The Collatz Conjecture or 3x+1 problem can be summarized as follows:
Take any positive integer n. If n is even, divide n by 2 to get n / 2. If n is odd, multiply n by 3 and add 1 to get 3n + 1. Repeat the process indefinitely. The conjecture states that no matter which number you start with, you will always reach 1 eventually.
Given a number n, return the number of steps required to reach 1.
Starting with n = 12, the steps would be as follows:
Resulting in 9 steps. So for input n = 12, the return value would be 9.
Go to the root of your PHP exercise directory, which is <EXERCISM_WORKSPACE>/php
.
To find the Exercism workspace run
% exercism debug | grep Workspace
Get PHPUnit if you don't have it already.
% wget --no-check-certificate https://phar.phpunit.de/phpunit.phar
% chmod +x phpunit.phar
Execute the tests:
% ./phpunit.phar collatz-conjecture/collatz-conjecture_test.php
An unsolved problem in mathematics named after mathematician Lothar Collatz https://en.wikipedia.org/wiki/3x_%2B_1_problem
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
<?php
require "collatz-conjecture.php";
class CollatzConjecture extends PHPUnit\Framework\TestCase
{
public function testZeroStepsForOne()
{
$this->assertEquals(0, steps(1));
}
public function testDivideIfEven()
{
$this->assertEquals(4, steps(16));
}
public function testEvenAndOddSteps()
{
$this->assertEquals(9, steps(12));
}
public function testLargeNumberOfEvenAndOddSteps()
{
$this->assertEquals(152, steps(1000000));
}
public function testZeroIsAnError()
{
$this->expectException(InvalidArgumentException::class);
$this->expectExceptionMessage('Only positive numbers are allowed');
steps(0);
}
public function testNegativeValueIsAnError()
{
$this->expectException(InvalidArgumentException::class);
$this->expectExceptionMessage('Only positive numbers are allowed');
steps(-1);
}
}
<?php
function steps(int $num) : int {
if ($num < 1) {
throw new InvalidArgumentException('Only positive numbers are allowed');
}
$result = 0;
while ($num !== 1) {
$num = $num % 2 ? $num * 3 + 1 : $num / 2;
$result++;
}
return $result;
}
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