Admissibility,
Monotonicity, Informedness
Three questions we ask about
heuristic algorithms?
- Is it guaranteed to find the shortest path?
- Is it better than another heuristic?
- Does it make steady progress toward a goal?
Admissibility
A search algorithm is admissible
if it is guaranteed to find a minimal path to a solution whenever such a
solution exists. Breadth first search is admissible, because it looks at
every state at level n before considering any state at level n+1.
The
A* (A star) Algorithm
Let's characterize a class of
admissible heuristic search strategies, using the evaluation function:
f(n) = g(n) + h(n)
g(n) represents the actual
distance at which the state n has been found in the graph, and h(n)
is the heuristic estimate of the distance from n to a goal state.
So f(n) represents an estimate of the total cost of the path from
the start, through n to the goal.
Let's define the evaluation function
f*:
f*(n) = g*(n) + h*(n)
where g*(n) is the
cost of the shortest path from the start to n and h*(n)
returns the actual cost of the shortest path from n to the goal.
So f* is the actual cost of the optimal path from the
start to the goal through n.
We call the function f*
an oracle -- an ideal evaluation function that can see the
shortest path from the start to the goal. Of course, oracles of this type don't
exist for most search problems. For real problems, we want an evaluation
function, f, that approaches f*
Algorithm A: In algorithm A, g(n), the cost of the current path
from start to n, is a reasonable estimate of g*:
g(n) >= g*(n)
g(n) and g*(n) will only be equal if the
search has found the optimal path to n. Similarly, in algorithm A, we
replace h*(n) with h(n). Although we can't actually
compute h*(n), we can often determine whether h(n) is
bounded by h*(n) -- that is, we can often determine that h(n)
is less than or equal to the actual cost of a minimal path (h*(n)).
Algorithm A: Algorithm A is the best-first search algorithm plus the
evaluation function f(n) = g(n) + h(n).
Algorithm A*: is algorithm A in which
h(n) <= h*(n)
for all states n. In other
words, if an algorithm uses an evaluation function that underestimates the
cost to the goal it is an A* algorithm.
Key Point: All A* algorithms are admissible.
Breadth-first Search is an A* algorithm in which
f(n) = g(n) + 0
In other words, bread-first search
uses a trivial estimate of the distance to the goal.
Route Finding Example: For route-finding problems, the straight-line distance from
city n to a goal city is a lower bound on the distance of an optimal
route from n to the goal.
Eight puzzle Example: The heuristic of number of tiles out-of-place is
certainly less than the actual number of moves to the goal state. So this
heuristic (combined with best-first search) is an admissible algorithm. So is
the heuristic sum of the distances of out-of-place tiles, because it too
underestimates the actual number of moves required to reach a goal state.
Monotonicity
This property asks if an algorithm
is locally admissible---that is, it always underestimates the cost
between any two states in the search space. Recall that A* does not
require that g(n) = g*(n). A heuristic function, h is monotone
if:
- For all states ni and nj, where nj is a descendant of ni, h(ni) - h(nj) <= cost(ni,nj).
- The heuristic evaluation of the goal state is 0: h(Goal) = 0.
In other words, for any two states
in the search space, a monotonic heuristic always underestimates the cost of
going from ni to nj. The heuristic is everywhere
admissible.
If best-first search is used with a
monotonic heuristic, you can skip checking for shortest path when a state is
encountered a second time. The second occurrence will not be shorter because
the heuristic finds the shortest path to any state the first time the state is
found. Thus, for a monotone heuristic, searching a graph is no different from
searching a tree.
Any monotonic heuristic is
admissible. For some sequence of states, s1,
s2, s3,...,sg, from start to goal, if the
heuristic underestimates the cost of going from s1 to s2
and from s2 to s3 and so on, then it underestimates the
cost of going from any state to the goal. So, it follows that h(n) <= cost(sn,
sg) = h*(n).
Informedness
For two A* heuristics, h1
and h2, if h1(n) <= h2(n), for all states n
in the search space, then heuristic h2 is more informed than
h1.
Eight Puzzle Heuristics: Breadth-first search is an A* heuristic with h1(n)
= 0, for all n. We have also noted that h2, the number of tiles out
of place is a lower bound for h*. So it follows that:
h1 <= h2 < h*(n)
Therefore, "number of tiles out
of place" is more informed than breadth-first heuristic. This should be
obvious.
The following figure illustrates the
difference between the "number of out-of-place tiles" heuristic and
breadth-first search. The heavy dark line shows the optimal solution path. The
states that show the numbers on the tiles are the states that would be the
portion of the space that is searched by the "tiles out-of-place"
heuristic. The rest of the states shown are the ones that would also be
examined by breadth-first search. You can see the significant pruning done by
the more informed heuristic.
Peseudocode
function
A*(start,goal)
closedset := the empty set // The set of nodes already evaluated.
openset := {start} // The set of tentative nodes to be
evaluated, initially containing the start node
came_from := the empty map // The map of navigated nodes.
g_score[start] := 0 // Cost from start along best known path.
// Estimated total cost from start to goal
through y.
f_score[start] := g_score[start] +
heuristic_cost_estimate(start, goal)
while openset is not empty
current := the node in openset having
the lowest f_score[] value
if current = goal
return reconstruct_path(came_from,
goal)
remove current from openset
add current to closedset
for each neighbor in neighbor_nodes(current)
if neighbor in closedset
continue
tentative_g_score := g_score[current]
+ dist_between(current,neighbor)
if neighbor not in openset or
tentative_g_score <= g_score[neighbor]
came_from[neighbor] := current
g_score[neighbor] :=
tentative_g_score
f_score[neighbor] := g_score[neighbor]
+ heuristic_cost_estimate(neighbor, goal)
if neighbor not in openset
add neighbor to openset
return failure
function reconstruct_path(came_from,
current_node)
if came_from[current_node] in set
p := reconstruct_path(came_from,
came_from[current_node])
return (p + current_node)
else
return current_node
Example

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