{"id":"chips-parity-reduction","text":"Moving chips by any even distance is free, so `min_cost_to_move_chips` reduces to `min(count_odd, count_even)` — move the smaller parity group across the boundary at cost 1 each.","truth_value":"IN","source":"entries/2026/06/06/minimum-cost-to-move-chips-to-the-same-position-solution.md","source_url":"","source_hash":"","justifications":[],"dependents":[],"metadata":{},"created_at":"","updated_at":"","reviewed_at":"","verified_at":"","retracted_at":"","explanation":{"steps":[{"node":"chips-parity-reduction","truth_value":"IN","reason":"premise"}]}}