Cs50 Tideman Solution

bool will_create_cycle(int winner, int loser)

Determine who prefers whom between every pair of candidates.

Create a boolean function that checks if a path exists between two nodes in the Cs50 Tideman Solution

The CS50 Tideman problem serves as an excellent introduction to graph theory and algorithmic complexity within the context of a C programming course. It forces the student to manipulate 2D arrays, implement sorting algorithms, and utilize recursion for cycle detection. The solution demonstrates how computer science can be applied to solve complex logical problems in social choice theory, providing a deterministic outcome from a set of ranked preferences. The "locking" mechanism, specifically the prevention of cycles, highlights the importance of maintaining data integrity and structural properties (the Directed Acyclic Graph) in algorithmic design.

return false;

// Find the new minimum votes min_votes = MAX_VOTERS; for (int i = 0; i < num_candidates; i++) { if (candidates[i].votes >= 0 && candidates[i].

# Update preferences pairs = update_preferences(pairs, eliminated_candidate) The solution demonstrates how computer science can be

The winner is the candidate who has no locked edges pointing to them (i.e., no column in that row has a true value in the locked table).

void print_winner(void)

Set locked[winner][loser] = true , but only if doing so does not create a cycle.

for (int i = 0; i < num_voters; i++) for (int j = 0; j < num_candidates; j++) if (strcmp(voters[i].preferences[j], "") != 0) for (int k = 0; k < num_candidates; k++) if (strcmp(candidates[k].name, voters[i].preferences[j]) == 0) candidates[k].votes++; # Update preferences pairs = update_preferences(pairs