Split total into 5 different numbers - php

Ok what i'm wanting to do is split a number ($row['count']) into 5, this is easy enough if you want equal numbers:
$sum = ($row['count'] / 5);
$fsum = floor($sum);
but I want each number to be different and still add up to total ie $row['count'] how can this be achieved?
Update:
If this helps its to be used to update 5 rows in a database:
$query = "SELECT * FROM foo";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
$sum = ($row['count'] / 5);
$fsum = floor($sum);
$id = $row['id'];
$update = "UPDATE foo SET foo1='$fsum', foo2='$fsum', foo3='$fsum', foo4='$fsum', foo5='$fsum' WHERE id='$id'";
mysql_query($update);
}// while
so ideally the $update query would be something like:
$update = "UPDATE foo SET foo1='$fsum1', foo2='$fsum2', foo3='$fsum3', foo4='$fsum4', foo5='$fsum5' WHERE id='$id'";

This is my take:
function randomize($sum, $parts) {
$part_no = count($parts);
$continnue_counter = 0;
while (count(array_unique($parts)) != $part_no) {
$changing = array_rand($parts, 2);
if (($parts[$changing[0]] - 1) == 0 || ($parts[$changing[1]] - 1) == 0) { // don't let them go under 1
++$continnue_counter;
// sometime one element get everything and others even out on 1
// just throw away everything you got so far and start over
if ($continnue_counter > 10) {
$parts = setup($sum, $part_no);
$continnue_counter = 0;
}
continue;
}
$continnue_counter = 0;
$signum = mt_rand(0, 100) % 2 ? 1 : -1;
$delta = $signum * mt_rand(1, min($parts[$changing[0]] - 1, $parts[$changing[1]] - 1)); // -1 to make sure they don't go under 0
$parts[$changing[0]] += $delta;
$parts[$changing[1]] -= $delta;
}
return $parts;
}
function setup($sum, $part_no) {
$parts = array_fill(0, $part_no, (int)($sum / $part_no));
// acount for the reminder of (int) cast
$reminder = $sum - array_sum($parts);
while ($reminder) {
$parts[array_rand($parts)] += 1;
--$reminder;
}
return $parts;
}
$part_no = 5;
$sum = 42;
$parts = randomize($sum, setup($sum, $part_no));
var_export($parts);
print array_sum($parts)
Update:
I've added a version that introduces a little more entropy.
Update2:
The more random one had a tendency to decrement everything to 1 except one part, added an explicit detection to deal with this. Still the algorithm behind it has unknown termination time.

