project-euler-solutions/C/p068.c

/* Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, and each line adding to nine.
 *
 *         4
 *          \
 *           \
 *            3
 *           /  \
 *          /    \
 *         1------2------6
 *        /
 *       /
 *      5
 *
 * Working clockwise, and starting from the group of three with the numerically lowest external node (4,3,2 in this example),
 * each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.
 *
 * It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.
 *
 * Total	Solution Set
 * 9	4,2,3; 5,3,1; 6,1,2
 * 9	4,3,2; 6,2,1; 5,1,3
 * 10	2,3,5; 4,5,1; 6,1,3
 * 10	2,5,3; 6,3,1; 4,1,5
 * 11	1,4,6; 3,6,2; 5,2,4
 * 11	1,6,4; 5,4,2; 3,2,6
 * 12	1,5,6; 2,6,4; 3,4,5
 * 12	1,6,5; 3,5,4; 2,4,6
 *
 * By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.
 *
 * Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum 16-digit string
 * for a "magic" 5-gon ring?
 *
 *             O
 *              \
 *               O
 *             /  \   O
 *            /    \ /
 *           O      O
 *          / \    /
 *         O   O--O---O
 *              \
 *               O
 */

#define _POSIX_C_SOURCE 199309L

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include "projecteuler.h"

int *eval_ring(int **ring, int n);
long int array_to_long(int *n, int len);
int compare(void *a, void *b);

int main(int argc, char **argv)
{
    int i;
    int *eval, **ring;
    long int n, max = 0;
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    if((ring = (int **)malloc(10*sizeof(int *))) == NULL)
    {
        fprintf(stderr, "Error while allocating memory\n");
        return 1;
    }

    for(i = 0; i < 10; i++)
    {
        if((ring[i] = (int *)malloc(sizeof(int))) == NULL)
        {
            fprintf(stderr, "Error while allocating memory\n");
            return 1;
        }
        *ring[i] = i + 1;
    }

    /* Check if the first permutation of values is a possible solution.*/
    eval = eval_ring(ring, 5);

    /* Convert the vector into an integer number.*/
    n = array_to_long(eval, 15);

    if(n > max)
    {
        max = n;
    }

    free(eval);

    /* Generate all possible permutations, for each one check if
     * it's a possible solution for the ring and save the maximum.*/
    while(next_permutation((void **)ring, 10, compare) != -1)
    {
        eval = eval_ring(ring, 5);

        n = array_to_long(eval, 15);

        if(n > max)
        {
            max = n;
        }

        free(eval);
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;

    printf("Project Euler, Problem 68\n");
    printf("Answer: %ld\n", max);

    printf("Elapsed time: %.9lf seconds\n", elapsed);

    return 0;
}

int compare(void *a, void *b)
{
    int *n1, *n2;

    n1 = (int *)a;
    n2 = (int *)b;

    return *n1 - *n2;
}

/* Function to evaluate the ring. The ring is represented as a vector of 2*n elements,
 * the first n elements represent the external nodes of the ring, the next n elements
 * represent the internal ring.*/
int *eval_ring(int **ring, int n)
{
    int i, j, magic_val, val;
    int *res;

    for(i = 1; i < n; i++)
    {
        /* We need to start from the lowest external node, so if
         * the first element in the vector is not the lowest of
         * the first n elements (the external elements), the configuration
         * is not a valid one.*/
        if(*ring[i] < *ring[0])
        {
            return NULL;
        }
    }

    if((res = (int *)malloc(3*n*sizeof(int))) == NULL)
    {
        fprintf(stderr, "Error while allocating memory\n");
        return NULL;
    }

    /* Each group of three must have the same value.*/
    magic_val = *ring[0] + *ring[n] + *ring[n+1];

    for(i = 0, j = 0; i < n; i++, j += 3)
    {
        /* We need to find the maximum 16-digit string, this is
         * possible only if the element "10" is used only once,
         * i.e. if it's one of the external nodes.*/
        if(*ring[n+i] == 10 || *ring[n+(i+1)%n] == 10)
        {
            return NULL;
        }

        /* Check that the value of the current three-element group
         * is the "magic" value.*/
        val = *ring[i] + *ring[n+i] + *ring[n+(i+1)%n];

        if(val != magic_val)
        {
            return NULL;
        }

        /* Save the current element group in the result string.*/
        res[j] = *ring[i];
        res[j+1] = *ring[n+i];
        res[j+2] = *ring[n+(i+1)%n];
    }

    return res;
}

/* Function to convert the vector into a long int.*/
long int array_to_long(int *n, int len)
{
    int i, k = 0;
    long int res = 0;

    if(n == NULL)
    {
        return 0;
    }

    for(i = len - 1; i >= 0; i--)
    {
        res += n[i] * pow(10, k);

        if(n[i] >= 10)
        {
            k += 2;
        }
        else
        {
            k++;
        }
    }

    return res;
}