* Palindrome.php
/**
* Largest palindrome product
* https://projecteuler.net/problem=4
* A palindromic number reads the same both ways.
* The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
* Find the largest palindrome made from the product of two 3-digit numbers.
-----------------------------------------------------------------------------------------
The palindrome can be written as:
abccba
Which then simpifies to:
100000a + 10000b + 1000c + 100c + 10b + a
And then:
100001a + 10010b + 1100c
Factoring out 11, you get:
11(9091a + 910b + 100c)
So the palindrome must be divisible by 11.
Seeing as 11 is prime, at least one of the numbers must be divisible by 11.
*/
<?php
class Palindrome {
/** @var int */
private $width;
public function __construct($width) {
$this->width = $width;
}
public static function pow(int $m, int $p) {
if ($p === 0) {
return 1;
}
$t = $m;
$res = 1;
for (; $p > 0; $p >>= 1) {
if ($p & 0x01 !== 0) {
$res *= $t;
}
$t *= $t;
}
return $res;
}
private function int2digitArray(int $num) : array {
$a = [];
while ($num > 0) {
$digit = $num % 10;
array_unshift($a, $digit);
$num = floor($num / 10);
}
return $a;
}
private function isPalindrome(array $a) : bool {
for ($i = 0, $j = count($a)-1; $i < $j; $i++, $j--) {
if ($a[$i] !== $a[$j]) {
return false;
}
}
return true;
}
public function intIsPalindrome(int $num) : bool {
$a = $this->int2digitArray($num);
return $this->isPalindrome($a);
}
public function largest(int &$maxI, int &$maxJ) {
$low = pow(10, $this->width-1);
$m = 10 * $low - 1;
$max = $product = 0;
// $i = 999, $j = 990
for ($i = $m; $i >= $low; $i--) {
for ($j = $m-9; $j >= $low; $j -= 11) {
// printf("i=%d, j=%d\n", $i, $j);
$product = $i * $j;
if ($product > $max) {
// echo $product . PHP_EOL;
if (self::intIsPalindrome($product)) {
$max = $product;
$maxI = $i;
$maxJ = $j;
}
}
}
}
return $max;
}
}
$p = new Palindrome(3);
$i = $j = 0;
echo $p->largest($i, $j).PHP_EOL; // 906609
printf("[%d, %d]\n", $i, $j); // [993, 913]