Related

Get lowest price on sum of combinations in given array

This code is working fine when the array length is 8 or 10 only. When we are checking this same code for more than 10 array length.it get loading not showing the results.
How do reduce my code. If you have algorithm please share. Please help me.
This program working flow:
$allowed_per_room_accommodation =[2,3,6,5,3,5,2,5,4];
$allowed_per_room_price =[10,30,60,40,30,50,20,60,80];
$search_accommodation = 10;
i am get subsets = [5,5],[5,3,2],[6,4],[6,2,2],[5,2,3],[3,2,5]
Show lowest price room and then equal of 10 accommodation; output like as [5,3,2];
<?php
$dp=array(array());
$GLOBALS['final']=[];
$GLOBALS['room_key']=[];
function display($v,$room_key)
{
$GLOBALS['final'][] = $v;
$GLOBALS['room_key'][] = $room_key;
}
function printSubsetsRec($arr, $i, $sum, $p,$dp,$room_key='')
{
// If we reached end and sum is non-zero. We print
// p[] only if arr[0] is equal to sun OR dp[0][sum]
// is true.
if ($i == 0 && $sum != 0 && $dp[0][$sum]) {
array_push($p,$arr[$i]);
array_push($room_key,$i);
display($p,$room_key);
return $p;
}
// If $sum becomes 0
if ($i == 0 && $sum == 0) {
display($p,$room_key);
return $p;
}
// If given sum can be achieved after ignoring
// current element.
if (isset($dp[$i-1][$sum])) {
// Create a new vector to store path
// if(!is_array(#$b))
// $b = array();
$b = $p;
printSubsetsRec($arr, $i-1, $sum, $b,$dp,$room_key);
}
// If given $sum can be achieved after considering
// current element.
if ($sum >= $arr[$i] && isset($dp[$i-1][$sum-$arr[$i]]))
{
if(!is_array($p))
$p = array();
if(!is_array($room_key))
$room_key = array();
array_push($p,$arr[$i]);
array_push($room_key,$i);
printSubsetsRec($arr, $i-1, $sum-$arr[$i], $p,$dp,$room_key);
}
}
// Prints all subsets of arr[0..n-1] with sum 0.
function printAllSubsets($arr, $n, $sum,$get=[])
{
if ($n == 0 || $sum < 0)
return;
// Sum 0 can always be achieved with 0 elements
// $dp = new bool*[$n];
$dp = array();
for ($i=0; $i<$n; ++$i)
{
// $dp[$i][$sum + 1]=true;
$dp[$i][0] = true;
}
// Sum arr[0] can be achieved with single element
if ($arr[0] <= $sum)
$dp[0][$arr[0]] = true;
// Fill rest of the entries in dp[][]
for ($i = 1; $i < $n; ++$i) {
for ($j = 0; $j < $sum + 1; ++$j) {
// echo $i.'d'.$j.'.ds';
$dp[$i][$j] = ($arr[$i] <= $j) ? (isset($dp[$i-1][$j])?$dp[$i-1][$j]:false) | (isset($dp[$i-1][$j-$arr[$i]])?($dp[$i-1][$j-$arr[$i]]):false) : (isset($dp[$i - 1][$j])?($dp[$i - 1][$j]):false);
}
}
if (isset($dp[$n-1][$sum]) == false) {
return "There are no subsets with";
}
$p;
printSubsetsRec($arr, $n-1, $sum, $p='',$dp);
}
$blockSize = array('2','3','6','5','3','5','2','5','4');
$blockvalue = array('10','30','60','40','30','50','20','60','80');
$blockname = array("map","compass","water","sandwich","glucose","tin","banana","apple","cheese");
$processSize = 10;
$m = count($blockSize);
$n = count($processSize);
// sum of sets in array
printAllSubsets($blockSize, $m, $processSize);
$final_subset_room = '';
$final_set_room_keys = '';
$final_set_room =[];
if($GLOBALS['room_key']){
foreach ($GLOBALS['room_key'] as $set_rooms_key => $set_rooms) {
$tot = 0;
foreach ($set_rooms as $set_rooms) {
$tot += $blockvalue[$set_rooms];
}
$final_set_room[$set_rooms_key] = $tot;
}
asort($final_set_room);
$final_set_room_first_key = key($final_set_room);
$final_all_room['set_room_keys'] = $GLOBALS['room_key'][$final_set_room_first_key];
$final_all_room_price['set_room_price'] = $final_set_room[$final_set_room_first_key];
}
if(isset($final_all_room_price)){
asort($final_all_room_price);
$final_all_room_first_key = key($final_all_room_price);
foreach ($final_all_room['set_room_keys'] as $key_room) {
echo $blockname[$key_room].'---'. $blockvalue[$key_room];
echo '<br>';
}
}
else
echo 'No Results';
?>
I'm assuming your task is, given a list rooms, each with the amount of people it can accommodate and the price, to accommodate 10 people (or any other quantity).
This problem is similar to 0-1 knapsack problem which is solvable in polynomial time. In knapsack problem one aims to maximize the price, here we aim to minimize it. Another thing that is different from classic knapsack problem is that full room cost is charged even if the room is not completely occupied. It may reduce the effectiveness of the algorithm proposed at Wikipedia. Anyway, the implementation isn't going to be straightforward if you have never worked with dynamic programming before.
If you want to know more, CLRS book on algorithms discusses dynamic programming in Chapter 15, and knapsack problem in Chapter 16. In the latter chapter they also prove that 0-1 knapsack problem doesn't have trivial greedy solution.

How to convert word to number using my function?

