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 `$i` to an array every iteration. Then you want to print the calculation with `implode()` and get the result with `array_sum()`. — Rizier123, Sep 22 '16 at 16:38
You may be interested in the [`range()`](https://secure.php.net/range) function. Doing `$nums = range(1, $num);` will give you an array of ints. — gen_Eric, Sep 22 '16 at 16:43
@MichielNuyts If you get stuck or confused post your code here and explain where you got stuck. — Rizier123, Sep 22 '16 at 16:44

Striezel · Accepted Answer · 2016-09-22T16:55:36.907

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 '16 at 16:55

answered Sep 22 '16 at 16:48

Striezel

3,693
7
23
37

You shouldn't close ` – Zac Crites Sep 22 '16 at 16:49
Also your arguments to `implode` are in the wrong order. – Zac Crites Sep 22 '16 at 16:52
@ZacCrites: Well, it does not matter for that example, and on the other hand it feels a little bit weird to open a ` – Striezel Sep 22 '16 at 16:53
As for the arguments: According to the [PHP documentation](http://php.net/manual/en/function.implode.php) 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. – Striezel Sep 22 '16 at 16:57
http://stackoverflow.com/questions/4410704/why-would-one-omit-the-close-tag – Zac Crites Sep 22 '16 at 16:58

Sherif · Answer 2 · 2016-09-22T17:27:51.017

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

@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. — Sherif, Sep 22 '16 at 17:23

score 1 · Answer 3 · answered Sep 22 '16 at 16:54

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 '16 at 16:54

Martavis Griffin

128
8

score 0 · Answer 4 · answered Sep 22 '16 at 16:58

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
    }
}

score -1 · Answer 5 · answered Sep 22 '16 at 16:48

-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 '16 at 16:48

Zac Crites

822
10
14

Dennis · Answer 6 · 2016-09-22T17:35:36.627

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

PHP Simple Calculation with an Array

6 Answers6

output

Generalization

output