Cs50 Tideman Solution 〈Easy ✪〉

// Store the candidate's index at the specified rank ranks[rank] = i; return true;

void add_pairs(void)

The Ranked Pairs method works by determining which candidate wins in head-to-head matchups. The algorithm follows a strict five-step process: Voters rank all candidates in order of preference.

Locked: Alice→Bob, Bob→Charlie. Testing Charlie→Alice: Start from loser (Alice? No! loser is the second arg? Careful.) Cs50 Tideman Solution

void print_winner(void)

For every candidate ranked higher, increment preferences[i][j] , where i is the preferred candidate and j is the less-preferred candidate.

# Update vote counts vote_counts = candidate: 0 for candidate in candidates for pair in pairs: if len(pair) > 0: vote_counts[pair[0]] += 1 // Store the candidate's index at the specified

// Reset pair_count to ensure it starts from 0 pair_count = 0; // Loop through all pairs of candidates for (int i = 0; i < candidate_count; i++)

Update a 2D array ( preferences[i][j] ) representing how many voters prefer candidate i over candidate j .

Your code sees: Charlie → ? → Alice? It checks: does Charlie beat anyone locked? Yes, Charlie beats nobody yet in locked. Wait — we check recursively. Testing Charlie→Alice: Start from loser (Alice

if (preferences[pairs[j].winner][pairs[j].loser] > preferences[pairs[max_index].winner][pairs[max_index].loser])

pairs[pair_count].winner = i; pairs[pair_count].loser = j; pair_count++;

Some students only check if preferences[i][j] > preferences[j][i] , forgetting that ties should be skipped entirely.