CS50 Solution pset3 tideman

tideman.c

#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

// Max number of candidates
#define MAX 9

// preferences[i][j] is number of voters who prefer i over j
int preferences[MAX][MAX];

// locked[i][j] means i is locked in over j
bool locked[MAX][MAX];
bool lock = true;

// Each pair has a winner, loser
typedef struct
{
    int winner;
    int loser;
}
pair;

// Array of candidates
string candidates[MAX];
pair pairs[MAX * (MAX - 1) / 2];

int pair_count;
int candidate_count;

// Function prototypes
bool vote(int rank, string name, int ranks[]);
void record_preferences(int ranks[]);
void add_pairs(void);
void sort_pairs(void);
int comparator(const void *a, const void *b);
void lock_pairs(void);
void print_winner(void);

int main(int argc, string argv[])
{
    // Check for invalid usage
    if (argc < 2)
    {
        printf("Usage: tideman [candidate ...]\n");
        return 1;
    }

    // Populate array of candidates
    candidate_count = argc - 1;
    if (candidate_count > MAX)
    {
        printf("Maximum number of candidates is %i\n", MAX);
        return 2;
    }
    for (int i = 0; i < candidate_count; i++)
    {
        candidates[i] = argv[i + 1];
    }

    // Clear graph of locked in pairs and the preferences array from garbage values
    for (int i = 0; i < candidate_count; i++)
    {
        for (int j = 0; j < candidate_count; j++)
        {
            locked[i][j] = false;
            preferences[i][j] = 0;
        }
    }

    pair_count = 0;
    int voter_count = get_int("Number of voters: ");

    // Query for votes
    for (int i = 0; i < voter_count; i++)
    {
        // ranks[i] is voter's ith preference
        int ranks[candidate_count];

        // Query for each rank
        for (int j = 0; j < candidate_count; j++)
        {
            string name = get_string("Rank %i: ", j + 1);

            if (!vote(j, name, ranks))
            {
                printf("Invalid vote.\n");
                return 3;
            }
        }

        record_preferences(ranks);

        printf("\n");
    }

    add_pairs();
    sort_pairs();
    lock_pairs();
    print_winner();
    return 0;
}

// Update ranks given a new vote
bool vote(int rank, string name, int ranks[])
{
    for (int i = 0; i < candidate_count; i++)
    {
        if (strcmp(name,  candidates[i]) == 0)
        {
            ranks[rank] = i;
            return true;
        }
    }
    return false;
}

// Update preferences given one voter's ranks
void record_preferences(int ranks[])
{
    for (int i = 0; i < candidate_count; i++)
    {
        for (int j = i + 1; j < candidate_count; j++)
        {
            preferences[ranks[i]][ranks[j]]++;
        }
    }
}

// Record pairs of candidates where one is preferred over the other
void add_pairs(void)
{
    for (int i = 0; i < candidate_count; i++)
    {
        for (int j = i + 1; j < candidate_count; j++)
        {
            if (preferences[i][j] > preferences[j][i])
            {
                pairs[pair_count].winner = i;
                pairs[pair_count].loser = j;
                pair_count++;
            }
            else if (preferences[i][j] < preferences[j][i])
            {
                pairs[pair_count].winner = j;
                pairs[pair_count].loser = i;
                pair_count++;
            }
        }
    }
}

// function used for sort
int comparator(const void *a, const void *b)
{
    pair *ab = (pair *)a;
    pair *ba = (pair *)b;

    // uses pointers to access the preferences and check how much a candidate wins over another
    return (preferences[ba->winner][ba->loser] - preferences[ab->winner][ab->loser]);
}

// Sort pairs in decreasing order by strength of victory
void sort_pairs(void)
{
    qsort(pairs, pair_count, sizeof(pair), comparator);
}

bool has_cycle(int winner, int loser)
{
    while (winner != -1 && winner != loser)
    {
        bool found = false;
        for (int i = 0; i < candidate_count; i++)
        {
            if (locked[i][winner])
            {
                found = true;
                winner = i;
            }
        }
        if (!found)
        {
            winner = -1;
        }
    }
    if (winner == loser)
    {
        return true;
    }
    return false;
}

// Lock pairs into the candidate graph in order, without creating cycles
void lock_pairs(void)
{
    //TODO
    for (int i = 0; i < pair_count; i++)
    {
        if (!has_cycle(pairs[i].winner, pairs[i].loser))
        {
            locked[pairs[i].winner][pairs[i].loser] = true;
        }
    }
}

// Print the winner of the election
void print_winner(void)
{
    //TODO
    for (int col = 0; col < MAX; col++)
    {
        bool found_source = true;
        for (int row = 0; row < MAX; row++)
        {
            if (locked[row][col] == true)
            {
                found_source = false;
                break;
            }
        }
        if (found_source)
        {
            printf("%s\n", candidates[col]);
            return;
        }
    }
    return;
}

Although this code passes all the checks made on CS50, it seems to me that the has_cycle function is incomplete. Meaning it misses some cases where there is a cycle. Let me show with an example:

Imagine there are 4 candidates: A, B, C and D (for simplicity). The ordered pairs are (represented as {winner, loser}):

1. {A, D}
2. {A, B}
3. {C, B}
4. {D, C}
5. {B, D}
6. {A, C}

If you draw this simple diagram, you'll see that the fifth pair, {B, D}, is a pair that cannot be drawn, because it would create the cycle B -> D -> C -> B.

Even though that is the case, the function has_cycle seems to miss this case, and incorrectly returns false (meaning there is no cycle formed by adding the pair {B, D}.

It seems to me that the proper way to write such a function would be by using recursion. Although I could not figure out how to write it yet.

Please let me know if I am missing something or if someone figured out how to write this function using recursion.

Hi,
I'm not sure about whether my answer would be suitable to solve your doubt regarding 6 votes for the 4 candidates thing, yet I have figured out the recursive version of has_cycle()...

---

bool has_cycle(int original_winner, int current_loser)
{
    if (current_loser == original_winner)
        return true;

    for (int i = 0; i < pair_count; ++i)
    {
        if (pairs[i].winner == current_loser && locked[pairs[i].winner][pairs[i].loser])
        {
            if (has_cycle(original_winner, pairs[i].loser))
                return true;
        }
    }
    return false;
}

We are given a winner and a loser, original_winner and current_loser respectively, I'd talk about the base case later.
We'll make a recursive call has_cycle(original_winner /* this doesn't change */, pair[i].loser) for each new pair in pairs, where the current_loser is the winner, if that pair already has a edge locked in.

Now in the recursive call to has_cycle, if the original_winner that we passed in while calling the function in lock_pairs() is the same as the current_loser in the recursive call, this means we have circled around, thus a cycle exists, and we return true.

If no cycle exists, it returns false.

Test for some randomized testcases at this website, where the graph has a red edge, and do a dry-run of the above recursive algorithm on that graph. That might help you understand it better.

Hope the answer helped :)

Thanks :)