Cs50 Tideman Solution -

We have an election. Voters rank candidates. The Tideman method (a ranked-pair voting system) works like this:

Your job is to write the functions for steps 2, 3, 4, and 5. But step 4—lock_pairs—is where most people get stuck.

This is the hardest part. You must detect if adding a new edge from winner to loser would create a cycle in the locked graph.

A cycle exists if there’s already a path from loser back to winner. Write a recursive function:

bool creates_cycle(int start, int current)
if (start == current)
        return true;
    for (int i = 0; i < candidate_count; i++)
        if (locked[current][i] && creates_cycle(start, i))
            return true;
    return false;

Before locking a pair (w, l), check: if (!creates_cycle(w, l)) then locked[w][l] = true.

A Comprehensive Analysis of the CS50 Solution

A pair struct stores winner and loser.
add_pairs loops through all (i, j) where preferences[i][j] > preferences[j][i] and adds that pair.

sort_pairs sorts pairs in decreasing order of victory margin:
preferences[pair.winner][pair.loser] - preferences[pair.loser][pair.winner]

Tip: Use a simple sorting algorithm (bubble sort works fine here) because the number of candidates is small (max 9).

Tideman is hard because it introduces graph theory without explicitly teaching it. But once you understand cycle detection as “can the loser reach the winner already,” the solution becomes clean and elegant.

Stick with it. Get the small test cases working (3 candidates). Then scale up. And remember — in CS50, the Tideman problem is marked as “more comfortable” for a reason. If you complete it, you have truly leveled up.

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

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:

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 Cs50 Tideman Solution

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

The CS50 Tideman problem is widely regarded by students as one of the most difficult challenges in the CS50x curriculum. It focuses on the Tideman (or Ranked Pairs) voting system, where winners are determined through a directed graph of candidate preferences. The Story of the "Lock Pairs" Battle

For many students, the "story" of solving Tideman is a journey from initial confidence to total frustration, ending in a hard-won "Aha!" moment.

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:

Understanding the Tideman Algorithm

The Tideman algorithm works as follows:

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].

CS50 Tideman problem set, the most challenging "feature" to develop is the lock_pairs

function. Its goal is to create a directed acyclic graph (DAG) by locking pairs of candidates in order of their strength of victory, provided that locking a pair does not create a cycle. The Core Logic: lock_pairs

To develop this feature, you must implement a helper function (often called

) that uses recursion or a search algorithm (like Depth-First Search) to check if adding an edge from Candidate A to Candidate B creates a path back to A. 1. Implement the Cycle Check We have an election

Create a boolean function that checks if a path exists between two nodes in the

// Returns true if adding an edge from 'start' to 'end' creates a cycle has_cycle(

// Base Case: If the target (end) can already reach the start, a cycle is found (start == end)

// 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)) ; Use code with caution. Copied to clipboard lock_pairs Iterate through your sorted

array. For each pair, call your cycle check before locking it in the adjacency matrix. lock_pairs( ; i < pair_count; i++)

// Check if locking this pair (winner -> loser) creates a cycle back to the winner

(!has_cycle(pairs[i].winner, pairs[i].loser)) locked[pairs[i].winner][pairs[i].loser] = ; Use code with caution. Copied to clipboard Key Considerations Sorting is Critical sort_pairs

function must correctly sort pairs in decreasing order of "strength of victory" (the number of voters who preferred the winner over the loser) before lock_pairs is called. : The simplest base case for the recursion is when the node of the current edge is the same as the node of the initial edge you are trying to lock. Graph Representation locked[i][j] 2D boolean array represents a directed edge from candidate to candidate

(or Ranked Pairs) algorithm is widely considered the most difficult problem in CS50x.

Its goal is to determine a winner in an election using a ranked-choice system that satisfies the Condorcet criterion

: if there is a candidate who would win every head-to-head matchup against every other candidate, that person must win the election. The Problem with Plurality

In a standard "winner-take-all" election, a candidate can win with of the vote even if

of the population dislikes them. Tideman solves this by having voters rank candidates (1st, 2nd, 3rd), then breaking those rankings down into every possible pair of opponents to see who is truly preferred by the majority. 1. Count the votes

The first step is to record voter preferences. We use a 2D array, preferences[i][j]

, where the value represents how many voters preferred candidate over candidate

. 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

To ensure the most "certain" victories are respected first, we sort the array in descending order based on the margin of victory

. This is the "Ranked" part of Ranked Pairs. If Candidate A beats B by 10 votes, and Candidate C beats D by only 2 votes, Candidate A's victory is processed first. 4. Lock the pairs (The "Cycle" Check) Your job is to write the functions for steps 2, 3, 4, and 5

This is the core challenge of Tideman. We iterate through our sorted pairs and "lock" them into a directed graph (using a boolean matrix locked[i][j] : You can only lock a pair if it does create a cycle. What is a cycle?

If A beats B, and B beats C, you cannot lock a victory for C over A. Doing so would create a "Rock-Paper-Scissors" loop where no one is at the top. : You must implement a recursive function (often called

) 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

The Tideman algorithm ensures a fair winner by ranking margins of victory and strictly forbidding cycles. The final winner is the candidate who remains "undefeated" in the locked preference graph, acting as the ultimate source of the electoral flow. used to detect cycles?

The CS50 Tideman problem set is widely considered the most difficult assignment in Harvard’s introductory computer science course. It tasks students with implementing the Tideman voting method (also known as Ranked Pairs), a system designed to find a "Condorcet winner"—a candidate who would win against any other candidate in a head-to-head matchup. 1. Record Voter Preferences

The first step is to process every voter's ballot. For each voter, the code must update a 2D preferences[i][j] array, where the value at index [i][j] represents the number of voters who preferred candidate i over candidate j. 2. Identify and Sort Matchups

Once all votes are in, the program identifies every possible head-to-head pair.

Identify: A pair is added to a pairs array if one candidate has more votes than the other.

Sort: The pairs are then sorted in descending order based on the "strength of victory" (the difference in votes between the winner and loser of that pair). 3. Build the Winner Graph

This is the most complex phase. The program iterates through the sorted pairs and "locks" them into a directed graph (using a locked[i][j] boolean matrix).

Cycle Detection: A pair is only locked if it does not create a cycle in the graph. For example, if A beats B and B beats C, you cannot lock a pair where C beats A, as this would create a loop where no clear winner exists.

Recursion: Most students use a recursive helper function to "trace" the path from the winner of the current pair to see if it eventually leads back to the loser. 4. Identify the Source

After all valid pairs are locked, the winner of the election is the "source" of the graph. This is the candidate who has zero arrows pointing toward them (no locked[i][j] is true where j is that candidate). Key Challenges & Academic Honesty

Complexity: Unlike earlier problems like Runoff or Cash, Tideman requires advanced logic for graph theory and recursion.

Academic Integrity: Because of its difficulty, students are frequently warned that looking up a full "Tideman solution" is considered a violation of academic honesty.

Here’s a concise, practical guide to implementing CS50’s Tideman (ranked pairs) solution in C, with key concepts, structure, and pitfalls.

The core unit of the Tideman algorithm is the pairwise comparison. A pair struct is defined to hold the indices of the winner and loser of a specific matchup:

typedef struct 
    int winner;
    int loser;
 pair;