package dev.bogwalk.batch0

import dev.bogwalk.util.maths.gcd
import dev.bogwalk.util.maths.lcm
import dev.bogwalk.util.maths.primeNumbers
import kotlin.math.abs
import kotlin.math.log2
import kotlin.math.pow

/**
 * Problem 5: Smallest Multiple
 *
 * https://projecteuler.net/problem=5
 *
 * Goal: Find the smallest positive number that can be evenly divided by each number in the range
 * [1, N].
 *
 * Constraints: 1 <= N <= 40
 *
 * e.g.: N = 3
 *       {1, 2, 3} evenly divides 6 to give quotient {6, 3, 2}
 */

class SmallestMultiple {
    /**
     * Repeatedly calculates the lcm of 2 values (via reduce()), starting from the largest and
     * stepping backwards until the middle of the range, as the smaller half of a range will
     * already be factors of the larger half.
     *
     * SPEED (BETTER) 4.6e+04ns for N = 40
     */
    fun lcmOfRange(n: Int): Long {
        val rangeMax = n.toLong()
        return (rangeMax downTo (rangeMax / 2 + 1)).reduce { acc, num ->
            abs(acc * num) / gcd(acc, num)
        }
    }

    /**
     * Solution mimics the one above but uses top-level helper that leverages BigInteger and its
     * built-in methods instead of Long.
     *
     * SPEED (WORST) 1.2e+05ns for N = 40
     */
    fun lcmOfRangeBI(n: Int): Long {
        val rangeMax = n.toLong()
        val range = (rangeMax downTo (rangeMax / 2 + 1)).toList().toLongArray()
        return lcm(*range)
    }

    /**
     * Uses prime numbers to calculate the lcm of a range, based on the formula:
     *
     *      p_i^a_i <= n
     *
     *      a_i * log(p_i) <= log(n)
     *
     *      a_i = floor(log(n) / log(p_i))
     *
     *      e.g. N = 6, primes < N = {2, 3, 5};
     *      the exponent of the 1st prime will be 2 as 2^2 < 6 but 2^3 > 6;
     *      the exponent of the 2nd prime will be 1 as 3^1 < 6 but 3^2 > 6;
     *      the exponent of the 3rd prime will be 1 as 5^1 < 6 but 5^2 > 6;
     *      therefore, lcm = 2^2 * 3^1 * 5^1 = 60.
     *
     * This is an adaptation of the prime factorisation method for calculating the LCM.
     *
     * SPEED (BEST) 1.9e+04ns for N = 40
     */
    fun lcmOfRangeUsingPrimes(n: Int): Long {
        var lcm = 1L
        val primes = primeNumbers(n)
        for (prime in primes) {
            lcm *= if (prime * prime <= n) {
                val exponent = (log2(n.toDouble()) / log2(prime.toDouble())).toInt()
                (prime.toDouble().pow(exponent)).toLong()
            } else {
                prime.toLong()
            }
        }
        return lcm
    }
}
package dev.bogwalk.batch0

import dev.bogwalk.util.maths.gcd
import dev.bogwalk.util.maths.lcm
import dev.bogwalk.util.maths.primeNumbers
import kotlin.math.abs
import kotlin.math.log2
import kotlin.math.pow

/**
 * Problem 5: Smallest Multiple
 *
 * https://projecteuler.net/problem=5
 *
 * Goal: Find the smallest positive number that can be evenly divided by each number in the range
 * [1, N].
 *
 * Constraints: 1 <= N <= 40
 *
 * e.g.: N = 3
 *       {1, 2, 3} evenly divides 6 to give quotient {6, 3, 2}
 */

class SmallestMultiple {
    /**
     * Repeatedly calculates the lcm of 2 values (via reduce()), starting from the largest and
     * stepping backwards until the middle of the range, as the smaller half of a range will
     * already be factors of the larger half.
     *
     * SPEED (BETTER) 4.6e+04ns for N = 40
     */
    fun lcmOfRange(n: Int): Long {
        val rangeMax = n.toLong()
        return (rangeMax downTo (rangeMax / 2 + 1)).reduce { acc, num ->
            abs(acc * num) / gcd(acc, num)
        }
    }

    /**
     * Solution mimics the one above but uses top-level helper that leverages BigInteger and its
     * built-in methods instead of Long.
     *
     * SPEED (WORST) 1.2e+05ns for N = 40
     */
    fun lcmOfRangeBI(n: Int): Long {
        val rangeMax = n.toLong()
        val range = (rangeMax downTo (rangeMax / 2 + 1)).toList().toLongArray()
        return lcm(*range)
    }

    /**
     * Uses prime numbers to calculate the lcm of a range, based on the formula:
     *
     *      p_i^a_i <= n
     *
     *      a_i * log(p_i) <= log(n)
     *
     *      a_i = floor(log(n) / log(p_i))
     *
     *      e.g. N = 6, primes < N = {2, 3, 5};
     *      the exponent of the 1st prime will be 2 as 2^2 < 6 but 2^3 > 6;
     *      the exponent of the 2nd prime will be 1 as 3^1 < 6 but 3^2 > 6;
     *      the exponent of the 3rd prime will be 1 as 5^1 < 6 but 5^2 > 6;
     *      therefore, lcm = 2^2 * 3^1 * 5^1 = 60.
     *
     * This is an adaptation of the prime factorisation method for calculating the LCM.
     *
     * SPEED (BEST) 1.9e+04ns for N = 40
     */
    fun lcmOfRangeUsingPrimes(n: Int): Long {
        var lcm = 1L
        val primes = primeNumbers(n)
        for (prime in primes) {
            lcm *= if (prime * prime <= n) {
                val exponent = (log2(n.toDouble()) / log2(prime.toDouble())).toInt()
                (prime.toDouble().pow(exponent)).toLong()
            } else {
                prime.toLong()
            }
        }
        return lcm
    }
}