Cs50 Tideman Solution [hot] May 2026
CS50 Tideman Solution: A Comprehensive Guide to the Voting System Problem
The CS50 Tideman problem is a popular problem in the CS50 course, a free online computer science course offered by Harvard University. The problem is part of the problem set 3, which focuses on algorithms and data structures. In this article, we will provide a comprehensive guide to solving the CS50 Tideman problem, also known as the "Voting System" problem.
What is the CS50 Tideman Problem?
The CS50 Tideman problem is a problem that requires you to implement a voting system based on the Tideman algorithm. The problem statement is as follows:
- In a voting system, there are
ncandidates, and each voter ranks the candidates in order of preference. - The Tideman algorithm is a ranked-choice voting system, where the winner is determined by iteratively eliminating the candidate with the fewest first-choice votes.
- The algorithm must satisfy the following constraints:
- The input consists of a list of
pairs, where each pair represents a voter and their ranked preferences. - The output is the winner of the election.
- The input consists of a list of
Understanding the Tideman Algorithm
The Tideman algorithm works as follows:
- Count First-Choice Votes: Count the number of first-choice votes for each candidate.
- Find the Candidate with the Fewest Votes: Find the candidate with the fewest first-choice votes.
- Eliminate the Candidate: Eliminate the candidate with the fewest first-choice votes.
- Update Preferences: Update the preferences of each voter by removing the eliminated candidate from their ranked list.
- Repeat: Repeat steps 2-4 until there is only one candidate left.
CS50 Tideman Solution in Python
Here is a Python solution to the CS50 Tideman problem:
def main():
# Get the number of candidates and voters
candidates = []
num_candidates = int(input("Enter the number of candidates: "))
for i in range(num_candidates):
candidate = input(f"Enter candidate i+1: ")
candidates.append(candidate)
num_voters = int(input("Enter the number of voters: "))
# Get the ranked preferences for each voter
pairs = []
for i in range(num_voters):
voter_preferences = []
print(f"\nEnter the ranked preferences for voter i+1:")
for j in range(num_candidates):
preference = input(f"Enter preference j+1: ")
voter_preferences.append(preference)
pairs.append(voter_preferences)
# Run the Tideman algorithm
winner = tideman(candidates, pairs)
if winner is not None:
print(f"\nThe winner is: winner")
else:
print("\nNo winner.")
def tideman(candidates, pairs):
# Count first-choice votes
vote_counts = candidate: 0 for candidate in candidates
for pair in pairs:
vote_counts[pair[0]] += 1
# Find the candidate with the fewest votes
min_votes = min(vote_counts.values())
min_vote_candidates = [candidate for candidate, count in vote_counts.items() if count == min_votes]
# Eliminate the candidate(s) with the fewest votes
eliminated_candidates = []
while len(min_vote_candidates) > 0:
eliminated_candidate = min_vote_candidates[0]
eliminated_candidates.append(eliminated_candidate)
candidates.remove(eliminated_candidate)
# Update preferences
pairs = update_preferences(pairs, eliminated_candidate)
# Update vote counts
vote_counts = candidate: 0 for candidate in candidates
for pair in pairs:
if len(pair) > 0:
vote_counts[pair[0]] += 1
# Find the candidate with the fewest votes
min_votes = min(vote_counts.values())
min_vote_candidates = [candidate for candidate, count in vote_counts.items() if count == min_votes]
# Return the winner
if len(candidates) == 1:
return candidates[0]
else:
return None
def update_preferences(pairs, eliminated_candidate):
updated_pairs = []
for pair in pairs:
updated_pair = [preference for preference in pair if preference != eliminated_candidate]
updated_pairs.append(updated_pair)
return updated_pairs
if __name__ == "__main__":
main()
CS50 Tideman Solution in C
Here is a C solution to the CS50 Tideman problem:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_CANDIDATES 9
#define MAX_VOTERS 9
#define MAX_NAME_LENGTH 50
// Structure to represent a candidate
typedef struct
char name[MAX_NAME_LENGTH];
int votes;
Candidate;
// Structure to represent a voter
typedef struct
