Cs50: Tideman Solution

Maya was the new programmer tasked with tabulating the votes. She had the first part down: counting each ballot to build a 2D array of preferences . It told her that Alice beat Bob (5 votes to 2), Bob beat Charlie (4 to 3), and Charlie beat Alice (3 to 2). A perfect, frustrating cycle.

He drew on the whiteboard:

"Show me your cycle detection," Kai said. Cs50 Tideman Solution

Maya’s heart sank. She had been checking loser → X → winner . But what about loser → X → Y → winner ?

// Returns true if adding edge winner->loser creates a cycle bool creates_cycle(int winner, int loser) { // If the loser can reach the winner through existing locked edges, // then adding winner->loser would complete a cycle. return dfs(loser, winner); } bool dfs(int current, int target) { if (current == target) return true; for (int i = 0; i < candidate_count; i++) { if (locked[current][i] && dfs(i, target)) return true; } return false; } Maya was the new programmer tasked with tabulating the votes

She stared at her lock_pairs function. It was midnight. Her screen showed the dreaded red “:(” from check50 .

Her friend, an old sysadmin named Kai, peered over her shoulder. "You're trying to lock every pair in order of strength, right?" A perfect, frustrating cycle

"You’re not just looking for a loop," Kai said. "You’re looking for a chain . Before you lock a new edge from winner to loser , ask yourself: is there any path from the loser back to the winner using the edges already locked? If yes, this new edge would complete the cycle. Skip it."

Every year, the village of Coderidge held an election for the Keeper of the Orchard. Unlike other villages, they used a complex ranked voting system designed by a long-dead mathematician named Tideman. The rule was simple: if there was a way to trace a circle of preference (A beats B, B beats C, C beats A), that circle was a paradox, and the weakest link in that circle must be ignored.

In a directed graph, adding an edge from A → B creates a cycle if and only if B can already reach A.

Maya pointed. "I wrote a recursive function creates_cycle(winner, loser) . It checks if the loser has any locked edges pointing to another candidate. Then it checks if that candidate points back to the original winner. If yes, it’s a cycle."