count 0 digits places in factorial result without memory exception - php

I want to find 0 digits after decimal results of factorial number without using as a string and in bulit functions .
For e.g.
Number Factoial result count of 0 occured after nums
1! = 1 => 0
5! = 120 => 1
10! = 3628800 => 2
15! = 1307674368000 => 3
As per last column i want to find out the total count of "0" at last number.Without having memory exception in php.
Thanks in advance

I provide you a function using substr. You can try this
function countLastZero($num){
$count = 0;
while(strlen($num) > 0){
$las = substr($num, -1);
if($las == 0){
$count++;
$num = substr($num, 0, strlen($num) - 1);
}else{
break;
}
}
return $count;
}
echo countLastZero(1307674368000); //output 3

Related

Setting probability of getting a random number 1 or 2

Is there a way to set a probability control on getting a random value either 1 or 2? Suppose if I want more than 70% of the time it should be 1 and only 30% should be the chance of getting 2 in rand(1,2). Is it possible?
Generate a number between 1 and 10, then if the number is greater than 7, return 2, else return 1
x = rand(1, 10);
if(x > 7){
return 2;
}
else {
return 1;
}
You will have 70% and 30% of chance to get the number you want
Why don't you generate a uniformly distributed random number and then filter the number with an if?
int x,r;
r=rand() % 100; // r in the range 0 to 99
if(r<=70){x=1;}
else{x=2;}
This should do the trick:
function yourrand($prob) {
$r = rand();
if ($r < $prob) { return 1; }
else { return 2; }
}

PHP - Get length of digits in a number

I would like to ask how I can get the length of digits in an Integer. For example:
$num = 245354;
$numlength = mb_strlen($num);
$numlength should be 6 in this example. Somehow I can't manage it to work?
Thanks
EDIT: The example code above --^ and its respective method mb_strlen(); works just fine.
Maybe:
$num = 245354;
$numlength = strlen((string)$num);
Accepted answer won't work with the big numbers. The better way to calculate the length of any number is to invoke floor(log10($num) + 1) with a check for 0.
$num = 12357;
echo $num !== 0 ? floor(log10($num) + 1) : 1; // prints 5
It has multiple advantages. It's faster, you don't do the casting of types, it works on big numbers, it works with different number systems like bin, hex, oct.
The equation does the logarithm with base 10 then makes the floor of it and adds 1.
This solution can work independently on the base, so if you want to calculate the length of binary or hex just change the base of the logarithm.
Working fiddle
The accepted solution presents a problem when evaluating negative numbers.
It works with a positive number:
$num = 245354;
$numlength = strlen((string)$num);
// Result: 6
But with a negative number, the (-) is added to the count:
$num = -245354;
$numlength = strlen((string)$num);
// Result: 7
Quick workaround:
$num = -245354;
$numlength = strlen((string)abs($num));
// Result: 6
More elegant way :)
ceil(log10($num));
You could also use some basic math!
$digits = (int)(log($num,10)+1)
<?php
$num = 123;
$num2 = 1234;
$num3 = 12345;
function digits($num){
return (int) (log($num, 10) + 1);
}
echo "\n $num: " . digits($num); // 123: 3
echo "\n $num2:" . digits($num2); // 1234: 4
echo "\n $num3:" . digits($num3); // 12345: 5
echo "\n";
Another way to find out the length of a number in digits would be to divide the integer part of the number to 10 until it becomes 0.
Example:
2021/10 = 202.1
202/10 = 20.2
20/10 = 2
2/10 = 0.2
Code:
function numberGetLength($number) {
$count = 0;
while (intval($number) > 0) {
$number = intval($number) / 10;
$count += 1;
}
return $count
}
Just using some version of (int)(log($num,10)+1) fails for 10, 100, 1000, etc. It counts the number 10 as 1 digit, 100 as two digits, etc. It also fails with 0 or any negative number.
If you must use math (and the number is non-negative), use:
$numlength = (int)(log($num+1, 10)+1);
Or for a math solution that counts the digits in positive OR negative numbers:
$numlength = ($num>=0) ? (int)(log($num+1, 10)+1) : (int)(log(1-$num, 10)+1);
But the strlen solution is just about as fast in PHP.
In PHP types are loosely set and guessed, if you want to see something as a string if it is an integer, float, and (i have not tried this) bool then #Gorjunav is the most correct answer.
Reset the variable as a string
$stringNum = (string) $num;
Then you can go anything string related you want with it! And vice-versa for changing a string to an int
$number = (int) $stringNum;
and so on...
count only integer value
`<?php
$n1 =12345;
$n2 =123454.55;
$n3 =12345564.557;
echo "The Number you Type: ".$n1."<br>";
$count = 0;
while ($n1 != 0)
{
$n1 = $n1 / 10;
$n1 = intval($n1);
++$count;
}
echo "The Digit in a Number: ".$count;
}
?>`
echo strlen((string) abs($num)); // using **abs** it'll work with negative integers as well
Tested in PHP 4.4.9 - 8.0.0
$array = array(-1, 0, -0, 1, 4, 9, 10, -10, 20, -20, 100, -100);
foreach( $array as $key => $num ){
echo $key."\t{$num}\t=>\t".($num !== 0 ? floor(log10(abs($num)) + 1) : 1)."\n";
}
/* Output:
0 -1 => 1
1 0 => 1
2 0 => 1
3 1 => 1
4 4 => 1
5 9 => 1
6 10 => 2
7 -10 => 2
8 20 => 2
9 -20 => 2
10 100 => 3
11 -100 => 3
*/
The following function work for either integers or floats (works with PHP7+):
function digitsCount($number): int
{
$number = abs($number);
$numberParts = explode(".", $number);
return
strlen($numberParts[0]) +
(strlen($numberParts[1] ?? 0));
}

