# AI Class: A* Search

October 16, 2011

The first week of the online Stanford AI class has gone by. The format of the presentation is pretty good, with lots of short lecture videos chained together with little quizzes embedded. It’s really pretty neat. There are some minor errors in some of the lectures, which is normal, though in this kind of online setting, it’s more difficult for the keener students to correct the lecturer as the lecture goes along, so some people are likely to spend a bit of time confused because of that. Some people have complained on Reddit about the relatively superficial level of the coverage of the lectures, but as far as I’m concerned, that’s what the book (and papers) are for. The course is called “Introduction to Artificial Intelligence”, after all.

## Motivation: Take me home, Romanian roads

The two things we covered this week were a very quick introduction to the field, then some stuff about search. Although there is a big pile of code available for download associated with Russell & Norvig’s AI book which would have made playing with some of this stuff easier, I decided to write some code of my own in Haskell, mostly for my own amusement. I wanted to implement A* search, with a clean interface for setting up problems.

The example used in the lectures, of finding driving routes on a simple road map of Romania, is a good simple problem to look at. I wanted the definition of this problem to look something like this:

import AStar

import Data.Maybe
import Data.List

data Rom = A | B | C | D | E | F | G | H | I | L | M |
N | O | P | R | S | T | U | V | Z
deriving (Eq, Ord, Show)

initial :: Rom
initial = A

goal :: Rom -> Bool
goal = (== B)

distances :: [((Rom, Rom), Int)]
distances = [((A, S), 140), ((A, T), 118), ((A, Z), 75), ((B, F), 211),
((B, G), 90), ((B, P), 101), ((B, U), 85), ((C, D), 120),
((C, P), 138), ((C, R), 146), ((D, M), 75), ((E, H), 86),
((F, S), 99), ((H, U), 98), ((I, N), 87), ((I, V), 92),
((L, M), 70), ((L, T), 111), ((O, S), 151), ((O, Z), 71),
((P, R), 97), ((R, S), 80), ((U, V), 142)]

euclidean :: [(Rom, Int)]
euclidean = [(A, 366), (B, 0), (C, 160), (D, 242), (E, 161), (F, 176),
(G, 77), (H, 151), (I, 226), (L, 244), (M, 241), (N, 234),
(O, 380), (P, 100), (R, 193), (S, 253), (T, 329), (U, 80),
(V, 199), (Z, 374)]

instance SearchState Rom where
actions x = map const $nub$ filter (/= x) $(map fst ps) ++ (map snd ps) where ps = map fst$ filter (\((a, b), _) -> a == x || b == x) distances

stepCost x y
| x < y = fromJust $lookup (x, y) distances | otherwise = fromJust$ lookup (y, x) distances

heuristic :: Rom -> Int
heuristic x = fromJust $lookup x euclidean Here, we define a data type Rom to represent Romanian cities, define an initial state and a goal function, set up a couple of data structures to record city-to-city road distances and Euclidean distances of cities from Bucharest (our goal), then define Rom as an instance of the type class SearchState which includes code to find the possible actions we can take from any state (i.e. the cities we can travel to directly from a given city) and the costs (i.e. distance) associated with going from one state to another. Finally, we define a heuristic, which just returns the Euclidean distance from a given city to the goal city. Given that definition, I wanted to be able to type something like astar initial goal heuristic to run an A* search for the goal state, starting at the initial state and using the given heuristic. We might also want to look at astar initial goal (const 0) to do a uniform cost search. I’d like to both get a result and a count of the number of search tree nodes expanded (which will allow me to see how much of a difference different heuristics make to the efficiency of the search). ## A* implementation To make this work, the AStar module is defined as follows. It’s basically one type class, some supporting scaffolding, and the astar search function. We need a few data structures to implement A* search efficiently, so we import, in particular, Data.Set for sets and Data.PSQueue for priority search queues (priority queues that also support fast membership testing): module AStar where import Data.Maybe import Data.List import Data.Ord import qualified Data.Set as S import qualified Data.PSQueue as PS import Debug.Trace The SearchState type class is defined as: class SearchState a where actions :: a -> [a -> a] pathCost :: [a] -> Int stepCost :: a -> a -> Int stepCost _ _ = 1 pathCost sts = sum$ zipWith stepCost (tail sts) (init sts)

First, for a given search state, we want to find the actions that we can perform from that state. The actions method of the type class does this, returning a list of functions that will transform one state to another – we can then just apply these functions to a state to get the result of any action. We also want to be able to calculate the costs of paths composed of states, either in one go, from a list of states, using pathCost, or step by step, which is what stepCost does. We define defaults for stepCost (each state change has a cost of 1) and pathCost (the cost of a path through a list of states is just the sum of the individual steps).

