Getting specific numbers in for loop - php
I want to get 3 int values in a for loop connected with each other. What I basically want is this:
000
001
002
003
010
011
...
323
330
331
I want to have each of the 3 numbers in a variable starting from 0 and max 3 when it gets higher than 3 it increases the number left to it by 1
For example
for($i = 0; $i <= $array; $i++){
echo $a . $b . $c . "<br />"; //Output would be the example I showed above
}
You can use base_convert to convert between arbitrary bases, in your case base 4.
for($i=0, $max = base_convert(333, 4, 10); $i < $max; ++$i) {
echo base_convert($i, 10 , 4);
}
To get 0-padded output, use printf with a format specifier:
printf('%03d', base_convert($i, 10 , 4));
for($i = 0; $i <= 63; $i++){
$c = $i % 4;
$b = ($i - $c)/4 % 4;
$a = (($i - $c)/4 - $b)/4;
echo $a . $b . $c . "<br />"; //Output would be the example I showed above
}
It's really just an explicit version of knittl's answer. The for loop has us step through every number from 0 thru 63, which happens to correspond with 0 through 333 in base 4. We then take $i, which is in base 10, and convert it to base 4 step by step. The % is the modulo operator - it returns the remainder after division. The least significant char is simply the remainder of $i/4, so we save that as $c. The next char is the 4 place (like the 10 place in base 10). So we subtract $c, which is already accounted for and divide by 4 and do the same thing.
for($a = 0; $a <= 3; $a++){
for($b = 0; $b <= 3; $b++){
for($c = 0; $c <= 3; $c++){
echo $a . $b . $c . "<br />";
}
}
}
Try this:
$count = 20; //how many numbers do you want
for($i =0; $i<$count; $i++) {
echo str_pad(base_convert($i, 10, 4),3,'0' , STR_PAD_LEFT) . '<br/>';
}
base_convert() converts every $i value from 10 base to 4.
str_pad() fill it with '0' to given length (3 here).
STR_PAD_LEFT means that zeros should be added on left side.
Related
Can someone help me return a value?
I have created a loop which returns a random number between two values. Cool.
But now I want the script to return the following value too: The number of unique numbers between two similar numbers.
Example:
4
5
8
22
45
3
85
44
4
55
15
23
As you see there is a double which is the four and there are 7 numbers inbetween. So I would like the script to echo these numbers two so in this case it should echo 7 but if there are more doubles in the list it should echo all the numbers between certain doubles.
This is what I have:
for ($x = 0; $x <= 100; $x++) {
$min=0;
$max=50;
echo rand($min,$max);
echo "<br>";
}
Can someone help me or guide me? I'm learning :)
Thanks!
So You need to seperate script for three parts:
getting randoms and save them to array (name it 'result'),
analyze them,
print (echo) results
Simply - instead of printing every step of loop, save them to array(), exit loop, analyze every item with other, example:
take i element of list
check is i+j element is the same
if is it the same - save j-i to second array() (name it 'ranges')
And after this, print two arrays (named by me as 'result' and 'ranges')
UPDATE:
Here's solution, hope You enjoy:
$result = array(); #variable is set as array object
$ranges = array(); #same
# 1st part - collecting random numbers
for ($x = 0; $x < 20; $x++)
{
$min=0;
$max=50;
$result[] = rand($min,$max); #here's putting random number to array
}
$result_size = count($result); #variable which is containg size of $result array
# 2nd part - getting ranges between values
for ($i = 0; $i < $result_size; $i++)
{
for ($j = 0; $j < $result_size; $j++)
{
if($i == $j) continue; # we don't want to compare numbers with itself,so miss it and continue
else if($result[$i] == $result[$j])
{
$range = $i - $j; # get range beetwen numbers
if($range > 0 ) # this is for miss double results like 14 and -14 for same comparing
{
$ranges[$result[$i]] = $range;
}
}
}
}
#3rd part - priting results
echo("RANDOM NUMBERS:<br>");
foreach($result as $number)
{
echo ("$number ");
}
echo("<br><br>RANGES BETWEEN SAME VALUES:<br>");
foreach($ranges as $number => $range)
{
echo ("For numbers: $number range is: $range<br>");
}
Here's sample of echo ($x is set as 20):
RANDOM NUMBERS:
6 40 6 29 43 32 17 44 48 21 40 2 33 47 42 3 22 26 39 46
RANGES BETWEEN SAME VALUES:
For numbers: 6 range is: 2
For numbers: 40 range is: 9
Here is your fish:
Put the rand into an array $list = array(); and $list[] = rand($min,$max); then process the array with two for loops.
