I've got a $identifier, $start_number and $end_number.
The start number is the number where the for loop should start
counting from
The end number is where the loop should stop counting
The identifier determinates how much is getting added to the start number
This for loops looks something like this:
$start_number = 102;
$end_number = 1051;
$identifier = 24;
for($i = $start_number; $i <= $end_number; $i += $identifier) {
//The first two times, add 1 to the identifier
//The second two times (we're at 4 now) add 5 to the identifier
//The third two times (were at 6 now) add 10 to the identifier
//The fourth two times (we're at 8 now) add 20 to the identifier
//etc...
}
I want it to add a dynamic number (which changes) to the $identifier each 2 times it loops, how do i do this?
Just keep track of where you are in your loop by using a counter. Then you can use the modulus operator to determine if it an even number iteration. You can add the appropriate value by using an array to store the values to add to $identifier with the count being the key to get your correct value.
$start_number = 102;
$end_number = 1051;
$identifier = 24;
$add = array(
2 => 1,
4 => 5,
6 => 10,
8 => 20
);
$count = 1;
for($i = $start_number; $i <= $end_number; $i += $identifier) {
if ($count % 2 === 0) {
$identifier += $add[$count];
}
$count++;
}
for($i = $identifier; $i <= $end_number; $i += $identifier) {
if($i%2 == 0){
//do your work
}
}
Related
For an endless number such as Pi, how would one go about finding the first occurrence of an exact sum of digits for a given number n.
For example. If n=20
Pi=3.14159265358979323846264338327950288419716939...
then the first occurrence is from digit 1 to digit 5 since:
1+4+1+5+9=20
if n=30, then the first occurrence is from digit 5 to digit 11
since 9+2+6+5+3+5=30
answer should have a working php demo
The answer to this is using sliding window that will maintain the sum. so maintain two pointers say i and j. Keep increasing j and adding the elements inside. when it crosses the desired sum increase i and decrease the element at i. Then keep increasing j until the sum is reached or the sum overflows so you repeat the above process.
Example sum = 30
141592653589793238 >> i=j=0 current_sum = 1
141592653589793238 >> i=0 j=6 current_sum=28
in the next iteration adding 5 will result in current_sum>30 so hence you increment i
141592653589793238 >> i=1 j=6 current_sum=27
141592653589793238 >> i=2 j=6 current_sum=23
141592653589793238 >> i=2 j=7 current_sum=28
Keep going in this manner and it will finally reach the window that is equal to the sum =30 . That should break you out of the loop and help you find the answer.
Method 1 (suggested by Ashwin Bhat)
This implementation uses two pivots. The sum of digits between $pivot_a and $pivot_b is computed. Depending on the value of the sum, we increment $pivot_b (if the sum is less) or $pivot_a (if the sum is greater). If the sum is equal to $n, break. The values of the pivots give the appropriate digit indices.
$pi = "314159265358979323846264338327950288419716939";
$n = 30;
$pivot_a = $pivot_b = 0;
$sum = 0;
for( ; $pivot_b < strlen($pi); ) {
if($sum < $n) {
$sum += $pi[$pivot_b++];
} elseif ($sum > $n) {
$sum -= $pi[$pivot_a++];
} else {
print('Solution found from digit '.$pivot_a.' to '.$pivot_b.'.');
exit;
}
}
print('No match was found.');
Method 2
This implementation uses one pivot only, from which it starts summing up the digits. If the sum happens to be greater than the desired value, it resets the sum to zero, shifts the pivot one position and starts the summing again.
$pi = "314159265358979323846264338327950288419716939";
$n = 30;
// Let's sum up all the elements from $pivot until we get the exact sum or a
// number greater than that. In the latter case, shift the $pivot one place.
$pivot = 0;
$sum = 0;
for($k=0 ; $sum != $n && $k < strlen($pi) ; $k++) {
$sum += $pi[$k];
print($pi[$k]);
if($sum > $n) {
print(' = '.$sum.' fail, k='.($pivot+1).PHP_EOL);
$sum = 0;
$k = $pivot++;
} elseif($sum < $n) {
print("+");
}
}
print(' = '.$n.' found from digit '.$pivot.' to '.$k.'.');
The implementation is not very effective but tries to explain the steps. It prints
3+1+4+1+5+9+2+6 = 31 fail, k=1
1+4+1+5+9+2+6+5 = 33 fail, k=2
4+1+5+9+2+6+5 = 32 fail, k=3
1+5+9+2+6+5+3 = 31 fail, k=4
5+9+2+6+5+3 = 30 found from digit 4 to 10.
Here's another approach. It builds an array of sums along the way and, on every iteration, attempts to add the current digit to the previous sums, and so on, while always only keeping the sums that are still relevant (< target).
