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Joined 2026-09-21 Message
Incoming links: List concatenation
Linear flattening of a binary tree 2026-10-24
For a leaf-labelled binary tree, define
go (One x) acc = x:acc and go (Two l r) acc = go l (go r acc), then apply go t []. The invariant in the title follows by structural induction and associativity of list concatenation. It proves preservation of leaf order. Each leaf supplies one output constructor and each internal node requires constant bookkeeping, giving linear time in the number of nodes. By contrast, flattening both subtrees and concatenating their results can copy prefixes of sizes on a left-branching tree, taking quadratic time. The accumulator is the already-known suffix in a difference list representation.Both versions preserve left-to-right leaf order:
flatten :: Tree a -> [a]
flatten (One x) = [x]
flatten (Two l r) = flatten l ++ flatten r
flattenFold :: Tree a -> [a]
flattenFold = foldTree (\x -> [x]) (++)The tree fold replaces each leaf by a singleton list and each internal node by list concatenation. The defining equations of
flattenFold are consequently those of flatten, proving their equality by structural induction.Let be the number of leaves; the total number of nodes is . List concatenation copies its left operand. For a tree whose left subtree contains leaves at each successive internal node, these copied lengths are . Their sum is , so the worst-case running time is .
A difference list representation avoids copying prefixes:
flattenFast :: Tree a -> [a]
flattenFast t = go t []
where
go (One x) acc = x:acc
go (Two l r) acc = go l (go r acc)The recursion invariant is
go t acc = flatten t ++ acc. The leaf case is immediate; the internal-node case uses associativity of list concatenation. Each node is visited once and each leaf contributes one output constructor, giving running time when the result is fully consumed. This is linear flattening of a binary tree and does not assume balance.Record the outcome at the first completed board and otherwise branch over every available move:
data Outcome = Winner Player | Draw deriving (Eq,Ord,Show)
finals :: Board -> [(Board,Outcome)]
finals b@(os,xs,es)
| won os = [(b,Winner O)]
| won xs = [(b,Winner X)]
| null es = [(b,Draw)]
| otherwise = concatMap (\q -> finals (play b q)) esThe recursion terminates because each recursive call decreases the number of empty positions by one. For correctness, induct on that number. A completed board contributes exactly its own outcome. Otherwise every legal continuation has a unique first available move, and the induction hypothesis covers exactly all completions after that move. List concatenation collects these branches without extending completed games. Repeated final positions are allowed here because the computation deliberately explores every legal move sequence.
For proper lazy lists, the generic condition is a finite first list, or an empty second list. A finite first list reaches its terminating constructor, after which list concatenation exposes the second one. An infinite first list never yields control to the second; it therefore cannot deliver occurrences originating there unless there are none. This concerns input occurrences, rather than an accidental overlap of the two inputs' value sets.
Repeated root removal gives the conventional list:The representation interpretation satisfies
toList :: List a -> [a]
toList rs
| null rs = []
| otherwise = head rs : toList (tail rs)toList (cons x rs) = x : toList rs; alternatively, structural induction on the number of stored elements proves that each step emits exactly the next preorder traversal value. The output has cons cells, and each step takes constant time, so fully consuming it costs time. No repeated list concatenation is needed.Join either two halves or successive pairs:By the defining property of
unhalve :: ([a],[a]) -> [a]
unhalve (xs,ys) = xs ++ ys
unpairs :: [(a,a)] -> [a]
unpairs = foldr (\(x,y) zs -> x:y:zs) []splitAt, concatenating the two halves restores the original list. For unpairs, structural induction on the even input proves that its first pair restores its first two elements and the recursive result restores the rest. Thus both requested function compositions are the identity. Fully consuming either result takes at most linear time; list concatenation only copies the left half once.The direct tree fold for an inorder traversal isBy structural induction, the child lists are ascending; the binary search tree property places every left value below the root and every right value above it. The displayed list concatenation therefore gives an ascending list containing every stored value exactly once.
flattenFold :: Tree a -> [a]
flattenFold = foldTree (\ls x rs -> ls ++ (x:rs)) []For a version with linear time complexity on every tree shape, use an accumulator:This also has a higher-order tree fold expression using difference lists:Its folded value is a function that prepends the inorder traversal to a supplied tail. The accumulator invariant is
flatten :: Tree a -> [a]
flatten t = onto t []
where
onto Empty zs = zs
onto (Tree l x r) zs = onto l (x : onto r zs)flattenLinear :: Tree a -> [a]
flattenLinear t =
foldTree (\l x r -> l . (x:) . r) id t []onto t zs = flattenFold t ++ zs. Each node contributes one constructor, giving time to consume the output. By contrast, the direct list-valued fold can take on a left-skewed tree because ++ repeatedly copies left traversals. Predicate prefix-suffix reconstruction 2026-10-24
Concatenating the initial list prefix accepted by a predicate with the remaining list suffix reconstructs the input, provided every test reached before the first false result terminates. Prove the finite case by structural induction: a true test copies the head, and a false test returns the untouched list suffix. An undefined reached test is an obstruction because reconstruction loses an exposed cons cell. An infinite all-true list prefix can still reconstruct the input, since list concatenation never demands its second argument there.