$min=0;
$max=50;
$list = array();
for ($x = 0; $x <= 100; ++$x) {
$list[] = rand($min,$max);
}
print "<pre>";print_r($list);print "</pre>";
$ranges = array();
$count = count($list);
for ($i = 0; $i < $count; ++$i) {
$a = $list[$i];
for ($j = $i+1; $j < $count; ++$j) {
$b = $list[$j];
if($a == $b) {
$ranges[] = $j-$i;
}
}
}
print "<pre>";print_r($ranges);print "</pre>";
generating same number with the given length
I have this math assignment that I should make into code.
I've tried all I thought of but I couldn't find a solution.
All this should be done without using php functions, only math operations.
You can use while, for, and such...
So I have number for example 9
Now I should create number of the length 9 which would be 999999999
If I had, for example, number 3, then the result should be 333.
Any ideas?
$gen = -1;
while($highest > 0) {
$gen = $highest + ($highest * 10);
$highest = $highest - 1;
}
echo $gen;
Here is a method that does not build a string; it uses pure math. (There will be many, many ways to do this task)
$x=9;
$result=0;
for($i=$x; $i; --$i){ // this looping expression can be structured however you wish potato-potatoe
$result+=$x*(10**($i-1)); // x times (10 to the power of (i-1))
}
echo $result;
// 999999999
*note: ** acts like pow() if you want to look it up.
Late edit: here is a clever, little loopless method (quietly proud). I am only calling range() and foreach() to demo; it is not an integral component of my method.
Demo: https://3v4l.org/GIjfG
foreach(range(0,9) as $n){
// echo "$n -> ",(integer)(1/9*$n*(10**$n)-($n/10)),"\n";
// echo "$n -> ",(1/9*$n*(10**$n)-(1/9*$n)),"\n";
// echo "$n -> ",(int)(1/9*10**$n)*$n,"\n";
// echo "$n -> ",(int)(10**$n/9)*$n,"\n";
echo "$n -> ",(10**$n-1)/9*$n,"\n";
}
Output:
0 -> 0
1 -> 1
2 -> 22
3 -> 333
4 -> 4444
5 -> 55555
6 -> 666666
7 -> 7777777
8 -> 88888888
9 -> 999999999
1/9 is the hero of this method because it generates .111111111(repeating). From this float number, I am using 10**$n to "shift" just enough 1s to the left side of the decimal point, then multiplying this float number by $n, then the float must be converted to an integer to complete.
Per #axiac's comment, the new hero is 10**$n-1 which generates a series of nines to the desired length (no float numbers). Next divide the nines by nine to generate a series of ones which becomes the perfect multiplier. Finally, multiply the series of ones and the input number to arrive at the desired output.
There are two operations you need to accomplish:
given a number $number, append the digit $n to it;
repeat operation #1 some number of times ($n times).
Operation #1 is easy:
$number = $number * 10 + $n;
Operation #2 is even easier:
for ($i = 0; $i < $n; $i ++)
What else do you need?
Initialization of the variable used to store the computed number:
$number = 0;
Put them in order and you get:
// The input digit
// It gives the length of the computed number
// and also its digits
$n = 8;
// The number we compute
$number = 0;
// Put the digit $n at the end of $number, $n times
for ($i = 0; $i < $n; $i ++) {
$number = $number * 10 + $n;
}
// That's all
If intval() is accepted:
$result = '';
$input = 9;
for($i=0; $i < $input; $i++){
$result .= $input;
}
$result = intval($result);
else:
$result = 0;
$input = 9;
for($i=0; $i < $input; $i++){
$factor = 1;
for($j = 0; $j < $i; $j++){
$factor *= 10;
}
$result += $input * $factor;
}
=>
9 + 90 + 900 + 9000 + 90000...
