Cs50 Tideman Solution [patched]

// Function to recount votes void recount_votes(voter_t *voters_prefs, int voters, candidate_t *candidates_list, int candidates) // Recount votes for (int i = 0; i < voters; i++) for (int j = 0; j < candidates; j++) if (candidates_list[voters_prefs[i].preferences[j] - 1].votes == 0) // Move to next preference voters_prefs[i].preferences[j] = -1; else break;

return -1;

This indicates that candidate 1 wins the election.

✅ Double-check the comparison - you want strongest victories first, so the pair with the highest preferences[winner][loser] should come first.

: Create a function like has_cycle(int start, int end) . If start == end , a loop exists; return true . Cs50 Tideman Solution

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

Here is a C solution to the CS50 Tideman problem:

#include <stdio.h> #include <stdlib.h> #include <string.h>

typedef struct int winner; int loser; pair; If start == end , a loop exists; return true

return false;

Scan columns in the 2D locked array. If a candidate's column contains exclusively false values, it means nobody has a locked victory over them. Print that candidate's name. Deep Dive: Solving the Recursion in lock_pairs

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

void record_preferences(int ranks[]) for (int i = 0; i < candidate_count; i++) for (int j = i + 1; j < candidate_count; j++) preferences[ranks[i]][ranks[j]]++; Use code with caution. 3. add_pairs() Print that candidate's name

Do you need help understanding how creates_cycle works? I can walk you through an example! Share public link

Compare preferences[i][j] with preferences[j][i] . If one value is higher, add that combination to the global pairs array as a new pair struct.

pair temp = pairs[i]; pairs[i] = pairs[max_index]; pairs[max_index] = temp;

: For every voter, use a nested loop to compare each candidate in their list to every candidate ranked : If candidate is ranked higher than candidate by a voter, increment preferences[A][B] Dev Genius 3. Create and Sort Pairs

This is the computationally complex aspect of the solution. The algorithm must create a directed graph by locking pairs one by one.