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How does pruning choice points in the code below make it more efficient (Prolog)?

In the code given below, there is the ! (cut) that prunes the choice point for efficiency. I am pretty certain that the reverse predicate and the agent_do_moves predicate are essential. solve_task(Task,Cost):- agent_current_position(oscar,P), solve_task_a(Task,[b(0,0,P)],[],R,Cost,_NewPos),!, % prune choice point for efficiency reverse(R,[_Init|Path]), agent_do_moves(oscar,Path). ...