PHP: for loop $i value turns zero if the number is greater
I want to turn the $i variable value to start counting from 1 if the given value is greater than 10:
here is what i am trying to achieve
<?php
$givenValue = 15; //number of x value
for ($i = 1; $i < $givenValue; $i++) {
if ($givenValue > 10){
$i = 1;
}
echo $i."<br>";
}
?>
This is how i want my result to look like
output: 1
output: 2
output: 3
output: 4
output: 5
output: 6
output: 7
output: 8
output: 9
output: 10
output: 1
output: 2
output: 3
output: 4
output: 5
in for loop body
Any help is welcome
You can use modulo calculation to get the result you want.
I also changed your if from $givenvalue to $i as $givenvalue will "always" be 10+.
$givenValue = 15; //number of x value
for ($i = 1; $i <= $givenValue; $i++) {
if ($i > 10){
Echo $i%10 . "\n";
}else{
echo $i . "\n";
}
}
https://3v4l.org/5afc5
Another option, if that is possible for you, is to start at zero and only use modulo calculation and add one to it to get the same result.
This also means I need to stop the loop at <$givenvalue as your original code shows.
$givenValue = 15; //number of x value
for ($i = 0; $i < $givenValue; $i++) {
Echo $i%10+1 . "\n";
}
https://3v4l.org/r0sgA
A method that uses less looping is to add 10 to the loop on each iteration and create the values using range().
Then add them to the array with array_merge, and output with implode.
$givenValue = 47; //number of x value
$breakpoint = 10;
$arr=[];
For($i = $breakpoint; $i< $givenValue;){
// Add new values from 1-$breakpoint in array
$arr = array_merge($arr, range(1,$breakpoint));
$i +=$breakpoint;
}
// Loop will exit before all values been collected
// Add the rest of the values
$arr = array_merge($arr, range(1,$givenValue-($i-10)));
// Echo the values in array
Echo implode("\n", $arr);
https://3v4l.org/jGsO4
Your code can be written like this:
<?php
$givenValue = 15; //number of x value
for ($i = 1; $i <= $givenValue; $i++)
{
if ($i > 10)
{
$i = 1;
$givenValue-=10;
}
echo "output: $i\n";
}
?>
http://sandbox.onlinephpfunctions.com/code/ed34d8dcd12a9a5a866b73338ad1209f55298519
You are resenting the counter, I would expect the behaviour you have. To do what you want add another counter to the mix
$j=1;
$givenValue = 15; //number of x value
for ($i = 1; $i <= $givenValue; $i++) {
if ($j > 10){
$j = 1;
}
echo $j."\n";
++$j;
}
You also had several missing ;
Output:
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
If you want to end on 5 you have to do 16 as the $givenValue or change it to <= less than or equal
See it here live
See what I have now, the $i variable counts to the $givenValue then the $j variable counts along side it, but with a range of 1-10 ( resets to 1 after 10 )
Decoding function to better understand
I am revisiting PHP and want to relearn where i lack and i found one problem, i am unable to understand the following code, where as it should output 6 according to the quiz, i got it from but i broke it down to simple pieces and commented out to better understand, according to me the value of $sum should be 4, but what i am doing wrong, maybe my breakdown is wrong?
$numbers = array(1,2,3,4);
$total = count($numbers);
//$total = 4
$sum = 0;
$output = "";
$i = 0;
foreach($numbers as $number) {
$i = $i + 1;
//0+1 = 1
//0+2 = 2
//0+3 = 3
//0+4 = 4
if ($i < $total) {
$sum = $sum + $number;
//1st time loop = 0 < 4 false
//2nd time loop = 0 < 1 false
//3rd time loop = 0 < 2 false
//5th time loop = 0 < 3 false
//6th time loop = 4 = 4 true
//$sum + $number
//0 + 4
//4
}
}
echo $sum;
This is very basic question and might get down vote but it is also a strong backbone for people who want to become PHP developer.
You don't understand the last part in the loop. It actually goes like this now:
if($i < $total) {
$sum = $sum + $number;
//1st time loop: $sum is 0. $sum + 1 = 1. $sum is now 1.
//2nd time loop: $sum is 1. $sum + 2 = 3. $sum is now 3.
//3rd time loop: $sum is 3. $sum + 3 = 6. $sum is now 6.