Finding out combinations of x amount of integers to sum a given number

I am trying to figure out how I can loop out possible combinations of a x amount of integers to sum a specifik number.
Let's say, I have number 7 and I need to figure out how I can sum that number with integers in pairs 3.
1+2+4 = 7
3+3+1 = 7
5+1+1 = 7
2+2+3 = 7
Repeated combinations of numbers doesn't interest me, e.g.:
1+2+4 = 7
2+4+1 = 7
4+2+1 = 7
Anyone got any ideas of how I should proceed to reach this result?
Thanks.
Here is the solution for your problem.
function printPartitions($target, $max, $s){
if($target === 0 )
echo $s;
else
{
if($max > 1)
{
printPartitions($target, $max-1, $s);
}
if($max <= $target)
{
printPartitions($target-$max, $max, $max . " " . $s);
}
}
}
printPartitions(5, 5, "<br/>");
You have to specify the $target Value, $max value.
e.g.
printPartitions(7, 7, "<br/>");
It will give you output like:
1 1 1 1 1 1 1
1 1 1 1 1 2
1 1 1 2 2
1 2 2 2
1 1 1 1 3
1 1 2 3
2 2 3
1 3 3
1 1 1 4
1 2 4
3 4
1 1 5
2 5
1 6
7
I've got a solution to my problem. I feel I should defientely share it here, if anyone would ever need it. My solutions is based on this post: https://stackoverflow.com/a/19067884/3293843
<?php
function sampling($chars, $size, $combinations = array()) {
# if it's the first iteration, the first set
# of combinations is the same as the set of characters
if (empty($combinations)) {
$combinations = $chars;
}
# we're done if we're at size 1
if ($size == 1) {
return $combinations;
}
# initialise array to put new values in
$new_combinations = array();
# loop through existing combinations and character set to create strings
foreach ($combinations as $combination) {
foreach ($chars as $char) {
$new_combinations[] = $combination .'#'. $char;
}
}
# call same function again for the next iteration
return sampling($chars, $size - 1, $new_combinations);
}
// example
$chars = array('1', '2', '3','4');
$target = 7;
$maxLengthOfIntegers = 3;
$output = sampling($chars, $maxLengthOfIntegers);
$repeatedEntries = array();
//presenting the output
foreach($output as $out){
$explodeOut = explode('#',$out);
sort($explodeOut);
if(array_sum($explodeOut) == $target){
$sortedPattern = implode('',$explodeOut);
if(!in_array($sortedPattern,$repeatedEntries)){
echo $sortedPattern.'<br/>';
$repeatedEntries[] = $sortedPattern;
}
}
}
?>
Thank you for your time and efforts.
Regards,
Jacob
you can try this algorithm
$ans = array();
for($i=1;$i<=5;$i++)
{
$i1 = 7-$i;
$i2 = intval($i1 - $i);
$value = $i."+".$i1."+".$i2;
$ans = array_unshift($ans,$value);
}
print_r($ans);
hope this helps.. PLease let me know

