I'm trying to make something like this to my variable data value...
$maxvalue = 0;
$basevalue = 0;
if($basevalue == 0) {$maxvalue = 0;}
else if ($basevalue == 1) {$maxvalue = 884;}
else if ($basevalue == 2) {$maxvalue = 1819;}
else if ($basevalue == 3) {$maxvalue = 2839;}
and so on.. i believe there is no exact computation on how the $maxvalue shifts as the basevalue increase. Can someone suggest me a simplier way to do this? thanks in advance!
$maxvalues = array(0, 884, 1819, 2839, ...);
$maxvalue = $maxvalues[$basevalue];
It looks like there's a pattern, almost like a faculty, but also with some other calculations. All numbers are multiples of 17. The following function returns the numbers you provided, so I think it might work for the higher numbers too:
function getMaxValue($base)
{
// Factor of $base = 51 + $base^2 + Factor($base - 1). You
// could solve that in a recursion, but a loop is generally better.
$factor = 0;
for ($i = 1; $i <= $base; $i++)
$factor += 51 + ($i * $i);
return $factor * 17;
}
// Test
for ($i = 0; $i < 100; $i++)
{
echo "$i -- " . getMaxValue($i) . "<br>\n";
}
Here's the solution that prevented me putting all of them in an array..
$maxvalue = 17/6*(2*($basevalue*$basevalue*$basevalue)+3
($basevalue*$basevalue)+307*$basevalue);
Thanks for all the help
Related
I need to get 50 random numbers out of range 1-100 without repeating. The current way i do is :
$array = array();
while (count($array) <= 50) {
$temp = random_int(1,100);
if (!in_array($temp, $array))
$array[] = $temp;
}
However, the looping is too many because I need to generate for more than 100,000 times.
Is there other ways that I can get a 50 random non-repeating numbers without looping ?
For example:
$number= range(1,100);
$array = array_slice(shuffle($number),0,50);
I can't use shuffle because it uses pseudo random number.
Is there other ways to achieve what I need, or ways that could shorten time.
pre fill a array of numbers and pick from them, and then remove it.
it prevents the unnecessary random generations you have
$numbers = [];
for ($i = 1; $i <= 100; $i++) {
$numbers[] = $i;
}
$randomNumbers = [];
for ($i = 1; $i <= 50; $i++) {
$r = rand(0, count($numbers) - 1);
$randomNumbers[] = $numbers[$r];
array_splice($numbers, $r, 1);
}
This would be my approach:
This gives you 50 numbers in any case, and they are defenitely different from each other. PLUS: you dont have to prefill some other array:
$start = microtime(true);
for($i = 0; $i <= 100000; $i++){
$arr = [];
while(sizeof($arr) < 50){
$num = rand(1, 100);
$arr[$num] = $num;
}
if(array_unique($arr) !== $arr || sizeof($arr) !== 50 ){
print("FAIL");
}
//print(array_unique($arr) == $arr ? "true" : "false");print("<br>");
//print(sizeof($arr));print("<br>");
//print_r(array_count_values ($arr));print("<br>");
//print_r($arr);print("<br>");
}
$time_elapsed_secs = microtime(true) - $start;
print($time_elapsed_secs);print("<br>");
Running this 100000 times takes about 0.4sec for me.
The actual generation is done in this part:
$arr = [];
while(sizeof($arr) < 50){
$num = rand(1, 100);
$arr[$num] = $num;
}
We can do in 2 steps:
$x = 0;
$arr = [];
while($x < 50){
$tmp = rand(1, 100);
if(!in_array($tmp, $arr)){
$arr[] = $tmp;
$x++;
}
}
I am trying to display possibilities for additions of specific numbers but have not been getting the right results.
