I have below a function (from a previous question that went unanswered) that creates an array with n amount of values. The sum of the array is equal to $max.
function randomDistinctPartition($n, $max) {
$partition= array();
for ($i = 1; $i < $n; $i++) {
$maxSingleNumber = $max - $n;
$partition[] = $number = rand(1, $maxSingleNumber);
$max -= $number;
}
$partition[] = $max;
return $partition;
}
For example: If I set $n = 4 and $max = 30. Then I should get the following.
array(5, 7, 10, 8);
However, this function does not take into account duplicates and 0s. What I would like - and have been trying to accomplish - is to generate an array with unique numbers that add up to my predetermined variable $max. No Duplicate numbers and No 0 and/or negative integers.
Ok, this problem actually revolves around linear sequences. With a minimum value of 1 consider the sequence:
f(n) = 1 + 2 + ... + n - 1 + n
The sum of such a sequence is equal to:
f(n) = n * (n + 1) / 2
so for n = 4, as an example, the sum is 10. That means if you're selecting 4 different numbers the minimum total with no zeroes and no negatives is 10. Now go in reverse: if you have a total of 10 and 4 numbers then there is only one combination of (1,2,3,4).
So first you need to check if your total is at least as high as this lower bound. If it is less there is no combination. If it is equal, there is precisely one combination. If it is higher it gets more complicated.
Now imagine your constraints are a total of 12 with 4 numbers. We've established that f(4) = 10. But what if the first (lowest) number is 2?
2 + 3 + 4 + 5 = 14
So the first number can't be higher than 1. You know your first number. Now you generate a sequence of 3 numbers with a total of 11 (being 12 - 1).
1 + 2 + 3 = 6
2 + 3 + 4 = 9
3 + 4 + 5 = 12
The second number has to be 2 because it can't be one. It can't be 3 because the minimum sum of three numbers starting with 3 is 12 and we have to add to 11.
Now we find two numbers that add up to 9 (12 - 1 - 2) with 3 being the lowest possible.
3 + 4 = 7
4 + 5 = 9
The third number can be 3 or 4. With the third number found the last is fixed. The two possible combinations are:
1, 2, 3, 6
1, 2, 4, 5
You can turn this into a general algorithm. Consider this recursive implementation:
$all = all_sequences(14, 4);
echo "\nAll sequences:\n\n";
foreach ($all as $arr) {
echo implode(', ', $arr) . "\n";
}
function all_sequences($total, $num, $start = 1) {
if ($num == 1) {
return array($total);
}
$max = lowest_maximum($start, $num);
$limit = (int)(($total - $max) / $num) + $start;
$ret = array();
if ($num == 2) {
for ($i = $start; $i <= $limit; $i++) {
$ret[] = array($i, $total - $i);
}
} else {
for ($i = $start; $i <= $limit; $i++) {
$sub = all_sequences($total - $i, $num - 1, $i + 1);
foreach ($sub as $arr) {
array_unshift($arr, $i);
$ret[] = $arr;
}
}
}
return $ret;
}
function lowest_maximum($start, $num) {
return sum_linear($num) + ($start - 1) * $num;
}
function sum_linear($num) {
return ($num + 1) * $num / 2;
}
Output:
All sequences:
1, 2, 3, 8
1, 2, 4, 7
1, 2, 5, 6
1, 3, 4, 6
2, 3, 4, 5
One implementation of this would be to get all the sequences and select one at random. This has the advantage of equally weighting all possible combinations, which may or may not be useful or necessary to what you're doing.
That will become unwieldy with large totals or large numbers of elements, in which case the above algorithm can be modified to return a random element in the range from $start to $limit instead of every value.
I would use 'area under triangle' formula... like cletus(!?)
Im really gonna have to start paying more attention to things...
Anyway, i think this solution is pretty elegant now, it applies the desired minimum spacing between all elements, evenly, scales the gaps (distribution) evenly to maintain the original sum and does the job non-recursively (except for the sort):
Given an array a() of random numbers of length n
Generate a sort index s()
and work on the sorted intervals a(s(0))-a(s(1)), a(s(1))-a(s(2)) etc
increase each interval by the
desired minimum separation size eg 1
(this necessarily warps their
'randomness')
decrease each interval by a factor
calculated to restore the series sum
to what it is without the added
spacing.
If we add 1 to each of a series we increase the series sum by 1 * len
1 added to each of series intervals increases sum by:
len*(len+1)/2 //( ?pascal's triangle )
Draft code:
$series($length); //the input sequence
$seriesum=sum($series); //its sum
$minsepa=1; //minimum separation
$sorti=sort_index_of($series) //sorted index - php haz function?
