package dev.bogwalk.batch4

import dev.bogwalk.util.maths.isHexagonalNumber
import dev.bogwalk.util.maths.isPentagonalNumber
import dev.bogwalk.util.maths.isTriangularNumber

/**
 * Problem 45: Triangular, Pentagonal, & Hexagonal
 *
 * https://projecteuler.net/problem=45
 *
 * Goal: Given a & b, find all numbers below N that are both types of numbers queried (as below).
 *
 * Constraints: 2 <= N <= 2e14
 *              a < b
 *              a,b in {3, 5, 6} -> {triangle, pentagonal, hexagonal}
 *
 * Triangle Number:
 *
 *      tN = n(n + 1) / 2
 *
 * Pentagonal Number:
 *
 *      pN = n(3n - 1) / 2
 *
 * Hexagonal Number:
 *
 *      hN = n(2n - 1)
 *
 * Some numbers can be all 3 type ->
 * e.g. T_1 = P_1 = H_1 = 1
 *      T_285 = P_165 = H_143 = 40755
 *
 * e.g.: N = 10_000, a = 3, b = 5
 *       result = {1, 210}
 *       N = 100_000, a = 5, b = 6
 *       result = {1, 40755}
 */

class TriPentHex {
    /**
     * HackerRank specific implementation will never request numbers that are both triangular and
     * hexagonal.
     *
     * Solution is optimised by generating the fastest growing number type first. Hexagonal
     * numbers jump to the limit faster than the other 2, and the pentagonal number sequence
     * grows faster than triangular numbers.
     *
     * SPEED (WORSE) 319.36ms for N = 2e14, pentagonal-hexagonal combo
     */
    fun commonNumbers(n: Long, a: Int, b: Int): List<Long> {
        val common = mutableListOf(1L)
        var i = 2L
        if (a == 3 && b == 5) {
            while (true) {
                val pentagonal = i * (3 * i - 1) / 2
                if (pentagonal >= n) break
                if (pentagonal.isTriangularNumber() != null) {
                    common.add(pentagonal)
                }
                i++
            }
        } else { // a == 5 && b == 6
            while (true) {
                val hexagonal = i * (2 * i - 1)
                if (hexagonal >= n) break
                if (hexagonal.isPentagonalNumber() != null) {
                    common.add(hexagonal)
                }
                i++
            }
        }
        return common
    }

    /**
     * For triangular-pentagonal combos, where T_x = P_y:
     *
     *      x(x + 1) / 2 = y(3y - 1) / 2, after completing the square becomes:
     *
     *      (6x - 1)^2 - 3(2y + 1)^2 = -2, a diophantine:
     *
     *      36x^2 - 12x - 12y^2 - 12y, solved for 2 integer variables:
     *
     *      T_(x+1) = -2T_x - P_y and
     *
     *      P_(y+1) = -3T_x - 2P_y - 1
     *
     * For pentagonal-hexagonal combos, where P_x = H_y:
     *
     *      x(3x - 1) / 2 = y(2y - 1), after completing the square becomes:
     *
     *      (6x - 1)^2 - 3(4y - 1)^2 = -2, a diophantine:
     *
     *      36x^2 - 12x - 48y^2 + 24y, solved for 2 integer variables:
     *
     *      P_(x+1) = 97P_x + 112H_y - 44 and
     *
     *      H_(y+1) = 84P_x + 97H_y - 38
     *
     * SPEED (BETTER) 3.7e4ns for N = 2e14, pentagonal-hexagonal combo
     */
    fun commonNumbersFormula(n: Long, a: Int, b: Int): List<Long> {
        val common = mutableListOf<Long>()
        // P_1 = H_1 = 1
        var x = 1
        var y = 1
        var num = 1L
        if (a == 3 && b == 5) {
            while (num < n) {
                // negative terms signify non-integer solutions of the diophantine
                if (x > 0L && y > 0L) common.add(num)
                x = (-2 * x - y).also {
                    y = -3 * x - 2 * y - 1
                }
                // could insert term into either triangular or pentagonal formula
                num = 1L * y * (y + 1) / 2
            }
        } else { // a == 5 && b == 6
            while (num < n) {
                if (x > 0 && y > 0) common.add(num)
                x = (97 * x + 112 * y - 44).also {
                    y = 84 * x + 97 * y - 38
                }
                num = 1L * y * (2 * y - 1)
            }
        }
        return common
    }

    /**
     * Project Euler specific implementation that finds the next number, after {1, 40755} that is
     * triangular, pentagonal, and hexagonal.
     *
     * All hexagonal numbers are a subset of triangular numbers made from an odd n,
     * as T_(2n - 1) == H_n, based on completing the squares below:
     *
     *      n(n + 1) / 2 = m(2m - 1)
     *
     *      n^2 + n = 4m^2 - 2m
     *
     *      (n + 0.5)^2 = 4(m - 0.25)^2
     *
     *      n = 2m - 1
     *
     * So this solution only needs to check for hexagonal numbers that are also pentagonal.
     */
    fun nextTripleType(): Long {
        var next: Long = 40755
        while (true) {
            next++
            if (
                next.isHexagonalNumber() != null &&
                next.isPentagonalNumber() != null
            ) break
        }
        return next
    }
}
package dev.bogwalk.batch4

import dev.bogwalk.util.maths.isHexagonalNumber
import dev.bogwalk.util.maths.isPentagonalNumber
import dev.bogwalk.util.maths.isTriangularNumber