<?php
$num3 = 20;
$set = null;
$balance = $num3;
$dig = mt_rand(1,5);
for($i = $balance; $i > 0; $i -= $dig ){
echo $dig.'<br>';
if($i < 1){
$set .= $i;
$balance = 0;
}
else{
$set .= $dig.'+';
$dig = mt_rand(1,5);
}
}
echo $set.'='.$num3;
?>
Here are some of the outputs:
2+5+1+4+5+3+=20
1+4+3+5+3+=20
3+1+1+2+3+4+4+1+3+=20
Appreciate any pointers. Thank in advance...
Ok, even though the requirement isn't completely clear, here's an approach:
(edit: demonstrating prevention of endless loop)
$max_loops = 1000;
$balance = 20;
$found = [];
while($balance > 0 && $max_loops-- > 0) {
$r = mt_rand(1, 5);
if ($balance - $r >= 0) {
$found[] = $r;
$balance -= $r;
}
}
echo implode(' + ', $found) . ' = '.array_sum($found);
Note: This code has a small risk of getting caught in an endless loop... though it's doubtful that it'll ever happen :)
Edit: Now the loop contains a hard-limit of 1000 iterations, after which the loop will end for sure...
To provoke an endless loop, set $balance = 7 and modify mt_rand(4, 5).
You can use a recursive function for this:
function randomAddends($target, $maxAddend = 5, $sum = 0, $addends = [])
{
// Return the array of addends when the target is reached
if ($target <= $sum) {
return $addends;
}
// generate a new random addend and add it to the array
$randomAddend = mt_rand(1, min($maxAddend, $target - $sum));
$addends[] = $randomAddend;
// make the recursive call with the new sum
return randomAddends($target, $maxAddend, $sum + $randomAddend, $addends);
}
I have a N*N Matrix.Now i want to know the diagonal difference of this Matrix.What will be the best approach of this solution?
I am trying with given approach:
Such as it is 3*3 Matrix say it is:
11 15 85
66 72 21
14 21 47
the diagonal simple formula will be:
firstD= (11+72+47) = 130
secondD = (85+72+14)= 171
diagonalDiff = |firstD - secondD| = |130-171| = 41
If I count every row such as first to find out firstD (First row's first value + Sec row's Sec value + Third row's third value+..).This is my thinking.
Can anyone tell me best approaches?
Try this:
$arr = array(
array(11, 15, 85),
array(66, 72, 21),
array(14, 21, 47),
);
$arrDiag = count($arr);
$firstD = 0;
$secondD = 0;
$i = 0;
for($j = 0; $j < $arrDiag; $j++){
$firstD += $arr[$i++][$j];
$secondD += $arr[$arrDiag - $i][$j];
}
echo abs($firstD - $secondD);//41
Model your matrix with a multi-dimensional array and iterate through it. The easiest way should be the following:
<?
$matrix = array(array(1,2,3),array(4,5,6),array(7,8,9)); //Insert or define your matrix here..
$n = count($matrix); //Size of matrix, thanks to VolkerK
$firstD = 0;
$lastD = 0;
for($i = 0; $i < $n; $i++){
$firstD += $matrix[$i][$i];
$lastD += $matrix[$i][$n-$i-1];
}
echo $firstD."\n";
echo $lastD;
Here is a pseudo code for your problem using one simple loop:
// $array - predefined 2 dimentional array with $N rows and $N columns
$result = 0;
for ($i=0;$i<$N;$i++) {
$result += ($array[$i,$i] - &array[$i,$N-$i-1]);
}
return echo abs($result);
that way you can do the calculation in one pass, and do a diff between two elements in each row insead of calculation the sum of each diagonal
Try this:
function diagonalDifference($arr) {
$left = 0;
$right = 0;
$i = 0;
foreach($arr as $ar){
$left+= $ar[0+$i];
$right+= $ar[count($ar) - (1+$i)];
$i++;
}
return abs($left - $right);
}
Try this
$result=0;
for($i=0;$i<=count($arr)-1;$i++){
$result= $result+($arr[$i][$i])-($arr[(count($arr)-1-$i)] [$i]);
}
return abs($result);
This is the code you need:
$first = 0;
$second = 0;
for($i = 0; $i < N; $i++) {
for($j = 0; $j < N; $j++) {
if($i == $j) {
$first += $matrix[$i][$j];
} else if($i + $j == N) {
$second += $matrix[$i][$j];
}
}
}
$diagonalDiff = abs($first - $second);
Where $matrix is a N*N array
Just using the function array_reduce:
function diagonalDifference($arr) {
$i = 0;
$n = count($arr);
return abs(array_reduce($arr,
function ($c, $str) use (&$i, $n ) {
$i++;
return $c + $str[$i-1] - $str[$n-$i];
}, 0));
}
demo
you can try this :
$first_diag=$second_diag=0;
$matrix=array(array(11,15,85),array(66,72,21),array(14,21,47));
foreach($matrix as $index=>$sub_array){
$first_diag +=$sub_array[$index];
$second_diag +=$sub_array[count($matrix)-1-$index];
}
print abs ($first_diag-$second_diag);
For one you can use a matrix library like Math_Matrix. But if this is the only operation you are gona need then it's best to write a generalized function for this using the same method you quoted yourself.