Generate a combination of numbers

I have a PHP page with two variables: $nbRank and $nbNumeric. Depending of these two variables, I want to generate an array containing all combinations existing. For example:
If $nbRank = 3 and $nbNumeric = 2 I would have:
0 0 0
0 0 1
0 0 2
0 1 0
0 1 1
0 1 2
0 2 0
0 2 1
0 2 2
1 0 0
1 0 1
1 0 2
1 1 0
1 1 1
1 1 2
1 2 0
1 2 1
1 2 2
2 0 0
2 0 1
2 0 2
2 1 0
2 1 1
2 1 2
2 2 0
2 2 1
2 2 2
So, I create different loop and formulas to get the final result, but it doesn't works. This is what I did :
$result = array();
$nbIntRank = 0;
$nbIntNumeric = 0;
$nbRank = array();
$nbNumeric = array();
$nb_rangs = 3;
$nb_chiffres = 2;
for ($i = 1; $i <= $nb_rangs; $i++){
$nbRank[$i] = 0;
}
$nbIntRank = count($nbRank);
for ($i = 0; $i <= $nb_chiffres; $i++){
$nbNumeric[$i] = $i;
}
$nbIntNumeric = count($nbNumeric);
$algo = ($nb_rangs * ($nb_chiffres + 1)) * ($nb_rangs * ($nb_chiffres + 1));
$nbLine = $algo / ($nb_rangs);
$occ = 0;
for ($i = 0; $i < $nbLine; $i++){
foreach ($nbRank as $nbrItem => $nbrValue){
$result[$i][] = $nbrValue;
$occ++;
}
}
echo '#############<br />';
echo '### DATAS ###<br />';
echo '#############<br /><br />';
echo '- Nb Elements : '.$algo.'<br />';
echo '- Nb Lines : '.$nbLine.'<br />';
echo '- Nb Valuable Occurency : '.$occ.'<br />';
echo '<br /><hr /><br />';
echo '##############<br />';
echo '### PARSER ###<br />';
echo '##############<br /><br />';
echo '<pre>';
var_dump($result);
echo '</pre>';
I managed to create my final array with empty values (81 values, in 27 lines of 3 elements) but it only contains 0.
You indicated you'll be fine with pseudo-code.. Sorry I cannot offer specific correction for your php code [if these answers appear - they might be more educating], but I'd chose a recursive solution for this problem.
In each level of the recursion, try all possibilities, and call the same function to find all combinations of one smaller size.
Pseudo-Code:
findCombinations(range,size,sol,resultList):
if (size ==0): #base clause
resultList.append(copy(sol)) #making a copy of sol and appending it as a solution
return
for each i in (0,range):
sol.append(i)
findCombinations(range,size-1,sol,resultList) #recursive invokation, with smaller size
sol.deleteLast() #clean up environment before next calls
Invoke with findCombinations(3,3,[],resultList) where [] is just empty list, and resultList will hold the list of combination when the algorithm is done. this invokation will get all combinations of size 3 with elemenets 0,1,2.
Complexity note:
The number of possibilities is growing exponentially [O(rangesize)], so if you try to invoke it with 20,20 for instance - it might take some [very long] time, for any solution.
$nbRank = 3;
$nbNumeric = 2;
foreach (range(0, base_convert(str_pad('', $nbRank, $nbNumeric), $nbNumeric+1, 10)) as $i) {
echo str_pad(base_convert($i, 10, $nbNumeric+1), 3, 0, STR_PAD_LEFT) . PHP_EOL;
}
Simple idea: What you want is every number from 0 to X with base $nbNumeric, thus we just convert the maximum number to base 10, iterate over it with the common 10-based operators, and convert it back to base $nbNumeric again.
Probably more readable, but in fact exactly the same
$nbRank = 3;
$nbNumeric = 2;
// Top is "base_convert(222, 3, 10);" and therefore the upper limit
$top = base_convert(str_pad('', $nbRank, $nbNumeric), $nbNumeric+1, 10);
for ($i = 0; $i <= $top; $i++) {
echo str_pad(base_convert($i, 10, $nbNumeric+1), 3, 0, STR_PAD_LEFT) . PHP_EOL;
}
Here's a recursive solution:
$nbRank = 3;
$nbNumeric = 2;
function getCombinations ($length, $min, $max, $aStartingCombinations)
{
if ($length == 1)
{
return range ($min, $max);
}
$final = array ();
foreach (getCombinations ($length - 1, $min, $max, $aStartingCombinations) as $combination)
{
for ($i = $min; $i <= $max; $i++)
{
$final [] = $combination . $i;
}
}
return $final;
}
print_r (getCombinations ($nbRank, 0, $nbNumeric, array ()));