To maintain a record of the search tree, we need a node type, holding a state, possibly a parent (Nothing if this is the initial node) and the cost at this node. Note that states can nominally appear in multiple nodes if a state is reached via multiple paths in the search graph (in fact, in the search function below, we only keep the best path to any particular state, so at any time, at most one node is associated with any state). We make our nodes an instance of Ord so that we can put them into data structures that require comparison. We set things up so that the comparison is done on the cost of the nodes.

data NodeInfo a = NodeInfo { state :: a,
parent :: Maybe (NodeInfo a),
cost :: Int } deriving (Eq, Show)

instance Eq a => Ord (NodeInfo a) where
compare = comparing cost

The main search function is a little complicated, but it has 7 main steps:

1. Initialise the search frontier with a node for the initial state (with no parent and zero cost).
2. If the goal has been reached, extract the path from the initial to the goal node from the search tree and return it, along with a count of the number of search tree nodes expanded to find the solution.
3. As long as there are nodes in the frontier, pick the lowest cost node for expansion (the frontier is represented using a priority search queue, so this minimum cost extraction is $O(log n)$ in time).
4. Add the node to be expanded to the “explored” set.
5. Determine the actions from the state of the node to be expanded, the result states of those actions and the costs of the possible result states (here, “cost” means the known path cost to the state of the node being expanded, plus the heuristic cost estimate for the result state).
6. Combine the existing frontier (minus the node being expanded) with the new states: any states not already in the frontier are added, and states that already exist in the frontier have their paths and costs updated if we have found a better path to the state than the one already recorded in the search tree.
7. Repeat from step 3.

astar :: (Show a, Eq a, SearchState a, Ord a) =>
a -> (a -> Bool) -> Maybe [a]
astar init goalTest = helper initFrontier S.empty
where -- Initial frontier contains just the single initial node (no
-- parent, zero cost).
initFrontier = PS.singleton init (NodeInfo init Nothing 0)

-- Tail recursive helper.
helper fr ex
| PS.null fr = Nothing   -- Fail if the frontier is empty.
| otherwise = if (goalTest s)
then Just (reverse $res s p) else helper fr'' ex' where -- Extract next state to expand and associated -- information. We produce trace output to show the -- states on the frontier. Just (s PS.:-> ni@(NodeInfo _ p c), fr') = PS.minView (traceShow (PS.keys fr) fr) -- Add to explored set. ex' = S.insert s ex -- Determine possible actions from this state and -- resulting state, filtering out any in the -- explored set. rs = filter (\r -> not (r S.member ex'))$
map (result s) (actions s)

-- Calculate full costs of possible result states.
cs = map (\r -> c + stepCost s r + heuristic r) rs

-- Find which of the new states are already in the
-- queue.
exis = map (\n -> PS.lookup n fr) rs

