Say I have the following Haskell tree type, where "State" is a simple wrapper:
data Tree a = Branch (State a) [Tree a]
| Leaf (State a)
deriving (Eq, Show)
I also have a function "expand :: Tree a -> Tree a" which takes a leaf node, and expands it into a branch, or takes a branch and returns it unaltered. This tree type represents an N-ary search-tree.
Searching depth-first is a waste, as the search-space is obviously infinite, as I can easily keep on expanding the search-space with the use of expand on all the tree's leaf nodes, and the chances of accidentally missing the goal-state is huge... thus the only solution is a breadth-first search, implemented pretty decent over here, which will find the solution 开发者_运维知识库if it's there.
What I want to generate, though, is the tree traversed up to finding the solution. This is a problem because I only know how to do this depth-first, which could be done by simply called the "expand" function again and again upon the first child node... until a goal-state is found. (This would really not generate anything other then a really uncomfortable list.)
Could anyone give me any hints on how to do this (or an entire algorithm), or a verdict on whether or not it's possible with a decent complexity? (Or any sources on this, because I found rather few.)
Have you looked at Chris Okasaki's "Breadth-First Numbering: Lessons from a Small Exercise in Algorithm Design"? The Data.Tree
module includes a monadic tree builder named unfoldTreeM_BF
that uses an algorithm adapted from that paper.
Here's an example that I think corresponds to what you're doing:
Suppose I want to search an infinite binary tree of strings where all the left children are the parent string plus "a", and the right children are the parent plus "bb". I could use unfoldTreeM_BF
to search the tree breadth-first and return the searched tree up to the solution:
import Control.Monad.State
import Data.Tree
children :: String -> [String]
children x = [x ++ "a", x ++ "bb"]
expand query x = do
found <- get
if found
then return (x, [])
else do
let (before, after) = break (==query) $ children x
if null after
then return (x, before)
else do
put True
return (x, before ++ [head after])
searchBF query = (evalState $ unfoldTreeM_BF (expand query) []) False
printSearchBF = drawTree . searchBF
This isn't terribly pretty, but it works. If I search for "aabb" I get exactly what I want:
|
+- a
| |
| +- aa
| | |
| | +- aaa
| | |
| | `- aabb
| |
| `- abb
|
`- bb
|
+- bba
|
`- bbbb
If this is the kind of thing you're describing, it shouldn't be hard to adapt for your tree type.
UPDATE: Here's a do-free version of expand
, in case you're into this kind of thing:
expand q x = liftM ((,) x) $ get >>= expandChildren
where
checkChildren (before, []) = return before
checkChildren (before, t:_) = put True >> return (before ++ [t])
expandChildren True = return []
expandChildren _ = checkChildren $ break (==q) $ children x
(Thanks to camccann for prodding me away from old control structure habits. I hope this version is more acceptable.)
I'm curious why you need the expand
function at all--why not simply construct the entire tree recursively and perform whatever search you want?
If you're using expand
in order to track which nodes are examined by the search, building a list as you go seems simpler, or even a second tree structure.
Here's a quick example that just returns the first result it finds, with the spurious Leaf
constructor removed:
data State a = State { getState :: a } deriving (Eq, Show)
data Tree a = Branch {
state :: State a,
children :: [Tree a]
} deriving (Eq, Show)
breadth ts = map (getState . state) ts ++ breadth (concatMap children ts)
search f t = head $ filter f (breadth [t])
mkTree n = Branch (State n) (map mkTree [n, 2*n .. n*n])
testTree = mkTree 2
Trying it out in GHCi:
> search (== 24) testTree
24
For contrast, here's a naive depth-first search:
depth (Branch (State x) ts) = x : (concatMap depth ts)
dSearch f t = head $ filter f (depth t)
...which of course fails to terminate when searching with (== 24)
, because the left-most branches are an endless series of 2s.
精彩评论