Cs50 Tideman Solution Now

Introduction

Tideman is a voting system implemented in the CS50 course, where voters rank candidates in order of preference. The goal of the Tideman solution is to determine the winner of an election based on the ranked ballots. In this report, we will outline the problem, provide a high-level overview of the solution, and walk through the implementation.

Problem Statement

The Tideman problem involves implementing a voting system that takes in a list of voters, candidates, and ranked ballots. The system must then determine the winner of the election based on the following rules:

1. The candidate with the fewest first-place votes is eliminated.
2. In the event of a tie, the candidate with the fewest second-place votes is eliminated, and so on.
3. The process continues until a candidate has more than half of the first-place votes.

Solution Overview

The Tideman solution involves the following steps:

1. Read Input: Read in the number of voters, candidates, and ranked ballots.
2. Initialize Data Structures: Initialize data structures to store the vote counts and candidate preferences.
3. Count First-Place Votes: Count the first-place votes for each candidate.
4. Check for Winner: Check if a candidate has more than half of the first-place votes. If so, declare them the winner.
5. Eliminate Candidate: Eliminate the candidate with the fewest votes.
6. Recount Votes: Recount the votes after eliminating a candidate.
7. Repeat: Repeat steps 4-6 until a winner is declared.

Implementation

The implementation involves the following functions: Cs50 Tideman Solution

How to Detect a Cycle (Without Recursion... Yet)

Many solutions use recursion. But recursion for cycle detection can be confusing when you are also learning C syntax.

Here is a clean, non-recursive (or minimally recursive) mental model:

The Rule: You can lock edge (A → B) if and only if there is no path from B back to A already in the graph.

Why? Because locking A → B when B → ... → A already exists completes the cycle.

3.3 Locking Pairs (lock_pairs)

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

• The Constraint: A pair cannot be locked if it would create a cycle in the graph. If a cycle is created, there is no "source" candidate, and the election cannot be won.
• The Logic:
  1. Iterate through the sorted pairs array.
  2. Temporarily set locked[pair.winner][pair.loser] to true.
  3. Check if this creates a cycle.
  4. If a cycle is detected, revert the lock (set to false). If not, keep it true.

The Complete Solution (Step-by-Step)

Let’s implement lock_pairs and its helper creates_cycle.

Step 5: The lock_pairs Function (The Hard Part)

This is where most students fail. We must lock pairs one by one, but before locking, check if the new edge would create a cycle. Introduction Tideman is a voting system implemented in

Definition of a cycle: There is already a path from the new loser to the new winner in the locked graph. If that path exists, adding winner→loser creates a cycle.

Strategy:

1. Start with an empty locked matrix (all false).
2. For each pair in sorted order:
   • Temporarily lock it (set locked[pair.winner][pair.loser] = true).
   • Check if this creates a cycle using a recursive depth-first search (DFS) starting from the new winner.
   • If a cycle exists, unlock it (false). Otherwise, keep it locked.

Cycle detection: Write a helper function bool creates_cycle(int winner, int loser) or bool has_cycle(int start, int current).

The classic solution uses a recursive function bool cycle(int end, int cycle_start) that tries to traverse from end to cycle_start following locked edges.

Elegant recursive approach:

// Returns true if there is a path from start to end in the locked graph
bool is_path(int start, int end)
if (start == end)
        return true;
for (int i = 0; i < candidate_count; i++)
if (locked[start][i] && is_path(i, end))
        return true;
return false;


void lock_pairs(void)
for (int i = 0; i < pair_count; i++)
int w = pairs[i].winner;
int l = pairs[i].loser; The candidate with the fewest first-place votes is
    // Check if locking (w -> l) creates a cycle
    // A cycle occurs if there is already a path from l to w
    if (!is_path(l, w))
locked[w][l] = true;
return;

Why this works:

• Adding w -> l creates a cycle if you can already go from l to w using previously locked edges.
• If l can reach w, then adding w -> l closes the loop.
• This is the most concise, correct solution.

3. Locking Pairs Without Cycles

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.