char preferences[MAX_CANDIDATES][MAX_NAME_LENGTH];
Voter;
int main()
// Get the number of candidates and voters
int num_candidates, num_voters;
printf("Enter the number of candidates: ");
scanf("%d", &num_candidates);
printf("Enter the number of voters: ");
scanf("%d", &num_voters);
// Get the names of the candidates
Candidate candidates[num_candidates];
for (int i = 0; i < num_candidates; i++)
printf("Enter candidate %d: ", i+1);
scanf("%s", candidates[i].name);
candidates[i].votes = 0;
// Get the ranked preferences for each voter
Voter voters[num_voters];
for (int i = 0; i < num_voters; i++)
printf("\nEnter the ranked preferences for voter %d:\n", i+1);
for (int j = 0; j < num_candidates; j++)
printf("Enter preference %d: ", j+1);
scanf("%s", voters[i].preferences[j]);
// Run the Tideman algorithm
char* winner = tideman(candidates, num_candidates, voters, num_voters);
if (winner != NULL)
printf("\nThe winner is: %s\n", winner);
else
printf("\nNo winner.\n");
return 0;
char* tideman(Candidate candidates[], int num_candidates, Voter voters[], int num_voters) {
// Count first-choice votes
for (int i = 0; i < num_candidates; i++)
candidates[i].votes = 0;
for (int i = 0; i < num_voters; i++)
for (int j = 0; j < num_candidates; j++)
if (strcmp(voters[i].preferences[j], "") != 0)
for (int k = 0; k < num_candidates; k++)
if (strcmp(candidates[k].name, voters[i].preferences[j]) == 0)
candidates[k].votes++;
break;
// Find the candidate with the fewest votes
int min_votes = MAX_VOTERS;
for (int i = 0; i < num_candidates; i++)
if (candidates[i].votes < min_votes)
min_votes = candidates[i].votes;
// Eliminate the candidate(s) with the fewest votes
int eliminated_candidates = 0;
while (eliminated_candidates < num_candidates - 1) {
// Find the candidate with the fewest votes
int min_vote_index = -1;
for (int i = 0; i < num_candidates; i++)
if (candidates[i].votes == min_votes)
min_vote_index = i;
break;
// Eliminate the candidate
eliminated_candidates++;
candidates[min_vote_index].votes = -1;
// Update preferences
for (int i = 0; i < num_voters; i++)
for (int j = 0; j < num_candidates; j++)
if (strcmp(voters[i].preferences[j], candidates[min_vote_index].name) == 0)
for (int k = j; k < num_candidates - 1; k++)
strcpy(voters[i].preferences[k], voters[i].preferences[k+1]);
strcpy(voters[i].preferences[num_candidates-1], "");
j--;
// Update vote counts
for (int i = 0; i < num_candidates; i++)
candidates[i].votes = 0;
for (int i = 0; i < num_voters; i++)
for (int j = 0; j < num_candidates; j++)
if (strcmp(voters[i].preferences[j], "") != 0)
for (int k = 0; k < num_candidates; k++)
if (strcmp(candidates[k].name, voters[i].preferences[j]) == 0)
candidates[k].votes++;
break;
// Find the new minimum votes
min_votes = MAX_VOTERS;
for (int i = 0; i < num_candidates; i++) {
if (candidates[i].votes >= 0 && candidates[i].
Once upon a time in the land of Cambridge, Massachusetts, a weary student named Alex sat before a glowing screen, haunted by the legendary monster of Problem Set 3: . The Call to Battle
Alex had conquered the simple Runoff election, but Tideman was a different beast. The CS50 curriculum demanded a more sophisticated winner—a candidate who could win head-to-head battles without creating an infinite loop of confusion. The Strategy To defeat the beast, Alex had to master several weapons: Cs50 Tideman Solution
The Preferences: Alex collected the ranks of every voter, tallying how many people preferred Alice over Bob.
The Pairs: Every possible matchup was listed. Alice vs. Bob, Bob vs. Charlie, Charlie vs. Alice.
The Sorting: Alex used a "Strength of Victory" spell, sorting the pairs so the most landslide wins came first. The Locked Trap
Then came the hardest part: locking the pairs. Alex had to draw arrows from winners to losers. But there was a curse—if an arrow created a "cycle" (where Alice beats Bob, Bob beats Charlie, and Charlie beats Alice), the arrow could not be drawn.
Alex spent three days staring at a "No Cycle" function, battling the dark magic of Recursion. "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
Once the arrows were locked, the answer was revealed. The winner was the Source—the one candidate who had arrows pointing at others but no arrows pointing at themselves.