function diagonalDiff($n){
$firstD = 0;
$secondD = 0;
for($i=0;$i<$n;$i++){
for($j=0;$j<$n;$j++){
if($i == $j) {
$first += $matrix[$i][$j];
} else if($i + $j == $n) {
$secondD += $matrix[$i][$j];
}
}
}
return abs($firstD-$secondD);
}
Should give you the answer you need for a matrix of a given size $n. (Only square Matrixes ofcourse :) )
Another better solution and it's easy to understand:
<?php
$arr = array(
array(11, 15, 85),
array(66, 72, 21),
array(14, 21, 47),
);
$arrDiag = count($arr);
$firstD = 0;
$secondD = 0;
$i = 0;
for($j = 0; $j < $arrDiag; $j++){
$i++;
$firstD += $arr[$j][$j];
$secondD += $arr[$arrDiag - $i][$j];
}
echo abs($firstD - $secondD);
?>
DEMO
Recently solved this question in Hacker Rank.
<?php
$m=array(array(3,4,-1,11,2),
array(-3,2,1,6,9),
array(3,4,6,5,-2),
array(1,9,9,7,-3),
array(12,9,16,7,-3));
echo count($m)."<br>";
$i=0;
$j=0;
$ek=0;
$k=0;
$n=1;
$es=count($m)-1;
for($i=0;$i<count($m);$i++)
{
for($j=0;$j<count($m);$j++)
{
echo $m[$i][$j]." ";
}
echo "<br>";
}
echo "<br>";
for($k=0;$k<count($m);$k++)
{
for($n=0;$n<count($m);$n++)
{
if($k==$n){
$ek=$ek+$m[$k][$n];
}
}
echo "<br>";
}
echo "<hr>";
$p=0;
for($k=0;$k<count($m);$k++)
{
echo $m[$k][$es-$p]."<br> ";
$p++;
}
?>
Sorry I used different variable names, I had to take this to my vs code.
$b = array(
array(1,2,5),
array(3,4,5),
array(3,4,5)
);
//So for each diag
echo $b[0][0] + $b[1][1] + $b[2][2]; //to sum the first diag
echo $b[0][2] + $b[1][1] + $b[2][0]; //to sum the second diag
//notice the pattern 00, 11, 22 vs 02,11, 20. hence why I have written the function below
function difference($b){
$d1 = 0;
$d2 = 0;
$count = count($b);
for($i=0; $i<$count; $i++){
$d1 += $b[$i][$i];
$d2 += $b[$i][$count-1-$i];
}
return abs($d1 - $d2);
}
I had the same issue, the instructions was given by the site, mislead me. However, I solved it in this way,
$n = count and store the length of input array
//define variables to store first diagonals and second diagonals
$fd = 0;
$sd = 0;
//loop through the 2D array with $i increment
$j = get an array from the main array
//in each iteration
$pd += get the values from the left to the right and add;
$sd += get the values from the right to the left and add;
}
find absolute difference between the sums using built in abs function and return it;
I think this would help someone
I need to calculate from a given array the number that is equal or higher and closest to a given number in PHP. Example:
Number to fetch:
6.85505196
Array to calculate:
3.11350000
4.38350000
4.04610000
3.99410000
2.86135817
0.50000000
Only correct combination should be:
3.99410000 + 2.86135817 = 6.85545817
Can somebody help me? It's been 3 hours I'm getting mad!
