Solution — Cs50 Tideman

Lock_Pairs and the Cycle CheckThis is the hardest part of the problem set. You must iterate through your sorted pairs and lock them into the locked[i][j] boolean matrix. However, before locking, you must check if adding that edge creates a cycle.

Drawing arrows from winners to losers to create a directed graph, while carefully skipping any arrow that creates a closed loop (cycle).

The core of the Tideman algorithm works by:

typedef struct

Happy coding, and may your locks always be cycle-free.

: Before locking a pair, the program must check if doing so would create a cycle (e.g., A beats B, B beats C, and C beats A). This is typically solved using

// Read in voter preferences for (int i = 0; i < *voters; i++) (*voters_prefs)[i].preferences = malloc(*candidates * sizeof(int)); for (int j = 0; j < *candidates; j++) scanf("%d", &(*voters_prefs)[i].preferences[j]); Cs50 Tideman Solution

My earlier simple example fails because the cycle is three edges. The recursion must start from loser and try to reach winner . But in our state:

: If the candidate name is found, set ranks[rank] = candidate_index and return true . Else false .

Alex spent three days staring at a "No Cycle" function, battling the dark magic of . "How do I know if I'm pointing back to where I started?" Alex cried out. After many mugs of coffee and failed check50 runs, the logic clicked. To see if an arrow from A to B would create a cycle, Alex had to check if B already had a path leading back to A. The Source of Victory Lock_Pairs and the Cycle CheckThis is the hardest

return;

if you are struggling with sort_pairs .

This logic ensures that for each ballot, every pair of candidates is correctly tallied based on the voter's order of preference. Drawing arrows from winners to losers to create

If a matchup results in a perfect tie, it is not added to the pairs array.

for (int i = 0; i < pair_count; i++)