-- Combine the existing states in the queue and the
-- new ones we've found.  Only ever keep one entry
-- in the queue for each state, the one with the
-- smallest cost.
toadd = catMaybes $zipWith3 combine rs cs exis combine r c Nothing = Just (r, NodeInfo r (Just ni) c) combine r c (Just (NodeInfo _ _ c')) | c < c' = Just (r, NodeInfo r (Just ni) c) | otherwise = Nothing -- Insert new states into new frontier. fr'' = foldl (\q (k, p) -> PS.insert k p q) fr' toadd -- Build the result by following the parent links -- through the search graph. res n Nothing = [init] res n (Just p) = n : res (state p) (parent p) This is a bit more verbose than the implementations in the Data.Graph module, but it works fine and is relatively efficient. Here’s what we get for the Romanian roads problem: *Main> astar initial goal (const 0) [A] [S,T,Z] [O,S,T] [L,O,S] [F,L,O,R] [F,L,R] [C,F,L,P] [C,F,M,P] [B,C,M,P] [B,C,D,P] [B,C,D] [B,D] [B] Just ([A,S,R,P,B],13) *Main> astar initial goal heuristic [A] [S,T,Z] [F,O,R,T,Z] [C,F,O,P,T,Z] [B,C,O,P,T,Z] [B,C,O,T,Z] Just ([A,S,R,P,B],6) *Main> First, we try the search using uniform cost search (i.e. a heuristic function of (const 0)). This gives us the right answer (as reported in the lectures), and expands 13 nodes in the search tree to find the solution. Then, using the Euclidean distance as a heuristic for the actual distance by road, we get the same answer (guaranteed since our heuristic is admissible, i.e. optimistic) by only expanding 6 search tree nodes. You can look at the order in which the nodes in the search tree are expanded and you’ll see that they correspond to the discussion in the lecture. ## Homework problem Let’s look at how to set up a couple of other problems. First, one of the homework problems dealt with a very simple setting: a 4 × 6 grid where we need to find a route from the top left to the bottom right moving only horizontally or vertically at each step, with a given heuristic function at each grid cell to estimate the remaining distance to the goal cell (numbers in the grid cells here show the heuristic value and the circles mark the initial and goal states): In our framework, this problem looks like this: import AStar data HWState = HWState Int Int deriving (Eq, Ord) instance Show HWState where show (HWState r c) = ("ABCD" !! (r-1)) : (show c) instance SearchState HWState where actions (HWState r c) = map (const . uncurry HWState)$
filter valid [ (r-1, c), (r+1, c), (r, c-1), (r, c+1) ]
where valid (r, c) = r >= 1 && r <= 4 && c >= 1 && c <= 6

initial :: HWState
initial = HWState 1 1

goal :: HWState -> Bool
goal (HWState 4 6) = True
goal (HWState _ _) = False

heuristic :: HWState -> Int
heuristic (HWState 4 _) = 1
heuristic (HWState _ 6) = 1
heuristic (HWState 3 _) = 2
heuristic (HWState _ 5) = 2
heuristic (HWState 2 _) = 3
heuristic (HWState _ 4) = 3
heuristic (HWState _ _) = 4

The problem state, defined by the HWState type, just gives the row and column where we are at this step, we define a Show instance for the state so we can see what’s going on, we define an initial state (top left) and a function to test whether a state is the goal state (easy here: are we at the bottom right?). Then we get to the meat of the code, which is an instance definition for the SearchState class (defined in the AStar module). Here, this just has a method to find out what possible actions we can take from a given state (moving to any of the adjacent cells, module edge effects). We use the default path cost definition from the SearchState type class. Finally, we have a function to calculate the suggested heuristic for a given state. This all works as expected (I’ve taken out the traceShow call, so we don’t see the states that are expanded at each step), although here the heuristic doesn’t seem to help at all:

*Main> astar initial goal (const 0)
Just ([A1,A2,A3,A4,A5,A6,B6,C6,D6],24)
*Main> astar initial goal heuristic
Just ([A1,A2,A3,A4,A5,A6,B6,C6,D6],24)
*Main>

## Sliding blocks puzzles

Let’s try a more substantial example. The sliding blocks puzzles of the 15-Puzzle kind are a classic search problem. Things are quite a bit more complicated here than in the two previous examples. The boards for these puzzles look like this (for the 3 × 3 “8-puzzle” version on the left and the 4 × 4 “15-puzzle” version on the right):

First we need some imports (we’ll use a map to help represent the board state, and we’ll need some random numbers to generate scrambled initial board states).

import Prelude hiding (Left, Right)
import qualified Data.Map as M
import System.Random

import AStar

Next, we define our board state. We need to know how big the board is (typically, we’ll consider the 3 × 3 “8-puzzle” and the 4 × 4 “15-puzzle”), and what tiles are in each position, which we represent as a map from a (row, column) tuple to an integer tile value. For convenience, we also record the row and column where the blank is.

type Board = M.Map (Int, Int) Int
data BoardState = BoardState { size :: Int,
board :: Board,
bpos :: (Int, Int) } deriving (Eq, Ord)

We often need to generate the list of all coordinate pairs for the board, so let’s have a helper function that does that for a given board size.

coords :: Int -> [(Int, Int)]
coords n = [(r, c) | r <- [1..n], c <- [1..n]]

Moves in these puzzles are most easily thought of in terms of moving the blank, rather than moving a tile. Any single move just moves the blank by one square, either up, down, left or right. Here, if we try to apply an invalid move to a board state (e.g. moving the blank off the left hand side of the board), we define the result to be an unchanged board. This is convenient when we come to generate scrambled boards, but it’s not a problem in the A* search, since we will never try to apply invalid moves.

