I need a function that randomizes an array similar to what shuffle does, with the difference that each element has different chances.
For example, consider the following array:
$animals = array('elephant', 'dog', 'cat', 'mouse');
elephant has an higher chance of getting on the first index than dog. Dog has an higher chance than cat and so on. For example, in this particular example elephant could have a chance of 40% in getting in the 1st position, 30% of getting on the 2nd position, 20% on getting on 3rd and 10% getting on last.
So, after the shuffling, the first elements in the original array will be more likely (but not for sure) to be in the first positions and the last ones in the last positions.
Normal shuffle may be implemented just as
dropping items randomly at some range
picking them up from left to right
We can adjust dropping step, drop every element not into whole range, but at some sliding window. Let N would be amount of elements in array, window width would be w and we'll move it at each step by off. Then off*(N-1) + w would be total width of the range.
Here's a function, which distorts elements' positions, but not completely at random.
function weak_shuffle($a, $strength) {
$len = count($a);
if ($len <= 1) return $a;
$out = array();
$M = mt_getrandmax();
$w = round($M / ($strength + 1)); // width of the sliding window
$off = ($M - $w) / ($len - 1); // offset of that window for each step.
for ($i = 0; $i < $len; $i++) {
do {
$idx = intval($off * $i + mt_rand(0, $w));
} while(array_key_exists($idx, $out));
$out[$idx] = $a[$i];
}
ksort($out);
return array_values($out);
}
$strength = 0 ~normal shuffle.
$strength = 0.25 ~your desired result (40.5%, 25.5%, 22%, 12% for elephant)
$strength = 1 first item will never be after last one.
$strength >= 3 array is actually never shuffled
Playground for testing:
$animals = array( 'elephant', 'dog', 'cat', 'mouse' );
$pos = array(0,0,0,0);
for ($iter = 0; $iter < 100000; $iter++) {
$shuffled = weak_shuffle($animals, 0.25);
$idx = array_search('elephant', $shuffled);
$pos[$idx]++;
}
print_r($pos);
Try to use this algorithm:
$animals = [ 'elephant', 'dog', 'cat', 'mouse' ]; // you can add more animals here
$shuffled = [];
$count = count($animals);
foreach($animals as $chance => $animal) {
$priority = ceil(($count - $chance) * 100 / $count);
$shuffled = array_merge($shuffled, array_fill(0, $priority, $animal));
}
shuffle($shuffled);
$animals = array_unique($shuffled);
You have an array, let's say of n elements. The probability that the i'th element will go to the j'th position is P(i, j). If I understood well, the following formula holds:
(P(i1, j1) >= P(i2, j2)) <=> (|i1 - j1| <= |j1 - i1|)
Thus, you have a Galois connection between the distance in your array and the shuffle probability. You can use this Galois connection to implement your exact formula if you have one. If you don't have a formula, you can invent one, which will meet the criteria specified above. Good luck.
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[] = $digit;
}
return null;
}
Demo: https://3v4l.org/9t3vf
Given n = 3 dogs and m = 3 pairs of enemies, a = [1, 2, 3] and b = [3, 3, 1], dog 1 is the enemy of dog 3, and dog 3 is the enemy of dogs 1 and 2. Because 3 is an enemy of both 1 and 2, it must be in its own container. dogs 1 and 2 can be together or separately. There are 4 possible groups: {1, 2} ,{1}, {2}, {3}. Note that the intervals are along the original line of dogs numbered consecutively from 1 to n, i.e. [1, 2, 3] in this case. The dogs cannot be reordered and dogs cannot be skipped, e.g. {2, 1} and {1, 3} are invalid.
So given the following:
case #1:
n = 5
m = 2
a = (1,2)
b = (3,5)
Result is: Total of 11 groups can be formed.
case #2
n = 8
m = 4
a = (2,3,4,3)
b = (8,5,6,4)
Result is: Total of 18 groups can be formed.
