Cs50 Tideman Solution May 2026

"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 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." Cs50 Tideman Solution

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. "You’re not just looking for a loop," Kai said

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. If yes, this new edge would complete the cycle

Her job was to "lock in" the strongest edges of victory to create a directed graph of the winner—without creating a cycle.

// 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; }