I created this function to converting numbers to words. And how I can convert words to number using this my function:
Simple function code:
$array = array("1"=>"ЯК","2"=>"ДУ","3"=>"СЕ","4"=>"ЧОР","5"=>"ПАНҶ","6"=>"ШАШ","7"=>"ҲАФТ","8"=>"ХАШТ","9"=>"НӮҲ","0"=>"НОЛ","10"=>"ДАҲ","20"=>"БИСТ","30"=>"СИ","40"=>"ЧИЛ","50"=>"ПАНҶОҲ","60"=>"ШАСТ","70"=>"ҲАФТОД","80"=>"ХАШТОД","90"=>"НАВАД","100"=>"САД");
$n = "98"; // Input number to converting
if($n < 10 && $n > -1){
echo $array[$n];
}
if($n == 10 OR $n == 20 OR $n == 30 OR $n == 40 OR $n == 50 OR $n == 60 OR $n == 70 OR $n == 80 OR $n == 90 OR $n == 100){
echo $array[$n];
}
if(mb_strlen($n) == 2 && $n[1] != 0)
{
$d = $n[0]."0";
echo "$array[$d]У ".$array[$n[1]];
}
My function so far converts the number to one hundred. How can I now convert text to a number using the answer of my function?
So, as #WillParky93 assumed, your input has spaces between words.
<?php
mb_internal_encoding("UTF-8");//For testing purposes
$array = array("1"=>"ЯК","2"=>"ДУ","3"=>"СЕ","4"=>"ЧОР","5"=>"ПАНҶ","6"=>"ШАШ","7"=>"ҲАФТ","8"=>"ХАШТ","9"=>"НӮҲ","0"=>"НОЛ","10"=>"ДАҲ","20"=>"БИСТ","30"=>"СИ","40"=>"ЧИЛ","50"=>"ПАНҶОҲ","60"=>"ШАСТ","70"=>"ҲАФТОД","80"=>"ХАШТОД","90"=>"НАВАД","100"=>"САД");
$postfixes = array("3" => "ВУ");
$n = "98"; // Input number to converting
$res = "";
//I also optimized your conversion of numbers to words
if($n > 0 && ($n < 10 || $n%10 == 0))
{
$res = $array[$n];
}
if($n > 10 && $n < 100 && $n%10 != 0)
{
$d = intval(($n/10));
$sd = $n%10;
$ending = isset($postfixes[$d]) ? $postfixes[$d] : "У";
$res = ($array[$d * 10]).$ending." ".$array[$sd];
}
echo $res;
echo "\n<br/>";
$splitted = explode(" ", $res);
//According to your example, you use only numerals that less than 100
//So, to simplify your task(btw, according to Google, the language is tajik
//and I don't know the rules of building numerals in this language)
if(sizeof($splitted) == 1) {
echo array_search($splitted[0], $array);
}
else if(sizeof($splitted) == 2) {
$first = $splitted[0];
$first_length = mb_strlen($first);
if(mb_substr($first, $first_length - 2) == "ВУ")
{
$first = mb_substr($first, 0, $first_length - 2);
}
else
{
$first = mb_substr($splitted[0], 0, $first_length - 1);
}
$second = $splitted[1];
echo (array_search($first, $array) + array_search($second, $array));
}
You didn't specify the input specs but I took the assumption you want it with a space between the words.
//get our input=>"522"
$input = "ПАНҶ САД БИСТ ДУ";
//split it up
$split = explode(" ", $input);
//start out output
$c = 0;
//set history
$history = "";
//loop the words
foreach($split as &$s){
$res = search($s);
//If number is 9 or less, we are going to check if it's with a number
//bigger than or equal to 100, if it is. We multiply them together
//else, we just add them.
if((($res = search($s)) <=9) ){
//get the next number in the array
$next = next($split);
//if the number is >100. set $nextres
if( ($nextres = search($next)) >= 100){
//I.E. $c = 5 * 100 = 500
$c = $nextres * $res;
//set the history so we skip over it next run
$history = $next;
}else{
//Single digit on its own
$c += $res;
}
}elseif($s != $history){
$c += $res;
}
}
//output the result
echo $c;
function search($s){
global $array;
if(!$res = array_search($s, $array)){
//grab the string length
$max = strlen($s);
//remove one character at a time until we find a match
for($i=0;$i<$max; $i++ ){
if($res = array_search(mb_substr($s, 0, -$i),$array)){
//stop the loop
$i = $max;
}
}
}
return $res;
}
Output is 522.