$sepsum=$minsepa*($length*($length+1))/2;
//sum of extra separation
$unsepfactor100=($seriesum*100)/($seriesum+sepsum);
//scale factor for original separation to maintain size
//(*100~ for integer arithmetic)
$px=series($sorti(0)); //for loop needs the value of prev serie
for($x=1 ; $x < length; $x++)
{ $tx=$series($sorti($x)); //val of serie to
$series($sorti($x))= ($minsepa*$x) //adjust relative to prev
+ $px
+ (($tx-$px)*$unsepfactor100)/100;
$px=$tx; //store for next iteration
}
all intervals are reduced by a
constant (non-random-warping-factor)
separation can be set to values other
than one
implementantions need to be carefuly
tweaked (i usualy test&'calibrate')
to accomodate rounding errors.
Probably scale everything up by ~15
then back down after. Intervals should survive if done right.
After sort index is generated, shuffle the order of indexes to duplicate values to avoid runs in the sequence of collided series.
( or just shuffle final output if order never mattered )
Shuffle indexes of dupes:
for($x=1; $x<$len; $x++)
{ if ($series($srt($x))==$series($srt($x-1)))
{ if( random(0,1) )
{ $sw= $srt($x);
$srt($x)= $srt($x-1);
$srt($x-1)= $sw;
} } }
A kind of minimal disturbance can be done to a 'random sequence' by just parting dupes by the minimum required, rather than moving them more than minimum -some 'random' amount that was sought by the question.
The code here separates every element by the min separation, whether duplicate or not, that should be kindof evenhanded, but overdone maybe. The code could be modified to only separate the dupes by looking through the series(sorti(n0:n1..len)) for them and calculating sepsum as +=minsep*(len-n) for each dupe. Then the adjustment loop just has to test again for dupe before applying adjustment.
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
The following is a programming task.
You are given a sequence of N integers. The task is to find the number of continuous sequences of integers such that their sum is zero.
For example if the sequence is:
2, -2, 6, -6, 8
There are 3 such sequences:
'2, -2'
'6, -6'
'2, -2, 6, -6'
I already have the following program written in PHP that reads the input from STDIN (first line containing the number of integers that follow.)
<?php
$n = fgets(STDIN) * 1;
$seq = array();
for ($i = 0; $i < $n; $i++) {
$seq[] = fgets( STDIN ) * 1;
}
$count = 0;
for( $i = 0; $i < $n; $i++)
{
$number = 0;
for( $j = $i; $j < $n; $j++)
{
$number += $seq[$j];
if( $number == 0 )
$count++;
}
}
echo 'count: ' . $count . PHP_EOL;
Input example
5
2
-2
6
-6
8
This works well for smaller sequences, but its efficiency is O(n^2).
What algorithm is appropriate - with possibly O(n) efficiency - for a sequence containing 100.000 integers?
Let's assume your data is stored in an array, and let it be arr.
Create an array sum, such that:
sum[i] = arr[0] + arr[1] + ... + arr[i]
And, in addition a single entry at the beginning with 0 (to handle a subarray that starts at the beginning and sums to zero)
Now, it is easy to see that for each two indices i,j such that i<j and sum[i]=sum[j], the continuous sequences arr[i+1]+arr[i+2]+...+arr[j] = 0.
By creating this array sum, you only have left to find how many duplicates are there. This cannot be done in O(n)1 (this is the element distinctness problem), but can be solved in O(nlogn) using sorting and then iterating and counting, which is still very fast for 100,000 entries.
Note, that if there are for example n duplicates of the number k in the array sum, there are Choose(n,2) = n(n-1)/2 continuous subsequences that are generated for these duplicates.
Example:
arr = [1,2,-2,5,6,-6,-5,8]
sum = [0,1,3,1,6,12,6,1,9]
sorted(sum) = [0,1,1,1,3,6,6,9,12]
There are 3 duplicates of 1 and 2 duplicates of 6, so you have total of:
Choose(3,2) + Choose(2,2) = 3*2/2 + 2/2 = 3+1 = 4
Which indeed match the 4 subsequences:
2,-2
2,-2,5,6,-6,-5
6,-6
5,6,-6,-5
(1) Without hashing, and then you will decay to O(n^2) worst case, but will benefit from O(n) average case, at the cost of O(n) extra space.
Since I can't reply to comments, this is a reply to amit's answer.
Maybe I have something wrong, but when applying your method to the original test case, we don't get the right answer:
input = [2, -2, 6, -6, 8]
sum = [2, 0, 6, 0, 8]
sorted(sum) = [0, 0, 2, 6, 8]
Since there are 2 duplicates of the number 0, this gives us (2*1)/2=1, which is not correct (correct answer would be 3).