Here's my code:
function countSubstrings($n, $a, $b) {
$tokenArr = array();
$x = 1;
while ($x <= $n){
$tokenArr[] = $x;
$x++;
}
$first = 0;
$last = $n - 1;
$outArr = array();
$pointer = 0;
/* generate groups left to right */
for ($i = $first; $i <= $last; $i++) {
$outArr[$pointer][] = $tokenArr[$i];
$tokenString = $tokenArr[$i];
$pointer++;
for ($j = $i + 1; $j <= $last; $j++) {
$tokenString .= $tokenArr[$j];
$outArr[$pointer] = str_split($tokenString);
$pointer++;
}
}
/* find the enemeies */
$intersects = array();
for($k = 0; $k < count($outArr); $k++){
if (count(array_intersect($outArr[$k], $a)) > 1 || count(array_intersect($outArr[$k], $b)) > 1) {
$intersects[] = $outArr[$k];
}
}
/* remove first and last items which are basically equal to $a and $b */
$intersects = array_slice($intersects, 1, -1);
/* remove the enemeies from generated groups */
foreach ($outArr as $keya => $valuea) {
if (in_array($valuea, $intersects)) {
unset($outArr[$keya]);
}
}
return count($outArr);
}
So far my code works in case: #1 but fails on #2.
The intersect logic seems to be incorrect to me as we have to check if the relationship formed by [a , b], for example, [1,2] exists in $outArr or not. Current check of count(array_intersect($outArr[$k], $a)) > 1 does not care about that. It rather checks if any element in $outArr[$k] is present in $a or not.
So, change the current logic from:
/* find the enemeies */
$intersects = array();
for($k = 0; $k < count($outArr); $k++){
if (count(array_intersect($outArr[$k], $a)) > 1 || count(array_intersect($outArr[$k], $b)) > 1) {
$intersects[] = $outArr[$k];
}
}
/* remove first and last items which are basically equal to $a and $b */
$intersects = array_slice($intersects, 1, -1);
to
$intersects = array();
foreach($a as $index => $val1){
$val2 = $b[$index];
foreach($outArr as $current_group){
if(in_array($val1,$current_group) && in_array($val2,$current_group)){ // check if both exist as they are enemies
$intersects[] = $current_group;
}
}
}
Demo: https://3v4l.org/Q2rnP
In the above code, we:
loop through all elements of $a and simultaneously with $b with the help of $index in foreach.
Check if for the current group in $outArr, whether both $a[$index](a.k.a $val1) and $b[$index](a.k.a $val2) exist in the group or not.
If both exist in current group, we put them under intersect as they are enemies. Your rest of the logic is correct.
Efficient Solution:
We have to exploit this line:
A group is defined as an interval (x, y) such that all dogs in the range from x to y form a group.
This means that we need to look at subarrays(as you correctly judged) instead of subsequences.
Now, we loop from 1 to N and if we find a number which has an enemy on the left, we can only form the next groups from that number + 1 onwards. Anything before them can't be included anyway since we are looking at subarrays.
For example, let's assume 5 is an enemy of 3 in a line of 1 to 5 and no other enemies are present. So, group formations would look like below.
Representation:
1 2 3 4 5
-1 -1 5 -1 3
|___|
|___|___|
|___|___|___|
|___|
|___|___|
|___|
|___| // the connection/group (4,5) remains and breaks everything before 4 since 3 is an enemy of 5 and we are looking for subarrays. So everything before 4 is disconnected anyway.
So, our next starting animal/dog to look from is 4.
For each enemy/animal, we maintain the nearest enemy on the left if present. If present, we update the next animal to look from for groups as proved above. In the below code, $prev_start is the variable that maintains the next animal to look from.