UPDATE: I finally finished my code as following:
$arr = array(3.1135, 4.3835, 4.0461, 3.9941, 2.86135817, 0.5);
$fetch = 6.85505196;
$bestsum = get_fee($arr, $fetch);
print($bestsum);
function get_fee($arr, $fetch) {
$bestsum = 999999999;
$combo = array();
$result = array();
for ($i = 0; $i<count($arr); $i++) {
combinations($arr, $i+1, $combo);
}
foreach ($combo as $idx => $arr) {
$sum = 0;
foreach ($arr as $value) {
$result[$idx] += $value;
}
if ($result[$idx] >= $fetch && $result[$idx] < $bestsum) $bestsum = $result[$idx];
}
return $bestsum;
}
function combinations($arr, $level, &$combo, $curr = array()) {
for($j = 0; $j < count($arr); $j++) {
$new = array_merge($curr, array($arr[$j]));
if($level == 1) {
sort($new);
if (!in_array($new, $combo)) {
$combo[] = $new;
}
} else {
combinations($arr, $level - 1, $combo, $new);
}
}
}
I hope the following example might help you. Please try this
<?php
$array = array(
"3.11350000",
"4.38350000",
"4.04610000",
"3.99410000",
"2.86135817",
"0.50000000"
);
echo "<pre>";
print_r($array);// it will print your array
for($i=0; $i<count($array); $i++)
{
$j=$i+1;
for($j;$j<count($array); $j++)
{
$sum = $array[$i] + $array[$j];
// echo $array[$i]. " + ".$array[$j]." = ".$sum."<br>"; //this will display all the combination of sum
if($sum >= 6.85505196 && ($sum <= round(6.85505196)) )//change the condition according to your requirement
{
echo "The correct combinations are:<br/><br/>";
echo "<b>". $array[$i]. " + ".$array[$j]." = ".$sum."<b>";
echo "<br/>";
}
}
echo "<br/>";
}
?>
We will get the result as below
Array
(
[0] => 3.11350000
[1] => 4.38350000
[2] => 4.04610000
[3] => 3.99410000
[4] => 2.86135817
[5] => 0.50000000
)
The correct combinations are:
4.04610000 + 2.86135817 = 6.90745817
3.99410000 + 2.86135817 = 6.85545817
You should do it in two steps:
a. Work out (or look up) an algorithm to do the job.
b. Implement it.
You don't say what you've managed in the three hours you worked on this, so here's a "brute force" (read: dumb) algorithm that will do the job:
Use a variable that will keep your best sum so far. It can start out as zero:
$bestsum = 0;
Try all single numbers, then all sums of two numbers, then all sums of three numbers, etc.: Every time you find a number that meets your criteria and is better than the current $bestsum, set $bestsum to it. Also set a second variable, $summands, to an array of the numbers you used to get this result. (Otherwise you won't know how you got the solution). Whenever you find an even better solution, update both variables.
When you've tried every number combination, your two variables contain the best solution. Print them out.
That's all. It's guaranteed to work correctly, since it tries all possibilities. There are all sorts of details to fill in, but you can get to work and ask here for help with specific tasks if you get stuck.
Thank you all for your help!
My code is working pretty cool when is needed to fetch one or two numbers (addition) only. But can't figure out how to add more combinations up to the total count of elements in my given array.
I mean if there are, let's say, 8 numbers in my array I want to try all possible combinations (additions to each other) as well.