/**
 * Problem 45: Triangular, Pentagonal, & Hexagonal
 *
 * https://projecteuler.net/problem=45
 *
 * Goal: Given a & b, find all numbers below N that are both types of numbers queried (as below).
 *
 * Constraints: 2 <= N <= 2e14
 *              a < b
 *              a,b in {3, 5, 6} -> {triangle, pentagonal, hexagonal}
 *
 * Triangle Number:
 *
 *      tN = n(n + 1) / 2
 *
 * Pentagonal Number:
 *
 *      pN = n(3n - 1) / 2
 *
 * Hexagonal Number:
 *
 *      hN = n(2n - 1)
 *
 * Some numbers can be all 3 type ->
 * e.g. T_1 = P_1 = H_1 = 1
 *      T_285 = P_165 = H_143 = 40755
 *
 * e.g.: N = 10_000, a = 3, b = 5
 *       result = {1, 210}
 *       N = 100_000, a = 5, b = 6
 *       result = {1, 40755}
 */

class TriPentHex {
    /**
     * HackerRank specific implementation will never request numbers that are both triangular and
     * hexagonal.
     *
     * Solution is optimised by generating the fastest growing number type first. Hexagonal
     * numbers jump to the limit faster than the other 2, and the pentagonal number sequence
     * grows faster than triangular numbers.
     *
     * SPEED (WORSE) 319.36ms for N = 2e14, pentagonal-hexagonal combo
     */
    fun commonNumbers(n: Long, a: Int, b: Int): List<Long> {
        val common = mutableListOf(1L)
        var i = 2L
        if (a == 3 && b == 5) {
            while (true) {
                val pentagonal = i * (3 * i - 1) / 2
                if (pentagonal >= n) break
                if (pentagonal.isTriangularNumber() != null) {
                    common.add(pentagonal)
                }
                i++
            }
        } else { // a == 5 && b == 6
            while (true) {
                val hexagonal = i * (2 * i - 1)
                if (hexagonal >= n) break
                if (hexagonal.isPentagonalNumber() != null) {
                    common.add(hexagonal)
                }
                i++
            }
        }
        return common
    }

    /**
     * For triangular-pentagonal combos, where T_x = P_y:
     *
     *      x(x + 1) / 2 = y(3y - 1) / 2, after completing the square becomes:
     *
     *      (6x - 1)^2 - 3(2y + 1)^2 = -2, a diophantine:
     *
     *      36x^2 - 12x - 12y^2 - 12y, solved for 2 integer variables:
     *
     *      T_(x+1) = -2T_x - P_y and
     *
     *      P_(y+1) = -3T_x - 2P_y - 1
     *
     * For pentagonal-hexagonal combos, where P_x = H_y:
     *
     *      x(3x - 1) / 2 = y(2y - 1), after completing the square becomes:
     *
     *      (6x - 1)^2 - 3(4y - 1)^2 = -2, a diophantine:
     *
     *      36x^2 - 12x - 48y^2 + 24y, solved for 2 integer variables:
     *
     *      P_(x+1) = 97P_x + 112H_y - 44 and
     *
     *      H_(y+1) = 84P_x + 97H_y - 38
     *
     * SPEED (BETTER) 3.7e4ns for N = 2e14, pentagonal-hexagonal combo
     */
    fun commonNumbersFormula(n: Long, a: Int, b: Int): List<Long> {
        val common = mutableListOf<Long>()
        // P_1 = H_1 = 1
        var x = 1
        var y = 1
        var num = 1L
        if (a == 3 && b == 5) {
            while (num < n) {
                // negative terms signify non-integer solutions of the diophantine
                if (x > 0L && y > 0L) common.add(num)
                x = (-2 * x - y).also {
                    y = -3 * x - 2 * y - 1
                }
                // could insert term into either triangular or pentagonal formula
                num = 1L * y * (y + 1) / 2
            }
        } else { // a == 5 && b == 6
            while (num < n) {
                if (x > 0 && y > 0) common.add(num)
                x = (97 * x + 112 * y - 44).also {
                    y = 84 * x + 97 * y - 38
                }
                num = 1L * y * (2 * y - 1)
            }
        }
        return common
    }

    /**
     * Project Euler specific implementation that finds the next number, after {1, 40755} that is
     * triangular, pentagonal, and hexagonal.
     *
     * All hexagonal numbers are a subset of triangular numbers made from an odd n,
     * as T_(2n - 1) == H_n, based on completing the squares below:
     *
     *      n(n + 1) / 2 = m(2m - 1)
     *
     *      n^2 + n = 4m^2 - 2m
     *
     *      (n + 0.5)^2 = 4(m - 0.25)^2
     *
     *      n = 2m - 1
     *
     * So this solution only needs to check for hexagonal numbers that are also pentagonal.
     */
    fun nextTripleType(): Long {
        var next: Long = 40755
        while (true) {
            next++
            if (
                next.isHexagonalNumber() != null &&
                next.isPentagonalNumber() != null
            ) break
        }
        return next
    }
}