Admissible heuristic

An admissible heuristic is used to estimate the cost of reaching the goal state in an informed search algorithm. In order for a heuristic to be admissible to the search problem, the estimated cost must always be lower than or equal to the actual cost of reaching the goal state. The search algorithm uses the admissible heuristic to find an estimated optimal path to the goal state from the current node. For example, in A* search the evaluation function (where ${\displaystyle n}$ is the current node) is:

${\displaystyle f(n)=g(n)+h(n)}$

where

${\displaystyle f(n)}$ = the evaluation function.

${\displaystyle g(n)}$ = the cost from the start node to the current node

${\displaystyle h(n)}$ = estimated cost from current node to goal.

${\displaystyle h(n)}$ is calculated using the heuristic function. With a non-admissible heuristic, the A* algorithm could overlook the optimal solution to a search problem due to an overestimation in ${\displaystyle f(n)}$.

${\displaystyle n}$ is a node

${\displaystyle h}$ is a heuristic

${\displaystyle h(n)}$ is cost indicated by ${\displaystyle h}$ to reach a goal from ${\displaystyle n}$

${\displaystyle h^{*}(n)}$ is the optimal cost to reach a goal from ${\displaystyle n}$

${\displaystyle h(n)}$ is admissible if, ${\displaystyle \forall n}$

${\displaystyle h(n)\leq h^{*}(n)}$

An admissible heuristic can be derived from a relaxed version of the problem, or by information from pattern databases that store exact solutions to subproblems of the problem, or by using inductive learning methods.

Two different examples of admissible heuristics apply to the fifteen puzzle problem:

• Hamming distance
• Manhattan distance

The Hamming distance is the total number of misplaced tiles. It is clear that this heuristic is admissible since the total number of moves to order the tiles correctly is at least the number of misplaced tiles (each tile not in place must be moved at least once). The cost (number of moves) to the goal (an ordered puzzle) is at least the Hamming distance of the puzzle.

The Manhattan distance of a puzzle is defined as:

${\displaystyle h(n)=\sum _{\text{all tiles}}{\mathit {distance}}({\text{tile, correct position}})}$

Consider the puzzle below in which the player wishes to move each tile such that the numbers are ordered. The Manhattan distance is an admissible heuristic in this case because every tile will have to be moved at least the number of spots in between itself and its correct position.[2]

The subscripts show the Manhattan distance for each tile. The total Manhattan distance for the shown puzzle is:

${\displaystyle h(n)=3+1+0+1+2+3+3+4+3+2+4+4+4+1+1=36}$

If an admissible heuristic is used in an algorithm that, per iteration, progresses only the one path that has lowest total expected cost of several candidate paths and terminates the moment any path reaches the goal accepting that path as shortest (for example in A* search algorithm), then this algorithm will terminate on the shortest path. To see why, simply consider that any path that the algorithm terminates on was only progressed because its total expected cost was lowest of the candidates. For an admissible heuristic, none of the candidates overestimate their costs so their true costs can only be greater to or equal to that of the accepted path. Finally, the total expected cost is the true cost for a path that reaches goal because the only admissible heuristic on reaching goal is zero.

As an example[3] of why admissibility can guarantee optimality, let us say we have costs as follows:(the cost above/below a node is the heuristic, the cost at an edge is the actual cost)

 0     10   0   100   0
START ----  O  ----- GOAL
 |                   |
0|                   |100
 |                   | 
 O ------- O  ------ O
100   1    100   1   100

So clearly we would start off visiting the top middle node, since the expected total cost, i.e. ${\displaystyle f(n)}$, is ${\displaystyle 10+0=10}$. Then the goal would be a candidate, with ${\displaystyle f(n)}$ equal to ${\displaystyle 10+100+0=110}$. Then we would clearly pick the bottom nodes one after the other, followed by the updated goal, since they all have ${\displaystyle f(n)}$ lower than the ${\displaystyle f(n)}$ of the current goal, i.e. their ${\displaystyle f(n)}$ is ${\displaystyle 100,101,102,102}$. So even though the goal was a candidate, we could not pick it because there were still better paths out there. This way, an admissible heuristic can ensure optimality.

However, note that although an admissible heuristic can guarantee final optimality, it is not necessarily efficient.

While all consistent heuristics are admissible, not all admissible heuristics are consistent.

For tree search problems, if an admissible heuristic is used, the A* search algorithm will never return a suboptimal goal node.

• Consistent heuristic
• Heuristic function
• Search algorithm