//4th time loop: it doesn't get here. $i (4) < $total (4)
//This is false, so it doesn't execute this block.
}
echo $sum; // Output: 6
I altered your script a little so that it will print out what it's doing as it goes. I find it useful to do this kind of thing if I'm having a hard time thinking through a problem.
$numbers = array(1,2,3,4);
$total = count($numbers);
$sum = 0;
$i = 0;
$j = 0;
foreach($numbers as $number) {
$i = $i + 1;
echo "Iteration $j: \$i +1 is $i, \$sum is $sum, \$number is $number";
if ($i < $total) {
$sum = $sum + $number;
echo ", \$i is less than \$total ($total), so \$sum + \$number is: $sum";
} else {
echo ", \$i is not less than \$total ($total), so \$sum will not be increased.";
}
echo '<br>'; // or a new line if it's CLI
$j++;
}
echo $sum;
Lets Explain
Your initial value of $i is 0 but when you start looping you increment it by 1, so the start value of $i is 1.
When checking the condition you did't use equal sign to check for the last value whether you start value is 1. So its clear that your loop must be run for 1 less of total.
$i = 0;
foreach($numbers as $number) {
$i += 1;
if ($i < $total)
$sum += $number;
}
echo $sum;
Analysis
Step: 1 / 4
The value of $number is: 1 And The value of $i is: 1
Step: 2 / 4
The value of $number is: 2 And The value of $i is: 2
Step: 3 / 4
The value of $number is: 3 And The value of $i is: 3
When the loop again go for a check the value of $i increased by 1 and at 4. So trying to match the condition if ($i < $total), where the value of $i and $total is equal, so it will return false. So the loop only run for 3 time.
Result
6
how to find the sum of all the multiples of 3 or 5 below 1000 in php, issue?
i have an small issue with the way this problem is resolved.
some would say: println((0 /: ((0 until 1000).filter(x => x % 3 == 0 || x % 5 == 0))) (_+_)) is the solution witch adds to 233168
my way was to do:
$maxnumber = 1000;
for ($i = 3; $i < $maxnumber; $i += 3)
{
$t += $i;
echo $i.',';
}
echo '<br>';
for ($j = 5; $j < $maxnumber; $j += 5)
{
$d += $j;
echo $j.',';
}
echo '<br>';
echo $t;
echo '<br>';
echo $d;
echo '<br>';
echo $t+$d;
this will give me :
3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,102,105,108,111,114,117,120,123,126,129,132,135,138,141,144,147,150,153,156,159,162,165,168,171,174,177,180,183,186,189,192,195,198,201,204,207,210,213,216,219,222,225,228,231,234,237,240,243,246,249,252,255,258,261,264,267,270,273,276,279,282,285,288,291,294,297,300,303,306,309,312,315,318,321,324,327,330,333,336,339,342,345,348,351,354,357,360,363,366,369,372,375,378,381,384,387,390,393,396,399,402,405,408,411,414,417,420,423,426,429,432,435,438,441,444,447,450,453,456,459,462,465,468,471,474,477,480,483,486,489,492,495,498,501,504,507,510,513,516,519,522,525,528,531,534,537,540,543,546,549,552,555,558,561,564,567,570,573,576,579,582,585,588,591,594,597,600,603,606,609,612,615,618,621,624,627,630,633,636,639,642,645,648,651,654,657,660,663,666,669,672,675,678,681,684,687,690,693,696,699,702,705,708,711,714,717,720,723,726,729,732,735,738,741,744,747,750,753,756,759,762,765,768,771,774,777,780,783,786,789,792,795,798,801,804,807,810,813,816,819,822,825,828,831,834,837,840,843,846,849,852,855,858,861,864,867,870,873,876,879,882,885,888,891,894,897,900,903,906,909,912,915,918,921,924,927,930,933,936,939,942,945,948,951,954,957,960,963,966,969,972,975,978,981,984,987,990,993,996,999
5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120,125,130,135,140,145,150,155,160,165,170,175,180,185,190,195,200,205,210,215,220,225,230,235,240,245,250,255,260,265,270,275,280,285,290,295,300,305,310,315,320,325,330,335,340,345,350,355,360,365,370,375,380,385,390,395,400,405,410,415,420,425,430,435,440,445,450,455,460,465,470,475,480,485,490,495,500,505,510,515,520,525,530,535,540,545,550,555,560,565,570,575,580,585,590,595,600,605,610,615,620,625,630,635,640,645,650,655,660,665,670,675,680,685,690,695,700,705,710,715,720,725,730,735,740,745,750,755,760,765,770,775,780,785,790,795,800,805,810,815,820,825,830,835,840,845,850,855,860,865,870,875,880,885,890,895,900,905,910,915,920,925,930,935,940,945,950,955,960,965,970,975,980,985,990,995
$t - 166833
$d - 99500
and total:
266333
why am i wrong?
