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!
///PROBLEM SOLVED. SEE MY ANSWER BELLOW.///
I have an array that randomly generates TRUE values throughout it. I
specified the number of TRUE values i want and it works like a charm.
I was toying around with it`s values and generating certain actions
based on wether or not the value is TRUE. Problem is i need to do this
again, this time using the TRUE values to form an array from which i
need a certain number of TRUE values.
I.E. I wanted 3 TRUE values out of 5. The code gave me the 3 TRUE
values on random iterations:
[1]=>1; [1]=>;
[2]=>1; [2]=>1;
[3]=>; [3]=>1;
[4]=>; [4]=>;
[5]=>1; [5]=>1;
This is all fine and dandy. Now i need TRUE values on 2 out of the 3
previous values. Consider i am taking 5 bites out of an apple. On 3
occasions i choke on it, from which in 2 cases i lose 2,3 teeth.
I want to print the outcome.
"Took a bite"
"Took a bite and choked"
"Took a bite"
"Took a bite , choked , lost 1 tooth and got 9 left" (i have a total of 10 teeth)
"Took a bite , choked , lost 2 teeth and got 7 left"
This is what i need to see printed after the 3/5 and 2/3 random
calculations occured. If i took a bite on the [2],[4] and [5]
iterations, i must lose tooth 2 times randomly on those exact
iterations.
Sorry for this example.
///PROBLEM SOLVED. SEE MY ANSWER BELLOW.///
2) Lastly, how can i store what happened after the code ran? Like "After you ate that apple, you choked 3 times lost 3 teeth and still got 7 left".
The way i did it does not work and returns "me loosing 1 tooth and having 9 left despite the fact that my calculated value is 7", not stacking the values of my ordeal.
I know these questions are silly, but i searched everywhere for information, read manuals and stuff and cannot put the pieces together...
The method used in the " random 3/5" case:
$spots = array();
while (count($spots) < number) {
$rand = rand(1,36);
if (!isset($spots[$rand])) {
$spots[$rand] = TRUE;
}
}
$spots = $spots + array_fill(1, number, FALSE);
ksort($spots); /// credits to Nick J.
Then did a foreach($spots as $k => $v)
Then did my code statements, starting with if ($v == 1), do code...
And gave me a random 3/5 list like the one i posted first.
Thank you in advance,
Vlad
Ok, i edited my answer as i have found the solution to problem number 1.
The catch was to scan the array of elements that are FALSE, unset them, work with that further on as it will randomly select elements from the TRUE elements of the original array.
Hope this comes in handy. I will post the code here:
$spots = array();
while (count($spots) < number) {
$rand = rand(1,constant);
if (!isset($spots[$rand])) {
$spots[$rand] = TRUE;
}
}
$spots = $spots + array_fill(1, constant, FALSE);
ksort($spots);
print_r($spots) ;
foreach($spots as $v)
{
if($v != 1)
{
unset($spots[array_search($v,$spots)]);
}
}
print_r($spots);
If example array:
$spots = array("1","2","3","4","5");
And i want 3 random numbers to be picked (TRUE = 1 , FALSE = nothing)
then the output is:
[1]=>1;
[2]=>1;
[3]=>;
[4]=>;
[5]=>1;
If i want 2 random numbers out of the 3 just got, then the final output is:
[1]=>1;
[2]=>1;
[5]=>1;
I have a range of whole numbers that might or might not have some numbers missing. Is it possible to find the smallest missing number without using a loop structure? If there are no missing numbers, the function should return the maximum value of the range plus one.
This is how I solved it using a for loop:
$range = [0,1,2,3,4,6,7];
// sort just in case the range is not in order
asort($range);
$range = array_values($range);
$first = true;
for ($x = 0; $x < count($range); $x++)
{
// don't check the first element
if ( ! $first )
{
if ( $range[$x - 1] + 1 !== $range[$x])
{
echo $range[$x - 1] + 1;
break;
}
}
// if we're on the last element, there are no missing numbers
if ($x + 1 === count($range))
{
echo $range[$x] + 1;
}
$first = false;
}
Ideally, I'd like to avoid looping completely, as the range can be massive. Any suggestions?
Algo solution
There is a way to check if there is a missing number using an algorithm. It's explained here. Basically if we need to add numbers from 1 to 100. We don't need to calculate by summing them we just need to do the following: (100 * (100 + 1)) / 2. So how is this going to solve our issue ?
