I got the answer fine, but when I run the following code,
$total = 0;
$x = 0;
for ($i = 1;; $i++)
{
$x = fib($i);
if ($x >= 4000000)
break;
else if ($x % 2 == 0)
$total += $x;
print("fib($i) = ");
print($x);
print(", total = $total");
}
function fib($n)
{
if ($n == 0)
return 0;
else if ($n == 1)
return 1;
else
return fib($n-1) + fib($n-2);
}
I get the warning that I have exceeded the maximum execution time of 30 seconds. Could you give me some pointers on how to improve this algorithm, or pointers on the code itself? The problem is presented here, by the way.
Let's say $i equal to 13. Then $x = fib(13)
Now in the next iteration, $i is equal to 14, and $x = fib(14)
Now, in the next iteration, $i = 15, so we must calculate $x. And $x must be equal to fib(15). Now, wat would be the cheapest way to calculate $x?
(I'm trying not to give the answer away, since that would ruin the puzzle)
Try this, add caching in fib
<?
$total = 0;
$x = 0;
for ($i = 1;; $i++) {
$x = fib($i);
if ($x >= 4000000) break;
else if ($x % 2 == 0) $total += $x;
print("fib($i) = ");
print($x);
print(", total = $total\n");
}
function fib($n) {
static $cache = array();
if (isset($cache[$n])) return $cache[$n];
if ($n == 0) return 0;
else if ($n == 1) return 1;
else {
$ret = fib($n-1) + fib($n-2);
$cache[$n] = $ret;
return $ret;
}
}
Time:
real 0m0.049s
user 0m0.027s
sys 0m0.013s
You'd be better served storing the running total and printing it at the end of your algorithm.
You could also streamline your fib($n) function like this:
function fib($n)
{
if($n>1)
return fib($n-1) + fib($n-2);
else
return 0;
}
That would reduce the number of conditions you'd need to go through considerably.
** Edited now that I re-read the question **
If you really want to print as you go, use the output buffer. at the start use:
ob_start();
and after all execution, use
ob_flush();
flush();
also you can increase your timeout with
set_time_limit(300); //the value is seconds... so this is 5 minutes.
Related
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.
I tried to find For a given positive integer Z, check if Z can be written as PQ, where P and Q are positive integers greater than 1. If Z can be written as PQ, return 1, else return 0
I tried lots with online solution,
Check if one integer is an integer power of another
Finding if a number is a power of 2
but it's not what i need , any hint or any tips?
Here's the naive method - try every combination:
function check($z) {
for($p = 2; $p < sqrt($z); $p++) {
if($z % $p > 0) {
continue;
}
$q = $p;
for($i = 1; $q < $z; $i++) {
$q *= $p;
}
if($q == $z) {
//print "$z = $p^$i";
return 1;
}
}
return 0;
}
Similarly, using php's built in log function. But it may not be as accurate (if there are rounding errors, false positives may occur).
function check($z) {
for($p = 2; $p < sqrt($z); $p++) {
$q = log($z,$p);
if($q == round($q)) {
return 1;
}
}
return 0;
}
I'm a beginner learning PHP. I have tried to make a loop that has a different behaviour for both even and odd numbers. I've been playing around with it for a while, yet I still can't get it to work. Has anyone got a solution?
$count = 0;
$mod = $count % 2;
while ($count < 10)
{
if ($mod == 0) {
echo "even, ";
} else {
echo "odd, ";
}
$count++;
}
A silly mistake, mod inside while() loop.
$count = 0;
while ($count < 10) {
$mod = $count % 2; //Here
if ($mod == 0) {
echo "even, ";
} else {
echo "odd, ";
}
$count++;
}
$count = 0;
$mod = $count %2;
Is were your problem is.
You have to use the modulus (%) operator inside the for loop. Also, there is no need to store the value from the use of the modulus operator at all, it can be compared directly inside the for-loop.
for ($count = 0; $count < 10, $count++) {
if ($count % 2 == 0) {
echo "even, ";
} else {
echo "odd, ";
}
}
You can also switch the while to a for like this.
Welcome to PHP.
Edit #1:
As you are getting a new value of $count every execution of the for-loop the old value if $count % 2 will be incorrect. It has to recalculate for every $count. First it checks if 0 is divisible by 2, then onto 1 and so forth. For every value of $count you have to check the divisibility.
In most programming languages you aren't computing a variable onto another, instead you are taking the value of the variable. Like $a = $b + $c; in that case, if you change the value of $b or $c it does not automatically update $a. Instead you have to call $a = $b + $c again. It is the same with % operator.
$count = 0;
while ($count < 10) {
$mod = $count % 2;
if ($mod == 0) {
echo "even, ";
} else {
echo "odd, ";
}
$count++;
}
use for loop instead of while loop
for($count=0;$count<10;$count++)
{
if(($count % 2) == 0)
echo "even,";
else
echo "odd,";
}
I'm trying to program my own Sine function implementation for fun but I keep getting :
Fatal error: Maximum execution time of 30 seconds exceeded
I have a small HTML form where you can enter the "x" value of Sin(x) your looking for and the number of "iterations" you want to calculate (precision of your value), the rest is PhP.