Some numbers are multiples of both 3 and 5. (Your algorithm adds these numbers to the total twice.)
Because 6 * 5 == 30 and 10 * 3 == 30, you're adding the some numbers up twice.
$sum = 0;
$i = 0;
foreach(range(0, 999) as $i) {
if($i % 3 == 0 || $i % 5 == 0) $sum += $i;
}
Because you double-count numbers that are multiple of both 3 and 5, i.e. multiples of 15.
You can account for this naively by subtracting all multiples of 15.
for ($j = 15; $j < $maxnumber; $j += 15)
{
$e += $j;
echo $j.',';
}
$total = $total - $d;
In your case, if it is 15, you will add the number twice.
Try this:
$t = 0;
$d = 0;
for ($i = 0; $i <= $maxnumber; $i++){
if ($i % 3 == 0)
$t+= $i;
else if ($i % 5 == 0)
$d += $i;
}
echo $t.'<br>'.$d;
I think that in your code, if a number is a multiple of 3 and 5, it is added twice. Take 15 for example. It's in your list of multiples of 3 and in the list of multiples of 5. Is this the behaviour you want?
One of the best approach to this solution (to achieve optimum time complexity), run an Arithmetic Progression series and find the number of terms in all series by using AP formula: T=a+(n-1)d, then find sum by : S=n/2[2*a+(n-1)d]
where : a=first term ,n=no. of term , d=common deference, T=nth term
The code solution below has been implemented to suit the question above - so the values 3 and 5 are hard-coded. However, the function can modified such that values are passed in as variable parameters.
function solution($number){
$val1 = 3;
$val2 = 5;
$common_term = $val1 * $val2;
$sum_of_terms1 = calculateSumofMulitples($val1,$number);
$sum_of_terms2 = calculateSumofMulitples($val2,$number);
$sum_of_cterms = calculateSumofMulitples($common_term,$number);
$final_result = $sum_of_terms1 + $sum_of_terms2 - $sum_of_cterms;
return $final_result;
}
function calculateSumofMulitples($val, $number)
{
//first, we begin by running an aithmetic prograssion for $val up to $number say N to get the no of terms [using T=a +(n-1)d]
$no_of_terms = (int) ($number / $val);
if($number % $val == 0) $no_of_terms = (int) ( ($number-1)/$val ); //since we are computing for a no of terms below N and not up to/exactly/up to N. if N divides val with no remainder, use no of terms = (N-1)/val
//second, we compute the sum of terms [using Sn = n/2[2a + (n-1)d]
$sum_of_terms = ($no_of_terms * 0.5) * ( (2*$val) + ($no_of_terms - 1) * $val );
// sum of multiples
return $sum_of_terms;
}
You can run a single loop checking whether the number is multiple of 3 OR 5:
for ($i = 0; $i < $maxnumber; $i++)
{
if($i%3 || $i%5){
$t += $i;
echo $i.',';}
}
I think the original code is not including numbers which are multiples of both 3 and 5 in the total: if the test for multiple of 3 matches, it takes that and goes on. If you total the multiples of 15 up to 1000, you get 33165, which is exactly the difference between your total, 266333, and the original total, 233168.
Here's my solution to the question:
<?php
$sum = 0;
$arr = [];
for($i = 1; $i < 1000; $i++){
if((int)$i % 3 === 0 || (int)$i % 5 === 0)
{
$sum += $i;
array_push($arr,$i);
}
}
echo $sum;
echo '<br>';
print_r($arr);//Displays the values meeting the criteria as an array of values