func New(initialState, finalState State) *graph.Problem {
return &graph.Problem{
Actions,
heuristic,
graph.NewCost(1<<31 - 1),
initialState,
func(s graph.State) bool { return s.(State) == finalState },
Result,
StepCost,
graph.NewCost(0)}
}
func New(from City) *graph.Problem {
return &graph.Problem{
Actions: func(s graph.State) []graph.Action {
rv := []graph.Action{}
for k := range edge[s.(City)] {
rv = append(rv, k)
}
return rv
},
Heuristic: func(s graph.State) graph.Cost { return graph.NewCost(toBucharest[s.(City)]) },
Infinity: graph.NewCost(1<<31 - 1),
InitialState: from,
IsGoal: func(s graph.State) bool { return s.(City) == Bucharest },
Result: func(s graph.State, a graph.Action) graph.State { return a.(City) },
StepCost: func(s graph.State, a graph.Action) graph.Cost {
return graph.NewCost(edge[s.(City)][a.(City)])
},
Zero: graph.NewCost(0),
}
}
// Function h returns a lower bound on cost of solving the puzzle.
// Each piece must move at least the manhatten distance to its correct position.
func heuristic(_s graph.State) graph.Cost {
manhatten := [9][9]int{
{0, 1, 2, 1, 2, 3, 2, 3, 4}, {1, 0, 1, 2, 1, 2, 3, 2, 3}, {2, 1, 0, 3, 2, 1, 4, 3, 2},
{1, 2, 3, 0, 1, 2, 1, 2, 3}, {2, 1, 2, 1, 0, 1, 2, 1, 2}, {3, 2, 1, 2, 1, 0, 3, 2, 1},
{2, 3, 4, 1, 2, 3, 0, 1, 2}, {3, 2, 3, 2, 1, 2, 1, 0, 1}, {4, 3, 2, 3, 2, 1, 2, 1, 0}}
ts := _s.(State)
c := 0
for i := uint(0); i < 9; i++ {
tile := ts.Get(i)
if tile != 0 {
c += manhatten[i][tile]
}
}
return graph.NewCost(c)
}
func StepCost(s graph.State, a graph.Action) graph.Cost {
return graph.NewCost(1)
}