Cs50 Tideman Solution < 2027 >

This function checks if there is a path from loser to winner.

// Returns true if locking winner->loser would create a cycle
bool creates_cycle(int winner, int loser)
// If loser directly points to winner, cycle is immediate
    if (loser == winner)
return true;
// Check all candidates to see if loser points to someone,
// and that someone eventually points to winner
for (int i = 0; i < candidate_count; i++)
if (locked[loser][i]) // If loser has an edge to i
     creates_cycle(i, winner))
return true;
return false;

How it works:

The Tideman voting method works like this:

In add_pairs, reset pair_count = 0 before adding new pairs.

The recursive is_path function is safe for up to 9 candidates. For larger sets, use an iterative DFS, but CS50’s MAX is 9.

Cycle Detection Logic:

bool cycle(int winner, int loser)
// If we have reached the winner again, we have a cycle
    if (loser == winner)
return true;
// Check all candidates
    for (int i = 0; i < candidate_count; i++)
// If the loser beats someone, check if that path leads back to the winner
        if (locked[loser][i])
if (cycle(winner, i))
return true;
return false;

Printing the Winner Logic:

void print_winner(void)
for (int i = 0; i < candidate_count; i++)
bool is_source = true;
        for (int j = 0; j < candidate_count; j++)
if (locked[j][i] == true)
is_source = false;
                break;
if (is_source)
printf("%s\n", candidates[i]);
            return;

The CS50 Tideman problem set is widely considered the most difficult assignment in Harvard’s introductory computer science course. It tasks students with implementing the Tideman voting method (also known as Ranked Pairs), a system designed to find a "Condorcet winner"—a candidate who would win against any other candidate in a head-to-head matchup. 1. Record Voter Preferences

The first step is to process every voter's ballot. For each voter, the code must update a 2D preferences[i][j] array, where the value at index [i][j] represents the number of voters who preferred candidate i over candidate j. 2. Identify and Sort Matchups

Once all votes are in, the program identifies every possible head-to-head pair.

Identify: A pair is added to a pairs array if one candidate has more votes than the other.

Sort: The pairs are then sorted in descending order based on the "strength of victory" (the difference in votes between the winner and loser of that pair). 3. Build the Winner Graph

This is the most complex phase. The program iterates through the sorted pairs and "locks" them into a directed graph (using a locked[i][j] boolean matrix). Cs50 Tideman Solution

Cycle Detection: A pair is only locked if it does not create a cycle in the graph. For example, if A beats B and B beats C, you cannot lock a pair where C beats A, as this would create a loop where no clear winner exists.

Recursion: Most students use a recursive helper function to "trace" the path from the winner of the current pair to see if it eventually leads back to the loser. 4. Identify the Source

After all valid pairs are locked, the winner of the election is the "source" of the graph. This is the candidate who has zero arrows pointing toward them (no locked[i][j] is true where j is that candidate). Key Challenges & Academic Honesty

Complexity: Unlike earlier problems like Runoff or Cash, Tideman requires advanced logic for graph theory and recursion.

Academic Integrity: Because of its difficulty, students are frequently warned that looking up a full "Tideman solution" is considered a violation of academic honesty.

This function is identical to the one in plurality. It should record the voter’s rank for each candidate.

Logic: If the candidate name is found, set ranks[rank] = candidate_index and return true. Else false. This function checks if there is a path

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;

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.

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.

We will write a helper that answers: "Starting from the loser, can I eventually reach the winner using existing locked edges?" How it works: The Tideman voting method works like this:

If yes → cycle → don’t lock. If no → safe to lock.