data MoveDir = Up | Down | Left | Right

move :: MoveDir -> BoardState -> BoardState
move dir (BoardState n m bpos@(brow, bcol)) = BoardState n m'' bpos'
where m'' = M.insert bpos' 0 m'
m' = M.insert bpos (m M.! bpos') m
bpos' = moveBlank dir
moveBlank Up = if (brow > 1) then (brow-1, bcol) else (brow, bcol)
moveBlank Down = if (brow < n) then (brow+1, bcol) else (brow, bcol)
moveBlank Left = if (bcol > 1) then (brow, bcol-1) else (brow, bcol)
moveBlank Right = if (bcol < n) then (brow, bcol+1) else (brow, bcol)

We add a Show instance for the board state so that we can see what’s going on.

instance Show BoardState where
show (BoardState n m _) =
unlines $map concat$ chunks n $map (format . (m M.!)) (coords n) where format x | x == 0 = " XX" | 1 <= x && x <= 9 = " " ++ show x | otherwise = " " ++ show x chunks _ [] = [] chunks n xs = (take n xs) : chunks n (drop n xs) Then, the SearchState instance is a bit of an anticlimax, since all we need to do is to generate the possible valid moves given a board state. We use the default implementation of path costs from SearchState, with each moving having the same cost. instance SearchState BoardState where actions (BoardState n m (brow, bcol)) = map move (lr ++ ud) where lr | bcol == 1 = [Right] | bcol == n = [Left] | otherwise = [Left, Right] ud | brow == 1 = [Down] | brow == n = [Up] | otherwise = [Up, Down] The “initial” state (we’ll actually start our searches from a scrambled state derived from this): initial :: Int -> BoardState initial n = BoardState n (M.fromList$ zip (coords n) ([1..n^2-1] ++ [0])) (n, n)

and the goal state is the same:

goal :: BoardState -> Bool
goal (BoardState n m _) = map (m M.!) (coords n) == ([1..n^2-1] ++ [0])

There are two possible heuristics that were mentioned in the lectures. The first just counts the number of tiles that are not in the correct position:

heuristic1 :: BoardState -> Int
heuristic1 (BoardState n m _) = sum $zipWith (\x y -> if x /= y then 1 else 0) (map (m M.!) (init$ coords n)) [1..n^2-1]

The second measures the Manhattan distance of each tile from its correct position and uses the sum of these distances for all the tiles as the heuristic value:

heuristic2 :: BoardState -> Int
heuristic2 (BoardState n m _) = M.foldrWithKey folder 0 m
where folder (r, c) v s = s + if (v > 0) then dist (r, c) (goodPos v) else 0
dist (r1, c1) (r2, c2) = abs (r1 - r2) + abs (c1 - c2)
goodPos v = ((v - 1) div n + 1, (v - 1) rem n + 1)

Finally, we need to be able to generate scrambled starting states. This is most easily done by generating a random list of moves and applying them to the initial board state. Since we ignore invalid moves in the move function above, this all works fine.

scramble :: Int -> BoardState -> IO BoardState
scramble n b = do
g <- getStdGen
return $foldl (flip move) b (makeMoves g) where makeMoves gen = map ([Up, Down, Left, Right] !!) (take n$ randomRs (0,3) gen :: [Int])

And then everything works as you might expect:

*Main> b0 <- scramble 20 (initial 3)
*Main> b0
5  1 XX
4  2  3
7  8  6

*Main> heuristic2 b0
6
*Main> astar b0 goal heuristic2
Just ([  5  1 XX
4  2  3
7  8  6
,  5  1  3
<< lines skipped >>
7 XX  8
,  1  2  3
4  5  6
7  8 XX
],54)
*Main>

Here, we generate a scrambled state for the 8-puzzle from the initial “perfect” state by applying 20 randomly selected moves. We display the scrambled initial state, the value of the Manhattan distance heuristic for this state (which gives us some idea of how much effort we’ll have to expend to find a solution), then run the A* search using the Manhattan distance heuristic. The result is a list of states (some of the output snipped) and a count of the number of search tree nodes expanded (54, in this case).

In a post in a day or two, I’ll look a little at how much of a difference using heuristics really makes in this problem…