We're going to get the first element of the array and the last one. We calculate the sum with this algo. We then use array_sum() to calculate the actual sum. If the results are the same, then there is no missing number. We could then "backtrack" the missing number by substracting the actual sum from the calculated one. This of course only works if there is only one number missing and will fail if there are several missing. So let's put this in code:
$range = range(0,7); // Creating an array
echo check($range) . "\r\n"; // check
unset($range[3]); // unset offset 3
echo check($range); // check
function check($array){
if($array[0] == 0){
unset($array[0]); // get ride of the zero
}
sort($array); // sorting
$first = reset($array); // get the first value
$last = end($array); // get the last value
$sum = ($last * ($first + $last)) / 2; // the algo
$actual_sum = array_sum($array); // the actual sum
if($sum == $actual_sum){
return $last + 1; // no missing number
}else{
return $sum - $actual_sum; // missing number
}
}
Output
8
3
Online demo
If there are several numbers missing, then just use array_map() or something similar to do an internal loop.
Regex solution
Let's take this to a new level and use regex ! I know it's nonsense, and it shouldn't be used in real world application. The goal is to show the true power of regex :)
So first let's make a string out of our range in the following format: I,II,III,IIII for range 1,3.
$range = range(0,7);
if($range[0] === 0){ // get ride of 0
unset($range[0]);
}
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
echo $str;
The output should be something like: I,II,III,IIII,IIIII,IIIIII,IIIIIII.
I've come up with the following regex: ^(?=(I+))(^\1|,\2I|\2I)+$. So what does this mean ?
^ # match begin of string
(?= # positive lookahead, we use this to not "eat" the match
(I+) # match I one or more times and put it in group 1
) # end of lookahead
( # start matching group 2
^\1 # match begin of string followed by what's matched in group 1
| # or
,\2I # match a comma, with what's matched in group 2 (recursive !) and an I
| # or
\2I # match what's matched in group 2 and an I
)+ # repeat one or more times
$ # match end of line
Let's see what's actually happening ....
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^
(I+) do not eat but match I and put it in group 1
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^
^\1 match what was matched in group 1, which means I gets matched
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^ ,\2I match what was matched in group 1 (one I in thise case) and add an I to it
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^^ \2I match what was matched previously in group 2 (,II in this case) and add an I to it
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^^^ \2I match what was matched previously in group 2 (,III in this case) and add an I to it
We're moving forward since there is a + sign which means match one or more times,
this is actually a recursive regex.
We put the $ to make sure it's the end of string
If the number of I's don't correspond, then the regex will fail.
See it working and failing. And Let's put it in PHP code:
$range = range(0,7);
if($range[0] === 0){
unset($range[0]);
}
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
if(preg_match('#^(?=(I*))(^\1|,\2I|\2I)+$#', $str)){
echo 'works !';
}else{
echo 'fails !';
}
Now let's take in account to return the number that's missing, we will remove the $ end character to make our regex not fail, and we use group 2 to return the missed number:
$range = range(0,7);
if($range[0] === 0){
unset($range[0]);
}
unset($range[2]); // remove 2
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
preg_match('#^(?=(I*))(^\1|,\2I|\2I)+#', $str, $m); // REGEEEEEX !!!
$n = strlen($m[2]); //get the length ie the number
$sum = array_sum($range); // array sum
if($n == $sum){
echo $n + 1; // no missing number
}else{
echo $n - 1; // missing number
}
Online demo
EDIT: NOTE
This question is about performance. Functions like array_diff and array_filter are not magically fast. They can add a huge time penalty. Replacing a loop in your code with a call to array_diff will not magically make things fast, and will probably make things slower. You need to understand how these functions work if you intend to use them to speed up your code.
This answer uses the assumption that no items are duplicated and no invalid elements exist to allow us to use the position of the element to infer its expected value.
This answer is theoretically the fastest possible solution if you start with a sorted list. The solution posted by Jack is theoretically the fastest if sorting is required.
In the series [0,1,2,3,4,...], the n'th element has the value n if no elements before it are missing. So we can spot-check at any point to see if our missing element is before or after the element in question.
So you start by cutting the list in half and checking to see if the item at position x = x
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
Yup, list[4] == 4. So move halfway from your current point the end of the list.
