The CS50 Tideman problem set requires implementing a "ranked pairs" voting system that guarantees a Condorcet winner if one exists. Solving it involves completing six primary functions: vote, record_preferences, add_pairs, sort_pairs, lock_pairs, and print_winner. Core Logic Overview
The algorithm follows three main phases to determine a winner:
Tallying: It examines every possible head-to-head matchup between candidates to see who is preferred.
Sorting: Matchups are ordered by "strength of victory," which is the margin by which the winner was preferred.
Locking: The strongest matchups are "locked" into a directed graph as edges (arrows), provided they do not create a cycle. Implementation Guide
Detailed walkthroughs and guides like those on Medium by Jacob and Jordan Rogers' Blog offer step-by-step logic. Key technical challenges include: CS50 Tideman - Dev Genius
The CS50 Tideman problem set challenges you to implement a Ranked Pairs voting system, which is designed to identify a Condorcet winner—a candidate who wins every head-to-head matchup.
The primary difficulty lies in constructing a directed graph of candidates and using recursion to ensure no cycles are created when "locking in" winning pairs. Core Logic: Tally, Sort, Lock The algorithm is broken down into three main phases:
Tally: Compare every pair of candidates to see who is preferred by more voters.
Sort: Order these pairs by their "strength of victory" (the number of voters preferring the winner) in descending order.
Lock: Add these pairs to a graph as directed edges (arrows) from winner to loser, only if doing so doesn't create a cycle. Phase 1: Recording Preferences
You must first populate a 2D preferences[i][j] array, where the value represents how many voters prefer candidate i over candidate j.
Vote Function: Validates a voter's choice by checking if the name exists in the candidates array and updates the ranks array.
Record Preferences: 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
After identifying all winner-loser pairs, you store them in a pairs array. Each pair struct contains a winner index and a loser index. CS50 Tideman - Dev Genius Cs50 Tideman Solution
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
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
The CS50 Tideman problem set is widely considered the most difficult challenge in the CS50 course. It requires implementing the Tideman voting system (ranked-choice voting), which involves complex graph theory and recursion to determine a winner while avoiding cycles. Core Problem Overview
The goal is to complete six functions that together simulate a "ranked-pair" election. The process follows these stages: The CS50 Tideman problem set requires implementing a
(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)
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? // Recursive Step: Check all candidates to see
This post is structured for someone who has struggled with the problem (as many do) and wants to understand why the solution works, not just what to type.
The winner is the candidate with no incoming locked edges (i.e., no locked[j][i] true for any j).
Loop through each candidate, and if none of the other candidates have a locked edge pointing to them, that candidate wins.
Before writing a single line of code, you must grasp the logic.
In a Tideman election:
Example: If candidate A beats B by 7 votes, and B beats C by 5 votes, but C beats A by 2 votes, a cycle exists (A→B→C→A). Tideman avoids this by skipping the weakest edge in the cycle.
A common mistake students make is sorting based only on the raw number of votes for the winner, rather than the margin of victory. However, the Tideman specification dictates sorting by victory strength (margin), which requires accessing both preferences[winner][loser] and preferences[loser][winner].
To detect a cycle, we use a recursive depth-first search (DFS). The function cycle_check(int start, int end) determines if adding an edge from start to end closes a loop.
start == end, we have returned to the node we began with. A cycle exists. Return true.locked[end][i] is true (meaning there is an arrow from the current endpoint to candidate i), recursively call the function with the start node and i as the new end node.false.The Tideman solution forces you to think algorithmically about graph theory, recursion, and sorting. It’s not about getting the code right on the first try, but about breaking the problem into small, testable pieces.
If you’re stuck, run check50 after completing each function, use printf debugging to see the state of preferences and pairs, and draw the cycle detection logic on paper.
Once you get it working, you’ll feel a deep sense of accomplishment — and a much stronger grasp of recursion and graph traversal.
Writing a "good" post about the CS50 Tideman problem usually means writing a "Problem Set Story." This is a popular format in the CS50 community (often seen on Medium, Dev.to, or Reddit) where you document your struggle and eventual triumph.
Because the course encourages academic honesty, a good post never pastes the full code solution. Instead, it explains the logic and the algorithm.
Here is a template and a drafted post you can use or adapt. It focuses on the hardest part of the problem: the sort_pairs and lock_pairs functions.
sort_pairs FunctionSort pairs in descending order of victory margin: margin = preferences[winner][loser] - preferences[loser][winner].
You can use any stable sorting algorithm. Bubble sort is fine for small candidate counts.
void sort_pairs(void)
for (int i = 0; i < pair_count - 1; i++)
for (int j = 0; j < pair_count - i - 1; j++)
int margin1 = preferences[pairs[j].winner][pairs[j].loser] - preferences[pairs[j].loser][pairs[j].winner];
int margin2 = preferences[pairs[j+1].winner][pairs[j+1].loser] - preferences[pairs[j+1].loser][pairs[j+1].winner];
if (margin1 < margin2)
pair temp = pairs[j];
pairs[j] = pairs[j+1];
pairs[j+1] = temp;
return;
Many students try to detect cycles by checking if locked[loser][winner] already exists. That is wrong. A cycle can be longer than two edges (A→B→C→A). Always use DFS.