The function either returns:
an array of 2 values representing the 0-based index interval within the digits,
or null if it couldn't find the target sum
Code:
function findFirstSumOccurrenceIndexes(string $digits, int $targetSum): ?array
{
$sums = [];
for ($pos = 0, $length = strlen($digits); $pos < $length; $pos++) {
$digit = (int)$digits[$pos];
if ($digit === $targetSum) {
return [$pos, $pos];
}
foreach ($sums as $startPos => $sum) {
if ($sum + $digit === $targetSum) {
return [$startPos, $pos];
}
if ($sum + $digit < $targetSum) {
$sums[$startPos] += $digit;
}
else {
unset($sums[$startPos]);
}
}
$sums[] = $digit;
}
return null;
}
Demo: https://3v4l.org/9t3vf
I have an events loop set up; and I also have an ads loop set up.
I want to inject each 'ad' into the events loop at random points. These loops/arrays have different set ups so can't push into the loop/array.
I had the below set up, which tended to work, but within ad.inc it was getting a random ad... whereas it should be getting the total count of ads and injecting them randomly into the events until that count is reached.
$count = 1;
$total = count($events);
$random = rand(3, $total);
foreach ($events as $event) {
include('./inc/events-item.inc');
if ($count == $random) {
include("./inc/ad.inc");
$random = rand($count, $total);
}
$count++;
}
For example, if my total events count is 30, and my total ads count is 4 then I should see the four ads randomly injected into the 30 events.
Any help?
Create array of all positions for ads. If you have 30 ads - there're 30 positions, from 0 to 29:
$positions = range(0, 29);
// now get 4 random elements from this array:
$rand_positions = array_rand($positions, 4);
// though `array_rand` returns array of keys
// in this case keys are the same as values
// iterate over your events and if counter equals
// to any value in $rand_positions - show ad
$i = 0;
foreach ($events as $event) {
include('./inc/events-item.inc');
if (in_array($i, $rand_positions, true)) {
include("./inc/ad.inc");
}
$i++;
}
You need to randomly select 4 (or however many ads you have) points between 30 (or however many entries you have) other points. Here is a possible solution.
// Set up counts
$entry_count = count($events);
$ad_count = 4;
// Create an array of entry indices, and an array of ad indices
$entry_indices = range(0, $entry_count - 1);
$ad_indices = array();
// Fill the ad indices with random elements from the entry indices
for ($i = 0; $i < $ad_count; $i++) {
$entry = rand(0, count($entry_indices));
array_push($ad_indices, $entry_indices[$entry]);
array_splice($entry_indices, $entry, 1);
}
// Sort it so we only need to look at the first element
sort($ad_indices);
// Iterate through the events
$count = 0;
foreach ($events as $event) {
include('./inc/events-item.inc');
// If we have any ad indices left, see if this entry is one of them
if (count($ad_indices) > 0 && $count == $ad_indices[0]) {
include("./inc/ad.inc");
array_shift($ad_indices);
}
$count++;
}
The problem statement is as following:
A particular kind of coding which we will refer to as "MysteryCode" is a binary system of encoding in which two successive values, differ at exactly one bit, i.e. the Hamming Distance between successive entities is 1. This kind of encoding is popularly used in Digital Communication systems for the purpose of error correction.
LetMysteryCodes(N)represent the MysteryCode list for N-bits.
MysteryCodes(1) = 0, 1 (list for 1-bitcodes,in this order)
MysteryCodes(2) = 00, 01, 11, 10 (list for 2-bitcodes,in this order)
MysteryCodes(3) =000, 001, 011, 010,110, 111, 101, 100 (list for 3-bitcodes,in this order)
There is a technique by which the list of (N+1) bitcodescan be generated from (N)-bitcodes.
Take the list of N bitcodesin the given order and call itList-N
Reverse the above list (List-N), and name the new reflected list: Reflected-List-N
Prefix each member of the original list (List-N) with 0 and call this new list 'A'
Prefix each member of the new list (Reflected-List-N) with 1 and call this new list 'B'
The list ofcodeswith N+1 bits is the concatenation of Lists A and B.
A Demonstration of the above steps: Generating the list of 3-bitMysteryCodesfrom 2-BitMysteryCodes
2-bit list ofcodes:00, 01, 11, 10
Reverse/Reflect the above list:10, 11, 01, 00
Prefix Old Entries with 0:000, 001, 011, 010
Prefix Reflected List with 1:110, 111, 101, 100
Concatenate the lists obtained in the last two steps:000, 001, 011, 010, 110, 111, 101, 100
Your Task
Your task is to display the last N "MysteryCodes" from the list of MysteryCodes for N-bits. If possible, try to identify a way in which this list can be generated in a more efficient way, than iterating through all the generation steps mentioned above.