PHP Random numbers

I would like to draw a random number from the interval 1,49 but I would like to add a number as an exception ( let's say 44 ) , I cannot use round(rand(1,49)) .So I decided to make an array of 49 numbers ( 1-49) , unset[$aray[44]] and apply array_rand
Now I want to draw a number from the interval [$left,49] , how can I do that using the same array that I used before ?The array now misses value 44.
The function pick takes an array as an argument with all the numbers you have already picked. It will then pick a number between the start and the end that IS NOT in that array. It will add this number into that array and return the number.
function pick(&$picked, $start, $end) {
sort($picked);
if($start > $end - count($picked)) {
return false;
}
$pick = rand($start, $end - count($picked));
foreach($picked as $p) {
if($pick >= $p) {
$pick += 1;
} else {
break;
}
}
$picked[] = $pick;
return $pick;
}
This function will efficiently get a random number that is not in the array AND WILL NEVER INFINITELY RECURSE!
To use this like you want:
$array = array(44); // you have picked 44 for example
$num = pick($array, 1, 49); // get a random number between 1 and 49 that is not in $array
// $num will be a number between 1 and 49 that is not in $arrays
How the function works
Say you are getting a number between 1 and 10. And you have picked two numbers (e.g. 2 and 6). This will pick a number between 1 and (10 minus 2) using rand: rand(1, 8).
It will then go through each number that has been picked and check if the number is bigger.
For Example:
If rand(1, 8) returns 2.
It looks at 2 (it is >= then 2 so it increments and becomes 3)
It looks at 6 (it is not >= then 6 so it exits the loop)
The result is: 3
If rand(1, 8) returns 3
It looks at 2 (it is >= then 2 so it increments and becomes 4)
It looks at 6 (it is not >= then 6 so it exits the loop)
The result is 4
If rand(1, 8) returns 6
It looks at 2 (it is >= then 2 so it increments and becomes 7)
It looks at 6 (it is >= then 6 so it increments and becomes 8)
The result is: 8
If rand(1, 8) returns 8
It looks at 2 (it is >= then 2 so it increments and becomes 9)
It looks at 6 (it is >= then 6 so it increments and becomes 10)
The result is: 10
Therefore a random number between 1 and 10 is returned and it will not be 2 or 6.
I implemented this a long time ago to randomly place mines in a 2-dimensional array (because I wanted random mines, but I wanted to guarantee the number of mines on the field to be a certain number)
Why not just check your exceptions:
function getRand($min, $max) {
$exceptions = array(44, 23);
do {
$rand = mt_rand($min, $max);
} while (in_array($rand, $exceptions));
return $rand;
}
Note that this could result in an infinite loop if you provide a min and max that force mt_rand to return an exception character. So if you call getRand(44,44);, while meaningless, will result in an infinite loop... (And you can avoid the infinite loop with a bit of logic in the function (checking that there is at least one non-exception value in the range $min to $max)...
The other option, would be to build the array with a loop:
function getRand($min, $max) {
$actualMin = min($min, $max);
$actualMax = max($min, $max);
$values = array();
$exceptions = array(44, 23);
for ($i = $actualMin; $i <= $actualMax; $i++) {
if (in_array($i, $exceptions)) {
continue;
}
$values[] = $i;
}
return $values[array_rand($values)];
}
The simplest solution would be to just search for a random number from min to max - number of exceptions. Then just add 1 to the result for each exception lower than the result.
function getRandom($min, $max)
{
$exceptions = array(23, 44); // Keep them sorted or you have to do sort() every time
$random = rand($min, $max - count($exceptions));
foreach ($exceptions as $ex)
{
if ($ex > $random) break;
++$random;
}
return $random;
}
Runtime should be O(1+n) with n being the number of exceptions lower than the result.

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