PHP Simple Calculation with an Array

Question

So i have a simple for loop to get this result from any given number (get).
1 + 2 + 3 + 4 = 10

$num = intval($_GET["number"]);
$total = 0;

for ($i = 1; $i <= $num; $i++) {

    echo $i;

    if ($i != $num) {
        echo " + ";
    } 
    $total += $i;
}
    echo " = " . $total;

Now I want to show the calculation of every step
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
And it should be done with an Array, but I can't seem to figure out the Algorithm. I think I'm overlooking something simple here.

Add the value $i to an array every iteration. Then you want to print the calculation with implode() and get the result with array_sum(). — Rizier123
– Rizier123, Commented Sep 22, 2016 at 16:38
You may be interested in the range() function. Doing $nums = range(1, $num); will give you an array of ints. — gen_Eric
– gen_Eric, Commented Sep 22, 2016 at 16:43

Striezel · Accepted Answer · 2016-09-22 16:55:36Z

1

Try something like this:

<?php
$num = intval($_GET["number"]);

//add all numbers to an array
$numbers = array();
for ($i = 1; $i <= $num; $i++)
{
  $numbers[] = $i;
  //show each array element with ' + ' in between the elements
  echo implode(' + ', $numbers);

  //show total sum
  echo " = " . array_sum($numbers) . "\n";
}
?>

Note that this does not work, if $_GET['number'] is zero or even below zero.

edited Sep 22, 2016 at 16:55

answered Sep 22, 2016 at 16:48

Striezel

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5 Comments

Zac Crites Over a year ago

You shouldn't close <?php tags since it can can cause whitespace to be accidentally written to the output buffer.

Zac Crites Over a year ago

Also your arguments to implode are in the wrong order.

Striezel Over a year ago

@ZacCrites: Well, it does not matter for that example, and on the other hand it feels a little bit weird to open a <?php tag but never close it.

Striezel Over a year ago

As for the arguments: According to the PHP documentation you are only half right. implode() accepts arguments in both orders, and that is why the script also worked as intended when I tested it. Anyway, I've changed the order to avoid confusion.

Zac Crites Over a year ago

stackoverflow.com/questions/4410704/…

Martavis Griffin · Accepted Answer · 2016-09-22 16:54:17Z

1

Here's the simplest way I can think...

$num = intval($_GET['number']);
$intArray = range(1,$num);

echo implode(" + ",$intArray)." = ".array_sum($intArray);

answered Sep 22, 2016 at 16:54

Martavis Griffin

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Comments

Sherif · Accepted Answer · 2016-09-22 17:27:51Z

1

You don't actually need a loop to do an arithmetic progression. An arithmetic progression like this can be calculated in constant time with the formula n * (n[-1] + n[1]) / 2.

For example the progression of 4, where n1 = 1, n2 = 2, n3 = 3, and n4 = 4 is simply 4 * (4 + 1) / 2 == 10.

function progression($n) {
    return $n * ($n + 1) / 2;
}

echo progression(4); // 10

However, to show the result of the progression at any given step you simply limit the upper-bound of that progression (i.e. $n).

$n = 4;
for ($i = 1; $i <= $n; $i++) {
    $operands = implode('+', range(1, $i));
    echo $operands . " = " . progression($i), "\n";
}

output

1 = 1
1+2 = 3
1+2+3 = 6
1+2+3+4 = 10

Generalization

This works for any linear arithmetic progression, regardless of the upper/lower bound. So for example the progression of 5 through 8 is still 4 * (5 + 8) / 2 which gives you 26.

So you can modify this function to a more general solution for any linear arithmetic progression as such.

function progression($size, $start = 1) {
    return $size * ($start + ($size + $start - 1)) / 2;
}

$n = 4;
$start = 5;
for ($i = $start; $i <= $n + $start - 1; $i++) {
    $operands = implode('+', range($start, $i));
    echo $operands . " = " . progression($i - $start + 1, $start), "\n";
}

output

5 = 5
5+6 = 11
5+6+7 = 18
5+6+7+8 = 26

edited Sep 22, 2016 at 17:27

answered Sep 22, 2016 at 16:51

Sherif

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2 Comments

Michiel Nuyts Over a year ago

Great explanation! Thanks

Sherif Over a year ago

@MichielNuyts Sure, no problem. If you found the answer helpful don't forget to upvote it or accept it so that others might find it more easily as well if they have the same/similar question.

nerdlyist · Accepted Answer · 2016-09-22 16:58:16Z

0

So assuming you are doing a range from the $_GET['number'] number then you can do something like (see comments in code for further explanation):

//This will create an array from 1 to number inclusive
$nums = range(1, $_GET['number']);
//The nums that have been used
$used = array();
//Now loop over that array
foreach($nums as $num){
    $used[] = $num; //Add this number to used
    if(count($used) > 1){//Dont care about first loop
        echo  implode(' + ', $used); // put all elements together by + sign
        echo ' = ' . array_sum($used) . "<br>"; //Show total plus a break
    }
}

answered Sep 22, 2016 at 16:58

nerdlyist

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Comments

Zac Crites · Accepted Answer · 2016-09-22 16:48:21Z

-1

<?php

$num = intval($_GET["number"]);
$terms = [1];

for ($i = 2; $i <= $num; $i++) {
    $terms[] = $i;
    $sum = array_sum($terms);
    echo implode(' + ', $terms) . ' = ' . $sum . PHP_EOL;
}

answered Sep 22, 2016 at 16:48

Zac Crites

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Comments

Dennis · Accepted Answer · 2016-09-22 17:35:36Z

-1

Going for maximum use of PHP array related functions:

$num = intval($_GET["number"]);
$array = range(1, $num);

for ($i = 2; $i <= $num; $i ++)
{
    $slice = array_slice($array, 0, $i);
    $total = array_sum($slice);
    echo implode(" + ", $slice) . " = " . $total . PHP_EOL;
}

Alternative with array_push

$num = intval($_GET["number"]);
$array = array(1);

for ($value = 2; $value <= $num; $value ++)
{
    array_push($array, $value);
    echo implode(" + ", $array) . " = " . array_sum($array) . PHP_EOL;
}

Output

1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10

edited Sep 22, 2016 at 17:35

answered Sep 22, 2016 at 17:06

Dennis

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