The maths are based of the "Series definition" of Sine on Wikipedia :
--> http://en.wikipedia.org/wiki/Sine#Series_definition
Here's my code :
<?php
function factorial($int) {
if($int<2)return 1;
for($f=2;$int-1>1;$f*=$int--);
return $f;
};
if(isset($_POST["x"]) && isset($_POST["iterations"])) {
$x = $_POST["x"];
$iterations = $_POST["iterations"];
}
else {
$error = "You forgot to enter the 'x' or the number of iterations you want.";
global $error;
}
if(isset($x) && is_numeric($x) && isset($iterations) && is_numeric($iterations)) {
$x = floatval($x);
$iterations = floatval($iterations);
for($i = 0; $i <= ($iterations-1); $i++) {
if($i%2 == 0) {
$operator = 1;
global $operator;
}
else {
$operator = -1;
global $operator;
}
}
for($k = 1; $k <= (($iterations-(1/2))*2); $k+2) {
$k = $k;
global $k;
}
function sinus($x, $iterations) {
if($x == 0 OR ($x%180) == 0) {
return 0;
}
else {
while($iterations != 0) {
$result = $result+(((pow($x, $k))/(factorial($k)))*$operator);
$iterations = $iterations-1;
return $result;
}
}
}
$result = sinus($x, $iterations);
global $result;
}
else if(!isset($x) OR !isset($iterations)) {
$error = "You forgot to enter the 'x' or the number of iterations you want.";
global $error;
}
else if(isset($x) && !is_numeric($x)&& isset($iterations) && is_numeric($iterations)) {
$error = "Not a valid number.";
global $error;
}
?>
My mistake probably comes from an infinite loop at this line :
$result = $result+(((pow($x, $k))/(factorial($k)))*$operator);
but I don't know how to solve the problem.
What I'm tring to do at this line is to calculate :
((pow($x, $k)) / (factorial($k)) + (((pow($x, $k))/(factorial($k)) * ($operator)
iterating :
+ (((pow($x, $k))/(factorial($k)) * $operator)
an "$iterations" amount of times with "$i"'s and "$k"'s values changing accordingly.
I'm really stuck here ! A bit of help would be needed. Thank you in advance !
Btw : The factorial function is not mine. I found it in a PhP.net comment and apparently it's the optimal factorial function.
Why are you computing the 'operator' and power 'k' out side the sinus function.
sin expansion looks like = x - x^2/2! + x^3/3! ....
something like this.
Also remember iteration is integer so apply intval on it and not floatval.
Also study in net how to use global. Anyway you do not need global because your 'operator' and power 'k' computation will be within sinus function.
Best of luck.
That factorial function is hardly optimal—for speed, though it is not bad. At least it does not recurse. It is simple and correct though. The major aspect of the timeout is that you are calling it a lot. One technique for improving its performance is to remember, in a local array, the values for factorial previously computed. Or just compute them all once.
There are many bits of your code which could endure improvement:
This statement:
while($iterations != 0)
What if $iterations is entered as 0.1? Or negative. That would cause an infinite loop. You can make the program more resistant to bad input with
while ($iterations > 0)
The formula for computing a sine uses the odd numbers: 1, 3, 5, 7; not every integer
There are easier ways to compute the alternating sign.
Excess complication of arithmetic expressions.
return $result is within the loop, terminating it early.
Here is a tested, working program which has adjustments for all these issues:
<?php
// precompute the factorial values
global $factorials;
$factorials = array();
foreach (range (0, 170) as $j)
if ($j < 2)
$factorials [$j] = 1;
else $factorials [$j] = $factorials [$j-1] * $j;
function sinus($x, $iterations)
{
global $factorials;
$sign = 1;
for ($j = 1, $result = 0; $j < $iterations * 2; $j += 2)
{
$result += pow($x, $j) / $factorials[$j] * $sign;
$sign = - $sign;
}
return $result;
}
// test program to prove functionality
$pi = 3.14159265358979323846264338327950288419716939937510582097494459230781640628620;
$x_vals = array (0, $pi/4, $pi/2, $pi, $pi * 3/2, 2 * $pi);
foreach ($x_vals as $x)
{
$y = sinus ($x, 20);
echo "sinus($x) = $y\n";
}
?>
Output:
sinus(0) = 0
sinus(0.78539816339745) = 0.70710678118655
sinus(1.5707963267949) = 1
sinus(3.1415926535898) = 3.4586691443274E-16
sinus(4.7123889803847) = -1
sinus(6.2831853071796) = 8.9457384260403E-15
By the way, this executes very quickly: 32 milliseconds for this output.
I need to find the greatest prime factor of a large number: up to 12 places (xxx,xxx,xxx,xxx). I have solved the problem, and the code works for small numbers (up to 6 places); however, the code won't run fast enough to not trigger a timeout on my server for something in the 100 billions.
I found a solution, thanks to all.
Code:
<?php
set_time_limit(300);
function is_prime($number) {
$sqrtn = intval(sqrt($number));
//won't work for 0-2
for($i=3; $i<=$sqrtn; $i+=2) {
if($number%$i == 0) {
return false;
}
}
return true;
}
$initial = 600851475143;
$prime_factors = array();
for($i=3; $i<=9999; $i++) {
$remainder = fmod($initial, $i);
if($remainder == 0) {
if(is_prime($i)) {
$prime_factors[] = $i;
}
}
}
//print_r($prime_factors);
echo "\n\n";
echo "<b>Answer: </b>". max($prime_factors);
?>
The test number in this case is 600851475143.
Your code will not find any prime factors larger than sqrt(n). To correct that, you have to test the quotient $number / $i also, for each factor (not only prime factors) found.
Your is_factor function
function is_factor($number, $factor) {
$half = $number/2;
for($y=1; $y<=$half; $y++) {
if(fmod($number, $factor) == 0) {
return true;
}
}
}
doesn't make sense. What's $y and the loop for? If $factor is not a divisor of $number, that will perform $number/2 utterly pointless divisions. With that fixed, reordering the tests in is_prime_factor will give a good speedup because the costly primality test needs only be performed for the few divisors of $number.
Here is a really simple and fast solution.
LPF(n)
{
for (i = 2; i <= sqrt(n); i++)
{
while (n > i && n % i == 0) n /= i;
}
return n;
}