As the terminal finally flashed green with the words All tests passed!, Alex didn't just find a solution; they had earned the right to call themselves a programmer. (CS50) TIDEMAN - PROBLEM SET 3 | SOLUTION
Here’s a concise, practical guide to implementing CS50’s Tideman (ranked pairs) solution in C, with key concepts, structure, and pitfalls.
Step 1: The Global Variables (Given by CS50)
// Number of candidates int candidate_count;
// locked[i][j] means i is locked over j bool locked[MAX][MAX];
Step 4: The print_winner Function (The Finishing Touch)
After locking, the winner is the candidate with no incoming edges (nobody has locked[j][i] == true for that i). CS50 Tideman Solution: A Comprehensive Guide to the
void print_winner(void)
for (int i = 0; i < candidate_count; i++)
bool is_source = true;
for (int j = 0; j < candidate_count; j++)
if (locked[j][i]) // If someone beats i
is_source = false;
break;
if (is_source)
printf("%s\n", candidates[i]);
return;
The Setup: It Starts Simple
The beginning is deceptively easy. You have to record preferences. If Alice beats Bob, Alice gets a point. Simple array work. I thought, "Hey, this isn't so bad."
Then came the graph.
Understanding the Tideman Algorithm
Before touching code, you must understand the three core stages:
- Record preferences – tally each pair of candidates to see who voters prefer.
- Sort pairs – order all possible candidate pairs by the strength of victory (largest margin first).
- Lock pairs without cycles – add pairs to a directed graph, skipping any pair that would create a cycle.
The winner is the candidate who ends up at the “source” of the locked graph (no incoming edges).
Why This Works (The “Aha” Moment)
The trick is realizing you don’t check cycles in the new edge itself. You check for existing paths.
Visual example:
- Locked edges: Alice → Bob, Bob → Charlie.
- Next pair: Charlie → Alice.
Your code sees: Charlie → ? → Alice? It checks: does Charlie beat anyone locked? Yes, Charlie beats nobody yet in locked. Wait — we check recursively.
Actually, step through:
creates_cycle(Charlie, Alice)- Does
locked[Charlie][any]? No. So loop ends. No cycle. Lock.
That’s wrong — wait! That would be a cycle (Charlie → Alice when Alice → Bob → Charlie already locked). Did we miss it?
Important correction: 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:
Locked: Alice→Bob, Bob→Charlie. Testing Charlie→Alice: Start from loser (Alice? No! loser is the second arg? Careful.) In a voting system, there are n candidates,
Let’s rename parameters for clarity:
bool creates_cycle(int from, int to) // We want to lock from->to
// If 'from' already beats 'to' indirectly
for (int i = 0; i < candidate_count; i++)
if (locked[from][i])
creates_cycle(i, to))
return true;
return false;
Then in lock_pairs:
if (!creates_cycle(winner, loser))
locked[winner][loser] = true;
Now test: Locked: Alice→Bob, Bob→Charlie. Check Charlie→Alice:
- From Charlie: any locked? No. Return false. Lock. → Bug! This should be a cycle!
The fix: We must check the reverse direction. A cycle occurs if there is a path from loser to winner using existing locked edges before adding winner→loser.
So the correct helper:
bool will_create_cycle(int winner, int loser) // If loser can reach winner already, then adding winner->loser creates cycle return can_reach(loser, winner);
bool can_reach(int from, int target) if (from == target) return true; for (int i = 0; i < candidate_count; i++) if (locked[from][i] && can_reach(i, target)) return true; return false;
Now test again: Locked: Alice→Bob, Bob→Charlie. Check Charlie→Alice: can_reach(Alice, Charlie)? No (Alice beats Bob, Bob beats Charlie, so yes! Alice can reach Charlie). Wait, that’s not what we check.
We check can_reach(loser, winner) = can_reach(Alice, Charlie)? Yes → cycle detected. Correct!
So the final correct version:
bool can_reach(int from, int target) if (from == target) return true; for (int i = 0; i < candidate_count; i++) if (locked[from][i] && can_reach(i, target)) return true; return false;bool will_create_cycle(int winner, int loser) return can_reach(loser, winner);
void lock_pairs(void) for (int i = 0; i < pair_count; i++) int w = pairs[i].winner; int l = pairs[i].loser; if (!will_create_cycle(w, l)) locked[w][l] = true;