In order to get nearest enemy on the left for each animal, we preprocess the enemy details as follows:
Preprocessing:
$enemies = array_combine(range(1,$n),array_fill(0,$n,-1)); // nothing tricky, just generates an array filled with sequential numbers as keys and sets it's value as -1
foreach($a as $index => $enemy_1){
$enemy_2 = $b[$index];
if($enemy_1 < $enemy_2){
$enemies[$enemy_2] = max($enemies[$enemy_2],$enemy_1);
}else if($enemy_2 < $enemy_1){
$enemies[$enemy_1] = max($enemies[$enemy_1],$enemy_2);
}
}
Computation:
$prev_start = 1;
$count = 0;
for($i=1;$i<=$n;++$i){
if($enemies[$i] !== -1){
$prev_start = max($enemies[$i] + 1,$prev_start);
}
$count += ($i - $prev_start + 1);
}
Since we preprocessed enemy details, we update $prev_start accordingly from where we have to start counting for groups again.
$count += ($i - $prev_start + 1); simply counts the number of groups(subarrays) to consider for counting.
Time complexity: O(m + n) where m is number of pairs and n is the number of dogs/animals.
Space complexity: O(n) where n is the number of dogs/animals.
Full Code:
<?php
function countSubarrays($n, $a, $b) {
$enemies = array_combine(range(1,$n),array_fill(0,$n,-1)); // nothing tricky, just generates an array filled with sequential numbers as keys and sets it's value as -1
foreach($a as $index => $enemy_1){
$enemy_2 = $b[$index];
if($enemy_1 < $enemy_2){
$enemies[$enemy_2] = max($enemies[$enemy_2],$enemy_1);
}else if($enemy_2 < $enemy_1){
$enemies[$enemy_1] = max($enemies[$enemy_1],$enemy_2);
}
}
$prev_start = 1;
$count = 0;
for($i=1;$i<=$n;++$i){
if($enemies[$i] !== -1){
$prev_start = max($enemies[$i] + 1,$prev_start);
}
$count += ($i - $prev_start + 1);
}
return $count;
}
Demo: https://3v4l.org/1W26C
QUESTION:
Given an array of integers, find the pair of adjacent elements that has the largest product and return that product.
Example:
https://app.codesignal.com/arcade/intro/level-2
For inputArray = [3, 6, -2, -5, 7, 3], the output should be
adjacentElementsProduct(inputArray) = 21.
7 and 3 produce the largest product.
Input/Output
Input:
inputArray: [3, 6, -2, -5, 7, 3]
Expected Output:
21
SOLUTION: My code that doesn't work:
function adjacentElementsProduct($inputArray) {
$total = 0;
$temp = 0;
$maxProduct = 0;
$var = 0;
if ($inputArray.count == 1) return 0;
for ($i = 0; $i < $inputArray[$inputArray.count-1]; $i++) {
if ($inputArray[i] + $inputArray[i+1] > $maxProduct) {
$maxProduct = $inputArray[i] * $inputArray[i+1];
}
}
return $maxProduct;
}
As with any programming task, the trick is to tackle it piece by piece. You tend to find your code is more readable when you break problems down into small components.
You need to:
Find the product of adjacent elements in an array
Find the largest product in that group of values
You can approach this without a large number of variables, nesting, etc.
function adjacentElementsProduct(array $inputs) {
$products = [];
for ($i = 1; $i < count($inputs); $i++) {
$products[] = $inputs[$i - 1] * $inputs[$i];
}
return max($products);
}
All we're doing is looping through the input array, starting with the second element. Calculating the product of the previous element and the current element then putting the result into an array of products. Finally we run that through max() which is going to handle finding the largest value for us.
Important to note: there's no validation taking place here. Can you trust that your array will only ever contain numerical values? Will it always contain at least two elements? If not you'll want to account for that.