More efficient or optimized solutions will receive higher credit.
Input Format
A single integer N.
Output Format
N lines, each of them with a binary number of N-bits. These are the last N elements in the list ofMysteryCodesfor N-bits.
Input Constraints 1 = N = 65
Sample Input 1
1
Sample Output 1
1
Explanation for Sample 1
Since N = 1, this is the (one) last element in the list ofMysteryCodesof 1-bit length.
Sample Input 2
2
Sample Output 2
11
10
Explanation for Sample 2 Since N = 2, these are the two last elements in the list ofMysteryCodesof 2-bit length.
Sample Input 3
3
Sample Output 3
111
101
100
$listN = 25;
$bits = array('0','1');
//check if input is valid or not
if(!is_int($listN))
{
echo "Input must be numeric!";
}
if($listN >= 1 && $listN <=65){
if($listN == 1){
echo '1'; exit;
}
ini_set('memory_limit', -1);
for($i=1; $i<=($listN - 1); $i++){
$reverseBits = array_reverse($bits);
$prefixBit = preg_filter('/^/', '0', $bits);
$prefixReverseBits = preg_filter('/^/', '1', $reverseBits);
$bits = array_merge($prefixBit, $prefixReverseBits);
unset($prefixBit, $prefixReverseBits, $reverseBits);
}
$finalBits = array_slice($bits, -$listN);
foreach($finalBits as $k=>$v){
echo $v."\n";
}
}
else{
echo "Invalid input!";
}
I have tried above solution, but didnt worked for input greater than 20.
for eg. If the input is 21, I got "Couldnt allocate memory" error.
It will be great if somebody figure out the optimized solutions...
The numbers follow a pattern which I transformed to below code.
Say given number is N
then create a N x N matrix and fill it's first column with 1's
and all other cells with 0's
Start from rightmost column uptil 2nd column.
For any column X start from bottom-most row and fill values like below:
Fill 2^(N - X + 1)/2 rows with 0's.
Fill 2^(N - X + 1) rows with 1's and then 0's alternatively.
Repeat step 2 till we reach topmost row.
Print the N x N matrix by joining the values in each row.
<?php
$listN = 3;
$output = [];
for ($i = 0; $i < $listN; $i++) {
$output[$i] = [];
for ($j = 0; $j < $listN; $j++) {
$output[$i][$j] = 0;
}
}
$output[$listN - 1][0] = 1;
for ($column = 1; $column < $listN; $column++) {
$zeroFlag = false;
for ($row = $listN - 1; $row >= 0;) {
$oneZero = 1;
if (!$zeroFlag) {
for ($k = 1; $k <= pow(2, $column) / 2 && $row >= 0; $k++) {
$output[$row][$listN - $column] = 0;
$row--;
$zeroFlag = true;
}
}
for ($k = 1; $k <= pow(2, $column) && $row >= 0; $k++) {
$output[$row][$listN - $column] = $oneZero;
$row--;
}
$oneZero = 0;
for ($k = 1; $k <= pow(2, $column) && $row >= 0; $k++) {
$output[$row][$listN - $column] = $oneZero;
$row--;
}
}
}
for ($i = 0; $i < $listN; $i++) {
$output[$i][0] = 1;
}
for ($i = 0; $i < $listN; $i++) {
print(join('', $output[$i]));
print("\n");
}
First of all, I apologize for my lack of English. I hope you do understand what I'm trying to explain here.
So basically I need to build a function that would limit the number of duplicate values inside an array.
The reason I need to do this is that I'm building a system that would divide numbers into groups and every group has to have the same amount of numbers.
EDIT: Random number represents the group number.
I've written a function do this but for some reason, it is not working properly.
function jagaTiimid($max, $liiget, $tArvLength, $tArv){
$tiimid = []; //Starting array
for($z=0;$z<$liiget;$z++){
$numbers = [];
$rn = randomNumber($tArvLength, $tArv, $numbers); //Generate a random number for a group, etc group 1, group 2, group 3
$mitu = countInArray($tiimid, $rn); //Check how many times that number has occured in array
if($mitu == $max){ //If it equals to maximum number of times then...
$rnUus = randomNumber($tArvLength, $tArv, $numbers); //generate a new random number
while($rnUus == $rn){
$numbers = [];
$rnUus = randomNumber($tArvLength, $tArv, $numbers);
} //loop until the new generated number doesn't equal to old rn.
$tiimid[] = $rnUus; //if it doesn't equal to $rn then push into array
}else{
$tiimid[] = $rn;
}
}
return $tiimid;
}
For some reason the number still occures more than it is suppose to.