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
Uh-oh, list[6] == 7. So somewhere between our last checkpoint and the current one, one element was missing. Divide the difference in half and check that element:
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
In this case, list[5] == 5
So we're good there. So we take half the distance between our current check and the last one that was abnormal. And oh.. it looks like cell n+1 is one we already checked. We know that list[6]==7 and list[5]==5, so the element number 6 is the one that's missing.
Since each step divides the number of elements to consider in half, you know that your worst-case performance is going to check no more than log2 of the total list size. That is, this is an O(log(n)) solution.
If this whole arrangement looks familiar, It's because you learned it back in your second year of college in a Computer Science class. It's a minor variation on the binary search algorithm--one of the most widely used index schemes in the industry. Indeed this question appears to be a perfectly-contrived application for this searching technique.
You can of course repeat the operation to find additional missing elements, but since you've already tested the values at key elements in the list, you can avoid re-checking most of the list and go straight to the interesting ones left to test.
Also note that this solution assumes a sorted list. If the list isn't sorted then obviously you sort it first. Except, binary searching has some notable properties in common with quicksort. It's quite possible that you can combine the process of sorting with the process of finding the missing element and do both in a single operation, saving yourself some time.
Finally, to sum up the list, that's just a stupid math trick thrown in for good measure. The sum of a list of numbers from 1 to N is just N*(N+1)/2. And if you've already determined that any elements are missing, then obvously just subtract the missing ones.
Technically, you can't really do without the loop (unless you only want to know if there's a missing number). However, you can accomplish this without first sorting the array.
The following algorithm uses O(n) time with O(n) space:
$range = [0, 1, 2, 3, 4, 6, 7];
$N = count($range);
$temp = str_repeat('0', $N); // assume all values are out of place
foreach ($range as $value) {
if ($value < $N) {
$temp[$value] = 1; // value is in the right place
}
}
// count number of leading ones
echo strspn($temp, '1'), PHP_EOL;
It builds an ordered identity map of N entries, marking each value against its position as "1"; in the end all entries must be "1", and the first "0" entry is the smallest value that's missing.
Btw, I'm using a temporary string instead of an array to reduce physical memory requirements.
I honestly don't get why you wouldn't want to use a loop. There's nothing wrong with loops. They're fast, and you simply can't do without them. However, in your case, there is a way to avoid having to write your own loops, using PHP core functions. They do loop over the array, though, but you simply can't avoid that.
Anyway, I gather what you're after, can easily be written in 3 lines:
function highestPlus(array $in)
{
$compare = range(min($in), max($in));
$diff = array_diff($compare, $in);
return empty($diff) ? max($in) +1 : $diff[0];
}
Tested with:
echo highestPlus(range(0,11));//echoes 12
$arr = array(9,3,4,1,2,5);
echo highestPlus($arr);//echoes 6
And now, to shamelessly steal Pé de Leão's answer (but "augment" it to do exactly what you want):
function highestPlus(array $range)
{//an unreadable one-liner... horrid, so don't, but know that you can...
return min(array_diff(range(0, max($range)+1), $range)) ?: max($range) +1;
}
How it works:
$compare = range(min($in), max($in));//range(lowest value in array, highest value in array)
$diff = array_diff($compare, $in);//get all values present in $compare, that aren't in $in
return empty($diff) ? max($in) +1 : $diff[0];
//-------------------------------------------------
// read as:
if (empty($diff))
{//every number in min-max range was found in $in, return highest value +1
return max($in) + 1;
}
//there were numbers in min-max range, not present in $in, return first missing number:
return $diff[0];
That's it, really.
Of course, if the supplied array might contain null or falsy values, or even strings, and duplicate values, it might be useful to "clean" the input a bit:
function highestPlus(array $in)
{
$clean = array_filter(
$in,
'is_numeric'//or even is_int
);
$compare = range(min($clean), max($clean));
$diff = array_diff($compare, $clean);//duplicates aren't an issue here
return empty($diff) ? max($clean) + 1; $diff[0];
}
Useful links:
The array_diff man page
The max and min functions
Good Ol' range, of course...