Here is how I would do it
$inputArray = [3, 6, -2, -5, 7, 3];
function adjacentElementsProduct($inputArray) {
$max = 0;
for($i = 0; $i < (sizeof($inputArray) - 1); $i++){
$b = $i+1;
if($inputArray[$i] > 0 && $inputArray[$b] > 0){
$max = (($inputArray[$i] * $inputArray[$b]) > $max) ? ($inputArray[$i] * $inputArray[$b]) : $max;
}
}
return $max;
}
echo adjacentElementsProduct($inputArray); // Outputs 21
function adjacentElementsProduct($inputArray) {
$res = [];
for($j=0;$j<count($inputArray);$j++){
$res[] = $inputArray[$j]*$inputArray[$j+1];
}
return (max($res) < 0) ? 0 : max($res);
}
As with any programming task, the trick is to tackle it piece by piece. You tend to find your code is more readable when you break problems down into small components.
You need to:
Find the product of adjacent elements in an array
Find the largest product in that group of values
In PHP its as below:
function adjacentElementsProduct($inputArray) {
$res = [];
for($j=1;$j<count($inputArray);$j++){
$res[] = $inputArray[$j-1]*$inputArray[$j];
}
return max($res);
}
$a = [3, 6, -2, -5, 7, 3]
echo adjacentElementsProduct($a); //21
I need to generate x amount of random odd numbers, within a given range.
I know this can be achieved with simple looping, but I'm unsure which approach would be the best, and is there a better mathematical way of solving this.
EDIT: Also I cannot have the same number more than once.
Generate x integer values over half the range, and for each value double it and add 1.
ANSWERING REVISED QUESTION: 1) Generate a list of candidates in range, shuffle them, and then take the first x. Or 2) generate values as per my original recommendation, and reject and retry if the generated value is in the list of already generated values.
The first will work better if x is a substantial fraction of the range, the latter if x is small relative to the range.
ADDENDUM: Should have thought of this approach earlier, it's based on conditional probability. I don't know php (I came at this from the "random" tag), so I'll express it as pseudo-code:
generate(x, upper_limit)
loop with index i from upper_limit downto 1 by 2
p_value = x / floor((i + 1) / 2)
if rand <= p_value
include i in selected set
decrement x
return/exit if x <= 0
end if
end loop
end generate
x is the desired number of values to generate, upper_limit is the largest odd number in the range, and rand generates a uniformly distributed random number between zero and one. Basically, it steps through the candidate set of odd numbers and accepts or rejects each one based how many values you still need and how many candidates still remain.
I've tested this and it really works. It requires less intermediate storage than shuffling and fewer iterations than the original acceptance/rejection.
Generate a list of elements in the range, remove the element you want in your random series. Repeat x times.
Or you can generate an array with the odd numbers in the range, then do a shuffle
Generation is easy:
$range_array = array();
for( $i = 0; $i < $max_value; $i++){
$range_array[] .= $i*2 + 1;
}
Shuffle
shuffle( $range_array );
splice out the x first elements.
$result = array_slice( $range_array, 0, $x );
This is a complete solution.
function mt_rands($min_rand, $max_rand, $num_rand){
if(!is_integer($min_rand) or !is_integer($max_rand)){
return false;
}
if($min_rand >= $max_rand){
return false;
}
if(!is_integer($num_rand) or ($num_rand < 1)){
return false;
}
if($num_rand <= ($max_rand - $min_rand)){
return false;
}
$rands = array();
while(count($rands) < $num_rand){
$loops = 0;
do{
++$loops; // loop limiter, use it if you want to
$rand = mt_rand($min_rand, $max_rand);
}while(in_array($rand, $rands, true));
$rands[] = $rand;
}
return $rands;
}
// let's see how it went
var_export($rands = mt_rands(0, 50, 5));
Code is not tested. Just wrote it. Can be improved a bit but it's up to you.
This code generates 5 odd unique numbers in the interval [1, 20]. Change $min, $max and $n = 5 according to your needs.