What am I missing? Thanks
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($result);
I would like to draw a random number from the interval 1,49 but I would like to add a number as an exception ( let's say 44 ) , I cannot use round(rand(1,49)) .So I decided to make an array of 49 numbers ( 1-49) , unset[$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.
I'm fairly new to PHP - programming in general. So basically what I need to accomplish is, create an array of x amount of numbers (created randomly) whose value add up to n:
Let's say, I have to create 4 numbers that add up to 30. I just need the first random dataset. The 4 and 30 here are variables which will be set by the user.
Essentially something like
x = amount of numbers;
n = sum of all x's combined;
// create x random numbers which all add up to n;
$row = array(5, 7, 10, 8) // these add up to 30
Also, no duplicates are allowed and all numbers have to be positive integers.
I need the values within an array. I have been messing around with it sometime, however, my knowledge is fairly limited. Any help will be greatly appreciated.
First off, this is a really cool problem. I'm almost sure that my approach doesn't even distribute the numbers perfectly, but it should be better than some of the other approaches here.
I decided to build the array from the lowest number up (and shuffle them at the end). This allows me to always choose a random range that will allows yield valid results. Since the numbers must always be increasing, I solved for the highest possible number that ensures that a valid solution still exists (ie, if n=4 and max=31, if the first number was picked to be 7, then it wouldn't be possible to pick numbers greater than 7 such that the sum of 4 numbers would be equal to 31).
$n = 4;
$max = 31;
$array = array();
$current_min = 1;
while( $n > 1 ) {
//solve for the highest possible number that would allow for $n many random numbers
$current_max = floor( ($max/$n) - (($n-1)/2) );
if( $current_max < $current_min ) throw new Exception( "Can't use combination" );
$new_rand = rand( $current_min, $current_max ); //get a new rand
$max -= $new_rand; //drop the max
$current_min = $new_rand + 1; //bump up the new min
$n--; //drop the n
$array[] = $new_rand; //add rand to array
}
$array[] = $max; //we know what the last element must be
shuffle( $array );
EDIT: For large values of $n you'll end up with a lot of grouped values towards the end of the array, since there is a good chance you will get a random value near the max value forcing the rest to be very close together. A possible fix is to have a weighted rand, but that's beyond me.
I'm not sure whether I understood you correctly, but try this:
$n = 4;
$max = 30;
$array = array();
do {
$random = mt_rand(0, $max);
if (!in_array($random, $array)) {
$array[] = $random;
$n--;
}
} while (n > 0);
sorry i missed 'no duplicates' too
-so need to tack on a 'deduplicator' ...i put it in the other question
To generate a series of random numbers with a fixed sum:
make a series of random numbers (of largest practical magnitude to hide granularity...)
calculate their sum
multiply each in series by desiredsum/sum
(basicaly to scale a random series to its new size)
Then there is rounding error to adjust for:
recalculate sum and its difference
from desired sum
add the sumdiff to a random element
in series if it doesnt result in a
negative, if it does loop to another
random element until fine.
to be ultratight instead add or
subtract 1 bit to random elements
until sumdiff=0
Some non-randomness resulting from doing it like this is if the magnitude of the source randoms is too small causing granularity in the result.
I dont have php, but here's a shot -
$n = ; //size of array
$targsum = ; //target sum
$ceiling = 0x3fff; //biggish number for rands
$sizedrands = array();
$firstsum=0;
$finsum=0;
//make rands, sum size
for( $count=$n; $count>0; $count--)
{ $arand=rand( 0, $ceiling );
$sizedrands($count)=$arand;
$firstsum+=$arand; }
//resize, sum resize
for( $count=$n; $count>0; $count--)
{ $sizedrands($count)=($sizedrands($count)*$targsum)/$firstsum;
$finsum+=$sizedrands($count);
}
//redistribute parts of rounding error randomly until done
$roundup=$targsum-$finsum;
$rounder=1; if($roundup<0){ $rounder=-1; }
while( $roundup!=0 )
{ $arand=rand( 0, $n );
if( ($rounder+$sizedrands($arand) ) > 0 )
{ $sizedrands($arand)+=$rounder;
$roundup-=$rounder; }
}
Hope this will help you more....
Approch-1
$aRandomarray = array();
for($i=0;$i<100;$i++)
{
$iRandomValue = mt_rand(1000, 999);
if (!in_array($iRandomValue , $aRandomarray)) {
$aRandomarray[$i] = $iRandomValue;
}
}
Approch-2
$aRandomarray = array();
for($i=0;$i<100;$i++)
{
$iRandomValue = mt_rand(100, 999);
$sRandom .= $iRandomValue;
}
array_push($aRandomarray, $sRandom);