Basically how it shouldn't end up is.
As you can see, one group(group 2) occurs more times than other group but it should be equal for both groups.
EDIT: CountInArray();
function countInArray($array, $what) {
$count = 0;
for ($i = 0; $i < count($array); $i++) {
if ($array[$i] === $what) {
$count++;
}
}
return $count;
}
When the first random pick hits a number that is already used $liiget times, the inner loop kicks in, but it does not check whether the newly generated random number already occurs $liiget times.
For efficiency I would keep track of the number of times a number has been used. Also, you could benefit from a safety net, in case there really is no number any more that would not exceed the maximum recurrence.
It is not necessary to have a nested loop. The code would look like this:
function jagaTiimid($max, $liiget, $tArvLength, $tArv){
$tiimid = []; //Starting array
$counts = []; // Helper for quick count
$tries = 0; // Counter to avoid infinite looping
while (count($tiimid) < $liiget && $tries++ < 100) {
$numbers = [];
$rn = randomNumber($tArvLength, $tArv, $numbers); //Generate a random number for a group, etc group 1, group 2, group 3
if (!isset($counts[$rn])) $counts[$rn] = 0; // initialise on first occurence
if ($counts[$rn] < $max) {
$tiimid[] = $rn; // add it to the result
$counts[$rn]++; // ... and adjust the count
$tries = 0; // reset the safety
}
}
return $tiimid;
}
replace while($rnUus == $rn) with while(countInArray($tiimid, $rnUus) >= $max)
– Ilya Bursov
So, I want to distribute evenly lists across 3 columns. The lists cannot be broken up or reordered.
At the moment, I have 5 lists each containing respectively 4, 4, 6, 3 and 3 items.
My initial approach was:
$lists = [4,4,6,3,3];
$columns = 3;
$total_links = 20;
$items_per_column = ceil($total_links/$columns);
$current_column = 1;
$counter = 0;
$lists_by_column = [];
foreach ($lists as $total_items) {
$counter += $total_items;
$lists_by_column[$current_column][] = $total_items;
if ($counter > $current_column*$links_per_column) {
$current_column++;
}
}
Results in:
[
[4],
[4,6],
[3,3]
]
But, I want it to look like this:
[
[4,4],
[6],
[3,3]
]
I want to always have the least possible variation in length between the columns.
Other examples of expected results:
[6,4,4,6] => [[6], [4,4], [6]]
[4,4,4,4,6] => [[4,4], [4,4], [6]]
[10,4,4,3,5] => [[10], [4,4], [3,5]]
[2,2,4,6,4,3,3,3] => [[2,2,4], [6,4], [3,3,3]]
Roughly what you need to do is loop over the number of columns within your foreach(). That will distribute them for you.
$numrows = ceil(count($lists) / $columns);
$thisrow = 1;
foreach ($lists as $total_items) {
if($thisrow < $numrows){
for($i = 1; $i <= $columns; $i++){
$lists_by_column[$i][] = $total_items;
}
}else{
//this is the last row
//find out how many columns need to fit.
//1 column is easy, it goes in the first column
//2 columns is when you'll need to skip the middle one
//3 columns is easy because it's full
}
$thisrow++;
}
This will be an even distribution, from left to right. But you actually want a modified even distribution that will look symmetrical to the eye. So within the foreach loop, you'll need to keep track of 1.) if you're on the last row of three, and 2.) if there are 2 remainders, to have it skip col2 and push to col3 instead. You'll need to set that up to be able to play around with it,...but you're just a couple of logic gates away from the land of milk and honey.
So, I ended up using this code:
$lists = [4,4,6,3,3];
$columns = 3;
$total_links = 20;
$items_per_column = ceil($total_links/$columns);
$current_column = 1;
$lists_by_column = [];
for ($i = 0; $i < count($lists); $i++) {
$total = $lists[$i];
$lists_by_column[$current_column][] = $lists[$i];
//Loop until reaching the end of the column
while ($total < $items_per_column && $i+1 < count($lists)) {
if ($total + $lists[$i+1] > $items_per_column) {
break;
}
$i++;
$total += $lists[$i];
$lists_by_column[$current_column][] = $lists[$i];
}
//When exiting the loop the last time we need another break
if (!isset($lists[$i+1])) {break;}
//If the last item goes onto the next column
if (abs($total - $items_per_column) < abs($total + $lists[$i+1] - $items_per_column)) {
$current_column++;
//If the last item goes onto the current column
} else if ($total + $lists[$i+1] > $items_per_column) {
$i++;
$lists_by_column[$current_column][] = $lists[$i];
$current_column++;
}
}