possible loop to find key
echo $limits_and_values[$pick]; // this is your random (weighted) value
This approach is brilliant because isset() is very fast and the maximum number of isset() calls in the while loop can only be as many as the largest weight (not to be confused with limit) in the array.
FOR YOUR CASE, THIS APPROACH WILL FIND THE VALUE IN 10 ITERATIONS OR LESS!
Here is my Demo that will accept a weighted array (like $values_and_weights), then in just four lines:
Restructure the array,
Generate a random number,
Find the correct value, and
Display it.
I need to total the number of clicks over 10 links on my page and then figure out the percentage of people that clicked each. This is easy division, but how do I make sure that I get a round 100% at the end.
I want to use the below code, but am worried that a situation could arise where the percentages do not tally to 100% as this function simply removes the numbers after the period.
function percent($num_amount, $num_total) {
$count1 = $num_amount / $num_total;
$count2 = $count1 * 100;
$count = number_format($count2, 0);
echo $count;
}
Any advice would be much appreciated.
Instead of calculating one percentage in your function you could pass all your results as an array and process it as a whole. After calculating all the percentages and rounding them make a check to see if they total 100. If not, then adjust the largest value to force them all to total 100. Adjusting the largest value will make sure your results are skewed as little as possible.
The array in my example would total 100.02 before making the adjustment.
function percent(array $numbers)
{
$result = array();
$total = array_sum($numbers);
foreach($numbers as $key => $number){
$result[$key] = round(($number/$total) * 100, 2);
}
$sum = array_sum($result);//This is 100.02 with my example array.
if(100 !== $sum){
$maxKeys = array_keys($result, max($result));
$result[$maxKeys[0]] = 100 - ($sum - max($result));
}
return $result;
}
$numbers = array(10.2, 22.36, 50.10, 27.9, 95.67, 3.71, 9.733, 4.6, 33.33, 33.33);
$percentages = percent($numbers);
var_dump($percentages);
var_dump(array_sum($percentages));
Output:-
array (size=10)
0 => float 3.51
1 => float 7.69
2 => float 17.22
3 => float 9.59
4 => float 32.86
5 => float 1.28
6 => float 3.35
7 => float 1.58
8 => float 11.46
9 => float 11.46
float 100
This will also work with an associative array as the function parameter. The keys will be preserved.
These figures could now be presented in a table, graph or chart and will always give you a total of 100%;
What you want to do is this.
Total the number of clicks across the board, then divide each number by the total.
For example:
1134
5391
2374
2887
In this case, four buttons, with a total of 11786 clicks, so:
1134 / 11786 = 0.09621....
5391 / 11786 = 0.45740....
2374 / 11786 = 0.20142....
2887 / 11786 = 0.24495....
Then for each division, round the result to 'two decimal points', so the first result:
0.09621.... becomes 0.10
because the 3rd point is 5 or above, it would remain at 0.09 if the 3rd point was below 5.
Once you have all of the results rounded, multiply each by 100 then add them up.
The ending result will always be 100.
Should warn you however that depending on how you use each individual percentage, when you round them, any result less that 0.05 will become 0%, unless you keep the value before you round it so you can declare it as a percentage less than 1.
I think you want to use ceil() or round() .
Since these are floating point numbers, there is room for error. Be careful how you round, and be sure that you don't independently calculate the last remaining percentages. Simply subtract the total of what you have from 1 or 100.
Make sure you dont calculate separate sides of the equation, sum one side, then subtract the other from 1 or 100 or however you are handling your percentages.
I run into this quite a bit and have a hack for it.
$percentages = array(
'1' => 87.5,
'2' => 12.5,
'3' => 0,
'4' => 0,
'5' => 0
);
If you round those percentages for output, you will end up with 88% and 13% (101%)
round($percentages['1']);
round($percentages['2']);
// 88
// 13
So here is the code I use to fix it.
$checkTotal = array_sum($percentages);
$max = max(array_keys($percentages));
if ($checkTotal > 100) {
$percentages[$max] = $percentages[$max] - 1;
}
if ($checkTotal < 100) {
$percentages[$max] = $percentages[$max] + 1;
}
If it is 100, do nothing.
If it is less than 100, add 1 to equal 100
If it is over 100, subtract 1 to equal 100
i am trying to make a very simple game where i have 7 positions which are all hidden and within those there a 3 winning positions. I can pick randomly 3 times. I need to display whether the pick is a winning or not after every pick and store the result in a base.
Currently my thought where to generate an array of winning numbers on the first pick and then pick random number and check if it is in the winning array.
But i have a feeling that there is much more efficient way to do so.
Would appreciate if you would use PHP for coding examples, but pseudo code will do as-well.
EDIT
i am looking for the way to solve this without populating array with winning positions. maybe there is a way to do this with weights or probability percents.
Something like on first pick i have 3/7*100 percent chance to win. save result to base.
on second pick i have either 3/6*100 or 2/6*100 percent chance to win based weather i won in previous pick which i get from base.
Revised answer: this example does not require you to store the complete state of the game in a variable; instead, you just need to store the try count and won count:
$won = 0;
for($try = 0; $try < 3; $try++) {
$slots = array_fill(0, 7 - $try, 0); // blank slots
$lucky = array_fill(0, 3 - $won, 1); // lucky slots
$pot = array_replace($slots, $lucky); // make some of the slots lucky
$win = $pot[array_rand($pot)]; // randomly pick a slot
$won += $win == 1; // update won count
echo sprintf("Try %d: win=%d, total wins=%d\n", $try + 1, $win, $won);
}
Original answer:
$pot = array( // pot is (an associative) array; 0 = blank, 1 = win
"pos_1" => 0,
"pos_2" => 0,
"pos_3" => 0,
"pos_4" => 0,
"pos_5" => 0,
"pos_6" => 0,
"pos_7" => 0
);
$win = array_rand($pot, 3); // picks three indexes from the pot randomly
foreach($win as $w) {
$pot[$w] = 1; // set winning indicator
}
print_r($pot);
Output: array containing state of the pots.
Array
(
[pos_1] => 0
[pos_2] => 1
[pos_3] => 0
[pos_4] => 1
[pos_5] => 1
[pos_6] => 0
[pos_7] => 0
)
You can just save the positions of the winning numbers. This way you can always check their values using the [] operator for arrays. After all, you just pick the positions and not the numbers.
Update:
This way you even don't need to hide numbers. It's quite possible to have some more abstract "winning things" - characters, words, structures. However, it is important that you do not alter your array of hidden "things" in any way or at least update the stored winning positions accordingly if they stay the same. If that's not the case you'd naturally need to update the saved winning positions.
<?php
$arr = array(true, true, false, false, false, false, false);
shuffle($arr);
function pick($arr, $index) {
return isset($arr[$index]) && $arr[$index] === true;
}
var_dump($arr);
var_dump(pick($arr, 3));
var_dump(pick($arr, 5));
var_dump(pick($arr, 1));
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.