<?php
function odd_filter($x)
{
if (($x % 2) == 1)
{
return true;
}
return false;
}
// seed with microseconds
function make_seed()
{
list($usec, $sec) = explode(' ', microtime());
return (float) $sec + ((float) $usec * 100000);
}
srand(make_seed());
$min = 1;
$max = 20;
//number of random numbers
$n = 5;
if (($max - $min + 1)/2 < $n)
{
print "iterval [$min, $max] is too short to generate $n odd numbers!\n";
exit(1);
}
$result = array();
for ($i = 0; $i < $n; ++$i)
{
$x = rand($min, $max);
//not exists in the hash and is odd
if(!isset($result{$x}) && odd_filter($x))
{
$result[$x] = 1;
}
else//new iteration needed
{
--$i;
}
}
$result = array_keys($result);
var_dump($result);
What I need to do to generate a sequence of non-repeating integers within a given range that meets the specific criteria that I have?
Here are the criteria:
Use only the numbers between 1 and MAX (let's say 9).
Numbers cannot repeat within the sequence except:
2a. Two of the first 5 numbers from the sequence must be repeated.
2b. These two numbers must be repeated at random points within the last 5 places in the final sequence (the last 5 includes the repeats).
For example:
SET: 1,2,3,4,5,6,7,8,9
Random Sequence (with repeats):
2,4,6,9,3,1,5,2,8,7,3
r, , , ,r, , ,x, , ,x
Here I have indicated the numbers that were randomly selected to be repeated (out of the first 5 in the random sequence) with an r and the insertion points where they were randomly placed (into the last 5 of the final sequence) with an x.
Any help in figuring this out is much appreciated. Actual use will be a bit more complicated than this, but I know what I will need to do once I can get this far.
Edit
To clarify a little more, I have 1-20, and I need a 22 digit random sequence. Every number must be used, two will be used twice as discussed in my original post. I chose 10 above to simplify a little. I should be able to adapt the logic you've all given.
I assume when you say "non-repeating" you mean "distinct" (unique) as opposed to "eventually becomes periodic" (as in "the digits of pi do not repeat")
Generate n distinct integers in your range.
Pick two from the first 5. Call these a and b.
Remove the last 3 from the list.
Insert a at position 0, 1, 2, or 3 in the sublist.
Insert b at position 0, 1, 2, 3, or 4 in the sublist.
Add the sublist back to the end of the list.
Removal of the sublist is not necessary but makes it easier to conceptualize.
Not obvious what to do if n+2 is less than 10. In particular, this algorithm may crash for n < 5 and return the wrong result for n=7.
If I understand you correctly, you have 1 to N random numbers that must be used in a 10-set permutation with some specific criteria about repeats. In php, I suggest this (not counting php-internals) O(n) solution:
//Generate a full list of keys
$source = range(1, MAX);
//NOTE: if MAX < 10, you must pad the array
//Get a random group of 10 of the keys
$input = array_rand(array_flip($source), 10);
//Shuffle (can be done later as well; this is the randomization).
//array_rand() does not change order.
shuffle($input);
//Select the first of 5 that must be repeated in the last 5
$one = rand(0, 4);
$onev = $input[$one];
//Remove this array key to prevent collisions with the second of 5
$input = array_diff($input, array($onev));
//Select a random index in the last 5 to be replaced with $one
$rep = rand(5, 9);
$repv = $input[$rep];
//Remove this array key to prevent collisions with the other to-be-replaced
$input = array_diff($input, array($repv));
//Acquire the new keys list of input now that two elements have been removed
$keys = array_slice(array_keys($input), 0, 3);
//Select the second-of-5 to replace in the last 5. No worry of collision now.
$two = array_rand($keys, 1);
$two = $keys[$two];
//Select the second from the last-of-5 to be replaced by $two
//No worry of collision because the other index is removed.
$keys = array_slice(array_keys($input), 4, 8);
$rept = array_rand($keys, 1);
$rept = $keys[$rept];
//Replace one of the last-of-five with one of the first-of-five
$input[$rept] = $input[$two];
//Restore removed keys as well as perform replacement of other last-of-five
$input[$one] = $onev;
$input[$rep] = $onev;
//re-randomize based on shuffle
ksort($input);
No loops, no conditionals.
