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'm trying to distribute 100% to total numbers (not equally), it can be done manually but I'm looking for a automatically way in PHP. I had to open calculator and get it done for manual.
What I'm trying to achieve is the result similar to this:
$value = 10000;
$total_numbers = 9
$a1 = $value*0.2;
$a2 = $value*0.175;
$a3 = $value*0.15;
$a4 = $value*0.125;
$a5 = $value*0.1;
$a6 = $value*0.08;
$a7 = $value*0.07;
$a8 = $value*0.05;
$a9 = $value*0.04;
So as you can see, the first variables have more quantity than the later ones, but if you add these, it will be 1 which is 100%. So lets say I have total_numbers=20 then I'll have to re-write it and get a calculator and do it the hard way to accomplish my goal. Is there any way this can be done automatically with a function where I can just tell the total number and it can distribute it to proportions or something?
The first one will always be bigger than rest, then second one bigger than rest but smaller than first, third one being greater than rest but small than first and second, and so on.
function distributeValue($value, $num) {
$parts = $num * ($num + 1) / 2;
$values = [];
for ($i = $num; $i > 1; --$i) {
$values[] = round($value * $i / $parts);
}
$values[] = $value - array_sum($values);
return $values;
}
var_dump(distributeValue(10000, 9));
This works by calculating the $numth triangle number (the number you get by adding all the numbers from 1 to $num) and dividing the total value up into this number of parts.
It then starts by taking $num parts, then $num-1 parts and so on.
Since it's rounding the numbers, the last step is to take the total minus all the other values which is around one part. If you are fine with getting floats instead of ints out, then you can remove the $values[] = $value - array_sum($values); line and change the condition of the for loop to $i > 0.
I've searched through a number of similar questions, but unfortunately I haven't been able to find an answer to this problem. I hope someone can point me in the right direction.
I need to come up with a PHP function which will produce a random number within a set range and mean. The range, in my case, will always be 1 to 100. The mean could be anything within the range.
For example...
r = f(x)
where...
r = the resulting random number
x = the mean
...running this function in a loop should produce random values where the average of the resulting values should be very close to x. (The more times we loop the closer we get to x)
Running the function in a loop, assuming x = 10, should produce a curve similar to this:
+
+ +
+ +
+ +
+ +
Where the curve starts at 1, peeks at 10, and ends at 100.
Unfortunately, I'm not well versed in statistics. Perhaps someone can help me word this problem correctly to find a solution?
interesting question. I'll sum it up:
We need a funcion f(x)
f returns an integer
if we run f a million times the average of the integer is x(or very close at least)
I am sure there are several approaches, but this uses the binomial distribution: http://en.wikipedia.org/wiki/Binomial_distribution
Here is the code:
function f($x){
$min = 0;
$max = 100;
$curve = 1.1;
$mean = $x;
$precision = 5; //higher is more precise but slower
$dist = array();
$lastval = $precision;
$belowsize = $mean-$min;
$abovesize = $max-$mean;
$belowfactor = pow(pow($curve,50),1/$belowsize);
$left = 0;
for($i = $min; $i< $mean; $i++){
$dist[$i] = round($lastval*$belowfactor);
$lastval = $lastval*$belowfactor;
$left += $dist[$i];
}
$dist[$mean] = round($lastval*$belowfactor);
$abovefactor = pow($left,1/$abovesize);
for($i = $mean+1; $i <= $max; $i++){
$dist[$i] = round($left-$left/$abovefactor);
$left = $left/$abovefactor;
}
$map = array();
foreach ($dist as $int => $quantity) {
for ($x = 0; $x < $quantity; $x++) {
$map[] = $int;
}
}
shuffle($map);
return current($map);
}
You can test it out like this(worked for me):
$results = array();
for($i = 0;$i<100;$i++){
$results[] = f(20);
}
$average = array_sum($results) / count($results);
echo $average;
It gives a distribution curve that looks like this:
I'm not sure if I got what you mean, even if I didn't this is still a pretty neat snippet:
<?php
function array_avg($array) { // Returns the average (mean) of the numbers in an array
return array_sum($array)/count($array);
}
function randomFromMean($x, $min = 1, $max = 100, $leniency = 3) {
/*
$x The number that you want to get close to
$min The minimum number in the range
$max Self-explanatory
$leniency How far off of $x can the result be
*/
$res = [mt_rand($min,$max)];
while (true) {
$res_avg = array_avg($res);
if ($res_avg >= ($x - $leniency) && $res_avg <= ($x + $leniency)) {
return $res;
break;
}
else if ($res_avg > $x && $res_avg < $max) {
array_push($res,mt_rand($min, $x));
}
else if ($res_avg > $min && $res_avg < $x) {
array_push($res, mt_rand($x,$max));
}
}
}
$res = randomFromMean(22); // This function returns an array of random numbers that have a mean close to the first param.
?>
If you then var_dump($res), You get something like this:
array (size=4)
0 => int 18
1 => int 54
2 => int 22
3 => int 4
EDIT: Using a low value for $leniency (like 1 or 2) will result in huge arrays, since testing, I recommend a leniency of around 3.
How can I generate fix smaller random numbers from a large number. Addition of these smaller numbers must be equal to large number. Suppose I want to generate 400 random number and addition of these smaller number = e.g. 1,000,000. every number should be unique and have any value assign to it. Like Number 1=1000 and number 2 may contain only 5. But total of all the number must be a large number. Is there any algorithm to do this kind of operation in php?
function array_generate_sum($n, $total)
{
$sum = 0;
$arr = array();
for( ; $n >= 0; $n--)
{
$current = $n == 0 ? $total - $sum : mt_rand(1, $total - $sum - $n);
$sum += $current;
$arr[] = $current;
}
return $arr;
}
// Generate an array of 5 values whose sum is 30
array_generate_sum(5, 30);
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)));