Cs50 Tideman Solution File

if (will_create_cycle(winner, i)) return true;

strcpy(voters[i].preferences[num_candidates-1], ""); j--;

For those uninitiated, the Tideman problem (pset3) is the infamous "boss fight" of the early weeks. It takes the concepts of plurality and runoff elections and cranks the difficulty up to 11 by introducing a ranked-choice voting system that uses a directed graph.

: Iterates through the ranks array. For every candidate at a higher rank (earlier index), you increment their preference count against every candidate at a lower rank (later index). Phase 2: Sorting Pairs Cs50 Tideman Solution

The core of the Tideman algorithm works by:

. As you iterate through each voter's ranked ballot, you update this matrix for every possible pair. 2. Create the pairs Next, we identify every pair of candidates where one candidate beat the other (i.e., preferences[i][j] > preferences[j][i] ). Each pair is stored in a struct containing: : The candidate with more votes. : The candidate with fewer votes. : The strength of the victory ( 3. Sort the pairs

The ultimate goal is to determine the winner of an election where voters rank candidates in order of preference. For every candidate at a higher rank (earlier

# Update preferences pairs = update_preferences(pairs, eliminated_candidate)

: The preferences[i][j] 2D array represents the number of voters who prefer candidate i over candidate j .

for (int j = 0; j < pair_count - i - 1; j++) for (int i = 0

int preferences[MAX][MAX]; bool locked[MAX][MAX]; string candidates[MAX]; int pair_count; int candidate_count;

Ensure you sort descending . Reversing the comparison sign is a common typo.

This function fills the pairs array with matchups where one candidate beats another. Compare preferences[i][j] with preferences[j][i] .

Then came the graph.

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