Think of the locked array as a directed graph. Each candidate is a node. When you lock a pair (winner → loser), you draw an arrow from the winner to the loser.
The recursive is_path approach is the clearest expression of cycle detection. It directly mirrors the definition: "Adding edge X→Y creates a cycle if Y can already reach X."
Here is a comprehensive solution to the CS50 Tideman problem:
Nested loops checking preferences[i][j] > preferences[j][i] . If true, set pairs[pair_count].winner = i , pairs[pair_count].loser = j , and increment pair_count . 4. void sort_pairs(void) Cs50 Tideman Solution
Tideman's algorithm works by:
To develop this feature, you must implement a helper function (often called
// Recursive Step: Check all candidates to see if 'end' has an edge to them ; i < candidate_count; i++) (locked[end][i]) (has_cycle(start, i)) ; Think of the locked array as a directed graph
Scan the locked matrix to find the "source" of the graph.
You can use a simple Selection Sort or Bubble Sort since the candidate count is small (maximum of 9).
We need to populate the global pairs array. A pair exists if preferences[i][j] > preferences[j][i] . If equal (tie), skip. The recursive is_path approach is the clearest expression
Increment preferences[i][j] , where i is the preferred candidate and j is the less-preferred candidate. 3. Add Pairs The add_pairs function populates the pairs array.
// Returns true if locking winner->loser would create a cycle bool creates_cycle(int winner, int loser)
To check for a cycle, you must write a recursive helper function (typically called has_cycle ). This function checks if the current loser can trace a path back to the current winner through already locked edges.
) that checks: "If I point an arrow from Winner to Loser, can I eventually follow a path of existing arrows from Loser back to Winner?" If the answer is yes, you skip that pair. 5. Identify the winner Once all non-cyclical pairs are locked, the winner is the of the graph. In graph theory, the source is the node with zero edges pointing towards it . In the code, you look for a candidate such that for all candidates locked[j][i] ✅ Summary