Function check_cycle(current_loser, target_winner): If current_loser is equal to target_winner: Return True (Cycle detected!) For every candidate in the contest: If current_loser is locked over this candidate: If check_cycle(this candidate, target_winner) is True: Return True Return False (Safe to lock) Use code with caution.
return;
void print_winner(void) for (int i = 0; i < candidate_count; i++) bool has_source = true; for (int j = 0; j < candidate_count; j++) // If anyone locks to 'i', 'i' is not the source if (locked[j][i]) has_source = false; break; if (has_source) printf("%s\n", candidates[i]); return; Use code with caution. Summary of Key Concepts
The first step is to record voter preferences. We use a 2D array, preferences[i][j] Cs50 Tideman Solution
If you are currently taking CS50, you probably just felt a cold shiver run down your spine seeing the word "Tideman."
The pairs array has an index that is independent of candidate indices. Don't use i or j to access the pairs array directly—maintain a separate counter.
To successfully write the solution, you must complete six specific functions inside the provided CS50 distribution code. 1. vote Don't use i or j to access the
bool creates_cycle(int from, int to) // We want to lock from->to
Many students try to detect cycles by checking if locked[loser][winner] already exists. That is wrong. A cycle can be longer than two edges (A→B→C→A). Always use DFS.
return;
Actually, step through:
It looked something like this mentally:
We need to populate the global pairs array. A pair exists if preferences[i][j] > preferences[j][i] . If equal (tie), skip. vote bool creates_cycle(int from, int to) // We
