Cs50 Tideman Solution Fixed

bool vote(int rank, string name, int ranks[]) for (int i = 0; i < candidate_count; i++) if (strcmp(candidates[i], name) == 0) ranks[rank] = i; return true; return false; Use code with caution. 2. record_preferences(ranks)

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# Eliminate the candidate(s) with the fewest votes eliminated_candidates = [] while len(min_vote_candidates) > 0: eliminated_candidate = min_vote_candidates[0] eliminated_candidates.append(eliminated_candidate) candidates.remove(eliminated_candidate) Cs50 Tideman Solution

This is the hardest part of CS50 Tideman. You must iterate through your sorted pairs and lock them into the locked[i][j] boolean matrix. However, you cannot lock a pair if it creates a cycle.

Ensure your sorting algorithm properly handles pairs with equal margins without breaking the structure. bool vote(int rank, string name, int ranks[]) for

Happy coding, and may your Tideman election always find its Condorcet winner

strcpy(voters[i].preferences[num_candidates-1], ""); j--; Can’t copy the link right now

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

If the name matches, update the ranks array to store the candidate's index.

int winner = pairs[i].winner; int loser = pairs[i].loser;