A word of warning on this solution. I wouldn't use it for a large set of numbers. If I were doing this same solution for a much larger set, I would use array_splice to drop chosen members from the array. As you get a much larger space, finding an unused number in your range becomes quite expensive, and demands a better solution than the brute force method below.
This will build half of your target set. You will call it twice, once for each half.
function build_half($min, $max, $num_elements, $arr = array() ){
while( count($arr) <= $num_elements)
{
$candidate = rand($min, $max);
if( !in_array($candidate, $arr))
{
array_push($arr, $candidate);
}
}
return $arr;
}
This will grab $this_many elements from the array.
function random_grab($arr, $this_many){ // don't try this on the subway
$nums_to_repeat = array();
// catch some edge cases...
if( $this_many > count($arr) )
{
return FALSE;
}
else if( $this_many == count($arr) )
{
return shuffle($arr);
}
while( count($nums_to_repeat) <= $this_many)
{
$rand_key = rand(0, count($arr) - 1);
if( ! in_array($arr[$rand_key], $nums_to_repeat))
{
array_push($nums_to_repeat, $arr[$rand_key]);
}
}
return $nums_to_repeat;
}
This is a fairly specialized case, but could be made more general by allowing the offset floor and ceiling to be passed in as parameters. For your problem they would be 5 and 9, so we just derive them directly.
function random_insert_2nd_half($target, $source){
$offsets_consumed = array();
$num_elements = count($target);
while( count($source) > 0 )
{
$offset = rand( ($num_elements/2), $num_elements - 1);
if( ! in_array( $offset, $offsets_consumed)
{
$arr[$offset] = array_pop($nums_to_repeat);
}
}
}
Ok so after having done all that, let's put it to work.
// Generate the first half of the array
$my_array = $repeated_nums = array();
$my_array = build_half(1, 10, 5);
// then grab the 2 random numbers from that first half.
$repeated_nums = random_grab($my_array, 2);
// So now we have our random numbers and can build the 2nd half of the array.
// we'll just repeat the call to the first function.
$my_array = build_half(1, 10, 5, $my_array);
// Then swap out two of the values in the second half.
$my_array = random_insert_2nd_half($my_array, $repeated_nums);
// at this point $my_array should match what you are looking for.
Hope this gets you on your way:
$max = 20; // max value
$repeats = 2; // numbers to be repeated
$nums = range(1, $max);
shuffle($nums);
$halfPoint = ceil($max / 2);
$firstHalf = array_slice($nums, 0, $halfPoint);
$repeaters = array_intersect_key($firstHalf, array_flip(array_rand($firstHalf, $repeats)));
$secondHalf = array_merge(array_slice($nums, $halfPoint), $repeaters);
shuffle($secondHalf);
$result = array_merge($firstHalf, $secondHalf);
var_dump(join(',', $result));
To generate distinct numbers within a range you can use something like this:
$arr_num = array();
while(count($arr_num)<=7)
{
$num = rand(1, 9);
if (!in_array($num, $arr_num))
{
$arr_num[] = $num;
}
}
$arr_num now has 8 distinct elements. Pick five elements of the array:
for ($i=0; $i<=4; $i+=1)
{
$new_arr[$i] = $arr_num[$i];
}
Now pick two numbers from $new_arr numbers:
$r1 = array_rand($new_arr);
$r2 = array_rand($new_arr);
Now you can insert these numbers into the original array at two of the last random positions. Hope it helped!
$max = 15;
$array = array(1, $max);
for($x = 1; $x <= $max; $x++)
{ $array[$x] = rand(1, $max); }
$firstDup = $array[rand(1,5)];
$secondDup = $firstDup;
do { $firstDup = $array[rand(1,5)];
} while($firstDup == $secondDup);
do { $array[rand($max-5,$max)] = $firstDup;
} while(!in_array($firstDup,array_slice($array,$max-5,5)));
do { $array[rand($max-5,$max)] = $secondDup;
} while(!in_array($secondDup,array_slice($array,$max-5,5)));