While there are many sorting algorithms you could use (like bubble sort or selection sort), a selection sort is a reliable choice for this problem. Here's a standard approach:
return 0;
// Structure to represent a voter typedef struct voter int *preferences; voter_t;
This public link is valid for 7 days and shares a thread, including any personal information you added. This link or copies made by others cannot be deleted. If you share with third parties, their policies apply. Can’t copy the link right now. Try again later. Cs50 Tideman Solution
Check C→A: Start = C (loser), end = A (winner). Path? B→C? No, C→ nothing. Actually: Wait — We check from loser (C) to winner (A). DFS from C: C→? No outgoing in locked yet. So no path. Lock it? But locking C→A would create A→B→C→A (cycle). Let’s trace:
for (int i = 0; i < candidate_count; i++) if (locked[loser][i]) if (creates_cycle(winner, i)) return true;
return false;
The problem specification guarantees a unique winner, simplifying the print_winner function. However, robust code would handle scenarios where multiple sources might exist (though the CS50 problem ensures this does not happen with the provided test data).
The Tideman problem is challenging precisely because it requires understanding graph theory concepts and recursive algorithms. However, by breaking down the problem into these six functions and understanding what each one accomplishes, you can build a complete solution that correctly implements the ranked pairs voting method.
: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)]) While there are many sorting algorithms you could
If the recursive loop finishes without finding a cycle, it is safe to set locked[winner][loser] = true . 6. Print Winner Once the graph is locked, you must identify the source.
Ordering these head-to-head matchups from the largest margin of victory to the smallest.
: Because of its difficulty, students are frequently warned that looking up a full "Tideman solution" is considered a violation of academic honesty . If you share with third parties, their policies apply
If preferences[i][j] > preferences[j][i] , initialize a new pair struct with i as the winner and j as the loser. Increment the global pair_count . Do nothing in the case of an exact tie. 4. Sort Pairs