PHP Simple Calculation with an Array - php

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.

Answer 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.

Answer 2

Here's the simplest way I can think...
$num = intval($_GET['number']);
$intArray = range(1,$num);
echo implode(" + ",$intArray)." = ".array_sum($intArray);

Answer 3

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

Answer 4

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
}
}

Answer 5

<?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;
}

Answer 6

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