My actual code is:
$bestsum = 1000000;
for ($i = 0; $i < count($txinfo["vout"]); $i++) {
if ($txinfo["vout"][$i]["value"] >= $spent && $txinfo["vout"][$i]["value"] < $bestsum) {
$bestsum = $txinfo["vout"][$i]["value"];
}
}
for($i = 0; $i < count($txinfo["vout"]); $i++) {
$j = $i + 1;
for($j; $j < count($txinfo["vout"]); $j++) {
$sum = $txinfo["vout"][$i]["value"] + $txinfo["vout"][$j]["value"];
if($sum >= $spent && $sum < $bestsum) {
$bestsum = $sum;
}
}
}
$fee = bcsub($bestsum, $spent, 8);
print("Fee: ".$fee);
New updated code.
<?php
$x = 6.85505196;
$num = array(3.1135, 4.3835, 4.0461, 3.9941, 2.86135817, 0.5);
asort($num); //sort the array
$low = $num[0]; // lowest value in the array
$maxpossible = $x+$low; // this is the maximum possible answer, as we require the number that is equal or higher and closest to a given number
$num = array_values($num);
$iterations = $x/$num[0]; // possible combinations loop, to equate to the sum of the given number using the lowest number
$sum=$num;
$newsum = $sum;
$k=count($num);
for($j=0; $j<=$iterations; $j++){
$l = count($sum);
for($i=0; $i<$l; $i++){
$genSum = $sum[$j]+$sum[$i];
if($genSum <= $maxpossible){
$newsum[$k] = $genSum;
$k++;
}
}
$newsum = array_unique($newsum);
$newsum = array_values($newsum);
$k = count($newsum);
$sum = $newsum;
}
asort($newsum);
$newsum = array_values($newsum);
for($i=0; $i<count($newsum); $i++){
if($x<=$newsum[$i]){
echo "\nMaximum Possible Number = ".$newsum[$i];
break;
}
}
?>
Can you tell me how the invert function for the following PHP function is?
<?php
function id2secure($old_number) {
$alphabet_en = '1357902468acegikmoqsuwybdfhjlnprtvxz-_';
$new_number = '';
while ($old_number > 0) {
$rest = $old_number%38;
if ($rest >= 38) { return FALSE; }
$new_number .= $alphabet_en[$rest];
$old_number = floor($old_number/38);
}
$new_number = strrev($new_number);
return $new_number;
}
echo id2secure(172293);
?>
Thank you very much in advance!
This is secure :) Took me a few minutes to crack it. Here you go,
function secure2id($new_number) {
$alphabet_en = '1357902468acegikmoqsuwybdfhjlnprtvxz';
$old_number = 0;
$new_number = strrev($new_number);
$len=strlen($new_number);
$n=0;
$base=1;
while($n<$len){
$c=$new_number[$n];
$index = strpos($alphabet_en, $c);
if ($index === false)
break;
$old_number += $base * $index;
$base *= 38;
$n++;
}
return $old_number;
}
Haven't tested this code, but it might work:
<?php function secure2id($sr)
{
$s = strrev($sr);
$alpha = '1357902468acegikmoqsuwybdfhjlnprtvxz';
$alpha2num = array();
$n = strlen($alpha);
for($i = 0; $i < $n; $i++)
{
$alpha2num[$alpha[$i]] = $i;
}
$rez = 0;
$n = strlen($s);
$b = 1;
for($i = 0; $i < $n; $i++)
{
$rez += $b * $alpha2num[$s[$i]];
$b *= 38;
}
return $rez;
} ?>
`
Are you asking how to convert base 38 to base 10? The numerical algorithm is this:
Let N be the new number in base 10. Set N to zero to start with.
Let X be the original number in base 38.
Multiply N by 38.
Let D be the most significant (leftmost) digit of X.
Let T be the value of D in base 10.
Add T to N.
Remove D from X (so the number X is now 1 digit shorter).
If X is empty, goto 10.
Goto 3.
Return N, which is now the fully converted base 10 number.
Now that you understand the math, it should be fairly straightforward to convert these steps to a PHP function.