The array_filter function
The array_map function might be worth a look
Just as array_sum might be
$range = array(0,1,2,3,4,6,7);
// sort just in case the range is not in order
asort($range);
$range = array_values($range);
$indexes = array_keys($range);
$diff = array_diff($indexes,$range);
echo $diff[0]; // >> will print: 5
// if $diff is an empty array - you can print
// the "maximum value of the range plus one": $range[count($range)-1]+1
echo min(array_diff(range(0, max($range)+1), $range));
Simple
$array1 = array(0,1,2,3,4,5,6,7);// array with actual number series
$array2 = array(0,1,2,4,6,7); // array with your custom number series
$missing = array_diff($array1,$array2);
sort($missing);
echo $missing[0];
$range = array(0,1,2,3,4,6,7);
$max=max($range);
$expected_total=($max*($max+1))/2; // sum if no number was missing.
$actual_total=array_sum($range); // sum of the input array.
if($expected_total==$actual_total){
echo $max+1; // no difference so no missing number, then echo 1+ missing number.
}else{
echo $expected_total-$actual_total; // the difference will be the missing number.
}
you can use array_diff() like this
<?php
$range = array("0","1","2","3","4","6","7","9");
asort($range);
$len=count($range);
if($range[$len-1]==$len-1){
$r=$range[$len-1];
}
else{
$ref= range(0,$len-1);
$result = array_diff($ref,$range);
$r=implode($result);
}
echo $r;
?>
function missing( $v ) {
static $p = -1;
$d = $v - $p - 1;
$p = $v;
return $d?1:0;
}
$result = array_search( 1, array_map( "missing", $ARRAY_TO_TEST ) );
Background;
to create a dropdown menu for a fun gambling game (Students can 'bet' how much that they are right) within a form.
Variables;
$balance
Students begin with £3 and play on the £10 table
$table(there is a;
£10 table, with a range of 1,2,3 etc to 10.
£100 table with a range of 10,20,30 etc to 100.
£1,000 table with a range of 100, 200, 300, 400 etc to 1000.)
I have assigned $table to equal number of zeros on max value,
eg $table = 2; for the £100 table
Limitations;
I only want the drop down menu to offer the highest 12 possible values (this could include the table below -IMP!).
Students are NOT automatically allowed to play on the 'next' table.
resources;
an array of possible values;
$a = array(1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,10,20,30,40,50,60,70,80,90,100,200,300,400,500,600,700,800,900,1000);
I can write a way to restrict the array by table;
(the maximum key for any table is (9*$table) )//hence why i use the zeroes above (the real game goes to $1 billion!)
$arrayMaxPos = (9*$table);
$maxbyTable = array_slice($a, 0, $arrayMaxPos);
Now I need a way to make sure no VALUE in the $maxbyTable is greater than $balance.
to create a $maxBet array of all allowed bets.
THIS IS WHERE I'M STUCK!
(I would then perform "array_slice($maxBet, -12);" to present only the highest 12 in the dropdown)
EDIT - I'd prefer to NOT have to use array walk because it seems unnecessary when I know where i want the array to end.
SECOND EDIT Apologies I realised that there is a way to mathematically ascertain which KEY maps to the highest possible bid.
It would be as follows
$integerLength = strlen($balance);//number of digits in $balance
$firstDigit = substr($balance, 0, 1);
then with some trickery because of this particular pattern
$maxKeyValue = (($integerlength*9) - 10 + $firstDigit);
So for example;
$balance = 792;
$maxKeyValue = ((3*9) - 10 + 7);// (key[24] = 700)
This though works on this problem and does not solve my programming problem.
Optional!
First of all, assuming the same rule applies, you don't need the $a array to know what prices are allowed on table $n
$table = $n; //$n being an integer
for ($i = 1; $i <= 10; $i++) {
$a[] = $i * pow(10, $n);
}
Will generate a perfectly valid array (where table #1 is 1-10, table #2 is 10-100 etc).
As for slicing it according to value, use a foreach loop and generate a new array, then stop when you hit the limit.
foreach ($a as $value) {
if ($value > $balance) { break; }
$allowedByTable[] = $value;
}
This will leave you with an array $allowedByTable that only has the possible bets which are lower then the user's current balance.
Important note
Even though you set what you think is right as options, never trust the user input and always validate the input on the server side. It's fairly trivial for someone to change the value in the combobox using DOM manipulation and bet on sums he's not supposed to have. Always check that the input you're getting is what you expect it to be!