leveling up PHP algorithm

I want some algorithm to make a PHP level up system I already made a system to level up people but what if people gained a huge amount of exp so they need to level up multiple times example of my code:
if(isset($_POST['id']) && isset($_POST['amount'])) {
$amount = $_POST['amount'];
$user = $_POST['id'];
$exp;$etnl;$newExp;$level;
`//select exp and etnl from DB
$result = DB::getInstance() ->query("SELECT exp,etnl,level FROM exp WHERE id = $user");
$result->setFetchMode(PDO::FETCH_ASSOC);
$row = $result->fetch();
$exp = $row['exp'];
$etnl = $row['etnl'];
$level = $row['level']
//c where to level up user or to add exp to his ep bar
if($amount + $exp > $etnl) {
$etnl = $etnl * 2;
$leftOvers = ($amount + $exp) - $etnl;
$newExp = $leftOvers;
$level = $level + 1;`
$result = DB::getInstance()->prepare("UPDATE exp SET exp=?,level=?,etnl=? WHERE id=?");
$result->execute(array($amount,$level,$etnl,$user));
} else {
$result = DB::getInstance()->prepare("UPDATE exp SET exp=? WHERE id=?");
$result->execute(array($amount,$user));
}
}`
`
but this system fails if the exp could level up a person 2 times any help or any new code? BTW etnl stands for exp till next level thank you
If you don't want to manually type out a level table like the other answer suggets you can also try doing your process with a recursive function:
$current_exp = 0;
$current_level = 1;
$to_next_level = 1000;
function did_level( $amount = 0 ) {
global $current_exp;
global $curent_level;
global $to_next_level;
$tmp_amount = $amount + $current_exp;
$tmp_amount_remainder = $tmp_amount - $to_next_level;
if ($tmp_amount >= $to_next_level) {
$current_level++;
$to_next_level *= 2;
$current_exp = $tmp_amount;
if ($tmp_amount_remainder > 0) {
// Level us up again.
did_level( $tmp_amount_remainder );
} else {
// Store the new level, to next level, and current exp
}
} else {
// Store the new exp amount
}
}
You have to get a table of levels which is basically a list of [exp, level] pairs. For example,
$level_list = [
['exp' => 0, 'level' => 1],
['exp' => 100, 'level' => 2],
['exp' => 1000, 'level' => 3],
// etc.
];
Then, you iterate the table starting from the end:
$i = count($level_list);
while($i > 0) {
$i--;
if ($user_exp >= $level_list[i]['exp']) {
// this is the level we need
break;
}
}
$target_level = $level_list[$i]['level'];
$target_exp = $level_list[$i]['exp'];

PHP MySQL read from a table to use in a graph

I have a graph that auto fills based on a few values. The values are the goal (total) the current (where it is at now) and in the total height of the graph. I am trying to pull the goal and current out of the database format it into money and it should fill the graph to the point it needs to be.
I had it working with a URL get which was ?current=584559096&goal=1000000000 and I just used $_GET. For easier maintenance for others around I want to pull it from a database which can be updated a post.
The code I have so far is:
<? php function formatMoney($number, $fractional=false) {
if ($fractional) {
$number = sprintf('%.0f', $number);
}
while (true) {
$replaced = preg_replace('/(-?\d+)(\d\d\d)/', '$1,$2', $number);
if ($replaced != $number) {
$number = $replaced;
} else {
break;
}
}
return $number;
}
$goal = $row['goal'];
$current = $row['current'];
$percent = round($current / $goal * 100);
$totalheight = 216;
$ourheight = $totalheight * $percent / 100;
$halfofarrowheight = 12;
$arrowheight = $totalheight-$ourheight-$halfofarrowheight;
if($arrowheight < 0)
$arrowheight = 0;
mysql_close($con);
?>
Here is my query
mysql_select_db("databasename", $con);
$result = mysql_query("select * from `dbname`.`tablename");
This is my error Warning: Division by zero in E:\inetpub\wwwroot\test.website.net\web\includes\globalfoot.php on line 36
Line 36 is the $percent = round($current / $goal * 100);
So, I been fooling with this thing for 4 days now trying to just get the numbers out of the goal column and current column format them to money and have the chart fill. For test sake lets say the goal is 1000000000000 and the current is 5000000000.
$row['goal'] = $goal;
$row['current'] = $current;
Please forgive me if I haven't understood your code, but shouldn't the above be:
$goal=$row['goal'];
$current=$row['current'];
Your code should look like this:
mysql_select_db("databasename", $con);
$result = mysql_query("select * from `dbname`.`tablename");
if (!$result || !($row = mysql_fetch_assoc($result))) {
echo "DB error: ".mysql_error();
exit;
}
$goal = $row['goal'];
$current = $row['current'];
// ...
Always, always double-check your variables when doing division.
$percent = ($goal > 0 || $goal < 0) ? round($current / $goal * 100) : 0;

