CODEWITHSUNDEEP
phparrays
Michiel Nuyts
Michiel Nuyts

Reputation: 51

PHP Simple Calculation with an Array

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.

Upvotes: 4

Views: 169

Answers (6)

Dennis
Dennis

Reputation: 8101

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

Upvotes: -1

Sherif
Sherif

Reputation: 11943

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

Upvotes: 1

nerdlyist
nerdlyist

Reputation: 2847

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
    }
}

Upvotes: 0

Striezel
Striezel

Reputation: 3758

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.

Upvotes: 1

Martavis Griffin
Martavis Griffin

Reputation: 128

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

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

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

Upvotes: 1

Zac Crites
Zac Crites

Reputation: 872

<?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;
}

Upvotes: -1

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