// 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++)
Here is a clean, non-recursive (or minimally recursive) mental model: Cs50 Tideman Solution
# Perform pairwise comparisons for i in range(len(candidates)): for j in range(i + 1, len(candidates)): # Determine which candidate wins the comparison winner = compare_candidates(candidates[i], candidates[j], voter_preferences) if winner == candidates[i]: win_counts[candidates[i]] += 1 else: win_counts[candidates[j]] += 1 // Check all candidates to see if loser
:param candidate1: Name of the first candidate :param candidate2: Name of the second candidate :param voter_preferences: List of voter preferences, where each preference is a list of candidate names in ranked order :return: The candidate that is preferred by more voters """ # Count the number of voters who prefer each candidate count1 = sum(1 for preference in voter_preferences if candidate1 in preference and candidate2 not in preference[:preference.index(candidate1)]) count2 = sum(1 for preference in voter_preferences if candidate2 in preference and candidate1 not in preference[:preference.index(candidate2)]) i++) Here is a clean
eliminate_candidate(candidates_list, candidates, eliminated);