Maya submitted her solution. And in the real election that followed, Alice became Keeper of the Orchard—not because she was the strongest in every head-to-head match, but because when paradoxes arose, the village had a coder wise enough to know which locks to leave open. Don't just check for a two-step loop. Use depth-first search to see if the loser has any path to the winner in the existing locked graph. If yes, skip the pair. That’s the entire secret of Tideman.
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.
In a directed graph, adding an edge from A → B creates a cycle if and only if B can already reach A.
Maya ran check50 . Green smiles across the board. She leaned back. Cs50 Tideman Solution
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.
"Show me your cycle detection," Kai said.
"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." Maya submitted her solution
Kai chuckled. "That's not just Tideman, Maya. That's life. Don't create cycles. Always check if the person you're stepping on has a hidden path back to you."
// 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; }
"Yes," Maya sighed. "I sort the pairs. Strongest first. Alice over Bob? Lock it. Bob over Charlie? Lock it. Charlie over Alice? Don't lock it because it creates a cycle. But my cycle detection is wrong." Use depth-first search to see if the loser
Her job was to "lock in" the strongest edges of victory to create a directed graph of the winner—without creating a cycle.
He drew on the whiteboard:
She stared at her lock_pairs function. It was midnight. Her screen showed the dreaded red “:(” from check50 .
Maya’s heart sank. She had been checking loser → X → winner . But what about loser → X → Y → winner ?