IMEI validation function

Does anybody know a PHP function for IMEI validation?
Short solution
You can use this (witchcraft!) solution, and simply check the string length:
function is_luhn($n) {
$str = '';
foreach (str_split(strrev((string) $n)) as $i => $d) {
$str .= $i %2 !== 0 ? $d * 2 : $d;
}
return array_sum(str_split($str)) % 10 === 0;
}
function is_imei($n){
return is_luhn($n) && strlen($n) == 15;
}
Detailed solution
Here's my original function that explains each step:
function is_imei($imei){
// Should be 15 digits
if(strlen($imei) != 15 || !ctype_digit($imei))
return false;
// Get digits
$digits = str_split($imei);
// Remove last digit, and store it
$imei_last = array_pop($digits);
// Create log
$log = array();
// Loop through digits
foreach($digits as $key => $n){
// If key is odd, then count is even
if($key & 1){
// Get double digits
$double = str_split($n * 2);
// Sum double digits
$n = array_sum($double);
}
// Append log
$log[] = $n;
}
// Sum log & multiply by 9
$sum = array_sum($log) * 9;
// Compare the last digit with $imei_last
return substr($sum, -1) == $imei_last;
}
Maybe can help you :
This IMEI number is something like this: ABCDEF-GH-IJKLMNO-X (without “-” characters)
For example: 350077523237513
In our example ABCDEF-GH-IJKLMNO-X:
AB is Reporting Body Identifier such as 35 = “British Approvals Board of Telecommunications (BABT)”
ABCDEF is Type Approval Code
GH is Final Assembly Code
IJKLMNO is Serial Number
X is Check Digit
Also this can help you : http://en.wikipedia.org/wiki/IMEI#Check_digit_computation
If i don't misunderstood, IMEI numbers using Luhn algorithm . So you can google this :) Or you can search IMEI algorithm
Maybe your good with the imei validator in the comments here:
http://www.php.net/manual/en/function.ctype-digit.php#77718
But I haven't tested it
Check this solution
<?php
function validate_imei($imei)
{
if (!preg_match('/^[0-9]{15}$/', $imei)) return false;
$sum = 0;
for ($i = 0; $i < 14; $i++)
{
$num = $imei[$i];
if (($i % 2) != 0)
{
$num = $imei[$i] * 2;
if ($num > 9)
{
$num = (string) $num;
$num = $num[0] + $num[1];
}
}
$sum += $num;
}
if ((($sum + $imei[14]) % 10) != 0) return false;
return true;
}
$imei = '868932036356090';
var_dump(validate_imei($imei));
?>
IMEI validation uses Luhn check algorithm. I found a link to a page where you can validate your IMEI. Furthermore, at the bottom of this page is a piece of code written in JavaScript to show how to calculate the 15th digit of IMEI and to valid IMEI. I might give you some ideas. You can check it out here http://imei.sms.eu.sk/index.html
Here is a jQuery solution which may be of use: https://github.com/madeinstefano/imei-validator
good fun from kasperhartwich
function validateImei($imei, $use_checksum = true) {
if (is_string($imei)) {
if (ereg('^[0-9]{15}$', $imei)) {
if (!$use_checksum) return true;
for ($i = 0, $sum = 0; $i < 14; $i++) {
$tmp = $imei[$i] * (($i%2) + 1 );
$sum += ($tmp%10) + intval($tmp/10);
}
return (((10 - ($sum%10)) %10) == $imei[14]);
}
}
return false;
}

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