Cs50 Tideman Solution Here

Tideman is hard because it introduces graph theory without explicitly teaching it. But once you understand cycle detection as “can the loser reach the winner already,” the solution becomes clean and elegant.

Stick with it. Get the small test cases working (3 candidates). Then scale up. And remember — in CS50, the Tideman problem is marked as “more comfortable” for a reason. If you complete it, you have truly leveled up.

Happy coding, and may your locks always be cycle-free.

Writing a "good" post about the CS50 Tideman problem usually means writing a "Problem Set Story." This is a popular format in the CS50 community (often seen on Medium, Dev.to, or Reddit) where you document your struggle and eventual triumph.

Because the course encourages academic honesty, a good post never pastes the full code solution. Instead, it explains the logic and the algorithm. Cs50 Tideman Solution

Here is a template and a drafted post you can use or adapt. It focuses on the hardest part of the problem: the sort_pairs and lock_pairs functions.

The hardest part is lock_pairs() — locking a pair if it doesn’t create a cycle.

Think of locked pairs as directed edges from winner to loser.

A cycle occurs if:

Adding edge A → B would create a path from B back to A using already locked edges.

So to check if locking (winner, loser) is safe:

Sort pairs in descending order of victory margin: margin = preferences[winner][loser] - preferences[loser][winner].

You can use any stable sorting algorithm. Bubble sort is fine for small candidate counts. Tideman is hard because it introduces graph theory

void sort_pairs(void)
for (int i = 0; i < pair_count - 1; i++)
for (int j = 0; j < pair_count - i - 1; j++)
int margin1 = preferences[pairs[j].winner][pairs[j].loser] - preferences[pairs[j].loser][pairs[j].winner];
            int margin2 = preferences[pairs[j+1].winner][pairs[j+1].loser] - preferences[pairs[j+1].loser][pairs[j+1].winner];
            if (margin1 < margin2)
pair temp = pairs[j];
                pairs[j] = pairs[j+1];
                pairs[j+1] = temp;
return;

This paper explores the algorithmic logic and implementation required to solve the "Tideman" problem from Harvard University’s CS50: Introduction to Computer Science course. The Tideman voting method (also known as Ranked Pairs) is a Condorcet method used to determine election results where voters rank candidates. This paper details the data structures used to represent the voting graph, the critical process of "locking" pairs to prevent cycles, and the specific programming strategies used to implement the sort_pairs, lock_pairs, and print_winner functions.

If you want, I can:

This post is structured for someone who has struggled with the problem (as many do) and wants to understand why the solution works, not just what to type.

#include <stdio.h>
#include <stdlib.h>
// Structure to represent a candidate
typedef struct candidate 
    int id;
    int votes;
 candidate_t;
// Structure to represent a voter
typedef struct voter 
    int *preferences;
 voter_t;
// Function to read input
void read_input(int *voters, int *candidates, voter_t **voters_prefs) 
    // Read in the number of voters and candidates
    scanf("%d %d", voters, candidates);
// Allocate memory for voters and candidates
    *voters_prefs = malloc(*voters * sizeof(voter_t));
    candidate_t *candidates_list = malloc(*candidates * sizeof(candidate_t));
// Read in voter preferences
    for (int i = 0; i < *voters; i++) 
        (*voters_prefs)[i].preferences = malloc(*candidates * sizeof(int));
        for (int j = 0; j < *candidates; j++) 
            scanf("%d", &(*voters_prefs)[i].preferences[j]);
// Function to count first-place votes
void count_first_place_votes(voter_t *voters_prefs, int voters, candidate_t *candidates_list, int candidates) 
    // Initialize vote counts to 0
    for (int i = 0; i < candidates; i++) 
        candidates_list[i].votes = 0;
// Count first-place votes
    for (int i = 0; i < voters; i++) 
        for (int j = 0; j < candidates; j++) 
            if (j == 0) 
                candidates_list[voters_prefs[i].preferences[j] - 1].votes++;
// Function to check for winner
int check_for_winner(candidate_t *candidates_list, int candidates) 
    // Check if any candidate has more than half of the first-place votes
    for (int i = 0; i < candidates; i++) 
        if (candidates_list[i].votes > candidates / 2) 
            return i + 1;
return -1;
// Function to eliminate candidate
void eliminate_candidate(candidate_t *candidates_list, int candidates, int eliminated) 
    // Decrement vote counts for eliminated candidate
    for (int i = 0; i < candidates; i++) 
        if (candidates_list[i].id == eliminated) 
            candidates_list[i].votes = 0;
// Function to recount votes
void recount_votes(voter_t *voters_prefs, int voters, candidate_t *candidates_list, int candidates) 
    // Recount votes
    for (int i = 0; i < voters; i++) 
        for (int j = 0; j < candidates; j++) 
            if (candidates_list[voters_prefs[i].preferences[j] - 1].votes == 0) 
                // Move to next preference
                voters_prefs[i].preferences[j] = -1;
             else 
                break;
int main() 
    int voters, candidates;
    voter_t *voters_prefs;
    read_input(&voters, &candidates, &voters_prefs);
candidate_t *candidates_list = malloc(candidates * sizeof(candidate_t));
    for (int i = 0; i < candidates; i++) 
        candidates_list[i].id = i + 1;
count_first_place_votes(voters_prefs, voters, candidates_list, candidates);
int winner = check_for_winner(candidates_list, candidates);
    while (winner == -1) 
        // Eliminate candidate with fewest votes
        int eliminated = -1;
        int min_votes = voters + 1;
        for (int i = 0; i < candidates; i++) 
            if (candidates_list[i].votes < min_votes) 
                min_votes = candidates_list[i].votes;
                eliminated = candidates_list[i].id;
eliminate_candidate(candidates_list, candidates, eliminated);
recount_votes(voters_prefs, voters, candidates_list, candidates);
count_first_place_votes(voters_prefs, voters, candidates_list, candidates);
winner = check_for_winner(candidates_list, candidates);
printf("The winner is: %d\n", winner);
return 0;